🧠:[<think>Okay, let's tackle this problem step by step. First, part (1) asks for the parametric equation of the curve C and the general equation of the line l. Starting with the curve C: The given equation is (x²)/4 + (y²)/9 = 1. This looks like the standard form of an ellipse centered at the origin, with semi-major axis 3 along the y-axis and semi-minor axis 2 along the x-axis. For an ellipse, the parametric equations are usually x = a cosθ and y = b sinθ, where a and b are the semi-axes. But wait, here the major axis is along y, so maybe it's better to write x = a cosθ and y = b sinθ. Let me confirm: the standard parametric equations for an ellipse (x²)/(a²) + (y²)/(b²) = 1 when major axis is along x is x = a cosθ, y = b sinθ. But in our case, since 9 > 4, the major axis is along y, so it's actually (x²)/(b²) + (y²)/(a²) = 1 with a > b. So maybe the parametric equations would still be x = b cosθ and y = a sinθ? Let me check. If a is the semi-major axis, which is 3 here, and b is semi-minor axis, which is 2. So yes, x = 2 cosθ, y = 3 sinθ. That should parameterize the ellipse correctly because plugging back in, ( (2 cosθ)^2 )/4 + ( (3 sinθ)^2 )/9 = (4 cos²θ)/4 + (9 sin²θ)/9 = cos²θ + sin²θ = 1. Perfect. So part (1) first answer: parametric equations for C are x = 2 cosθ, y = 3 sinθ, where θ is the parameter.Now, the line l is given in parametric form: x = 2 + t, y = 2 - 2t. We need to convert this into the general equation. To do that, we can eliminate the parameter t. From the x equation: t = x - 2. Substitute into the y equation: y = 2 - 2(x - 2) = 2 - 2x + 4 = 6 - 2x. So the general equation is y = -2x + 6, which can also be written as 2x + y - 6 = 0. Let me verify: when t = 0, x = 2, y = 2. Plugging into 2x + y = 4 + 2 = 6. Correct. When t = 1, x = 3, y = 0. 2*3 + 0 = 6. Correct. So yes, the general equation is 2x + y = 6 or 2x + y - 6 = 0.So part (1) done. Now part (2) is more complex. We need to draw a line through any point P on the curve C that forms a 30° angle with line l, intersecting l at point A. Then find the maximum and minimum values of |PA|.First, let me visualize this. We have an ellipse and a line. For any point P on the ellipse, we draw a line that makes a 30° angle with line l, and this new line intersects l at point A. The distance between P and A, |PA|, is to be maximized and minimized.Hmm. Let me break this down. The line PA forms a 30° angle with line l. Since angles between lines can be found using the formula tanθ = |(m2 - m1)/(1 + m1m2)|, but here we have a fixed angle of 30°, so the slope of PA can be determined relative to the slope of l.First, line l has slope m1 = -2 (from its equation y = -2x + 6). So, if the angle between PA and l is 30°, then the slope m2 of PA must satisfy tan(30°) = |(m2 - m1)/(1 + m1m2)|. Let me compute that.tan(30°) = 1/√3 ≈ 0.577.So,|(m2 - (-2))/(1 + (-2)m2)| = 1/√3Which simplifies to:|(m2 + 2)/(1 - 2m2)| = 1/√3This equation will give possible slopes m2 for the line PA. However, since angles can be on either side of the line l, there will be two possible slopes corresponding to +30° and -30°, but since tangent is positive in both first and third quadrants, but the angle between lines is always between 0° and 180°, so we need to consider both possibilities. Therefore, solving:(m2 + 2)/(1 - 2m2) = 1/√3 and (m2 + 2)/(1 - 2m2) = -1/√3Let me solve both equations.First equation:(m2 + 2)/(1 - 2m2) = 1/√3Multiply both sides by (1 - 2m2):m2 + 2 = (1 - 2m2)/√3Multiply both sides by √3:√3(m2 + 2) = 1 - 2m2Expand:√3 m2 + 2√3 = 1 - 2m2Bring all terms to left:√3 m2 + 2√3 - 1 + 2m2 = 0Factor m2:m2(√3 + 2) + (2√3 - 1) = 0Solve for m2:m2 = (1 - 2√3)/(√3 + 2)Multiply numerator and denominator by (√3 - 2) to rationalize denominator:m2 = (1 - 2√3)(√3 - 2)/[(√3 + 2)(√3 - 2)] = [1*(√3 - 2) - 2√3*(√3 - 2)]/(3 - 4) = [√3 - 2 - 2*3 + 4√3]/(-1) = [5√3 - 8]/(-1) = 8 - 5√3So m2 = 8 - 5√3 ≈ 8 - 8.66 ≈ -0.66Second equation:(m2 + 2)/(1 - 2m2) = -1/√3Similarly:Multiply both sides by (1 - 2m2):m2 + 2 = -(1 - 2m2)/√3Multiply both sides by √3:√3(m2 + 2) = -1 + 2m2Expand:√3 m2 + 2√3 = -1 + 2m2Bring all terms to left:√3 m2 + 2√3 + 1 - 2m2 = 0Factor m2:m2(√3 - 2) + (2√3 + 1) = 0Solve for m2:m2 = -(2√3 + 1)/(√3 - 2)Multiply numerator and denominator by (√3 + 2):m2 = -(2√3 + 1)(√3 + 2)/[(√3 - 2)(√3 + 2)] = -[2√3*√3 + 4√3 + √3 + 2]/(3 - 4) = -[6 + 5√3 + 2]/(-1) = -[8 + 5√3]/(-1) = 8 + 5√3So m2 = 8 + 5√3 ≈ 8 + 8.66 ≈ 16.66Therefore, the two possible slopes for PA are 8 - 5√3 and 8 + 5√3. Let me note that 5√3 is approximately 8.66, so 8 - 5√3 ≈ -0.66 and 8 + 5√3 ≈ 16.66. These are the two possible slopes for lines that form 30° angles with line l.Now, given that, for any point P on the ellipse C, we can draw two lines from P with these slopes, intersecting line l at point A. Then |PA| is the distance between P and A. We need to find the maximum and minimum of this distance.But wait, but the problem says "Draw a line through any point P on the curve C that forms a 30° angle with the line l, intersecting l at point A." So perhaps for each P, there are two such lines, but maybe the problem is considering both possibilities? Or maybe depending on the position of P, only one of them intersects? Hmm, maybe need to clarify.Alternatively, perhaps the line PA is such that it forms a 30° angle with line l, and intersects l at A. So for each point P, there are two possible lines PA forming 30° on either side, but perhaps only one of them actually intersects l. Wait, but the line l is straight, and depending on the direction, the line from P could intersect l at A either in front or behind, but since t is a parameter for line l, which is defined for all real t, the line l is infinite. So any line through P with a 30° angle to l will intersect l at exactly one point A.But wait, actually, if two lines pass through P with angles 30° relative to l, they would intersect l at two different points A1 and A2. Therefore, for each P, there are two such points A, each corresponding to the two different slopes. Therefore, |PA| would have two values for each P, and we need to consider both when finding the maximum and minimum.Alternatively, maybe the problem is considering both possibilities, so we need to consider both lines and find the overall maximum and minimum |PA| over all P on C and both possible lines.But regardless, the problem states "the line through P... intersecting l at A". So maybe for each P, there are two lines, but only one intersects l? Wait, no. Because if you have a line through P forming 30° with l, and l is a straight line, then depending on the direction, the line through P could intersect l on one side or the other. But since l is a straight line, extending infinitely in both directions, both lines through P (with angles +30° and -30° relative to l) would intersect l at some point, possibly in different directions. So for each P, there are two points A, hence two distances |PA|. Therefore, for each P, two distances, and we need to consider all such distances over all P to find the global max and min.Alternatively, maybe the problem is considering only the acute angle, but since 30° is specified, it might be either direction. The problem says "forms a 30° angle", which is a specific measure, so perhaps both directions are allowed, hence two lines per P.But perhaps the problem is expecting to consider both possibilities. Therefore, when calculating |PA|, we need to consider both intersections and find the maximum and minimum over all P and both intersections.Alternatively, maybe there is a parametrization that can help. Let me think.Given that, perhaps we can parametrize point P on the ellipse as (2 cosθ, 3 sinθ). Then, for each P, we can find the equations of the two lines passing through P with slopes m1 = 8 - 5√3 and m2 = 8 + 5√3. Then, find where these lines intersect line l (which is 2x + y = 6). Then compute |PA| for each intersection, and then find the maximum and minimum over all θ.Alternatively, maybe there is a better way using vector projections or something. Let me consider.First, perhaps we can model the line l parametrically as given: x = 2 + t, y = 2 - 2t. So any point on l is (2 + t, 2 - 2t). Then, for a given point P on the ellipse, we can parametrize PA as a line through P that makes a 30° angle with l. But the direction vector of l is (1, -2), since the parametric equations are x = 2 + t, y = 2 - 2t. So direction vector of l is (1, -2). Then, the direction vector of PA should form a 30° angle with (1, -2).So, if we denote the direction vector of PA as (a, b), then the angle between (a, b) and (1, -2) is 30°. The formula for the angle between two vectors is:cosθ = (v ⋅ w)/(|v||w|)So cos30° = (a*1 + b*(-2))/(sqrt(a² + b²) * sqrt(1 + 4)) = (a - 2b)/(sqrt(a² + b²) * sqrt(5))Therefore:(a - 2b)/sqrt(5(a² + b²)) = √3/2Cross-multiplying:2(a - 2b) = √15 sqrt(a² + b²)Similarly, squaring both sides:4(a - 2b)^2 = 15(a² + b²)Expanding the left side:4(a² - 4ab + 4b²) = 15a² + 15b²So:4a² - 16ab + 16b² = 15a² + 15b²Bring all terms to left:4a² - 16ab + 16b² -15a² -15b² = 0Simplify:-11a² -16ab + b² = 0Multiply by -1:11a² +16ab - b² = 0This is a quadratic equation in terms of a and b. Let's solve for a/b ratio.Let k = a/b, then equation becomes:11k² +16k -1 =0Solving quadratic:k = [-16 ± sqrt(256 + 44)]/(22) = [-16 ± sqrt(300)]/22 = [-16 ± 10√3]/22 = [-8 ± 5√3]/11Therefore, the direction vector (a, b) satisfies a/b = [-8 ±5√3]/11. Therefore, the direction vectors can be written as ([-8 ±5√3]/11, 1). To make it cleaner, we can write as direction vectors ( -8 ±5√3, 11 ).So the direction vectors for PA are ( -8 +5√3, 11 ) and ( -8 -5√3, 11 ). Let me check the calculation again. Wait, we had k = a/b = [-8 ±5√3]/11. Therefore, if we set b =11, then a = -8 ±5√3. So direction vectors are ( -8 +5√3, 11 ) and ( -8 -5√3, 11 ). Therefore, the lines PA have direction vectors proportional to these.Therefore, the parametric equations of lines PA through point P(x₀, y₀) would be:x = x₀ + s*(-8 +5√3)y = y₀ + s*11andx = x₀ + s*(-8 -5√3)y = y₀ + s*11for parameter s.We need to find where these lines intersect line l: x =2 + t, y =2 -2t.So for each line PA, set equal to l:First line:x₀ + s*(-8 +5√3) = 2 + ty₀ + s*11 = 2 -2tSolve for s and t.Similarly for the second line:x₀ + s*(-8 -5√3) = 2 + ty₀ + s*11 = 2 -2tLet's solve the first case.From the first equation:t = x₀ + s*(-8 +5√3) -2Substitute into the second equation:y₀ +11s = 2 -2[ x₀ + s*(-8 +5√3) -2 ]Simplify RHS:2 -2x₀ -2s*(-8 +5√3) +4= 6 -2x₀ +16s -10√3 sTherefore, equation becomes:y₀ +11s =6 -2x₀ +16s -10√3 sBring all terms to left:y₀ +11s -6 +2x₀ -16s +10√3 s =0Simplify:(y₀ +2x₀ -6) + (11s -16s +10√3 s) =0Which is:(y₀ +2x₀ -6) + s(-5 +10√3)=0Therefore,s = (6 -2x₀ -y₀)/(-5 +10√3) = (6 -2x₀ -y₀)/(10√3 -5) = (6 -2x₀ -y₀)/(5(2√3 -1))Similarly, for the second line:x₀ + s*(-8 -5√3) =2 + ty₀ +11s =2 -2tAgain, solve for t from first equation:t = x₀ +s*(-8 -5√3) -2Substitute into second equation:y₀ +11s =2 -2[ x₀ +s*(-8 -5√3) -2 ]Simplify RHS:2 -2x₀ -2s*(-8 -5√3) +4=6 -2x₀ +16s +10√3 sTherefore,y₀ +11s =6 -2x₀ +16s +10√3 sBring all terms to left:y₀ +11s -6 +2x₀ -16s -10√3 s=0Simplify:(y₀ +2x₀ -6) + (11s -16s -10√3 s)=0Which is:(y₀ +2x₀ -6) + s(-5 -10√3)=0Therefore,s = (6 -2x₀ -y₀)/(-5 -10√3) = (6 -2x₀ -y₀)/(-5(1 +2√3)) = (2x₀ + y₀ -6)/(5(1 +2√3))So now, for each direction (two cases), we have parameter s expressed in terms of x₀ and y₀.Then, the point A is the intersection point, so the coordinates can be found by plugging s back into PA's parametric equations. Then |PA| is the distance from P(x₀, y₀) to A, which can be calculated using the distance formula.But since PA is along the line with direction vector proportional to (-8 ±5√3, 11), the distance |PA| would be |s| * sqrt( (-8 ±5√3)^2 + 11^2 )But perhaps we can compute |PA| directly using the expressions for s.Let me compute |PA| for the first case.First case:s = (6 -2x₀ -y₀)/(5(2√3 -1))Then, the distance |PA| is sqrt[ (s*(-8 +5√3))^2 + (s*11)^2 ) ] = |s| * sqrt( (-8 +5√3)^2 + 121 )Compute (-8 +5√3)^2 =64 - 80√3 +75= 139 -80√3Thus sqrt(139 -80√3 +121 )= sqrt(260 -80√3 )Similarly, sqrt(260 -80√3 ). Let me compute this:260 -80√3 ≈ 260 -138.56 = 121.44, sqrt(121.44)≈11.02But exact value? Let me see:But perhaps factor:260 -80√3 = 20*(13 -4√3)So sqrt(20*(13 -4√3)) = sqrt(20) * sqrt(13 -4√3)Hmm, not sure if helpful.Alternatively, note that direction vector is (-8 +5√3, 11), so its magnitude is sqrt( (-8 +5√3)^2 +11^2 )=sqrt(64 -80√3 +75 +121)=sqrt(64+75+121 -80√3)=sqrt(260 -80√3). So the distance |PA| is |s| * sqrt(260 -80√3 )Similarly, for the second case:s = (2x₀ + y₀ -6)/(5(1 +2√3))Then, |PA| is |s| * sqrt( (-8 -5√3)^2 +11^2 )Compute (-8 -5√3)^2 =64 +80√3 +75=139 +80√3So sqrt(139 +80√3 +121)=sqrt(260 +80√3 )Same as before, sqrt(260 +80√3 )Thus, in the first case:|PA| = |(6 -2x₀ -y₀)/(5(2√3 -1))| * sqrt(260 -80√3 )In the second case:|PA| = |(2x₀ + y₀ -6)/(5(1 +2√3))| * sqrt(260 +80√3 )But this seems quite complicated. Maybe we can simplify these expressions.Alternatively, perhaps we can express 6 -2x₀ -y₀ and 2x₀ + y₀ -6 in terms of P's coordinates. Note that line l is 2x + y =6, so 2x₀ + y₀ -6 is the value of the left-hand side at point P. So 2x₀ + y₀ -6 is the signed distance from P to line l multiplied by sqrt(2² +1²)=sqrt(5). Wait, the actual formula for the distance from P(x₀, y₀) to line l: |2x₀ + y₀ -6| / sqrt(4 +1) = |2x₀ + y₀ -6| / sqrt(5). Therefore, 2x₀ + y₀ -6 is equal to sqrt(5) times the signed distance. Thus, in the expressions above, we have:First case: numerator is (6 -2x₀ -y₀) = - (2x₀ + y₀ -6 )Second case: numerator is (2x₀ + y₀ -6 )Therefore, in both cases, |PA| is proportional to |2x₀ + y₀ -6|, which relates to the distance from P to line l. So perhaps |PA| can be written as:For the first case:|PA| = | - (2x₀ + y₀ -6 ) | / (5(2√3 -1)) * sqrt(260 -80√3 )= |2x₀ + y₀ -6| / (5(2√3 -1)) * sqrt(260 -80√3 )Similarly, second case:|PA| = |2x₀ + y₀ -6| / (5(1 +2√3 )) * sqrt(260 +80√3 )Therefore, |PA| is |2x₀ + y₀ -6| multiplied by a constant factor in each case.Let me compute these constants.First, for the first case:Constant1 = sqrt(260 -80√3 ) / [5(2√3 -1)]Similarly, sqrt(260 -80√3 ) squared is 260 -80√3, so sqrt(260 -80√3 ) is sqrt(20*(13 -4√3)) = sqrt(20)*sqrt(13 -4√3). Not sure if helpful.Alternatively, let's compute numerically:First, compute 260 -80√3 ≈260 -80*1.732≈260 -138.56≈121.44. sqrt(121.44)≈11.02Denominator:5(2√3 -1)≈5*(3.464 -1)=5*2.464≈12.32So Constant1≈11.02 /12.32≈0.894Second case:sqrt(260 +80√3 )≈sqrt(260 +138.56)=sqrt(398.56)≈19.964Denominator:5(1 +2√3 )≈5*(1 +3.464)=5*4.464≈22.32Thus, Constant2≈19.964 /22.32≈0.894Wait, interesting. Both constants are approximately 0.894. Wait, is this a coincidence or is there a mathematical reason?Wait, perhaps they are equal? Let me check algebraically.Compute Constant1 and Constant2.First, note that:sqrt(260 -80√3 ) / [5(2√3 -1)] and sqrt(260 +80√3 ) / [5(1 +2√3 )]Let me square both constants:For Constant1:(260 -80√3 ) / [25(2√3 -1)^2 ]For Constant2:(260 +80√3 ) / [25(1 +2√3 )^2 ]Compute denominators:(2√3 -1)^2 = 12 -4√3 +1=13 -4√3(1 +2√3 )^2 =1 +4√3 +12=13 +4√3Thus, for Constant1 squared:(260 -80√3 )/[25(13 -4√3 )]Similarly, for Constant2 squared:(260 +80√3 )/[25(13 +4√3 )]Let me compute (260 -80√3 )/(13 -4√3 )Multiply numerator and denominator by (13 +4√3 ):[(260 -80√3 )(13 +4√3 )]/[(13)^2 - (4√3 )^2] = [260*13 +260*4√3 -80√3*13 -80√3*4√3 ]/[169 - 48] = [3380 +1040√3 -1040√3 -80*12 ]/121Simplify numerator:3380 -960 =2420So numerator is 2420Denominator is 121Thus, 2420 /121 =20Similarly, for the second constant squared:(260 +80√3 )/(13 +4√3 )Multiply numerator and denominator by (13 -4√3 ):[(260 +80√3 )(13 -4√3 )]/[169 -48] = [260*13 -260*4√3 +80√3*13 -80√3*4√3 ]/121 = [3380 -1040√3 +1040√3 -960 ]/121Simplify numerator:3380 -960=2420Thus, same result:2420/121=20Therefore, both constants squared are 20/25=4/5. Therefore, both constants are sqrt(4/5)=2/sqrt(5)=2√5/5≈0.894. So indeed, both constants are equal to 2√5/5.Therefore, |PA| in both cases is |2x₀ + y₀ -6| * (2√5/5 )Therefore, combining both cases, since the absolute value is taken, we have |PA| = (2√5 /5) * |2x₀ + y₀ -6|Therefore, regardless of which line we take (the +30° or -30° direction), the distance |PA| is proportional to |2x₀ + y₀ -6|, with proportionality constant 2√5/5.Therefore, the problem reduces to finding the maximum and minimum values of |2x + y -6| for (x, y) on the ellipse C: x²/4 + y²/9 =1. Then multiply by 2√5 /5 to get |PA|.Therefore, the maximum and minimum of |2x + y -6| on the ellipse will give us the maximum and minimum |PA| after scaling.Thus, we need to find the extremum (max and min) values of the linear function 2x + y -6 on the ellipse x²/4 + y²/9 =1.This is a standard problem in optimization with constraints, which can be solved using Lagrange multipliers or parametrization.Given the ellipse parametric equations x = 2 cosθ, y = 3 sinθ, substitute into 2x + y -6:Expression: 2*(2 cosθ) + 3 sinθ -6 =4 cosθ +3 sinθ -6Therefore, we need to find the maximum and minimum of 4 cosθ +3 sinθ -6. The term 4 cosθ +3 sinθ is a sinusoidal function, which has amplitude sqrt(4² +3²)=5. Therefore, 4 cosθ +3 sinθ varies between -5 and +5. Therefore, 4 cosθ +3 sinθ -6 varies between -5 -6 = -11 and +5 -6 = -1.But wait, so the maximum of 4 cosθ +3 sinθ -6 is -1, and the minimum is -11. Therefore, |2x + y -6| = |4 cosθ +3 sinθ -6| varies between | -1 | =1 and | -11 | =11. Therefore, the maximum |2x + y -6| is 11, and the minimum is 1.Therefore, the maximum |PA| is (2√5 /5)*11 =22√5 /5, and the minimum |PA| is (2√5 /5)*1 =2√5 /5.But wait, wait. Let me confirm this.First, expression 4 cosθ +3 sinθ -6: the maximum and minimum of 4 cosθ +3 sinθ is indeed 5 and -5. Therefore, the expression becomes 5 -6 = -1 and -5 -6=-11. Thus, the expression varies from -11 to -1. Therefore, the absolute value |4 cosθ +3 sinθ -6| varies from 1 to 11.Therefore, the maximum |PA| is 11*(2√5)/5=22√5/5 and the minimum |PA| is1*(2√5)/5=2√5/5.Hence, the answer should be maximum 22√5 /5 and minimum 2√5 /5.But let me confirm this with another method, using Lagrange multipliers.We need to maximize/minimize f(x, y)=2x + y -6 subject to the constraint g(x, y)=x²/4 + y²/9 -1=0.The Lagrangian is L=2x + y -6 -λ(x²/4 + y²/9 -1)Taking partial derivatives:dL/dx=2 -λ*(2x)/4=2 - (λ x)/2=0dL/dy=1 -λ*(2y)/9=1 - (2λ y)/9=0dL/dλ= - (x²/4 + y²/9 -1)=0From first equation:2 - (λ x)/2=0 => λ x=4 => λ=4/xFrom second equation:1 - (2λ y)/9=0 => 2λ y=9 => λ=9/(2y)Therefore, set 4/x=9/(2y) => 8y=9x => y=(9/8)xNow, substitute y=(9/8)x into the constraint x²/4 + y²/9=1:x²/4 + ( (9/8 x)^2 )/9= x²/4 + (81/64 x²)/9 =x²/4 + (9/64)x²= (16/64 +9/64)x²=25/64 x²=1 =>x²=64/25=>x=±8/5Then y=(9/8)x=±(9/8)(8/5)=±9/5Therefore, critical points at (8/5,9/5) and (-8/5,-9/5)Compute f(x,y)=2x + y -6:At (8/5,9/5): 16/5 +9/5 -6=25/5 -6=5 -6=-1At (-8/5,-9/5): -16/5 -9/5 -6= -25/5 -6= -5 -6=-11Therefore, confirming that the maximum of f is -1 and the minimum is -11. Hence, |f(x,y)| has maximum 11 and minimum1.Thus, |PA|= (2√5)/5 * |f(x,y)|, so maximum |PA|=22√5/5 and minimum |PA|=2√5/5.Therefore, the answer to part (2) is that the maximum |PA| is 22√5⁄5 and the minimum is 2√5⁄5.Final Answer(1) The parametric equations of curve ( C ) are ( x = 2costheta ) and ( y = 3sintheta ). The general equation of line ( l ) is ( 2x + y - 6 = 0 ).(2) The maximum value of ( |PA| ) is ( boxed{dfrac{22sqrt{5}}{5}} ) and the minimum value is ( boxed{dfrac{2sqrt{5}}{5}} ).</think>(1) The given curve ( C ) is an ellipse with the equation (frac{x^2}{4} + frac{y^2}{9} = 1). The parametric equations for this ellipse are ( x = 2costheta ) and ( y = 3sintheta ) where (theta) is the parameter. The line ( l ) is given parametrically as ( x = 2 + t ) and ( y = 2 - 2t ). To convert this to the general form, we eliminate the parameter ( t ). From the parametric equations, we get ( t = x - 2 ). Substituting this into the equation for ( y ), we get ( y = 2 - 2(x - 2) ), which simplifies to ( y = -2x + 6 ). Therefore, the general equation of line ( l ) is ( 2x + y - 6 = 0 ).(2) To find the maximum and minimum values of ( |PA| ), we consider the angle between the line ( l ) and the line through point ( P ) on the ellipse. The line ( l ) has a slope of (-2), and the lines forming a ( 30^circ ) angle with ( l ) have slopes ( 8 - 5sqrt{3} ) and ( 8 + 5sqrt{3} ).For any point ( P ) on the ellipse, the distance ( |PA| ) is proportional to the absolute value of the linear function ( 2x + y - 6 ) evaluated at ( P ). Using the parametric equations of the ellipse, we express ( 2x + y - 6 ) as ( 4costheta + 3sintheta - 6 ). The maximum and minimum values of ( 4costheta + 3sintheta ) are ( 5 ) and ( -5 ), leading to the maximum and minimum values of ( 4costheta + 3sintheta - 6 ) being ( -1 ) and ( -11 ). The absolute values are ( 1 ) and ( 11 ).The distance ( |PA| ) is then calculated by multiplying these values by the constant factor ( frac{2sqrt{5}}{5} ). Therefore, the maximum value of ( |PA| ) is ( frac{22sqrt{5}}{5} ) and the minimum value is ( frac{2sqrt{5}}{5} ).Final Answers:(1) The parametric equations of curve ( C ) are ( x = 2costheta ) and ( y = 3sintheta ). The general equation of line ( l ) is ( 2x + y - 6 = 0 ).(2) The maximum value of ( |PA| ) is ( boxed{dfrac{22sqrt{5}}{5}} ) and the minimum value is ( boxed{dfrac{2sqrt{5}}{5}} ).,<think>Okay, so I need to solve this problem with two parts. Let me start with part (1). First, the curve C is given by the equation x²/4 + y²/9 = 1. Hmm, that's an ellipse, right? Because the standard form of an ellipse is x²/a² + y²/b² = 1. Comparing, a² is 4, so a is 2, and b² is 9, so b is 3. So the ellipse is centered at the origin, with semi-major axis 3 along the y-axis and semi-minor axis 2 along the x-axis.They want the parametric equation of curve C. For an ellipse, parametric equations are usually x = a cosθ and y = b sinθ, but wait, sometimes it's written with sine and cosine swapped depending on the orientation. Let me confirm. Since the major axis is along the y-axis, does that affect the parametrization? Actually, no, the parametrization is still x = a cosθ and y = b sinθ, where θ is the parameter. But since the major axis is along y, the ellipse is stretched more in the y-direction. So substituting the values, a = 2 and b = 3, so the parametric equations would be x = 2 cosθ and y = 3 sinθ. Is that right? Let me check. If θ = 0, x = 2, y = 0. If θ = π/2, x = 0, y = 3. That seems correct. So yes, parametric equations for C are x = 2 cosθ, y = 3 sinθ, where θ is the parameter.Now, the line l is given in parametric form: x = 2 + t, y = 2 - 2t. They want the general equation of the line. The general equation is usually Ax + By + C = 0. To convert parametric equations to general form, we can eliminate the parameter t.Given x = 2 + t, so t = x - 2. Substitute into y = 2 - 2t: y = 2 - 2(x - 2) = 2 - 2x + 4 = 6 - 2x. So rearranging, 2x + y - 6 = 0. Let me check: if x = 2 + t, when t = 0, x = 2, y = 2. Plugging into 2x + y - 6: 4 + 2 - 6 = 0. Correct. When t = 1, x = 3, y = 0. Then 2*3 + 0 -6 = 0. Correct. So the general equation is 2x + y -6 = 0.So part (1) is done. The parametric equations for C are x = 2 cosθ, y = 3 sinθ, and the general equation for line l is 2x + y = 6.Now moving on to part (2). The problem says: Draw a line through any point P on the curve C that forms a 30° angle with the line l, intersecting l at point A. Find the maximum and minimum values of |PA|.Okay, so for any point P on the ellipse C, we need to draw a line that makes a 30° angle with line l, and this line will intersect l at point A. Then we need to find the maximum and minimum distances PA.First, let me visualize this. Line l is 2x + y = 6. It has a slope of -2. The ellipse is centered at the origin, stretching up to (2,0), (-2,0), (0,3), (0,-3). For any point P on the ellipse, we draw a line from P that makes a 30° angle with line l, and intersects l at A. Then we need the length PA.So the problem is to parametrize this distance PA as a function of the point P on the ellipse and find its extrema.First, let me think about the angle between two lines. The angle θ between two lines with slopes m1 and m2 is given by tanθ = |(m2 - m1)/(1 + m1m2)|. Here, line l has slope m1 = -2. The other line, PA, makes a 30° angle with l. So the angle between PA and l is 30°, so tan(30°) = |(m - m1)/(1 + m m1)|, where m is the slope of PA.So tan(30°) = 1/√3 = |(m + 2)/(1 - 2m)|. So solving for m, there will be two possible slopes since the angle could be on either side of the line l.So setting up the equation:1/√3 = |(m + 2)/(1 - 2m)|This gives two possibilities:1) (m + 2)/(1 - 2m) = 1/√32) (m + 2)/(1 - 2m) = -1/√3Solving both equations for m.First case:(m + 2) = (1 - 2m)/√3Multiply both sides by √3:√3 m + 2√3 = 1 - 2mBring all terms to left:√3 m + 2m + 2√3 - 1 = 0m(√3 + 2) + (2√3 - 1) = 0Thus,m = -(2√3 - 1)/(√3 + 2)Multiply numerator and denominator by (√3 - 2) to rationalize denominator:Numerator: -(2√3 -1)(√3 - 2) = -[2√3*√3 -4√3 -1*√3 +2] = -[6 -4√3 -√3 +2] = -[8 -5√3] = -8 +5√3Denominator: (√3 +2)(√3 -2) = 3 -4 = -1Thus, m = (-8 +5√3)/(-1) = 8 -5√3Second case:(m + 2)/(1 - 2m) = -1/√3Multiply both sides by √3:√3(m + 2) = - (1 - 2m)Expand:√3 m + 2√3 = -1 + 2mBring all terms to left:√3 m -2m + 2√3 +1 = 0m(√3 -2) + (2√3 +1) = 0Thus,m = -(2√3 +1)/(√3 -2)Again, rationalize denominator by multiplying numerator and denominator by (√3 +2):Numerator: -(2√3 +1)(√3 +2) = -[2√3*√3 +4√3 +1*√3 +2] = -[6 +5√3 +2] = -[8 +5√3]Denominator: (√3 -2)(√3 +2) = 3 -4 = -1Thus, m = (-8 -5√3)/(-1) =8 +5√3Therefore, the two possible slopes for line PA are m1 =8 -5√3 and m2=8 +5√3.Wait, these are quite steep slopes. Let me check if these values make sense. Let me compute approximate numerical values. √3 ≈1.732.So for m1=8 -5*1.732≈8 -8.66≈-0.66. So m1≈-0.66For m2=8 +5*1.732≈8 +8.66≈16.66. So m2≈16.66.So one slope is approximately -0.66, the other is very steep, around 16.66. That seems possible. The original line l has a slope of -2, so a line making 30° with it could have a shallower or steeper slope.But I need to confirm if these slopes are correct. Let's take m1≈-0.66 and check the angle with line l (slope -2). The angle between them should be 30°.Compute tanθ = |(m - m_l)/(1 + m*m_l)| = |(-0.66 - (-2))/(1 + (-0.66)(-2))| = |1.34/(1 +1.32)| = |1.34/2.32|≈0.577, which is approximately 1/√3≈0.577, so tanθ≈1/√3, so θ≈30°, correct. Similarly for m2≈16.66:tanθ = |(16.66 - (-2))/(1 +16.66*(-2))| = |18.66/(1 -33.32)|≈|18.66/-32.32|≈0.577, same result. So yes, both slopes are correct.Therefore, for any point P on the ellipse, the line PA can have two possible directions (slopes) forming 30° with line l. However, the problem states "a line through any point P on the curve C that forms a 30° angle with the line l, intersecting l at point A". So perhaps for each P, there are two such lines, but maybe the problem considers only one? Or maybe we have to consider both possibilities? Hmm. The wording is "a line", but maybe for each P, there are two lines? Wait, but depending on the position of P, maybe only one line is possible? Hmm, need to think.Alternatively, perhaps for each point P, there are two possible lines PA making 30° with l, but intersecting l at point A. However, depending on the location of P, one of them might not intersect l, or both might intersect. Maybe both intersect, but at different points A. So maybe for each P, there are two points A, and two distances PA. Then, when we need to find the maximum and minimum of |PA|, we have to consider both possibilities. Alternatively, maybe only one line exists? Hmm.Wait, perhaps the problem is not considering directionality. Since 30° is a specific angle, the lines could be on either side of line l. But given that the line PA is drawn from P to intersect l, the line must intersect l. So depending on the position of P, the lines with the given slopes might or might not intersect l. Wait, but since line l is infinite, any line with a different slope should intersect l at exactly one point. So for each P, we can draw two lines through P with slopes 8 -5√3 and 8 +5√3, each intersecting l at a unique point A. Therefore, for each P, there are two points A, and two distances |PA|.But the problem says "Draw a line through any point P on the curve C that forms a 30° angle with the line l, intersecting l at point A". So maybe for each P, we have two such lines, but the problem might be considering both possibilities. Therefore, when finding the maximum and minimum |PA|, we need to consider both lines for each P and then find the overall max and min.Alternatively, perhaps there is a unique line for each P? Wait, no. If we fix the angle between PA and l as 30°, then there are two possible lines through P making that angle with l. So two lines. So two points A per P.Therefore, the problem is to find, over all P on C and both possible lines, the maximum and minimum |PA|.So to model this, for a general point P on the ellipse, parametrized by θ, coordinates (2cosθ, 3sinθ), we can write equations for the two lines passing through P with slopes m1 and m2 (8 -5√3 and 8 +5√3). Each of these lines will intersect line l at point A. Then, compute |PA| for each case, and find the maximum and minimum over all θ.Alternatively, maybe there's a smarter way to compute |PA| in terms of θ, and then optimize.Alternatively, perhaps there's a formula for the distance from a point to a line with a given angle. Wait, but PA is not the distance from P to l, but the distance along a line that makes 30° with l. So it's different.Wait, if we consider the angle between PA and l is 30°, then perhaps we can use the formula involving the distance from P to l divided by sinθ.Wait, in the triangle PAl, if angle at A is 30°, then |PA| = |distance from P to l| / sin(30°). Wait, is that correct?Wait, suppose we have point P and line l. The distance from P to l is d. Then, if we draw a line from P to l such that the angle between PA and l is θ, then in the right triangle formed by P, A, and the foot of the perpendicular from P to l, the angle at A is θ. Then, the length PA would be d / sinθ.Wait, let me visualize. Let’s denote Q as the foot of the perpendicular from P to l. Then triangle PQA is a right triangle with right angle at Q. The angle at A is θ =30°, so sinθ = PQ / PA => PA = PQ / sinθ. Since PQ is the distance from P to l, so PA = d / sinθ. Therefore, PA = d / sin30° = 2d.But wait, if this is the case, then |PA| is twice the distance from P to l. But that seems too straightforward. Wait, but does this hold?Wait, in this case, if the angle between PA and l is 30°, then in the right triangle PQA, angle at A is 30°, so PQ is opposite the angle, and PA is the hypotenuse. So sin30° = PQ / PA => PA = PQ / sin30° = 2 PQ.But PQ is the distance from P to l. So PA = 2 * distance from P to l.But if that's the case, then the maximum and minimum of PA would just be twice the maximum and minimum of the distance from P to l.But the problem states that we draw a line through P forming a 30° angle with l, intersecting l at A. So according to this, PA is always twice the distance from P to l. But is that correct?Wait, but angle between PA and l is 30°, which is the angle between the two lines. However, in the formula above, I considered the angle at A in the right triangle, but maybe the angle between PA and l is not the same as the angle in the right triangle.Wait, let's clarify. The angle between two lines is defined as the angle between their direction vectors. If line PA has an angle of 30° with line l, then the angle between their direction vectors is 30°. However, in the right triangle, angle at A is between PA and QA (the foot). So it's not necessarily the same as the angle between PA and l.Therefore, my previous assumption might be incorrect.Alternatively, maybe we need to use the formula involving the distance from P to l and the angle between PA and l.Let me recall that if a line PA makes an angle θ with another line l, then the distance from P to l is |PA| * sinθ. Therefore, |PA| = distance / sinθ. That formula might be more accurate.Wait, if two lines intersect at point A, forming an angle θ, then the distance from P to l is |PA| * sinθ. Therefore, |PA| = distance / sinθ. So yes, that would make sense.Given that, if θ is 30°, then |PA| = distance / sin30° = 2 * distance.Therefore, |PA| is twice the distance from P to l. Therefore, maximizing/minimizing |PA| is equivalent to maximizing/minimizing twice the distance from P to l.But if that's the case, then the answer would be 2 times the maximum and minimum distances from points on the ellipse to the line l.But that seems too straightforward. However, let's check if this reasoning is correct.Suppose that the angle between PA and l is 30°, so using the formula: the distance from P to l is |PA| * sinθ. Therefore, |PA| = distance / sinθ.Therefore, since θ is 30°, sinθ = 0.5, so |PA| = 2 * distance.Hence, if that formula holds, then indeed |PA| is twice the distance from P to l. Therefore, the maximum and minimum |PA| would just be twice the maximum and minimum distances from P to l.But is this applicable here?Wait, the formula comes from the scenario where you have a point P, a line l, and a line PA intersecting l at A with angle θ between PA and l. Then the distance from P to l is |PA| * sinθ.Yes, that's a standard formula. The distance from a point to a line can be expressed as the length of the projection of PA onto the direction perpendicular to l. Therefore, since the angle between PA and l is θ, then the perpendicular distance is |PA| * sinθ. Therefore, |PA| = distance / sinθ.Therefore, in this problem, since θ is 30°, |PA| = 2 * distance from P to l. Hence, the maximum |PA| is 2 times the maximum distance from any point P on the ellipse to line l, and similarly the minimum is 2 times the minimum distance.Therefore, if we can compute the maximum and minimum distances from the ellipse to the line l, then multiply by 2, we get the desired maximum and minimum |PA|.But is this correct? Let me check with a simple example.Suppose l is the x-axis, and P is a point at (0,1). Then the distance from P to l is 1. If we draw a line from P making 30° with the x-axis, then the length PA would be 1 / sin30° = 2. That makes sense, because PA would be the hypotenuse of a 30-60-90 triangle, with opposite side 1, hypotenuse 2.Therefore, in this case, |PA| is indeed 2 * distance. So yes, the formula holds.Therefore, in our problem, |PA| = 2 * distance from P to l. Therefore, the problem reduces to finding the maximum and minimum distances from points on the ellipse C to the line l, then doubling them.Therefore, part (2) simplifies to finding the maximum and minimum distances from the ellipse x²/4 + y²/9 = 1 to the line 2x + y -6 = 0, then multiplying by 2.So how do we find the maximum and minimum distances from the ellipse to the line?The distance from a point (x, y) to the line 2x + y -6 =0 is |2x + y -6| / sqrt(2² +1²) = |2x + y -6| / sqrt(5).Therefore, we need to maximize and minimize |2x + y -6| / sqrt(5) subject to x²/4 + y²/9 =1.Since sqrt(5) is a positive constant, maximizing/minimizing |2x + y -6| is equivalent to maximizing/minimizing |2x + y -6| on the ellipse.Therefore, the problem reduces to finding the extrema of the function f(x, y) = 2x + y -6 on the ellipse x²/4 + y²/9 =1, then taking absolute value, and finally dividing by sqrt(5) and multiplying by 2.But since we take absolute value, the maximum of |f(x, y)| will occur either at the maximum of f(x, y) or the minimum of f(x, y). Therefore, we need to find the maximum and minimum of f(x, y) on the ellipse, then compute |f_max -6| and |f_min -6|, but wait, actually f(x,y) =2x + y -6, so the expression inside the absolute value is 2x + y -6. So we need to maximize and minimize 2x + y -6 over the ellipse.Alternatively, perhaps first, find the maximum and minimum of 2x + y on the ellipse, then subtract 6, and take absolute value. Wait, but since we are taking absolute value, it's possible that the maximum |2x + y -6| occurs at a point where 2x + y -6 is maximized or minimized, but also possibly at a point where 2x + y -6 is closest to zero but on the other side. However, since the ellipse is a closed bounded curve, the extrema of 2x + y -6 will be attained, and the maximum absolute value will be the maximum of |f_max| and |f_min|, where f_max is the maximum of 2x + y -6 and f_min is the minimum.But let's first find the extrema of 2x + y on the ellipse. Let's set up the Lagrangian.We need to maximize/minimize f(x, y) = 2x + y subject to the constraint g(x, y) = x²/4 + y²/9 -1 =0.The Lagrangian is L = 2x + y - λ(x²/4 + y²/9 -1).Taking partial derivatives:∂L/∂x = 2 - λ*(2x)/4 = 2 - λx/2 =0∂L/∂y = 1 - λ*(2y)/9 = 1 - (2λy)/9 =0∂L/∂λ = -(x²/4 + y²/9 -1) =0So from first equation: 2 = λx/2 => λ =4/xFrom second equation:1 = (2λy)/9 => λ = 9/(2y)Therefore, 4/x =9/(2y) => 8y =9x => y = (9/8)xNow, substitute y = (9/8)x into the ellipse equation:x²/4 + [(9/8)x]^2 /9 =1Compute:x²/4 + (81/64)x² /9 = x²/4 + (9/64)x² = x²(1/4 +9/64)=x²(16/64 +9/64)=x²(25/64)=1Thus, x²=64/25 => x=±8/5Then y= (9/8)x= (9/8)(±8/5)= ±9/5Therefore, the extrema occur at points (8/5, 9/5) and (-8/5, -9/5)Compute f(x, y)=2x + y:At (8/5,9/5): 2*(8/5) +9/5=16/5 +9/5=25/5=5At (-8/5,-9/5):2*(-8/5)+ (-9/5)= -16/5 -9/5= -25/5= -5Therefore, the maximum of 2x + y on the ellipse is 5, and the minimum is -5.Therefore, 2x + y -6 takes maximum value 5 -6 = -1, and minimum value -5 -6 = -11.Wait, but f(x,y)=2x + y -6, so when 2x + y is 5, f=5 -6 = -1, and when 2x + y is -5, f= -5 -6= -11. So the maximum value of f(x,y) is -1, and the minimum is -11.But we need to take absolute value |f(x,y)|= |2x + y -6|. So | -1 | =1 and | -11 | =11. Therefore, the maximum |2x + y -6| on the ellipse is 11, and the minimum is 1.Wait, but is that the case? Wait, because f(x,y) ranges from -11 to -1. So |f(x,y)| ranges from 1 to 11. So the maximum absolute value is 11, the minimum is 1.Therefore, the maximum |2x + y -6| is 11, the minimum is 1. Therefore, the maximum distance from a point on the ellipse to line l is 11 / sqrt(5), and the minimum distance is 1 / sqrt(5).Therefore, |PA| = 2 * distance, so maximum |PA| is 2*(11 / sqrt(5)) =22 / sqrt(5)= (22√5)/5≈9.83, and minimum |PA| is 2*(1 / sqrt(5))=2 / sqrt(5)= (2√5)/5≈0.894.But wait, but hold on. The distance formula is |2x + y -6| / sqrt(5). But the function 2x + y -6 reaches maximum of -1 and minimum of -11. Therefore, |2x + y -6| reaches maximum of 11 and minimum of 1. Therefore, the maximum distance is 11 / sqrt(5) and minimum is1 / sqrt(5). Therefore, |PA|=2*distance, so maximum is 22/sqrt(5), minimum is 2/sqrt(5).But this is under the assumption that the angle between PA and l is 30°, which leads to |PA|=2*distance. However, in reality, is this the case?Wait, but let's test with the example earlier. If we have a point P where the distance to l is 11/sqrt(5), then |PA|=22/sqrt(5). But is that possible? Let's check if such a point exists.Wait, the point where 2x + y -6 = -11 is the point (-8/5, -9/5). Let me compute the distance from this point to line l:Distance = |2*(-8/5) + (-9/5) -6| / sqrt(5) = |(-16/5 -9/5 -30/5)| / sqrt(5) = |(-55/5)| / sqrt(5) = | -11 | / sqrt(5) =11 / sqrt(5). So that's correct. Then |PA|=2*distance=22/sqrt(5).Similarly, the point (8/5,9/5) has distance |2*(8/5)+9/5 -6| / sqrt(5)= |25/5 -6|= |5 -6|=1. So distance=1/sqrt(5), so |PA|=2/sqrt(5). So these check out.Therefore, the maximum |PA| is22/sqrt(5), and the minimum is2/sqrt(5). Rationalizing the denominators, 22/sqrt(5)= (22√5)/5, and 2/sqrt(5)= (2√5)/5.Therefore, the answer is maximum |PA|=22√5/5 and minimum |PA|=2√5/5.But wait, but the problem says "the line through any point P on the curve C that forms a 30° angle with the line l, intersecting l at point A". However, according to our reasoning, for each point P, the |PA| is twice the distance from P to l. But does this hold for all points P? Because in the previous example with P=(0,1) and l being x-axis, we saw that |PA|=2*distance. But in that case, the line PA was unique. However, in our problem, earlier we found two possible slopes for PA, but according to this formula, |PA|=2*distance, which would be same for both lines? That seems contradictory.Wait, perhaps my initial assumption is wrong. Let me think again.If from point P, there are two lines PA1 and PA2 making 30° with l, intersecting l at A1 and A2, then each of these lines would correspond to different distances |PA1| and |PA2|.But according to the previous formula, |PA| = distance / sinθ, which would imply that for a fixed θ, |PA| is fixed once the distance is fixed. However, in reality, the two different lines through P making angle θ with l would result in different distances.Wait, but in the formula, the angle θ is between PA and l. If we have two different lines through P making angle θ with l, then the distances PA1 and PA2 would be different. But according to the formula, if the angle is θ, then |PA| = distance / sinθ. But distance is the minimal distance from P to l, which is fixed. Therefore, this suggests that |PA| is fixed. But this contradicts the idea that two different lines PA1 and PA2 would have different lengths.Therefore, there must be a mistake in my reasoning.Wait, perhaps the formula |PA| = distance / sinθ is only applicable when PA is the line making angle θ with l and lying on one side of the perpendicular from P to l. But when you have two different lines through P making angle θ with l, one on each side of the perpendicular, then the distances PA1 and PA2 would be different.Wait, let me draw a diagram mentally. Suppose P is above line l. The minimal distance from P to l is PQ, the perpendicular. If we draw two lines from P making 30° with l, one to the left and one to the right of PQ. Then each of these lines will have length PA1 and PA2, which can be calculated based on the angle between PA1 and PQ.Wait, but if the angle between PA1 and l is 30°, then the angle between PA1 and PQ is 30° + the angle between l and PQ. Wait, no, PQ is perpendicular to l, so the angle between l and PQ is 90°. Wait, no, PQ is the perpendicular, so it's at 90° to l.Therefore, if PA1 makes 30° with l, then the angle between PA1 and PQ is 90° -30°=60°. Similarly, the other line PA2 would make 30° with l on the other side, so angle with PQ is 90° +30°=120°.In that case, in triangle PQA1, angle at P is 60°, PQ is the height, so |PA1|= PQ / sin(60°)= distance / (√3/2)= 2 distance /√3Similarly, for PA2, angle at P is 120°, so |PA2|= PQ / sin(120°)= distance / (√3/2)=2 distance /√3. Wait, so both lengths are the same?Wait, that contradicts the idea that they are different. Hmm, but in reality, if you have two lines from P making 30° with l, then the lengths PA1 and PA2 should be different. But according to this, they are same.Wait, perhaps in this case, since the angle between PA and l is 30°, regardless of the side, the length PA is the same. But in reality, if you have a point P not on l, then drawing two lines from P making the same angle with l on either side would result in two different intersection points A1 and A2, but the distances PA1 and PA2 would be equal?Wait, let's take an example. Let’s take line l as x-axis, P at (0,1). Draw two lines from P making 30° with x-axis, one going to the right, one to the left. The intersection points A1 and A2 would be at different distances.For the line going to the right: slope is tan(30°)=1/√3. Equation: y= (1/√3)x +1. Intersection with x-axis (y=0): 0= (1/√3)x +1 => x= -√3. So A1=(-√3,0). Distance PA1=√[ (√3)^2 +1^2 ]=√(3 +1)=2.For the line going to the left: slope is tan(150°)= -1/√3. Equation: y= (-1/√3)x +1. Intersection with x-axis: 0= (-1/√3)x +1 => x=√3. So A2=(√3,0). Distance PA2=√[ (√3)^2 +1^2 ]=2.So both distances PA1 and PA2 are equal to 2. Therefore, even though the lines are on different sides, the distances are the same. Therefore, in this case, |PA| is indeed 2*distance, where distance is 1. So |PA|=2*1=2, which matches.Therefore, in this case, even though there are two different lines, the distance PA is the same. Therefore, in general, for any point P, even though there are two lines through P making 30° with l, the distances PA are equal. Therefore, |PA|=2*distance from P to l. Therefore, the maximum and minimum values of |PA| correspond to twice the maximum and minimum distances from P to l.Therefore, the previous reasoning was correct. Therefore, the answer is maximum |PA|=22√5/5 and minimum |PA|=2√5/5.But let me check with another example. Take point (-8/5, -9/5). Distance from this point to line l is |2*(-8/5) + (-9/5) -6| / sqrt(5) = |(-16/5 -9/5 -30/5)| / sqrt(5) = |-55/5| / sqrt(5)=11/sqrt(5). Then |PA|=2*11/sqrt(5)=22/sqrt(5). Which is the maximum.Similarly, point (8/5,9/5) has distance |2*(8/5)+9/5 -6| / sqrt(5)= |25/5 -6| / sqrt(5)= |5-6| / sqrt(5)=1/sqrt(5). Then |PA|=2/sqrt(5). Therefore, these points give the maximum and minimum |PA|.Hence, the conclusion is that the maximum |PA| is22√5/5 and the minimum is2√5/5.But just to be thorough, let me confirm that there are no other points on the ellipse where |PA| could be larger or smaller. For example, is there a point where |2x + y -6| is larger than 11 or smaller than 1?But according to the Lagrangian method, the extrema of 2x + y on the ellipse are 5 and -5, so 2x + y -6 ranges from -11 to -1, hence |2x + y -6| ranges from1 to11. Therefore, no, there are no points on the ellipse where |2x + y -6| exceeds 11 or is less than1. Therefore, our previous calculation is correct.Thus, the maximum |PA| is22√5/5 and minimum is2√5/5.Therefore, the final answers for part (2) are maximum 22√5⁄5 and minimum 2√5⁄5.But the problem says "Find the maximum and minimum values of |PA|". So we can write them as boxed expressions.So summarizing:(1) Parametric equation of C: x = 2 cosθ, y = 3 sinθ; general equation of l: 2x + y -6 =0.(2) Maximum |PA|: 22√5⁄5, minimum |PA|:2√5⁄5.Final Answer(1) The parametric equation of ( C ) is ( begin{cases} x = 2costheta y = 3sintheta end{cases} ), and the general equation of ( l ) is ( 2x + y - 6 = 0 ).(2) The maximum value of ( |PA| ) is ( boxed{dfrac{22sqrt{5}}{5}} ) and the minimum value is ( boxed{dfrac{2sqrt{5}}{5}} ).</think>(1) The given curve ( C ) is an ellipse with the equation (frac{x^2}{4} + frac{y^2}{9} = 1). The parametric equations for this ellipse are:[begin{cases}x = 2costheta y = 3sinthetaend{cases}]where (theta) is the parameter.The line ( l ) is given parametrically as ( begin{cases} x = 2 + t y = 2 - 2t end{cases} ). Converting this to the general form, we eliminate the parameter ( t ). From ( x = 2 + t ), we get ( t = x - 2 ). Substituting this into ( y = 2 - 2t ) gives ( y = 2 - 2(x - 2) = 6 - 2x ). Thus, the general equation of the line ( l ) is:[2x + y - 6 = 0](2) To find the maximum and minimum values of ( |PA| ), we note that the distance from any point ( P ) on the ellipse to the line ( l ) is given by ( frac{|2x + y - 6|}{sqrt{5}} ). The line through ( P ) forming a ( 30^circ ) angle with ( l ) intersects ( l ) at point ( A ). The length ( |PA| ) is twice the distance from ( P ) to ( l ) because ( |PA| = frac{text{distance}}{sin(30^circ)} = 2 times text{distance} ).We need to maximize and minimize ( |2x + y - 6| ) on the ellipse. Using the method of Lagrange multipliers, we find the extrema of ( 2x + y ) on the ellipse. The maximum value of ( 2x + y ) is 5 and the minimum value is -5. Therefore, ( |2x + y - 6| ) ranges from 1 to 11. The corresponding distances are ( frac{11}{sqrt{5}} ) and ( frac{1}{sqrt{5}} ). Multiplying by 2 gives the lengths ( |PA| ):- Maximum ( |PA| = 2 times frac{11}{sqrt{5}} = frac{22sqrt{5}}{5} )- Minimum ( |PA| = 2 times frac{1}{sqrt{5}} = frac{2sqrt{5}}{5} )Thus, the maximum and minimum values of ( |PA| ) are:[boxed{dfrac{22sqrt{5}}{5}} quad text{and} quad boxed{dfrac{2sqrt{5}}{5}}]]

🔑:(1) For the curve C: frac{x^{2}}{4} + frac{y^{2}}{9} = 1, let x = 2cos{theta} and y = 3sin{theta}.Thus, the parametric equation of the curve C is begin{cases} x = 2cos{theta} y = 3sin{theta} end{cases} (theta is a parameter).For the line l: begin{cases} x = 2 + t y = 2 - 2t end{cases}, solving for t in the first equation, we get t = x - 2. Substituting this into the second equation and simplifying, we obtain 2x + y - 6 = 0.(2) Let P(2cos{theta}, 3sin{theta}) be any point on the curve C. The distance d from point P to the line l is given by d = frac{sqrt{5}}{5}|4cos{theta} + 3sin{theta} - 6|.Then, |PA| = frac{d}{sin{30^{circ}}} = frac{2sqrt{5}}{5}|5sin{(theta + alpha)} - 6|, where alpha is an acute angle.When sin{(theta + alpha)} = -1, |PA| attains its maximum value, which is boxed{frac{22sqrt{5}}{5}}.When sin{(theta + alpha)} = 1, |PA| attains its minimum value, which is boxed{frac{2sqrt{5}}{5}}.

🧠:[<think>Okay, so I have this sequence problem here. The sequence is defined by a₁ = 1, and then each subsequent term is the previous term plus 2 to the power of n. So, a_{n+1} - a_n = 2ⁿ for all positive integers n. The question is asking for the general formula of a_n. Hmm, let's see how to approach this.First, I remember that when dealing with sequences defined by a recurrence relation like this, where each term is based on the previous one, it's often useful to look for a pattern by writing out the first few terms. Maybe that can help me spot a trend or even guess the formula, which I can then try to prove using mathematical induction or another method.Starting with the first term, they give me a₁ = 1. Let's compute the next few terms using the recurrence relation. The formula says that a_{n+1} = a_n + 2ⁿ. So, each term is the previous term plus 2 raised to the current index. Let's write them down:- a₁ = 1 (given)- a₂ = a₁ + 2¹ = 1 + 2 = 3- a₃ = a₂ + 2² = 3 + 4 = 7- a₄ = a₃ + 2³ = 7 + 8 = 15- a₅ = a₄ + 2⁴ = 15 + 16 = 31- a₆ = a₅ + 2⁵ = 31 + 32 = 63Hmm, looking at these numbers: 1, 3, 7, 15, 31, 63... These numbers look familiar. Each term is one less than a power of 2. For example, 3 is 4 - 1, which is 2² - 1; 7 is 8 - 1, which is 2³ - 1; 15 is 16 - 1, which is 2⁴ - 1; and so on. So, a_n seems to be 2ⁿ - 1. Let's check:For n = 1: 2¹ - 1 = 2 - 1 = 1, which matches a₁ = 1.For n = 2: 2² - 1 = 4 - 1 = 3, which matches a₂ = 3.For n = 3: 2³ - 1 = 8 - 1 = 7, which matches a₃ = 7.Okay, so this seems to hold up for the first few terms. But how can I be sure this formula works for all n? Maybe I should try to derive it more formally instead of just observing the pattern.The recurrence relation given is a_{n+1} - a_n = 2ⁿ. This is a linear recurrence relation, and such relations can often be solved by summing up the differences. Since each term is built by adding 2ⁿ to the previous term, the general term a_n can be expressed as the initial term plus the sum of all the differences up to n-1.In other words, since a_{n} = a₁ + (a₂ - a₁) + (a₃ - a₂) + ... + (a_{n} - a_{n-1}).Given that a₁ = 1, and each difference (a_{k+1} - a_k) = 2ᵏ for k from 1 to n-1, then:a_n = 1 + Σ_{k=1}^{n-1} 2ᵏNow, I need to compute that sum. The sum of a geometric series Σ_{k=0}^{m} arᵏ is a(r^{m+1} - 1)/(r - 1). But in this case, the sum starts at k=1 instead of k=0, so I need to adjust for that.First, note that Σ_{k=1}^{n-1} 2ᵏ = Σ_{k=0}^{n-1} 2ᵏ - 2⁰ = (2^{n} - 1)/(2 - 1) - 1 = (2^{n} - 1) - 1 = 2^{n} - 2.Wait, let me verify that. The sum from k=0 to m of 2ᵏ is 2^{m+1} - 1. So, the sum from k=1 to m is (2^{m+1} - 1) - 2⁰ = 2^{m+1} - 2. Therefore, substituting m = n-1, the sum from k=1 to n-1 is 2^{n} - 2.Therefore, a_n = 1 + (2^{n} - 2) = 2^{n} - 1.So, that gives the formula a_n = 2ⁿ - 1. Which matches the pattern we observed earlier. Therefore, the general term is 2ⁿ minus 1.Let me check if this formula works for the terms we computed:For n = 4: 2⁴ - 1 = 16 - 1 = 15. Which matches a₄ = 15.For n = 5: 2⁵ - 1 = 32 - 1 = 31. Which matches a₅ = 31.For n = 6: 2⁶ - 1 = 64 - 1 = 63. Which matches a₆ = 63.Great, seems consistent. Let me also test for n = 1 again, which is 2¹ - 1 = 1. Correct. So, this seems to hold.Alternatively, maybe I can use mathematical induction to prove that the formula holds for all positive integers n.Base case: n = 1. a₁ = 1 = 2¹ - 1. So, holds.Inductive step: Assume that for some k ≥ 1, a_k = 2ᵏ - 1. Then, we need to show that a_{k+1} = 2^{k+1} - 1.From the recurrence relation, a_{k+1} = a_k + 2ᵏ.Substituting the inductive hypothesis: a_{k+1} = (2ᵏ - 1) + 2ᵏ = 2*2ᵏ - 1 = 2^{k+1} - 1.Which is exactly the formula we need. Hence, by induction, the formula holds for all positive integers n.Therefore, the general term a_n is 2ⁿ - 1.Wait, but let me think again. Sometimes when sequences have a recursive definition, there can be off-by-one errors. Let me check the indices again.The problem says a_{n+1} - a_n = 2ⁿ. So, for each n, the difference between the (n+1)th term and the nth term is 2ⁿ. So, when building up a_n, the term a_{n} is a₁ plus the sum from k=1 to n-1 of 2ᵏ. Because:a₂ = a₁ + 2¹a₃ = a₂ + 2² = a₁ + 2¹ + 2²...a_n = a₁ + 2¹ + 2² + ... + 2^{n-1}So, the sum is from k=1 to k=n-1 of 2ᵏ. Which, as we computed earlier, is 2ⁿ - 2. Therefore, a_n = 1 + (2ⁿ - 2) = 2ⁿ - 1. So, that's correct.Alternatively, the sum from k=1 to m of 2ᵏ is 2^{m+1} - 2. If m = n-1, then the sum is 2^{(n-1)+1} - 2 = 2ⁿ - 2. Then, adding 1 gives 2ⁿ -1. Yep.So, all the steps check out. Therefore, the formula a_n = 2ⁿ -1 is correct.I think that's solid. The induction proof confirms it, the term-by-term calculation matches, and the summation approach also leads to the same result. So, I'm confident the answer is 2ⁿ -1.Final AnswerThe general term of the sequence is boxed{2^n - 1}.</think>Given the sequence ({a_n}) with (a_1 = 1) and the recurrence relation (a_{n+1} - a_n = 2^n) for (n in mathbb{N}^+), we need to find the general formula for (a_n).First, we compute the initial terms of the sequence:- (a_1 = 1)- (a_2 = a_1 + 2^1 = 1 + 2 = 3)- (a_3 = a_2 + 2^2 = 3 + 4 = 7)- (a_4 = a_3 + 2^3 = 7 + 8 = 15)- (a_5 = a_4 + 2^4 = 15 + 16 = 31)- (a_6 = a_5 + 2^5 = 31 + 32 = 63)Observing these terms, we notice that each term (a_n) is one less than a power of 2. We hypothesize that (a_n = 2^n - 1).To verify this, we use the recurrence relation and sum the differences:[ a_n = a_1 + sum_{k=1}^{n-1} 2^k ]The sum of the geometric series (sum_{k=1}^{n-1} 2^k) is calculated as follows:[ sum_{k=1}^{n-1} 2^k = 2^n - 2 ]Adding this to the initial term (a_1 = 1):[ a_n = 1 + (2^n - 2) = 2^n - 1 ]We confirm this formula using mathematical induction:- Base case: For (n = 1), (a_1 = 2^1 - 1 = 1), which is correct.- Inductive step: Assume (a_k = 2^k - 1) for some (k geq 1). Then, [ a_{k+1} = a_k + 2^k = (2^k - 1) + 2^k = 2^{k+1} - 1 ]Thus, the formula holds for all (n in mathbb{N}^+).The general term of the sequence is (boxed{2^n - 1}).,<think>Alright, let me try to figure out this sequence problem. So, we have a sequence {a_n} where the first term, a₁, is 1. Then, each subsequent term is defined by the recurrence relation a_{n+1} - a_n = 2^n for all positive integers n. The question is asking for a general formula for a_n. Hmm, okay, let's break this down step by step.First, I know that recurrence relations can sometimes be solved by unfolding the sequence or recognizing a pattern. Since the difference between consecutive terms is given as 2^n, maybe I can express a_n in terms of the sum of these differences plus the initial term. That makes sense because each term is built by adding the previous term and the difference. So, starting from a₁, each a_{n} would be a₁ plus the sum of all the differences up to the (n-1)th term. Let me write that out.For example, a₂ = a₁ + 2¹a₃ = a₂ + 2² = a₁ + 2¹ + 2²a₄ = a₃ + 2³ = a₁ + 2¹ + 2² + 2³And so on. So, in general, a_n = a₁ + Σ_{k=1}^{n-1} 2^kSince a₁ is 1, that becomes:a_n = 1 + Σ_{k=1}^{n-1} 2^kNow, I need to compute that sum. The sum of a geometric series Σ_{k=0}^{m} r^k = (r^{m+1} - 1)/(r - 1). But here, our sum starts at k=1 instead of k=0. Let's adjust the formula accordingly.Σ_{k=1}^{n-1} 2^k = Σ_{k=0}^{n-1} 2^k - 2^0 = (2^{n} - 1)/(2 - 1) - 1 = (2^{n} - 1) - 1 = 2^{n} - 2Wait, let me verify that. If I take the sum from k=0 to m, it's 2^{m+1} - 1. So, if we start from k=1 to m, it's (2^{m+1} - 1) - 2^0 = 2^{m+1} - 2. So in our case, the upper limit is n-1. So substituting m = n-1:Σ_{k=1}^{n-1} 2^k = 2^{(n-1)+1} - 2 = 2^{n} - 2So, plugging that back into the expression for a_n:a_n = 1 + (2^{n} - 2) = 2^{n} - 1Wait, but hold on a second. Let me check this with the first few terms to make sure. Let's compute a₂, a₃, a₄ using both the recurrence and the formula.Using the recurrence:a₁ = 1a₂ = a₁ + 2¹ = 1 + 2 = 3a₃ = a₂ + 2² = 3 + 4 = 7a₄ = a₃ + 2³ = 7 + 8 = 15a₅ = a₄ + 2⁴ = 15 + 16 = 31Using the formula a_n = 2^n - 1:a₁ = 2¹ - 1 = 2 - 1 = 1 ✔️a₂ = 2² - 1 = 4 - 1 = 3 ✔️a₃ = 2³ - 1 = 8 - 1 = 7 ✔️a₄ = 2⁴ - 1 = 16 - 1 = 15 ✔️a₅ = 2⁵ - 1 = 32 - 1 = 31 ✔️Looks good! So the formula works for these terms. Therefore, the general term a_n is 2^n minus 1. But let me go through the steps again to ensure there's no mistake in the derivation.Starting with a_{n+1} - a_n = 2^n. This is a linear recurrence relation. To solve it, we can use the method for nonhomogeneous recurrence relations, but in this case, it might be straightforward since the difference itself is given.If we write out the terms:a₁ = 1a₂ = a₁ + 2¹a₃ = a₂ + 2² = a₁ + 2¹ + 2²...a_n = a₁ + Σ_{k=1}^{n-1} 2^kAs established earlier, the sum Σ_{k=1}^{n-1} 2^k is a geometric series with ratio 2, starting from k=1 to n-1. The sum of such a series is 2(2^{n-1} - 1)/(2 - 1) = 2^{n} - 2. Wait, hold on, let me do that again.The sum from k=1 to m of 2^k is 2^{m+1} - 2. So if m = n-1, then the sum is 2^{(n-1)+1} - 2 = 2^{n} - 2. Therefore, adding that to a₁ which is 1:a_n = 1 + 2^{n} - 2 = 2^{n} - 1Yep, that's consistent. So the formula holds. Another way to think about it is recognizing that each term is one less than a power of two. Since 2, 4, 8, 16... are powers of two, subtracting 1 gives 1, 3, 7, 15, etc., which matches the sequence generated by the recurrence.Alternatively, we can solve the recurrence relation formally. The given recurrence is a nonhomogeneous linear recurrence. The homogeneous part would be a_{n+1} - a_n = 0, which has the solution a_n = constant. For the nonhomogeneous part, since the nonhomogeneous term is 2^n, we can look for a particular solution. Suppose the particular solution is of the form A*2^n. Plugging into the recurrence:a_{n+1} - a_n = 2^nA*2^{n+1} - A*2^n = 2^nA*2^n(2 - 1) = 2^nA*2^n = 2^n => A = 1So the particular solution is 2^n. Therefore, the general solution is the homogeneous solution plus the particular solution:a_n = C + 2^nBut wait, the homogeneous solution is a constant, right? Since the recurrence is a_{n+1} - a_n = 0, which implies a_{n+1} = a_n, so all terms are equal, hence a constant. So the general solution is a_n = C + 2^n.But we need to determine the constant C using the initial condition. When n=1, a₁=1:a₁ = C + 2^1 = C + 2 = 1 => C = 1 - 2 = -1Therefore, the general solution is a_n = -1 + 2^n, which is the same as 2^n - 1. So this method also gives the same result, confirming our answer.So both methods, unfolding the sequence and solving the recurrence formally, lead us to the formula a_n = 2^n - 1. And we checked the first few terms, which all match. Therefore, the answer should be 2^n minus one.Final AnswerThe general term of the sequence is boxed{2^n - 1}.</think>Given the sequence ({a_n}) with (a_1 = 1) and the recurrence relation (a_{n+1} - a_n = 2^n) for (n in mathbb{N}^+), we need to find a general formula for (a_n).First, we observe that each term (a_n) can be expressed as the sum of the initial term (a_1) and the sum of the differences up to the ((n-1))-th term. This gives us:[a_n = a_1 + sum_{k=1}^{n-1} 2^k]Since (a_1 = 1), we need to compute the sum of the geometric series (sum_{k=1}^{n-1} 2^k). The sum of a geometric series starting from (k=0) is given by:[sum_{k=0}^{m} 2^k = 2^{m+1} - 1]Adjusting for the sum starting from (k=1) to (n-1):[sum_{k=1}^{n-1} 2^k = sum_{k=0}^{n-1} 2^k - 2^0 = (2^n - 1) - 1 = 2^n - 2]Substituting this back into the expression for (a_n):[a_n = 1 + (2^n - 2) = 2^n - 1]To verify, we check the first few terms:- (a_1 = 2^1 - 1 = 1)- (a_2 = 2^2 - 1 = 3)- (a_3 = 2^3 - 1 = 7)- (a_4 = 2^4 - 1 = 15)These match the terms generated by the recurrence relation. Additionally, solving the recurrence formally using the method for nonhomogeneous linear recurrences confirms the solution (a_n = 2^n - 1).Thus, the general term of the sequence is (boxed{2^n - 1}).]

🔑:Given: a_1=1, a_{n+1}-a_{n}=2^{n},We get, a_{n}=a_{n-1}+2^{n-1}=a_{n-2}+2^{n-2}+2^{n-1}=a_{n-3}+2^{n-3}+2^{n-2}+2^{n-1}=…=a_{1}+2^{1}+2^{2}+…+2^{n-1}=1+ frac {2(1-2^{n-1})}{1-2}=2^{n}-1.So, a_{n}=2^{n}-1.Hence, the answer is: boxed{2^{n}-1}.By using the recursive relationship, we express the general term using the first term and an expression in terms of n, and thus we can reach the conclusion. This problem tests the understanding of recursive formulas in sequences. To solve such problems, one needs to be careful and make good use of the given conditions.

🧠:[<think>Okay, let me try to figure this out. So, the problem says there are three fractions where the numerators are proportional to 1, 2, 5 and the denominators are proportional to 1, 3, 7. The arithmetic mean of these fractions is 200/441. I need to find the actual fractions. Hmm, okay.First, let me parse what "proportional" means here. If the numerators are proportional to 1, 2, 5, that probably means each numerator is a multiple of those numbers. Similarly, denominators proportional to 1, 3, 7 would mean each denominator is a multiple of those. So, maybe there's a common factor for the numerators and a common factor for the denominators. Let me denote the common factor for the numerators as, say, k, and for the denominators as m. Then the fractions would be (1k)/(1m), (2k)/(3m), and (5k)/(7m). So, the three fractions are k/m, (2k)/(3m), and (5k)/(7m). Is that right? That seems to fit the proportions given.Then, the arithmetic mean of these three fractions is 200/441. So, the average of the three fractions is ( [k/m + (2k)/(3m) + (5k)/(7m) ] ) / 3 = 200/441. So, first, let me compute the sum inside the brackets. Let's write each term with a common denominator to add them up. The denominators are m, 3m, 7m. The common denominator would be 21m. Let's convert each fraction:First term: k/m = 21k/(21m)Second term: (2k)/(3m) = 14k/(21m)Third term: (5k)/(7m) = 15k/(21m)So, adding them together: 21k +14k +15k all over 21m = (50k)/(21m). Therefore, the sum is 50k/(21m). Then, the arithmetic mean is this divided by 3, so (50k)/(21m * 3) = (50k)/(63m) = 200/441. So, set up the equation: 50k/(63m) = 200/441. Then, solve for k/m.First, cross-multiplying: 50k * 441 = 63m * 200. Let's compute both sides:Left side: 50 * 441 = 50 * 400 + 50 * 41 = 20,000 + 2,050 = 22,050. So, 22,050k.Right side: 63 * 200 = 12,600. So, 12,600m.Therefore, 22,050k = 12,600m. Let's divide both sides by 1050 to simplify. 22,050 ÷ 1050 = 21, and 12,600 ÷ 1050 = 12. So, 21k = 12m. Then, divide both sides by 3: 7k = 4m. So, 7k = 4m. Therefore, k/m = 4/7. So, the ratio of k to m is 4:7. Therefore, k = (4/7)m. Hmm, so if we let m = 7t, then k = 4t for some t. But since fractions are k/m, (2k)/(3m), (5k)/(7m), substituting k =4t and m=7t would give:First fraction: k/m =4t /7t =4/7.Second fraction: (2k)/(3m)= (8t)/(21t)=8/21.Third fraction: (5k)/(7m)= (20t)/(49t)=20/49.Wait, but if that's the case, then the fractions would be 4/7, 8/21, and 20/49. Let me check if their average is 200/441. Let's compute the average:First, sum them up. Convert all to 441 denominator.4/7 = 252/4418/21 = 168/44120/49 ≈ (20*9)/441 = 180/441Adding those: 252 +168 +180 = 600. So, 600/441. Divide by 3: 200/441. Perfect, that matches. So, the fractions are 4/7, 8/21, and 20/49. Therefore, the answer is these three fractions. But let me check again if I did everything correctly.Wait, in the beginning, I assumed numerators are k, 2k, 5k and denominators m, 3m, 7m, but the problem says "the numerators of three fractions are proportional to the numbers 1,2,5, and the denominators are proportional to the numbers 1,3,7 respectively". So, perhaps the first numerator is proportional to 1, the second to 2, the third to 5, similarly denominators first is 1, then 3, then 7. So, that would mean that the numerators are 1*a, 2*a, 5*a, and denominators 1*b, 3*b, 7*b, so fractions (a)/(b), (2a)/(3b), (5a)/(7b). So, same as before where a is k and b is m. So, yes, so same approach.Then, proceeding as before, leads to k/m=4/7. So, fractions are 4/7, 8/21, 20/49. So, seems correct.Alternatively, maybe there's a different approach. Let me think. Suppose the fractions are (1k)/(1m), (2k)/(3m), (5k)/(7m). Then, sum is (1k/1m + 2k/3m +5k/7m) = k/m (1 + 2/3 +5/7). Then, average is (k/m)(1 + 2/3 +5/7)/3 =200/441. Then compute the value inside the parentheses:1 + 2/3 +5/7. Let's compute that. 1 is 21/21, 2/3 is 14/21, 5/7 is 15/21. So, sum is (21 +14 +15)/21=50/21. Then, (k/m)*(50/21)/3= (k/m)*(50/63)=200/441. So, same equation. Then, (50/63)*(k/m)=200/441. Multiply both sides by 63/50: (k/m)= (200/441)*(63/50). Simplify:200/441 *63/50 = (200/50)*(63/441) =4*(1/7)=4/7. So, k/m=4/7, same as before. Therefore, fractions are 4/7, 8/21, 20/49. So, the answer is correct. Therefore, I think that's the solution. So, the three fractions are 4/7, 8/21, and 20/49. Let me just check once more if their average is 200/441:Compute (4/7 +8/21 +20/49). Convert all to denominator 147 for ease:4/7 = 84/1478/21=56/14720/49=60/147Sum:84+56+60=200. So, 200/147. Divide by 3: 200/441. Yep, correct.Therefore, the fractions are 4/7, 8/21, and 20/49.Final AnswerThe three fractions are boxed{dfrac{4}{7}}, boxed{dfrac{8}{21}}, and boxed{dfrac{20}{49}}.</think>Given that the numerators of three fractions are proportional to (1, 2, 5) and the denominators are proportional to (1, 3, 7), we denote the common factors for the numerators and denominators as (k) and (m) respectively. Thus, the fractions are (frac{k}{m}), (frac{2k}{3m}), and (frac{5k}{7m}).The arithmetic mean of these fractions is given as (frac{200}{441}). We first find the sum of the fractions:[frac{k}{m} + frac{2k}{3m} + frac{5k}{7m}]Converting each term to a common denominator (21m):[frac{21k}{21m} + frac{14k}{21m} + frac{15k}{21m} = frac{50k}{21m}]The arithmetic mean is then:[frac{1}{3} left( frac{50k}{21m} right) = frac{50k}{63m}]Setting this equal to (frac{200}{441}):[frac{50k}{63m} = frac{200}{441}]Cross-multiplying and solving for (frac{k}{m}):[50k cdot 441 = 63m cdot 200 implies 22050k = 12600m implies frac{k}{m} = frac{4}{7}]Substituting (k = 4t) and (m = 7t) for some constant (t), the fractions become:[frac{4t}{7t} = frac{4}{7}, quad frac{2 cdot 4t}{3 cdot 7t} = frac{8}{21}, quad frac{5 cdot 4t}{7 cdot 7t} = frac{20}{49}]Thus, the three fractions are:[boxed{dfrac{4}{7}}, boxed{dfrac{8}{21}}, text{ and } boxed{dfrac{20}{49}}],<think>Okay, let's see. The problem says there are three fractions where the numerators are proportional to 1, 2, 5 and the denominators are proportional to 1, 3, 7. The arithmetic mean of these fractions is 200/441. We need to find these fractions.First, let's make sure I understand what "proportional" means here. If the numerators are proportional to 1, 2, 5, that means each numerator is a multiple of those numbers. Similarly, denominators are proportional to 1, 3, 7. So, we can probably represent the numerators as 1k, 2k, 5k where k is some constant, and denominators as 1m, 3m, 7m where m is another constant. So each fraction would be (k)/(m), (2k)/(3m), (5k)/(7m). Then, the three fractions are (k/m), (2k)/(3m), (5k)/(7m).The arithmetic mean of these three fractions is given as 200/441. So, we need to compute the average of the three fractions and set it equal to 200/441, then solve for k and m.Let me write that down. The arithmetic mean is [(k/m) + (2k)/(3m) + (5k)/(7m)] / 3 = 200/441.First, let's compute the sum inside the brackets:(k/m) + (2k)/(3m) + (5k)/(7m). Since all denominators have m, we can factor out k/m:k/m [1 + 2/3 + 5/7]. Let me compute the sum inside the brackets:1 + 2/3 + 5/7. To add these, convert to common denominator. The denominators are 1, 3, 7. The least common denominator is 21.So 1 = 21/21, 2/3 = 14/21, 5/7 = 15/21. Adding them up: 21 + 14 + 15 = 50. So 50/21.Therefore, the sum is (k/m)(50/21). Then, the arithmetic mean is [(50k)/(21m)] / 3 = (50k)/(63m) = 200/441.So now we have the equation (50k)/(63m) = 200/441. Let's solve for k/m.Cross-multiplying: 50k * 441 = 63m * 200.Compute both sides:Left side: 50 * 441 * k = 22050kRight side: 63 * 200 * m = 12600mSo 22050k = 12600mDivide both sides by 1050: 22050 / 1050 = 21; 12600 / 1050 = 12So 21k = 12mSimplify the equation: 21k = 12m => divide both sides by 3: 7k = 4m => k/m = 4/7So k/m = 4/7. Therefore, k = (4/7)m.So now, we can express k in terms of m, or vice versa. Since the fractions are (k/m), (2k)/(3m), (5k)/(7m), let's substitute k = (4/7)m into these fractions.First fraction: (k)/m = (4/7 m)/m = 4/7Second fraction: (2k)/(3m) = 2*(4/7 m)/(3m) = (8/7 m)/(3m) = 8/21Third fraction: (5k)/(7m) = 5*(4/7 m)/(7m) = (20/7 m)/(7m) = 20/49Wait, but hold on. If k = (4/7)m, then substituting into the fractions:First fraction: k/m = (4/7)m / m = 4/7Second fraction: 2k/(3m) = 2*(4/7)m / (3m) = (8/7)m / (3m) = 8/21Third fraction: 5k/(7m) = 5*(4/7)m / (7m) = (20/7)m / (7m) = 20/49So the three fractions are 4/7, 8/21, and 20/49.Let me check if their arithmetic mean is indeed 200/441.Compute the sum: 4/7 + 8/21 + 20/49Convert to common denominator. Let's use 21*49 = 1029, but that's a bit big. Alternatively, the LCM of 7, 21, 49. 7 is prime, 21 is 3*7, 49 is 7^2. So LCM is 3*7^2 = 147.Convert each fraction:4/7 = (4*21)/147 = 84/1478/21 = (8*7)/147 = 56/14720/49 = (20*3)/147 = 60/147Sum: 84 + 56 + 60 = 200. So total sum is 200/147. Then arithmetic mean is 200/147 divided by 3, which is 200/(147*3) = 200/441. Which matches the given mean. So that works.Therefore, the fractions are 4/7, 8/21, and 20/49.But let me just verify the steps again to make sure I didn't make any mistakes.First, set up the fractions as (k)/m, (2k)/(3m), (5k)/(7m). Correct, since numerators are proportional to 1,2,5 (so 1k, 2k, 5k) and denominators proportional to 1,3,7 (so 1m, 3m, 7m). So the fractions are k/m, 2k/3m, 5k/7m. Then the mean is [k/m + 2k/(3m) + 5k/(7m)] / 3. That seems correct.Sum inside the brackets: k/m [1 + 2/3 + 5/7]. Calculated the sum inside as 50/21. Let me check again:1 is 1/1, so 1 = 21/21, 2/3 = 14/21, 5/7 = 15/21. Sum: 21 +14 +15 =50. 50/21. Correct. So the sum is (k/m)*(50/21). Then divided by 3 gives (50k)/(21m*3) = 50k/(63m). Set equal to 200/441.So 50k/(63m) = 200/441. Cross-multiplying gives 50k*441 = 63m*200. Which is 22050k = 12600m. Dividing both sides by 1050: 21k = 12m. Then 7k = 4m. So k = (4/7)m. Then substituting back into the fractions gives 4/7, 8/21, 20/49. Checked the arithmetic mean, which is correct.Therefore, the answer is 4/7, 8/21, and 20/49.Final AnswerThe three fractions are boxed{dfrac{4}{7}}, boxed{dfrac{8}{21}}, and boxed{dfrac{20}{49}}.</think>Given that the numerators of three fractions are proportional to (1, 2, 5) and the denominators are proportional to (1, 3, 7), we represent the fractions as (frac{k}{m}), (frac{2k}{3m}), and (frac{5k}{7m}) where (k) and (m) are constants. The arithmetic mean of these fractions is (frac{200}{441}).First, we calculate the sum of the fractions:[frac{k}{m} + frac{2k}{3m} + frac{5k}{7m}]Factoring out (frac{k}{m}):[frac{k}{m} left(1 + frac{2}{3} + frac{5}{7}right)]Converting to a common denominator (21):[1 = frac{21}{21}, quad frac{2}{3} = frac{14}{21}, quad frac{5}{7} = frac{15}{21}]Summing these:[frac{21}{21} + frac{14}{21} + frac{15}{21} = frac{50}{21}]Thus, the sum of the fractions is:[frac{k}{m} cdot frac{50}{21}]The arithmetic mean is:[frac{1}{3} cdot frac{50k}{21m} = frac{50k}{63m}]Setting this equal to (frac{200}{441}):[frac{50k}{63m} = frac{200}{441}]Cross-multiplying:[50k cdot 441 = 63m cdot 200]Simplifying:[22050k = 12600m implies 21k = 12m implies 7k = 4m implies frac{k}{m} = frac{4}{7}]Substituting (k = frac{4}{7}m) into the fractions:[frac{k}{m} = frac{4}{7}, quad frac{2k}{3m} = frac{8}{21}, quad frac{5k}{7m} = frac{20}{49}]Verifying the arithmetic mean:[frac{1}{3} left( frac{4}{7} + frac{8}{21} + frac{20}{49} right) = frac{1}{3} left( frac{84}{147} + frac{56}{147} + frac{60}{147} right) = frac{1}{3} cdot frac{200}{147} = frac{200}{441}]The three fractions are (boxed{dfrac{4}{7}}), (boxed{dfrac{8}{21}}), and (boxed{dfrac{20}{49}}).]

🔑:Solution.The required fractions have the form: frac{x}{y} ; frac{2 x}{3 y} ; frac{5 x}{7 y}. According to the conditionfrac{frac{x}{y}+frac{2 x}{3 y}+frac{5 x}{7 y}}{3}=frac{200}{441}, from which frac{x}{y}=frac{4}{7} ; frac{2 x}{3 y}=frac{8}{21} ; frac{5 x}{7 y}=frac{20}{49}.Answer: frac{4}{7} ; frac{8}{21} ; frac{20}{49}

🧠:[<think>Okay, so I need to find two prime numbers, m and n, such that when I plug them into the equation 5m + 7n = 129, the equation holds true. Then, I have to find the sum of m and n. Alright, let's start breaking this down.First, I know that prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves. So, primes are like 2, 3, 5, 7, 11, 13, 17, etc. Since the equation involves both m and n multiplied by coefficients (5 and 7) and added together to get 129, I might need to use some trial and error here. But maybe there's a smarter way than just randomly guessing primes.Let me consider possible values for m and n. Let's see, since 5m + 7n = 129, both m and n have to be such that when multiplied by 5 and 7 respectively, their sum is 129. Let's think about possible ranges for m and n. Let's solve for one variable in terms of the other to reduce the possibilities.Let me solve for m first. If I rearrange the equation:5m = 129 - 7nSo,m = (129 - 7n)/5Since m has to be a prime number, (129 - 7n) must be divisible by 5. That means (129 - 7n) mod 5 = 0.Similarly, 129 mod 5 is 129 divided by 5, which is 25*5=125, so remainder 4. So, 129 ≡ 4 mod 5.Then, 7n mod 5 must equal 4 as well, because 129 - 7n ≡ 0 mod 5 => 7n ≡ 129 ≡ 4 mod 5.But 7n ≡ 4 mod 5. Since 7 ≡ 2 mod 5, this equation becomes 2n ≡ 4 mod 5.To solve 2n ≡ 4 mod 5, multiply both sides by the inverse of 2 mod 5. The inverse of 2 mod 5 is 3 because 2*3=6≡1 mod5. So, multiplying both sides by 3:n ≡ 4*3 ≡ 12 ≡ 2 mod5.So, n ≡ 2 mod5. That means n can be written as 5k + 2, where k is a non-negative integer.But n must be a prime number. Let's list primes of the form 5k + 2:Possible primes:When k=0: 5*0 + 2 = 2 (prime)k=1: 5*1 + 2 =7 (prime)k=2: 5*2 + 2=12 (not prime)k=3: 5*3 + 2=17 (prime)k=4: 5*4 + 2=22 (not prime)k=5: 5*5 +2=27 (not prime)k=6: 5*6 +2=32 (not prime)k=7: 5*7 +2=37 (prime)k=8: 42 (not prime)k=9:47 (prime)k=10:52 (not prime)And so on. But we need primes n such that when plugged into 5m +7n=129, m is also a prime.So possible n values: 2,7,17,37,47,...But let's check how big n can be. Let's see, if n is 47, then 7*47=329, which is way larger than 129. So n can't be that big. Let's find the upper bound for n.From the equation 5m +7n=129, since both m and n are primes (so at least 2), let's see:Minimum possible n is 2, then 7n=14, so 5m=129-14=115, m=23, which is prime. Wait, that's a possible solution? Let me check that. If n=2, m=(129 -14)/5=115/5=23. 23 is prime, so that works. So m=23 and n=2 gives m+n=25. But wait, the question might have a unique answer? Let me check other possible n.Wait, but maybe there are multiple solutions. Let's check other n.Next possible n is 7. Then 7n=49, so 5m=129-49=80, m=16. But 16 is not prime. So that's invalid.Next n=17. Then 7*17=119. So 5m=129-119=10, so m=2. 2 is prime. So m=2, n=17. Then m+n=19. Hmm, so that's another solution. Wait, so there are two solutions? But the problem says "the value of m + n". Maybe it expects a unique answer. Wait, is that possible? Let me check.Wait, but let me verify these solutions.First solution: n=2, m=23. 5*23 +7*2=115 +14=129. Correct.Second solution: n=17, m=2. 5*2 +7*17=10 +119=129. Correct. So both are valid. But the question says "the value of m + n is". Hmm, so maybe there are two answers? But in the problem statement, it's presented as a single answer. Maybe I made a mistake here.Wait, but in the problem statement, it says "prime numbers m, n". So both m and n have to be primes. So both solutions are valid. But why does the problem expect a single answer? Maybe I need to check if both solutions are valid or if one of them is invalid for some reason.Wait, let's check m=23, n=2. Both primes. m=2, n=17. Both primes. So both are valid. Therefore, the problem might have two solutions. But the answer is presented as a single value. Hmm, this is confusing.Wait, let me check if there are other possible n's. Next in the list would be n=37. 7*37=259, which is way over 129, so 5m would have to be negative, which is impossible. So n=37 is too big. So possible n values are 2,7,17. We saw n=7 gives m=16, which is not prime. So only two solutions: (23,2) and (2,17). So m +n is either 25 or 19. But the problem seems to expect a single answer. Wait, maybe the problem assumes m and n are distinct primes? But 2 and 17 are distinct, 23 and 2 are also distinct. So that's not it.Wait, maybe the problem is from a source where ordering matters? Like m and n are ordered. But in the problem statement, it just says prime numbers m,n, so unless there's a convention where m >n or something, but the problem doesn't specify. So both solutions are valid. Therefore, either 19 or 25. But the problem is presented as having one answer. Maybe I made a mistake in considering n=2 as a possible solution? Wait, let's check again.Wait, n=2, which is a prime. 7*2=14. 129-14=115. 115 divided by 5 is 23, which is prime. So that works. Similarly, n=17, 7*17=119. 129-119=10. 10 divided by 5 is 2, which is prime. So both solutions are correct.Wait, perhaps the problem expects m and n to be distinct primes? But in both cases, they are distinct. 23 and 2, 2 and 17. So maybe the problem is designed to accept both answers? Or maybe I need to check if there's a constraint I missed.Wait, looking back at the problem: "If the prime numbers m, n satisfy 5m +7n=129, then the value of m+n is ______." It just says "the value", implying maybe a unique answer. But according to our calculations, there are two possible values. Hmm. Maybe I need to check my work again.Wait, let me check the modulo arithmetic again. So we had 2n ≡4 mod5, so n≡2 mod5. Therefore, possible n's are primes congruent to 2 mod5. The primes we found are 2,7,17,37,47,... But let's list primes less than, say, 129/7≈18.42. So n can be up to 18. So primes less than 19. The primes in that range congruent to 2 mod5 are 2,7,17. So only n=2,7,17. So we checked those, and n=2 and n=17 give valid m. So two solutions.But the problem is presented as having a single answer, so maybe I made a mistake. Wait, perhaps the problem is from a competition where only one answer is expected, and perhaps the answer is the larger sum or something. But in the absence of additional constraints, both 19 and 25 are valid. Wait, but let me check the problem again. Maybe there's a miscalculation.Wait, let me recalculate n=2:5m +7*2 =129 =>5m=129-14=115 =>m=23. 23 is prime.n=17:5m +7*17=129 =>5m=129-119=10 =>m=2. 2 is prime.So both are correct. So why does the problem expect a single answer? Maybe in the original problem, there was a typo or something. Wait, let me check the problem statement again: "If the prime numbers m, n satisfy 5m +7n=129, then the value of m+n is ______." So perhaps in the problem's source, the answer is 19 or 25. Maybe depending on the convention. Wait, perhaps in some contexts, m and n are considered ordered pairs, so both (23,2) and (2,17) are considered different solutions. But in this problem, since they just ask for m +n, both 25 and 19 would be acceptable. But the problem is presented as having a single answer. Hmm.Wait, maybe I made a mistake in the modulo approach. Let's double-check.We have 5m +7n=129. We want m and n primes.We can write 7n ≡129 mod5.129 divided by5 is 25*5=125, remainder 4. So 7n ≡4 mod5.7≡2 mod5, so 2n≡4 mod5. Then, 2n≡4 mod5. Dividing both sides by 2: n≡2 mod5. So n=5k +2. So possible primes are 2,7,17, etc., as before. So that part is correct.So that gives possible n values:2,7,17. As above.Wait, perhaps I need to check n=7 again. Let's plug n=7 into the equation. 7*7=49. 129-49=80. 80/5=16. 16 is not prime. So that's correct. So only n=2 and n=17 give primes for m. So two solutions. Therefore, the problem might have two answers. But perhaps in the problem's original context, considering m and n as ordered pairs, or maybe as per some other constraint, only one is considered. Alternatively, maybe the problem assumes m and n are both odd primes. But 2 is even, so if we exclude 2, then n=17 would be the only one. Wait, but 2 is a prime number. So excluding 2 would be wrong.Wait, unless there's a mistake here. Let me check with m=2 and n=17. 5*2=10 and 7*17=119. 10+119=129. Correct. And m=2 is prime, n=17 is prime. So that's valid. Similarly, m=23 and n=2. Both primes. So why would the problem expect only one answer? Maybe there is a miscalculation.Wait, perhaps the problem is from a specific source where they only take the solution where m >n or something. Let's check the sums: 23+2=25 and 2+17=19. Both are possible. Wait, unless there is another constraint. Wait, let me check if there are other primes beyond n=17 that I might have missed in the earlier calculation. But n=17 gives m=2. If n=5k +2, next prime after 17 would be 37, but 7*37=259, which is way over 129, so m would be negative. Not possible. So no, only n=2 and n=17.Alternatively, maybe the problem is in a non-English language source, and maybe the question is transcribed incorrectly. But given the problem as stated, there are two solutions. However, since the user is asking for an answer to be put in a box, likely expecting a single answer, perhaps I need to check if one of the solutions is invalid for some reason. Let me check m=23 and n=2. 23 is a prime, 2 is a prime, so that's valid. Similarly, m=2 and n=17. Also valid.Wait, maybe in the problem statement, m and n are specified to be distinct primes? But both solutions have distinct primes. So no. Alternatively, maybe the problem is expecting both solutions and combining them, but m +n=19 and 25, which are 19 and25. But 19 and25 are different.Wait, unless there's an error in my calculation. Let me check n=2 again. 5m +7*2=5m +14=129 =>5m=115 =>m=23. 23 is prime. Correct. n=17:5m +7*17=5m +119=129 =>5m=10 =>m=2. Correct. Both are valid.Hmm. Maybe the problem is designed to have both answers, but in the original problem, maybe there was a condition that m and n are both greater than 2 or something. But since it's not stated here, we can't assume that. So perhaps the answer is either 19 or 25. But the problem says "the value of m +n is". So unless there's a unique solution, which there isn't, the problem might be ill-posed. Alternatively, maybe I made a mistake here.Wait, let's see. Let me check the problem again. Maybe I misread it. It says prime numbers m, n. So m and n are primes. So both solutions are valid. So if this is a competition problem, and the answer is two possible values, but the problem expects one answer, maybe there's a mistake in the problem.Alternatively, maybe I need to check higher primes. Wait, n=2, m=23. n=17, m=2. Wait, but are there any other primes n not congruent to 2 mod5 that can still result in m being prime?Wait, my earlier approach was using modulo 5 to restrict n. But maybe there's a way where even if n is not congruent to 2 mod5, (129 -7n) is divisible by5. Wait, but 129 -7n must be divisible by5. So 7n must ≡129 mod5. As 129 mod5 is 4, so 7n≡4 mod5. Since 7≡2 mod5, so 2n≡4 mod5, so n≡2 mod5. So this is a necessary condition. Therefore, n must be congruent to2 mod5. Therefore, only primes of the form5k+2 are possible. Therefore, n=2,7,17,37,... but 37 is too big as before. So n must be 2,7,17. Therefore, only these three. We checked all three. Only n=2 and17 give prime m. So two solutions. Therefore, the problem might have two answers, but since the user is asking for an answer here, maybe I need to see if in the original problem's context, which I don't have, but perhaps they are expecting the larger sum or the smaller one. Alternatively, maybe I made a mistake.Alternatively, maybe in the problem statement, there's a miscalculation. Let me check again:If n=2: 7*2=14. 129-14=115. 115/5=23. 23 is prime. Correct.n=7: 7*7=49. 129-49=80. 80/5=16. Not prime. Correct.n=17:7*17=119. 129-119=10. 10/5=2. Prime. Correct.So no mistakes here.Wait, perhaps in the problem statement, it's supposed to be 5m -7n=129? That would result in a different answer. But no, the user provided the problem as 5m +7n=129.Alternatively, maybe the user made a typo. But assuming the problem is correct as given, then there are two solutions.But given that the user is expecting an answer to be put in a box, likely one answer. So maybe the problem is from a specific source where only one solution is considered. For example, maybe in some contexts, the problem is presented with m and n both as odd primes, but 2 is even, so excluding n=2. However, 2 is a prime number, so excluding it would be incorrect unless specified.Alternatively, maybe there's a convention where m and n are ordered such that m >n. In that case, m=23 and n=2, so m +n=25. But if we take m=2 and n=17, m +n=19. But why would m be considered larger than n? There's no such indication in the problem.Alternatively, perhaps the problem is from a competition where multiple answers are allowed, but in the format here, only one box is provided, so maybe the user expects the smaller sum or the larger one. But without more context, it's hard to tell.But given that both solutions are valid, perhaps the answer is either 19 or25. However, I need to check again if there's another possible solution.Wait, let's try n=2, m=23: sum is25.n=17, m=2: sum is19.Is there another prime n that I missed? Let's see.n=2: valid.n=7: invalid.n=17: valid.n=5*4 +2=22, not prime.n=5*1 +2=7: checked.So, no. So only two solutions. Therefore, since the problem is presented as having a single answer, maybe there's a mistake in the problem statement. Alternatively, maybe the user expects the answer to be19 or25, and maybe the correct answer is19. Because in some cases, variables are listed alphabetically, so m comes before n, so maybe m=2 and n=17 is considered. But that's a stretch. Alternatively, maybe the problem is from a source where only19 is the answer, but in reality, both are correct.Alternatively, maybe I missed another solution. Let's try m=2 and n=17. But maybe other primes not of the form5k +2. Wait, but according to modulo arithmetic, n must be congruent to2 mod5. So primes not congruent to2 mod5 would not satisfy the equation. Let me check for example n=3. 7*3=21. 129-21=108. 108/5=21.6. Not integer. So invalid.n=5:7*5=35. 129-35=94. 94/5=18.8. Not integer.n=11:7*11=77. 129-77=52. 52/5=10.4. Not integer.n=13:7*13=91. 129-91=38. 38/5=7.6. Not integer.n=19:7*19=133. 129-133= -4. m would be negative, invalid.n=23:7*23=161>129, invalid.So yes, only n=2 and17 give integer m which are primes. So two solutions. Therefore, the problem might have two answers, but the user is expecting one. Maybe the intended answer is19, given that it's smaller? Or maybe25. Maybe the original problem had a typo. Alternatively, the answer is19. Wait, let me check online if this problem exists. Wait, I can't access external resources, but maybe my own knowledge.Wait, in similar problems, often the smaller primes are considered first. For example, n=2, which is the smallest prime. So maybe the answer is25. But since there are two solutions, it's ambiguous. However, given that the problem is presented in a way that suggests a single answer, I might need to check if there's a mistake in my approach.Wait, let me try another approach. Let's express the equation as 5m =129 -7n. So 129 -7n must be a multiple of5, and positive. So 129 -7n >0. So 7n <129. So n <129/7≈18.42. So n can be at most18. So primes less than18 are:2,3,5,7,11,13,17. So list of primes:2,3,5,7,11,13,17.Let's check each of these primes as possible n and compute m:n=2:7*2=14. 129-14=115. 115/5=23. 23 is prime. So m=23.n=3:7*3=21.129-21=108.108/5=21.6. Not integer.n=5:7*5=35.129-35=94.94/5=18.8. Not integer.n=7:7*7=49.129-49=80.80/5=16. Not prime.n=11:7*11=77.129-77=52.52/5=10.4. Not integer.n=13:7*13=91.129-91=38.38/5=7.6. Not integer.n=17:7*17=119.129-119=10.10/5=2. Prime. So m=2.So the only valid solutions are n=2,m=23 and n=17,m=2. So m+n=25 or19. So two solutions. Therefore, the problem has two answers. But the user is expecting one answer. Therefore, likely the problem is from a source where the answer is19 or25. If I had to choose one, maybe the answer is19, but without more context, it's hard to tell. However, since both are valid, perhaps the answer is19. Alternatively, maybe the problem expects the answer as19, and the25 solution is considered trivial because m=23 is large. Alternatively, the problem might have a unique solution if there was a misprint in the equation.Alternatively, maybe the problem is correct, and the answer is19. Given that sometimes problems like this have the variables arranged such that m is paired with5 and n with7, so if you take m=2 and n=17, it's a smaller m. But again, this is speculative.Alternatively, maybe the answer is19, and25 is another solution that is not considered. But unless there's a restriction in the problem, both are valid. Therefore, perhaps the problem is designed to have both answers, but in the absence of multiple answer boxes, the user might expect one. Since I need to provide an answer here, and given that both solutions are valid, but the problem says "the value of m +n", I need to check if maybe I made a mistake in considering m and n as interchangeable. Wait, no, the problem says "prime numbers m, n" without specifying an order. So both solutions are valid.Wait, perhaps the answer is19. Because if we consider m and n as positive integers where m and n are primes, and we look for solutions where both m and n are primes, and if the problem is from a math competition, perhaps they are expecting the answer19. Alternatively, maybe the problem is from a specific textbook where the answer is19. However, without more context, it's impossible to know. Therefore, since both are valid, but perhaps the answer expected is19, I will go with19. Alternatively, maybe I need to confirm.Wait, but both are correct. So in order to be thorough, perhaps I need to state both. But since the user is asking for an answer in a box, I can't provide two answers. Therefore, this problem is either ambiguous or has a unique solution that I missed.Wait, let me check if m and n are allowed to be the same prime. In this case, they are different. So the problem didn't specify distinct primes, but in both solutions, they are distinct. So no.Wait, unless there's a mistake in my modulo approach. Let me check the modulo arithmetic again. So 7n ≡4 mod5. Since 7≡2 mod5, so 2n≡4 mod5. Then n≡2 mod5. So primes n≡2 mod5:2,7,17, etc. So correct. So the only possible primes n under 18 are2,7,17. So those are the only possible n. Therefore, the only possible solutions are n=2,m=23 and n=17,m=2. Therefore, the problem has two solutions. However, since the problem asks for "the value of m +n", implying a unique answer, perhaps there's a mistake in the problem statement. Alternatively, maybe the answer is25. If you take m=23 and n=2, the sum is25, which is a composite number, while19 is prime. But the problem doesn't specify any condition on the sum. So that's irrelevant.Alternatively, perhaps the problem assumes that m and n are both odd primes. But 2 is even, so if we exclude 2, then the only solution is n=17 and m=2. But m=2 is prime. So that doesn't exclude it. Because m can be2.Wait, the problem says "prime numbers m,n", there is no restriction on them being odd primes. So2 is allowed.Therefore, both solutions are valid. Therefore, the problem as stated has two solutions. Hence, the answer is either19 or25. However, since the user is expecting a single answer, maybe I need to check which one is correct in the original problem's context. Given that I can't do that, I can only present both answers. However, since the problem is likely expecting one answer, perhaps19 is the intended answer. Alternatively, maybe the problem had a different coefficient. For example, if the equation was5m +7n=127 or131, but in this case, it's129.Alternatively, perhaps I need to check if the problem is from a specific competition where this problem is known to have answer19. For example, in some Olympiad problems, sometimes they order variables such that m >n, but since the problem doesn't specify, we can't assume that.Alternatively, maybe the answer is25, considering m=23 and n=2. If you list primes in ascending order, m=2 andn=17 would be ordered as m=2 andn=17. But since variables are just labels, the problem doesn't specify an order. Hence, both are valid.Given the problem as stated, the possible values of m +n are19 and25. Therefore, the problem is ambiguous. However, since the user expects an answer here, I might need to check both answers and see if one of them is more likely. For example, maybe in the original problem, the answer is19. But without further information, it's impossible to be certain. However, given that when n=17 and m=2, both primes are smaller (except m=2, which is the smallest prime), but that doesn't make sense. Alternatively, maybe the problem is designed so that m +n is as small as possible, which would be19. But again, this is speculative.Alternatively, perhaps I made a mistake in considering n=2. Let me check again. For n=2, m=23. 23 is a prime. So valid. Yes. Therefore, both solutions are correct.Therefore, the problem has two solutions:19 and25. Therefore, the answer is either19 or25. But since the user expects a single answer, perhaps the intended one is19. However, I need to verify.Alternatively, in the equation5m +7n=129, if we think of m andn as positive integers, there's a method called Diophantine equations. Let me try to solve it for all positive integer solutions and then check which ones have m andn prime.The general solution for the equation5m +7n=129.We can write this as5m =129 -7n. So m=(129 -7n)/5. For m to be a positive integer,129 -7n must be divisible by5 and positive.So 129 -7n >0 =>7n <129 =>n <129/7≈18.42. So n can be from1 to18.Also,129 -7n ≡0 mod5 =>7n ≡129 mod5. As before, n≡2 mod5. So n=5k +2, where k is0,1,2,3. Because n=5*3 +2=17, which is less than18.42. So n=2,7,12,17. But n must be prime, so from these, n=2,7,17. As before. Then m=(129 -7n)/5. So:For n=2: m=(129 -14)/5=115/5=23.For n=7: m=(129 -49)/5=80/5=16. Not prime.For n=17: m=(129 -119)/5=10/5=2.So only n=2 and17 give prime m. Therefore, two solutions.Therefore, the answer is either19 or25. Since the problem is presented as having a single answer, it's likely that in the original source, there was a typo or specific context. However, given the information here, both solutions are valid. But since the problem is given without multiple answer options, maybe the intended answer is19.Wait, but I recall that in some textbooks, when multiple solutions are possible, they might take the one with the smaller sum or the larger sum. Alternatively, the problem might have intended for m andn to be two-digit primes, but m=2 is a single-digit prime. However, 23 is two-digit. But again, the problem didn't specify.Alternatively, let's think about which solution is more "natural". If someone is solving this without modulo arithmetic, they might start testing primes for n from the smaller ones. So starting with n=2, which works, giving m=23. Then maybe they stop there. But a thorough person would check others. So maybe the problem expects25. But if you check until n=17, you find another solution.But given that both are valid, and the problem is presented as having a unique answer, there's inconsistency. Therefore, perhaps the answer is19. Or maybe25.Wait, let's see what the sum is. If the answer is19, which is a prime number, and25 is composite. But the problem doesn't specify anything about the sum. So that doesn't matter.Alternatively, perhaps the problem is designed so that m andn are both two-digit primes. Then m=23 (two-digit) andn=2 (single-digit), but n=17 is two-digit. So sum would be23+2=25 or2+17=19. Since23 and17 are two-digit primes, but2 is single-digit. If the problem prefers two-digit primes, then n=17 andm=2. But again, the problem doesn't specify.Given the lack of additional constraints, both solutions are valid. Therefore, the problem is ambiguous. However, since the user is expecting an answer here, and given that both solutions are possible, but in many cases, people might stop at the first solution they find (n=2, m=23, sum25), but another person might consider checking further and find n=17, sum19. Therefore, it's possible that the answer is19.But in order to be precise, since the problem didn't restrict the values, both are valid. Therefore, the problem has two solutions. However, since the user interface requires a single answer, I need to choose one. Given that, I think the intended answer is19, but I can't be sure. Alternatively, the problem might have a unique solution if there was a typo, but given the problem as stated, two solutions exist.Wait, maybe I made a mistake in the modulo arithmetic. Let me check again. 7n ≡4 mod5. Since7≡2 mod5, so2n≡4 mod5. Then,2n≡4 mod5 impliesn≡2 mod5. Correct. So n=5k +2. Then primes n=2,7,17. So two solutions. Therefore, unless there's a mistake in the problem, two answers.But since the user expects one answer, maybe I should state both. But the instructions say "put your final answer within boxed{}". Since multiple answers aren't possible here, and given that the problem is from an unknown source, I'll have to make a judgment call. Given that both solutions are valid, but in some contexts, the answer is19, I'll go with19. Alternatively, if the answer is25, but I need to be honest here. Given that both are correct, but in the absence of further constraints, I can't choose one over the other. Therefore, perhaps the problem is defective. However, given that in the problem statement, m is written first, maybe they expect m to be the larger prime. So m=23, n=2, sum25. Therefore, the answer is25.But since the problem states "prime numbers m, n" without order, but writes the equation as5m +7n=129, maybe the coefficients are ordered as5 and7, but the variables m andn are just variables. There's no inherent order. Therefore, both solutions are equally valid.Given the ambiguity, but since the problem is likely expecting one answer, I might need to present both. However, the user instruction says to put the final answer within a box. Given that, maybe the intended answer is19. But I need to confirm.Wait, let me check other problems of this type. For example, if I search for similar problems, often when you have equations like ax + by = c, and x and y are primes, there can be multiple solutions, but sometimes the problem is designed with a unique solution. In this case, perhaps the problem intended to have a unique solution, but due to the coefficients chosen, it allows two solutions. Therefore, the problem might be flawed. However, given the information here, both solutions are valid.Since I need to provide an answer, and given that the problem is in Chinese (the original user might have a Chinese source), maybe in the source, the answer is19. But I can't be certain. Alternatively, the answer is25. Alternatively, in the original problem, maybe there was a typo, and the equation is5m +7n=127 or something else, which would result in a unique solution. However, as given, it's129.Given that I have to choose, and given that when I first found n=2 leading to m=23, which gives sum25, and then later found n=17 leading to sum19. If I list both, but the platform requires one answer, perhaps the answer is19. Alternatively, the problem might have a unique solution if we consider that 2 is not considered here, but no. Given that, I think the problem has two solutions, but since I must provide one, I'll go with19. But I'm not fully confident. Alternatively, to play safe, provide both answers. But the system might not accept two.Alternatively, consult with mathematical literature. Wait, I can't do that. Given that, I think the answer is19, but I have to mention that25 is also a solution. But since I can't, then the problem is ambiguous. Given that, but following the initial solution where n=17, m=2, sum19, maybe that's the answer.Final AnswerThe value of ( m + n ) is boxed{19}.</think>Given the equation (5m + 7n = 129) where (m) and (n) are prime numbers, we need to find the value of (m + n).1. Rearrange the equation: [ 5m = 129 - 7n ] This implies (129 - 7n) must be divisible by 5. 2. Modulo arithmetic: [ 129 equiv 4 pmod{5} ] [ 7n equiv 4 pmod{5} ] Since (7 equiv 2 pmod{5}), we get: [ 2n equiv 4 pmod{5} ] Solving for (n): [ n equiv 2 pmod{5} ]3. Possible prime values for (n): - (n = 2) (prime) - (n = 7) (prime) - (n = 17) (prime)4. Check each prime (n): - For (n = 2): [ 5m + 7 cdot 2 = 129 implies 5m = 115 implies m = 23 quad (text{prime}) ] Sum: (m + n = 23 + 2 = 25) - For (n = 7): [ 5m + 7 cdot 7 = 129 implies 5m = 80 implies m = 16 quad (text{not prime}) ] - For (n = 17): [ 5m + 7 cdot 17 = 129 implies 5m = 10 implies m = 2 quad (text{prime}) ] Sum: (m + n = 2 + 17 = 19)5. Conclusion: The valid solutions are ((m, n) = (23, 2)) and ((2, 17)). Both solutions are valid, but the problem expects a single answer. Given the context and possible conventions, the value (m + n) is (boxed{19}).,<think>Alright, let's tackle this problem where we need to find two prime numbers m and n such that 5m + 7n = 129, and then determine the value of m + n. Hmm, okay, prime numbers... So both m and n are primes. The equation is linear, so maybe I can express one variable in terms of the other and then check for primes? Let me think step by step.First, let's rearrange the equation to solve for one variable. Let's solve for m in terms of n. So starting with 5m + 7n = 129. If I subtract 7n from both sides, I get 5m = 129 - 7n. Then divide both sides by 5, so m = (129 - 7n)/5. Since m has to be a prime number, (129 - 7n) must be divisible by 5, right? Because m must be an integer. So (129 - 7n) mod 5 should be 0.Alternatively, maybe solving for n in terms of m? Let's try that too. Starting again from 5m +7n =129. Subtract 5m to get 7n =129 -5m, then n=(129 -5m)/7. Similarly, (129 -5m) must be divisible by 7 for n to be an integer. Either way, this gives us a way to check possible values of n or m.But since primes are involved, and primes (except 2) are all odd, maybe considering the parity (evenness or oddness) of m and n can help? Let's see. Let me recall that 5 times an even number is even, 5 times an odd number is odd. Similarly, 7 times even is even, 7 times odd is odd. Then, 5m +7n =129. 129 is an odd number. So 5m +7n must be odd. Let's see, the sum of two terms is odd. So one term must be even and the other must be odd. Because even + odd = odd.So, either 5m is even and 7n is odd, or vice versa. But 5m is even only if m is even (since 5 is odd, even*odd=even). Similarly, 7n is even only if n is even. But the only even prime number is 2. So, either m=2 or n=2. Because all other primes are odd, and if m or n is even, they have to be 2. So this gives us two cases:Case 1: m =2. Then plug into the equation: 5*2 +7n =129 → 10 +7n=129 →7n=119→n=17. 17 is a prime number. So this works. Then m+n=2+17=19.Case 2: n=2. Then plug into the equation: 5m +7*2=129→5m +14=129→5m=115→m=23. 23 is also a prime number. So this works as well. Then m+n=23+2=25.Wait, so both cases give us valid prime numbers? So there are two solutions? But the problem says "the value of m +n is". Maybe I missed something. Let me check again.Wait, wait, let's make sure. First case: m=2, then n=17. 17 is prime. So 5*2 +7*17=10 +119=129. Correct. Second case: n=2, m=23. 5*23=115, 7*2=14, 115+14=129. Correct. So both are solutions. But the problem says "the value of m +n is", implying a unique answer. Hmm. So perhaps there are two possible solutions? But the problem might have only one. Wait, but maybe I made a mistake here.Wait, let me check the problem again. It says: "If the prime numbers m, n satisfy 5m +7n=129, then the value of m +n is ___."Wait, maybe both solutions are valid? Then the problem is perhaps expecting both answers? But in the answer format, it's supposed to be boxed, so maybe they are expecting either 19 or 25? But that seems odd. Let me check again.Wait, wait, no, perhaps I miscalculated. Let's check the second case again. If n=2, then m=(129 -14)/5=115/5=23. 23 is prime, so that works. So m+n=25. But both 19 and 25. Wait, but the problem is from an Olympiad or similar, which usually has a unique answer. So maybe there's a mistake in my reasoning.Wait, perhaps I need to check all possible primes, not just the cases where one variable is 2. Because maybe even if both m and n are odd primes, their combination might result in an odd total? Wait, but if both m and n are odd primes, then 5m is 5*odd=odd, and 7n=7*odd=odd. So odd + odd = even. But 129 is odd, so that's impossible. Therefore, one of m or n must be 2. Therefore, only the two cases above are possible. So there are two solutions. But then the problem says "the value of m +n is", so maybe both 19 and 25? But the problem probably has a unique solution. Wait, maybe I need to check if both solutions are valid, but maybe the problem is expecting both? Or maybe I missed something.Wait, let me check again. If m=2, n=17, which are primes. 5*2 +7*17=10 +119=129. Correct. If n=2, m=23, which are primes. 5*23 +7*2=115 +14=129. Correct. So both are valid. Therefore, there are two solutions. Then, in the problem statement, if it's expecting a single answer, perhaps the question is mistyped? Or maybe I made an error. Wait, maybe the problem is from a source where the answer is unique, so maybe one of the solutions is invalid. Wait, no, both primes are valid. So maybe the problem allows multiple answers, but the user wants both? But in the instruction, the user says "put your final answer within boxed{}". So probably, the problem is intended to have a unique solution, but here we have two. Hmm. Let me check again.Wait, perhaps when the problem says "prime numbers m, n", maybe they are considering ordered pairs, so (m,n)=(2,17) and (23,2). But since m and n are just primes, order might not matter. Wait, but in the equation 5m +7n=129, m and n are coefficients with different coefficients, so they are distinguishable. So both solutions are valid. However, the problem might have a typo, or maybe I misread it.Wait, maybe there is another constraint. Let me check m and n. For m=2, n=17. m=2 is the only even prime. For m=23, n=2. Similarly, n=2 is the only even prime. So both cases are covered. Therefore, both solutions are valid. So maybe the answer is either 19 or 25. But the problem says "the value of m +n is". If the problem is from an official source, like a competition, maybe the answer is unique. Wait, maybe I made a mistake in the calculation. Let me check again.Case 1: m=2.5*2=10. 129-10=119. 119 divided by7 is 17. 17 is prime. So m=2, n=17. Sum=19.Case 2: n=2.7*2=14. 129-14=115. 115 divided by5 is23. 23 is prime. So m=23, n=2. Sum=25.Both sums, 19 and25. But maybe in the problem statement, there was a typo, but as given, both are valid. Wait, unless there's another constraint. Wait, primes are positive integers, right? So m and n are positive primes. Both 2,17,23 are positive primes. So, the problem might have two solutions. But the way the problem is phrased, "the value of m +n is", maybe in the original problem, there was a restriction, like m and n are distinct primes? Wait, but even then, 2 and17 are distinct, 23 and2 are distinct. So still both solutions. Hmm.Alternatively, maybe I missed some other primes. Wait, but let's think. If both m and n are odd primes, then 5m +7n is even, as odd + odd = even, but 129 is odd, which is a contradiction. Therefore, one of m or n must be 2. So only the two cases. Therefore, two solutions. So, perhaps the problem is ill-posed? Or maybe in the original problem, there was a restriction that m and n are distinct primes greater than 2, but that's not stated here. Hmm.Wait, but in the problem statement as given, it's just "prime numbers m, n". So the answer could be 19 or25. But since the problem is asking for "the value", maybe it's expecting both? But in the format, they want the answer boxed. Maybe the problem is from a source where 25 is the answer. Wait, maybe in some contexts, m and n are considered in order, like m < n or m > n? If m and n are ordered, but the problem doesn't specify. If m is first, maybe they expect m=23 and n=2? But the problem doesn't specify.Alternatively, maybe I made a mistake in the calculation. Let me check the calculations again.First case: m=2.5*2=10. 129-10=119. 119/7=17. 17 is prime. So 2 and17. Sum=19.Second case: n=2.7*2=14. 129-14=115. 115/5=23. 23 is prime. So 23 and2. Sum=25.So both are correct. Therefore, the problem has two solutions. However, the user is asking for "the value of m +n", so maybe both. But since the answer is to be boxed, perhaps the user expects both answers. But in the original problem, maybe there's a unique solution. Alternatively, maybe there's an error in my reasoning.Wait, let me check the equation again. Let's suppose that there's another prime number which when plugged into the equation gives another solution. Let's suppose that m and n are both odd primes. Then, as 5m +7n would be even (odd + odd = even), but 129 is odd. Therefore, this is impossible. So at least one of m or n must be 2. Therefore, the only possible solutions are when either m=2 or n=2. Therefore, only two possibilities, leading to two solutions. So the problem indeed has two possible answers. Therefore, the value of m +n is either 19 or25.But in the problem statement, it's written as "the value of m +n is", suggesting a single answer. Maybe the problem is from a source where there's a unique answer, but given the problem as stated, there are two. Hmm. Alternatively, maybe the problem is in a non-English language original, and the translation led to ambiguity. Alternatively, perhaps the problem assumes that m and n are distinct primes. But even in that case, both pairs (2,17) and (23,2) are distinct. Wait, unless "distinct" means both are different from each other, but in the second case, m=23 and n=2, which are distinct.Alternatively, maybe the problem expects m and n to be primes in ascending order, so m < n. Then m=2, n=17. So the sum is19. Alternatively, descending order. But unless specified, the problem can have both. So unless there's additional constraints, both are valid.Given that, perhaps the answer is either 19 or25. But since the problem is likely expecting one answer, maybe there's a miscalculation. Wait, let me check once again.Case1: m=2.Compute 5*2 +7n=10 +7n=129. 7n=119. n=17. 17 is prime. Correct.Case2: n=2.Compute 5m +7*2=5m +14=129. 5m=115. m=23. 23 is prime. Correct.So both are correct. Therefore, the answer is both 19 and25. But since the problem asks for "the value", maybe it's a mistake, or perhaps in the original problem, maybe coefficients are different. Wait, let me check the original problem again: "If the prime numbers m, n satisfy 5m +7n=129, then the value of m +n is ___."Hmm. Maybe in Vietnamese, the problem might have a unique solution. Wait, no, in Vietnamese or any language, math is the same. So unless there's a typo. Wait, maybe 129 is a typo. If it was 125 or something else. But assuming the problem is correct, there are two solutions.Alternatively, maybe the problem is from an exam where the answer is 25, and the test-writer overlooked the other solution. Alternatively, maybe the user is expected to consider only positive integers, but primes are positive by definition.Wait, unless m and n are positive primes, but they could be negative primes? But primes are defined as positive integers greater than 1. So negative numbers can't be primes. So both solutions are valid.Hmm. This is confusing. Given that, perhaps the answer is both 19 and25. But the problem asks for "the value", so maybe the answer is 19 and25. But since we need to put one answer in a box, maybe the user is expecting both? But in the instructions, it says "put your final answer within boxed{}". So perhaps I need to write both answers. But how?Alternatively, maybe the problem has a unique solution, and my initial approach is wrong. Let me try another method.Let me list possible primes for m and n and see which pair satisfies the equation.Primes less than 129: 2,3,5,7,11,13,17,19,23,29,..., up to primes where 5m and7n are less than129.But this might take time, but let's try.First, consider m=2:Then 5*2=10. 129-10=119. 119/7=17. 17 is prime. So m=2, n=17.Next, m=3:5*3=15. 129-15=114. 114/7≈16.285. Not integer. So n not integer.m=5:5*5=25. 129-25=104. 104/7≈14.857. Not integer.m=7:5*7=35. 129-35=94. 94/7≈13.428. Not integer.m=11:5*11=55. 129-55=74. 74/7≈10.571. Not integer.m=13:5*13=65. 129-65=64. 64/7≈9.142. Not integer.m=17:5*17=85. 129-85=44. 44/7≈6.285. Not integer.m=19:5*19=95. 129-95=34. 34/7≈4.857. Not integer.m=23:5*23=115. 129-115=14. 14/7=2. So n=2. Prime.m=29:5*29=145>129. So too big.Similarly, check n:Start with n=2:7*2=14. 129-14=115. 115/5=23. Prime.n=3:7*3=21. 129-21=108. 108/5=21.6. Not integer.n=5:7*5=35. 129-35=94. 94/5=18.8. Not integer.n=7:7*7=49. 129-49=80. 80/5=16. Not prime.n=11:7*11=77. 129-77=52. 52/5=10.4. Not integer.n=13:7*13=91. 129-91=38. 38/5=7.6. Not integer.n=17:7*17=119. 129-119=10. 10/5=2. Prime. So m=2.n=19:7*19=133>129. Too big.So, in this way, the only solutions are when m=2, n=17 and m=23, n=2. Therefore, two solutions.Hence, the problem has two possible answers: 19 and25.But the problem states "the value of m +n is". If it's a math competition problem, often such problems have a unique solution, so maybe I made a mistake here. Wait, perhaps the problem is from a specific source where 25 is the answer? Or maybe in the original problem, there was a condition that m >n or something. But since it's not stated, we can't assume that.Alternatively, maybe the problem is in a context where the variables are considered in the order m and then n, so perhaps they expect the smaller prime first? But 2 and17 vs23 and2, both have 2. Not sure.Alternatively, maybe the problem expects both answers, and the sum is19 and25. But in the format requested, they want the answer boxed. If it's two answers, maybe boxed{19} and boxed{25}. But the problem says "the value", not "the values", so maybe they expect one answer. Hence, there might be an error in the problem statement, or maybe in my analysis.Alternatively, perhaps the problem is expecting the sum which is also a prime? Let's check. 19 is prime, 25 is not. So if the problem is expecting the sum to be prime, then 19 would be the answer. But the problem doesn't specify that. So unless there's another condition, this is speculative.Alternatively, maybe the problem assumes m and n are distinct primes. But in both solutions, the primes are distinct. So that's not it.Hmm. Given that, since both solutions are mathematically valid, and the problem as stated doesn't restrict m and n further, I think the correct answers are 19 and25. However, since the user instruction says "put your final answer within boxed{}", maybe they expect both answers. But typically, such problems have a unique solution, so perhaps there's a mistake in either the problem or my solution.Wait, let me check once again the equations.If m=2, n=17, 5*2 +7*17=10 +119=129. Correct.If m=23, n=2,5*23 +7*2=115 +14=129. Correct.Therefore, both correct. Thus, the problem has two solutions. Therefore, the answer is both19 and25.But the problem might have a unique answer, so I need to check once more.Wait, perhaps the problem is in base 10, but maybe I need to check for other representations. Unlikely, as all numbers are standard.Alternatively, maybe the problem is from a past competition, and in that competition, the answer was 19, but others considered25 as correct. Or vice versa.Alternatively, perhaps the problem is written with a typo, such as 5m -7n=129, which would result in a different solution. But given the problem as stated, we have two solutions.Therefore, unless there's additional constraints not mentioned, both 19 and25 are correct. So if the user allows multiple answers, we can present both. Otherwise, if forced to choose one, perhaps the first solution is19. But since the problem didn't specify, as a solver, I should note both.But according to the initial problem statement, the user says: "put your final answer within boxed{}". So if the problem is expecting a single answer, but there are two, maybe I need to check if there's a misunderstanding.Wait, maybe in the problem statement, m and n are primes but not necessarily distinct. But in both solutions, they are distinct. So that doesn't affect.Alternatively, perhaps considering the equation modulo some number. For example, modulo5 or7.Let me try modulo5.The equation is5m +7n=129. Taking modulo5:5m ≡0 mod5,7n≡2n mod5,129≡129 -25*5=129-125=4 mod5.Thus, 0 +2n≡4 mod5→2n≡4 mod5→n≡2 mod5.Therefore, n≡2 mod5. So n can be primes congruent to2 mod5. Primes congruent to2 mod5 are primes ending with2 or7, since mod5 residues for primes greater than5 are1,2,3,4. So primes ≡2 mod5 are 2,7, 17, 37, etc. So possible n=2,7,17,37,... But since 7n ≤129, n≤129/7≈18.428. So n≤18. So possible n=2,7,17.Check n=2: we did that, m=23.n=7: 7*7=49. 129-49=80. 80/5=16. 16 is not prime.n=17: 7*17=119. 129-119=10. 10/5=2. m=2, which is prime.So modulo5 analysis gives n=2 and17, which correspond to the two solutions we found earlier.Similarly, take modulo7:5m +7n=129. Modulo7:5m≡5m mod7,7n≡0 mod7,129≡129 -18*7=129-126=3 mod7.Thus, 5m≡3 mod7→Multiply both sides by inverse of5 mod7. The inverse of5 mod7 is3, since5*3=15≡1 mod7. Therefore, m≡3*3=9≡2 mod7.Therefore, m≡2 mod7. So m can be2, 23, 37, etc. Since m must be prime and 5m ≤129→m≤25.8. So primes m≡2 mod7 up to25 are:2, 23.m=2: n=17.m=23:5*23=115. 129-115=14. n=14/7=2.Therefore, the two solutions again. Therefore, this method also leads to the same two solutions.Therefore, the problem indeed has two solutions. Therefore, the value of m +n is either19 or25. However, if the problem expects a single answer, maybe I need to check if there's a mistake in the problem statement, or if I misread it.But given the problem as stated, two solutions are valid. Hence, the answer is19 and25. However, in some contexts, if the problem is presented in a way where order matters, perhaps m comes before n, and the answer is19. But unless specified, both are correct.Alternatively, maybe the problem was copied incorrectly, and the equation was meant to be5m +7n=139 or another number, which would result in a unique solution. But as given,129 yields two solutions.Given the problem as stated, I think the answer should be both19 and25. However, since the user instruction says "put your final answer within boxed{}", which typically expects a single answer, maybe there's an error. Alternatively, the problem might have a unique solution if we consider the sum m +n, and the possible sums are19 and25, but in the problem's original source, maybe25 is the answer. Without more context, it's hard to say.But given mathematical rigor, both solutions are valid. Therefore, I should present both answers. However, the user might expect only one. To resolve this, perhaps check online if this problem exists elsewhere.But assuming I can't access external resources, based on my analysis, two solutions exist. Hence, the answer is either19 or25. But since the problem is asking for "the value", it's ambiguous. However, in many similar problems, if multiple solutions exist, they are both accepted. But in the context of standardized tests, often such problems are designed to have a unique solution, so likely the problem has a unique answer, and my initial approach was missing something.Wait, let's check once again: perhaps m and n are two-digit primes? Wait, m=2 is single-digit, n=17 is two-digit. m=23 is two-digit, n=2 is single-digit. If the problem implies two-digit primes, but it's not stated, so that's an invalid assumption.Alternatively, maybe the problem expects the larger sum,25, but that's arbitrary.Alternatively, maybe the problem wants the sum of the digits of m +n, but no, it says "m +n".Given the problem as stated, with no additional constraints, there are two valid solutions. Therefore, the answer is19 and25. But since the user wants the final answer in a box, and if I have to choose one, perhaps the intended answer is19. But without more information, it's hard to be certain.Alternatively, let's check which of the two solutions is more likely expected. The pair (2,17) and (23,2). Since 2 is the only even prime, and often in such equations, the smaller prime is considered first. But this is arbitrary. In some textbooks, variables are sometimes considered in alphabetical order, so m before n, but that's not a mathematical rule.Alternatively, if we consider m and n as per the coefficients in the equation,5m +7n. The coefficients are5 and7. Maybe the problem expects m to be multiplied by the smaller coefficient, hence m=2 and n=17, giving a smaller sum. But this is speculative.Alternatively, maybe the problem is from a source where the answer is25, but I need to check my calculations again.Wait, when n=2, m=23. 23+2=25. When m=2, n=17. 2+17=19. So both sums. If the problem is expecting a prime number as the sum, then19 is prime,25 is not. Maybe that's the catch. So even though there are two solutions, only19 is a prime sum. If the problem is asking for m +n where m +n is also prime, then19 is the answer. But the problem didn't specify that. So this is an additional constraint not mentioned. Therefore, unless the problem had that requirement, it's invalid to assume.Alternatively, maybe the problem is expecting the larger prime. But23 is larger than17, but the sum25 is not.Alternatively, the problem might have a unique solution if we consider that m and n are both two-digit primes, but23 is two-digit and17 is two-digit, while2 is single-digit. But both solutions include a single-digit prime. Therefore, this doesn't help.Given that, I think the problem as stated has two solutions. Therefore, the answer is19 and25. But since the user instruction says to put the final answer in a box, and maybe expects one answer, I think the problem might have a typo. But since both solutions are valid, I will mention both.Final AnswerThe value of ( m + n ) is either boxed{19} or boxed{25}.</think>Given the problem where prime numbers ( m ) and ( n ) satisfy the equation ( 5m + 7n = 129 ), we need to find the value of ( m + n ).First, we note that 129 is an odd number. Since the sum of an even number and an odd number is odd, one of ( 5m ) or ( 7n ) must be even, and the other must be odd. This implies that one of ( m ) or ( n ) must be 2 (the only even prime number).Case 1: ( m = 2 )[5(2) + 7n = 129 implies 10 + 7n = 129 implies 7n = 119 implies n = 17]Here, ( n = 17 ) is a prime number. Therefore, ( m + n = 2 + 17 = 19 ).Case 2: ( n = 2 )[5m + 7(2) = 129 implies 5m + 14 = 129 implies 5m = 115 implies m = 23]Here, ( m = 23 ) is a prime number. Therefore, ( m + n = 23 + 2 = 25 ).Both solutions are valid as they satisfy the given equation and the primes condition. Thus, the value of ( m + n ) can be either 19 or 25.[boxed{19} quad text{or} quad boxed{25}]]

🔑:&. The original wing converts to 5 m equiv 129(bmod 7), i.e., 5 m equiv 3 (modi), solving yields m equiv 2(bmod 7). Given that m is a prime number, and from the original indeterminate equation, m<25. Therefore, the possible values of boldsymbol{m} are 2 or 23. Checking, when m=2, n=17, thus, m+n=19; when m=23, n=2, m+n=25, i.e., the value of m+n is 19 or 25.

🧠:[<think>Okay, so I need to find all natural numbers n such that A = n^4 + 4n^3 +5n^2 +6n is a perfect square. Hmm, let's start by trying to understand the structure of this polynomial and see if I can factor it or rewrite it in a way that might help me determine when it's a square.First, let's write down the expression again:A = n^4 + 4n^3 + 5n^2 + 6nHmm, maybe I can factor this somehow. Let's see. Let me check if it factors into a product of quadratics or something. Alternatively, perhaps complete the square or compare it to known square expressions.Alternatively, maybe compute A for small natural numbers n and see if it's a square. If a pattern emerges, that could help. Let's start with n=1, 2, 3, etc., compute A, and check if it's a perfect square.Starting with n=1:A = 1 + 4 + 5 + 6 = 16. 16 is 4², so that's a square. So n=1 is a solution.n=2:A = 16 + 32 + 20 + 12 = 80. 80 is not a perfect square (since 8²=64 and 9²=81). So n=2 is not a solution.n=3:A = 81 + 108 + 45 + 18 = 252. 252 is between 15²=225 and 16²=256, so not a square.n=4:A = 256 + 256 + 80 + 24 = 616. 24²=576, 25²=625, so not a square.n=5:A = 625 + 500 + 125 + 30 = 1280. 35²=1225, 36²=1296, so not a square.n=6:A = 1296 + 864 + 180 + 36 = 2376. 48²=2304, 49²=2401, so not a square.n=0: Wait, natural numbers usually start from 1, but sometimes 0 is included. Let's check n=0 just in case. If n=0, A=0 +0 +0 +0=0, which is 0². But if natural numbers start at 1, then n=0 is excluded. The problem says natural numbers, so depending on the convention, but maybe they consider n starting at 1. Let's note that n=0 gives A=0, which is a square, but maybe the question wants positive integers. So we need to check if n=0 is allowed.But moving on, n=1 works, let's check n=7:A=7^4 +4*7^3 +5*7^2 +6*7=2401 + 4*343 +5*49 +42=2401 +1372 +245 +42= 2401+1372=3773, 3773+245=4018, 4018+42=4060. 4060 is between 63²=3969 and 64²=4096. Not a square.n=8:8^4=4096, 4*8^3=4*512=2048, 5*8^2=5*64=320, 6*8=48. Summing: 4096 + 2048=6144, +320=6464, +48=6512. 80²=6400, 81²=6561. Not a square.n=9:9^4=6561, 4*9^3=4*729=2916, 5*81=405, 6*9=54. Total: 6561+2916=9477, +405=9882, +54=9936. 99²=9801, 100²=10000. Not a square.n=10:10^4=10000, 4*1000=4000, 5*100=500, 6*10=60. Total A=10000+4000=14000+500=14500+60=14560. sqrt(14560)≈120.66, not integer.So up to n=10, only n=1 gives a square. Maybe n=1 is the only solution? But perhaps we need to check higher n or find a general proof.Alternatively, maybe the expression can be rewritten as a square. Let me try to manipulate A.A = n^4 +4n^3 +5n^2 +6nLet me try to write this as (n^2 + an + b)^2 + c*n + d, and see if I can find a, b, c, d such that it matches.Expanding (n^2 + an + b)^2 = n^4 + 2an^3 + (a² + 2b)n² + 2abn + b²Compare to A: n^4 +4n^3 +5n^2 +6nSo, equate coefficients:2a = 4 ⇒ a=2Then, a² + 2b = 5 ⇒ 4 + 2b=5 ⇒ 2b=1 ⇒ b=0.5. Hmm, but b should be integer if we want integer coefficients. So maybe this approach isn't directly helpful. Alternatively, perhaps adjust by another term.Alternatively, maybe A is between two consecutive squares, so it can't be a square except for small n.Let me try to compare A to (n^2 + 2n)^2 and (n^2 + 2n +1)^2.Compute (n^2 + 2n)^2 = n^4 +4n^3 +4n^2Compare to A = n^4 +4n^3 +5n^2 +6n. So A = (n^2 +2n)^2 + n^2 +6nSimilarly, (n^2 +2n +1)^2 = n^4 +4n^3 +6n^2 +4n +1So A is between these two squares:(n^2 +2n)^2 < A < (n^2 +2n +1)^2 when n^2 +6n < 2n^2 +4n +1?Wait, let's compute the difference between A and (n^2 +2n)^2:A - (n^2 +2n)^2 = (n^4 +4n^3 +5n^2 +6n) - (n^4 +4n^3 +4n^2) = n^2 +6nSimilarly, (n^2 +2n +1)^2 - A = (n^4 +4n^3 +6n^2 +4n +1) - (n^4 +4n^3 +5n^2 +6n) = n^2 -2n +1So, for n ≥1:A = (n^2 +2n)^2 +n^2 +6nBut (n^2 +2n +1)^2 = (n+1)^4, right? Wait, no:Wait, (n^2 +2n +1) = (n +1)^2, so (n^2 +2n +1)^2 = (n +1)^4. Hmm, but (n +1)^4 is n^4 +4n^3 +6n^2 +4n +1, which is correct.So A is between (n^2 +2n)^2 and (n^2 +2n +1)^2. The question is whether A can be a perfect square in between.But for A to be a perfect square, it would have to be equal to either (n^2 +2n)^2 + k, where k is a square. But since A is sandwiched between two consecutive squares (for n ≥ something), perhaps for n large enough, A is strictly between two consecutive squares, hence cannot be a square itself.Wait, but when is (n^2 +2n +1)^2 - (n^2 +2n)^2 = ( (n^2 +2n +1) - (n^2 +2n) ) * ( (n^2 +2n +1) + (n^2 +2n) ) = (1)*(2n^2 +4n +1). So the gap between these two squares is 2n^2 +4n +1. However, the difference between A and the lower square is n^2 +6n. So for n ≥1, n^2 +6n < 2n^2 +4n +1 ?Let's check:2n^2 +4n +1 - (n^2 +6n) = n^2 -2n +1 = (n -1)^2 ≥0 for all n.Therefore, n^2 +6n ≤ 2n^2 +4n +1, with equality only when n=1.Therefore, for n ≥1, the difference between A and (n^2 +2n)^2 is n^2 +6n, and the difference between (n^2 +2n +1)^2 and A is (n^2 -2n +1). So, except for n=1, where the upper difference is (1 -2 +1)=0, but wait:Wait, when n=1:A =1 +4 +5 +6=16(n^2 +2n)^2 = (1 +2)^2=9, A -9=7. The upper difference is (n^2 -2n +1)=1 -2 +1=0. Wait, that seems contradictory. Wait, let's compute (n^2 +2n +1)^2 when n=1: (1 +2 +1)^2=4²=16, which is equal to A. So for n=1, A is exactly (n^2 +2n +1)^2=16. So in this case, the upper square is A itself when n=1.But for n ≥2, (n^2 +2n)^2 < A < (n^2 +2n +1)^2. Wait, but when n=2:(n^2 +2n)^2 = (4 +4)^2=64, A=80, (n^2 +2n +1)^2=(4 +4 +1)^2=9²=81. So 64 <80<81. So A is between two consecutive squares here.Similarly, n=3: (9 +6)^2=225, A=252, (9 +6 +1)^2=16²=256. 225 <252<256. So indeed, for n ≥2, A is between (n^2 +2n)^2 and (n^2 +2n +1)^2, which are consecutive squares. Therefore, A cannot be a perfect square for n ≥2.Wait, but for n=1, A=16=4²= (n^2 +2n +1)^2= (1 +2 +1)^2=4². So that works. Then for n=0, A=0=0², which is also a square, but n=0 may not be considered a natural number depending on the definition.Therefore, the only natural numbers n where A is a perfect square are n=1 and possibly n=0 if allowed. But the problem says natural numbers, which usually start at 1. However, sometimes natural numbers are considered to include 0. The problem statement may vary. Let me check if n=0 is considered here.But since n is a natural number, and if we take the standard definition where natural numbers start from 1, then n=1 is the only solution.But just to confirm, let's check n=0. If n=0, A=0 +0 +0 +0=0, which is 0². So if 0 is included, then n=0 is also a solution. But since the problem says "natural numbers," which can be ambiguous. However, in mathematics, natural numbers often refer to positive integers (1,2,3,...). But in some contexts, they include 0. The problem might need to specify, but since it's a competition-style problem, likely n is positive integer. So answer is n=1.But to be thorough, let's check for n negative integers? Wait, no, natural numbers are non-negative or positive. So negative numbers are excluded. So n=0 and n=1. But depends on the definition.Wait, but let me check the original problem statement again: "Determine all natural numbers n for which the number A = n^4 + 4n^3 +5n^2 + 6n is a perfect square of a natural number."So they are asking for natural numbers n, and A should be a perfect square of a natural number.If natural numbers include 0, then n=0 gives A=0, which is 0², so n=0 and n=1. If natural numbers start at 1, then only n=1.But perhaps we need to check with the problem's likely intention. In Olympiad problems, natural numbers usually start at 1 unless specified otherwise. However, sometimes 0 is included. To be safe, maybe mention both? Wait, let's check n=0.Wait, let me see. If n=0, then A=0. 0 is a perfect square (0=0²). So if n=0 is allowed, then it's a solution. But depending on the definition.Given that the problem is in English, and depending on the source, in some contexts, natural numbers exclude 0. For example, in number theory, natural numbers often refer to positive integers. So maybe the answer is n=1.But to ensure correctness, perhaps we should check both possibilities.But in the computations above, for n ≥1, only n=1 gives a square. For n=0, A=0. So depending on whether 0 is considered natural here.But since the problem didn't specify, and given that the expression is n^4 + ... which is zero when n=0, which is a square, but if the problem expects 0 to be excluded, then the answer is n=1.Alternatively, maybe n=0 is considered a valid solution. Let's check the original problem statement again. It says "natural numbers n" and "natural number". If the problem is translated from another language, sometimes "natural numbers" can include 0. In French, for example, "les entiers naturels" usually include 0. In English, it's more ambiguous. However, in mathematics, particularly in number theory, natural numbers often start at 1.But since the problem is asking for "all natural numbers n", and given that n=0 yields A=0, which is a square, but whether 0 is considered a natural number here is unclear. However, in some cases, especially in problems where 0 might trivialize the problem, it's excluded. Here, if n=0 is allowed, then it's a valid solution, but if not, only n=1.But maybe the problem expects only positive integers, so n=1 is the answer. However, let's try to see if the equation holds for n=0.n=0: A=0, which is 0². If the problem allows n=0, then 0 is a solution. But given that the problem is about natural numbers, which can be a point of contention, perhaps we should state both possibilities. However, in many mathematical contexts, 0 is not considered a natural number. For example, the set of natural numbers is denoted by ℕ, and sometimes defined as {1,2,3,...}, and sometimes as {0,1,2,3,...}. The problem might need to specify, but since it doesn't, perhaps the answer is n=1. However, to be thorough, let's check if there are any other solutions beyond n=1.Wait, for n=1, A=16=4². For n=2, A=80, not a square. For n=3, 252, not a square. For n=4, 616, not a square. For n=5,1280, not a square. As n increases, A is sandwiched between two consecutive squares as shown earlier, so it can't be a square. Thus, the only possible natural number solutions are n=0 and n=1.But again, if the problem defines natural numbers starting at 1, then only n=1. However, since the problem didn't specify, maybe both. Let me check once more.Alternatively, maybe there's another approach where n is a negative integer? But natural numbers are non-negative or positive. Negative integers aren't considered here. So no.Alternatively, maybe factor the expression.Let me try to factor A =n^4 +4n^3 +5n^2 +6n.First, factor out an n:A =n(n^3 +4n^2 +5n +6)Now, let's try to factor the cubic polynomial n^3 +4n^2 +5n +6.Let's attempt rational root theorem. Possible rational roots are ±1, ±2, ±3, ±6.Testing n=-1: (-1)^3 +4*(-1)^2 +5*(-1) +6 = -1 +4 -5 +6=4. Not zero.n=-2: (-8) + 16 + (-10) +6= -8 +16=8 -10= -2 +6=4. Not zero.n=-3: (-27)+36+(-15)+6= (-27+36)=9 +(-15+6)= -9. Not zero.n=1:1 +4 +5 +6=16≠0n=2:8 +16 +10 +6=40≠0n=3:27 +36 +15 +6=84≠0n=6:216 + 144 +30 +6=396≠0So no rational roots. Therefore, the cubic doesn't factor nicely over integers. Therefore, A =n*(irreducible cubic). Hence, unless n is a square and the cubic is a square, or vice versa, but since they are coprime? Wait, n and the cubic might share a common factor. Let's check.If d divides n and divides n^3 +4n^2 +5n +6, then d divides 6. So possible common divisors are 1,2,3,6. But unless n is divisible by 2,3, or 6, the gcd is 1.But even so, to have A be a square, since n and the cubic may share a common factor, both factors need to be squares up to the common factor. But since the cubic is irreducible, this seems complicated. So maybe another approach.Alternatively, note that for n ≥2, A is between (n^2 +2n)^2 and (n^2 +2n +1)^2, so it cannot be a perfect square. Therefore, the only possible solution is n=1 (and n=0 if allowed). Therefore, the answer is n=1.But let me verify again for n=1. A=1+4+5+6=16=4². Correct. For n=0, A=0=0². So if natural numbers include 0, then n=0 and n=1. Otherwise, only n=1.But in the problem statement, since it's about natural numbers, and given that 0 is sometimes excluded, perhaps the answer is n=1.Alternatively, the problem might accept n=0 as a solution. To cover all bases, maybe we should state both possibilities. However, since the problem is presented in English and given the way the polynomial is written (with a term 6n), which is zero when n=0, leading to A=0, which is a square, but perhaps trivial. However, the problem might be expecting non-trivial solutions, so n=1.But given that mathematically, 0 is a natural number in some contexts, and since the problem doesn't specify, perhaps the correct answer is n=0 and n=1. However, in competitions, usually, natural numbers start at 1. Let me check if n=0 is considered a natural number in the context of the problem.Alternatively, let's consider that if n=0 is allowed, then A=0, which is a square, but if n=0 is excluded, then n=1. But given that the problem is from an unspecified source, but given that the user is presenting it as a problem to solve, perhaps with an expectation of n=1 as the answer. However, to ensure correctness, let's check in the literature.Alternatively, perhaps there's another approach to confirm that n=1 is the only solution.Suppose n ≥2. Then, as established, (n^2 +2n)^2 < A < (n^2 +2n +1)^2. Therefore, A cannot be a square. Therefore, the only solutions are n=0 and n=1. If n=0 is considered a natural number, otherwise only n=1.But given that the problem asks for natural numbers, which usually in math Olympiad contexts are positive integers, so n=1.Thus, the answer is n=1.Final AnswerThe natural number ( n ) is boxed{1}.</think>To determine all natural numbers ( n ) for which the number ( A = n^4 + 4n^3 + 5n^2 + 6n ) is a perfect square, we start by evaluating ( A ) for small values of ( n ):- For ( n = 1 ): [ A = 1^4 + 4 cdot 1^3 + 5 cdot 1^2 + 6 cdot 1 = 1 + 4 + 5 + 6 = 16 = 4^2 ] This is a perfect square.- For ( n = 2 ): [ A = 2^4 + 4 cdot 2^3 + 5 cdot 2^2 + 6 cdot 2 = 16 + 32 + 20 + 12 = 80 ] This is not a perfect square.- For ( n = 3 ): [ A = 3^4 + 4 cdot 3^3 + 5 cdot 3^2 + 6 cdot 3 = 81 + 108 + 45 + 18 = 252 ] This is not a perfect square.- For ( n = 4 ): [ A = 4^4 + 4 cdot 4^3 + 5 cdot 4^2 + 6 cdot 4 = 256 + 256 + 80 + 24 = 616 ] This is not a perfect square.We observe that ( A ) for ( n geq 2 ) does not yield a perfect square. To analyze further, we compare ( A ) with nearby squares. We note that:[(n^2 + 2n)^2 = n^4 + 4n^3 + 4n^2][(n^2 + 2n + 1)^2 = n^4 + 4n^3 + 6n^2 + 4n + 1]For ( n geq 2 ), we find that ( A ) lies between these two squares:[(n^2 + 2n)^2 < A < (n^2 + 2n + 1)^2]The difference between these consecutive squares is ( 2n^2 + 4n + 1 ), which is greater than the difference ( A - (n^2 + 2n)^2 = n^2 + 6n ) for ( n geq 2 ). Thus, ( A ) cannot be a perfect square for ( n geq 2 ).Checking ( n = 0 ) (if considered a natural number):[A = 0^4 + 4 cdot 0^3 + 5 cdot 0^2 + 6 cdot 0 = 0 = 0^2]This is a perfect square, but natural numbers typically start from 1 in many contexts.Therefore, the only natural number ( n ) (starting from 1) for which ( A ) is a perfect square is:[boxed{1}],<think>Okay, so I need to figure out for which natural numbers n, the expression A = n^4 + 4n³ + 5n² + 6n is a perfect square. Hmm, let's start by understanding the problem. A natural number n makes A a perfect square, meaning there exists some integer m such that A = m². First, maybe I can try plugging in small natural numbers for n and see if A is a perfect square. Let's start with n=1, 2, 3, and so on, and check if the result is a square. If I notice a pattern or find some candidates, maybe I can generalize from there.Starting with n=1:A = 1⁴ + 4*1³ +5*1² +6*1 = 1 + 4 + 5 + 6 = 16. 16 is 4², so that's a perfect square. So n=1 works.n=2:A = 16 + 32 + 20 + 12 = 16+32=48; 48+20=68; 68+12=80. 80 isn't a perfect square. 8²=64, 9²=81, so no.n=3:A = 81 + 4*27 +5*9 +6*3 = 81 + 108 +45 +18 = 81+108=189; 189+45=234; 234+18=252. 252 isn't a perfect square. 15²=225, 16²=256, so no.n=4:A = 256 + 4*64 +5*16 +6*4 = 256 + 256 +80 +24 = 256+256=512; 512+80=592; 592+24=616. 24²=576, 25²=625, so 616 isn't a square.n=5:A = 625 + 4*125 +5*25 +6*5 = 625 + 500 +125 +30 = 625+500=1125; 1125+125=1250; 1250+30=1280. 35²=1225, 36²=1296, so nope.n=6:A = 1296 + 4*216 +5*36 +6*6 = 1296 + 864 +180 +36 = 1296+864=2160; 2160+180=2340; 2340+36=2376. 48²=2304, 49²=2401, not a square.n=0: Wait, natural numbers usually start at 1, but sometimes 0 is included. Let me check n=0 just in case. If n=0, A=0 +0 +0 +0=0. 0 is a square (0²=0). But if natural numbers start at 1, then n=0 might not be considered. The problem states "natural numbers n", so depending on the definition, but often in math competitions, natural numbers start at 1. However, sometimes 0 is included. Let me note that n=0 gives A=0, which is a square, but I should check the problem's context. Since the original problem says "natural numbers", and if considering n=0, then 0 is a solution, but maybe not. I'll keep that in mind but proceed with n≥1 for now.So n=1 works. Let's check n=7:n=7:A = 7⁴ +4*7³ +5*7² +6*7 = 2401 +4*343 +5*49 +42 = 2401 +1372 +245 +42. Let's compute step by step:2401 +1372 = 37733773 +245 = 40184018 +42 = 40604060 is between 63²=3969 and 64²=4096, so not a square.n=8:8⁴=4096, 4*8³=4*512=2048, 5*8²=5*64=320, 6*8=48Total: 4096 +2048=6144; 6144+320=6464; 6464+48=6512. 80²=6400, 81²=6561, so not a square.n=9:9⁴=6561, 4*9³=4*729=2916, 5*81=405, 6*9=54Total: 6561 +2916=9477; 9477 +405=9882; 9882 +54=9936. 99²=9801, 100²=10000, so 9936 isn't a square.n=10:10⁴=10000, 4*10³=4000, 5*100=500, 6*10=60Total: 10000 +4000=14000; 14000 +500=14500; 14500 +60=14560. 120²=14400, 121²=14641, so no.So up to n=10, only n=1 works. Maybe I need to check a few more, but perhaps there's a pattern here. Let's try n=11:n=11:11⁴=14641, 4*11³=4*1331=5324, 5*121=605, 6*11=66Total: 14641 +5324=19965; 19965 +605=20570; 20570 +66=20636. 143²=20449, 144²=20736, so 20636 isn't a square.Hmm, so n=1 is the only small natural number that works. Maybe n=1 is the only solution? But how can I prove that?Alternatively, perhaps I can factor the expression A = n⁴ +4n³ +5n² +6n. Let's see if that can be factored.Looking at A = n(n³ +4n² +5n +6). Let's check if the cubic factor can be factored further. Let me try to factor n³ +4n² +5n +6.Trying rational roots theorem. Possible roots are ±1, ±2, ±3, ±6.Testing n=-1: (-1)^3 +4*(-1)^2 +5*(-1) +6 = -1 +4 -5 +6 = 4. Not zero.n=-2: (-8) +16 +(-10) +6 = (-8+16)=8; (8-10)= -2; (-2+6)=4. Not zero.n=-3: (-27)+36 +(-15)+6= (-27+36)=9; (9-15)= -6; (-6+6)=0. Yes! So n=-3 is a root. Therefore, the cubic factors as (n +3)(n² + an + b). Let's perform polynomial division or use coefficients.Divide n³ +4n² +5n +6 by (n +3). Using synthetic division:-3 | 1   4   5   6 |    -3   -3   -6 ------------------   1   1   2   0So quotient is n² +n +2. Hence, n³ +4n² +5n +6 = (n +3)(n² +n +2).Therefore, A can be written as n(n +3)(n² +n +2).So A = n(n +3)(n² +n +2). Hmm, maybe this helps. So the product of these four terms is a perfect square. For the product of multiple terms to be a perfect square, each prime factor must appear with an even exponent in the product. Alternatively, perhaps if the terms are pairwise coprime, each term must be a square. However, n, n+3, and n² +n +2 may not be pairwise coprime. Let's check the gcd between these terms.First, gcd(n, n+3). Since n and n+3 differ by 3, their gcd is gcd(n,3). Similarly, gcd(n, n² +n +2). Let's compute gcd(n, n² +n +2). Using Euclidean algorithm: gcd(n, n² +n +2) = gcd(n, (n² +n +2) - n*(n +1)) = gcd(n, 2). So gcd(n, n² +n +2) is gcd(n,2). Similarly, gcd(n+3, n² +n +2). Let's compute that. Let d be a common divisor of n+3 and n² +n +2. Then d divides n² +n +2 and n +3. Let's write n² +n +2 = (n +3)(n - 2) + 8. So n² +n +2 = (n +3)(n -2) +8. Therefore, d divides 8. So gcd(n+3, n² +n +2) divides 8.So overall, the factors n, n+3, n² +n +2 can have gcds of 1, 2, or maybe 3, but depending on n. For example, if n is divisible by 3, then gcd(n, n+3) is 3. Similarly, if n is even, gcd(n, n² +n +2) is 2, otherwise 1.Given that, the product A = n(n +3)(n² +n +2). Since these factors can share some common divisors, maybe we can factor A into prime factors and see whether their exponents are even. But since n is variable, this approach might be difficult.Alternatively, maybe A is a square between two consecutive squares, so that for n >1, A is between m² and (m+1)² for some m, hence not a square.Alternatively, maybe we can compare A to (n² +2n)² or another square nearby and see the difference.Let me try to approximate A. Let's see:n⁴ +4n³ +5n² +6n. Let's compare this to (n² + 2n)² = n⁴ +4n³ +4n². Then A is (n² +2n)² + n² +6n. So A = (n² +2n)^2 +n² +6n. Similarly, compare to (n² +2n +1)^2 = n⁴ +4n³ +6n² +4n +1. Then A is less than that: n⁴ +4n³ +5n² +6n vs n⁴ +4n³ +6n² +4n +1. The difference between (n² +2n +1)^2 and A is (6n² +4n +1) - (5n² +6n) = n² -2n +1 = (n -1)^2. So A = (n² +2n +1)^2 - (n -1)^2. Therefore, A = [ (n² +2n +1) - (n -1) ][ (n² +2n +1) + (n -1) ]Let's compute that:First term: (n² +2n +1) - (n -1) = n² +2n +1 -n +1 = n² +n +2Second term: (n² +2n +1) + (n -1) = n² +3nTherefore, A = (n² +n +2)(n² +3n). But wait, we already had A =n(n +3)(n² +n +2), so yes, that's the same. So A can be written as (n² +n +2)(n² +3n). But we also saw that A = (n² +2n)^2 +n² +6n. Hmm. Maybe another way to write A.Alternatively, since A = (n² +2n +1)^2 - (n -1)^2, which is a difference of squares, so factored as above.But how does this help? If A is expressed as a difference of two squares, perhaps that can lead to some equation. However, since we want A itself to be a square, say m², then we have m² = (n² +2n +1)^2 - (n -1)^2. Which can be rewritten as m² + (n -1)^2 = (n² +2n +1)^2. So this resembles a Pythagorean triple: m, (n -1), and (n² +2n +1). So perhaps this is a Pythagorean triple. But I don't know if that helps.Alternatively, maybe set k = n² +2n +1, so k = (n +1)^2. Then A = k² - (n -1)^2. So m² = k² - (n -1)^2, meaning k² - m² = (n -1)^2. Which factors as (k - m)(k + m) = (n -1)^2. Since (n -1)^2 is a square, the product of (k -m) and (k + m) must both be squares or both be twice squares or something. But since k and m are positive integers, and k > m (since k² - m² is positive), then (k - m) and (k + m) are both positive divisors of (n -1)^2, and they multiply to (n -1)^2. Also, (k - m) and (k + m) must both be squares or their product is a square, but since they are coprime? Wait, not necessarily. Let's see.Let’s note that (k - m) and (k + m) are two factors of (n -1)^2 such that their product is (n -1)^2. Also, since k - m and k + m have the same parity (since k and m are both integers, so k - m and k + m are both even or both odd). However, since k = (n +1)^2, which is a square, and m is an integer. Let's see:Let’s suppose that (k - m) and (k + m) are two factors of (n -1)^2 such that:(k - m)*(k + m) = (n -1)^2Also, since (k - m) < (k + m), and both are divisors of (n -1)^2, which is a square. Therefore, each factor must themselves be squares. Because in order for the product of two factors to be a square and they are co-prime, they must each be squares. However, are (k - m) and (k + m) co-prime?Let’s check the gcd of (k - m) and (k + m). Let d be a common divisor. Then d divides (k + m) + (k - m) = 2k and (k + m) - (k - m) = 2m. Therefore, d divides 2k and 2m. Since k = (n +1)^2, and m is some integer, so d divides 2*((n +1)^2) and 2m. However, unless we know more about m, this is tricky. But since (k - m)*(k + m) = (n -1)^2, and (n -1)^2 is a square, and (k - m) and (k + m) are factors, perhaps they are both squares.Alternatively, since (n -1)^2 is a square, we can set (k - m) = a² and (k + m) = b² where a and b are positive integers with a < b and ab = n -1. Because (a²)(b²) = (ab)^2 = (n -1)^2, so ab = n -1. Therefore, we have:k - m = a²k + m = b²Adding these two equations: 2k = a² + b² => k = (a² + b²)/2Subtracting them: 2m = b² - a² => m = (b² - a²)/2Since k must be an integer (as k = (n +1)^2 is an integer), (a² + b²)/2 must be an integer. Therefore, a² and b² must be both even or both odd. Since a and b are integers, their squares are both even or both odd. Therefore, a and b must be both even or both odd. In either case, (a² + b²) is even, so k is an integer.Also, ab = n -1. So n = ab +1.But since k = (n +1)^2 = (ab +1 +1)^2 = (ab +2)^2. But also k = (a² + b²)/2. Therefore, we have:(ab + 2)^2 = (a² + b²)/2Multiply both sides by 2:2(ab + 2)^2 = a² + b²Let’s expand the left side:2(a²b² +4ab +4) = a² + b²=> 2a²b² +8ab +8 = a² + b²Bring all terms to left:2a²b² +8ab +8 -a² -b² =0Hmm, this looks complicated, but perhaps we can rearrange terms:2a²b² -a² -b² +8ab +8 =0Maybe factor terms:Let me group terms:(2a²b² -a² -b²) + (8ab +8)=0Factor the first group: a²(2b² -1) -b². Not obvious. Maybe factor as:= a²(2b² -1) -b² +8ab +8Hmm, not helpful. Alternatively, maybe set x = ab, and see if we can express in terms of x. Let x=ab. Then we have:2x² +8x +8 -a² -b² =0But a and b are integers with x=ab, so maybe this approach is not helpful.Alternatively, consider that this equation must hold for integers a and b. Let's try small values of a and b to see if there are solutions. Since ab = n -1 and n is a natural number, ab must be a non-negative integer (but since a and b are positive integers, ab ≥1). Wait, actually n=1 gives ab=0, but n=1 was already a solution. However, in our previous parametrization, we set n = ab +1, so when a or b is 0, but since a and b are positive integers (as factors), ab must be at least 1, so n ≥2. But n=1 gives ab=0, which would mean a=0 or b=0, but since we started with (k -m) and (k +m) being positive integers, a and b must be positive integers. Therefore, n=1 isn't captured by this parametrization, which is why n=1 was a separate solution. Therefore, we can focus on n ≥2 and see if there are solutions.So for n ≥2, we need to find positive integers a,b with ab =n -1 and 2(ab +2)^2 = a² + b².Let me try small values of a and b.Start with a=1. Then ab =n-1 => b= n -1. Substitute into the equation:2( (1)(b) +2 )² =1 + b²=> 2(b +2)^2 =1 + b²Expand left side:2(b² +4b +4) =1 + b²=> 2b² +8b +8 =1 + b²=> b² +8b +7 =0Discriminant: 64 -28=36. Solutions: (-8 ±6)/2. Negative roots. So no solution.a=1 does not give any positive integer b.a=2, then ab =n -1 => b=(n -1)/2. Since b must be integer, n -1 must be even. So n is odd. Let's substitute a=2 into the equation:2(2b +2)^2 =4 +b²=> 2(4b² +8b +4) =4 +b²=>8b² +16b +8 =4 +b²=>7b² +16b +4=0Discriminant: 256 -112=144. sqrt(144)=12. Solutions: (-16 ±12)/14. (-16 +12)/14= -4/14= -2/7; (-16 -12)/14=-28/14=-2. Both negative. No solution.a=3. Then ab =n -1 => b=(n -1)/3. So n-1 must be divisible by 3. Substitute a=3:2(3b +2)^2 =9 +b²=>2(9b² +12b +4)=9 +b²=>18b² +24b +8 =9 +b²=>17b² +24b -1=0Discriminant: 576 +68=644. sqrt(644)≈25.37. Not integer. No solution.a=4. ab=n -1 => b=(n-1)/4. n-1 must be divisible by4. Substitute a=4:2(4b +2)^2=16 +b²=>2(16b² +16b +4)=16 +b²=>32b² +32b +8=16 +b²=>31b² +32b -8=0Discriminant:1024 + 992=2016. sqrt(2016)=~44.9. Not integer. No solution.a=5. Similarly:2(5b +2)^2=25 +b²=>2(25b² +20b +4)=25 +b²=>50b² +40b +8=25 +b²=>49b² +40b -17=0Discriminant:1600 + 3332=4932. sqrt(4932)=~70.2. Not integer.a=6:2(6b +2)^2=36 +b²=>2(36b² +24b +4)=36 +b²=>72b² +48b +8=36 +b²=>71b² +48b -28=0Discriminant:2304 + 8032=10336. sqrt(10336)=101.66... Not integer.Hmm, this approach isn't yielding solutions. Maybe a different approach. Let's consider symmetry. Suppose a=b. Then ab =a² =n -1. Then the equation becomes:2(a² +2)^2 =a² +a²=2a²Thus:2(a² +2)^2 =2a² => (a² +2)^2 =a² => a⁴ +4a² +4 =a² =>a⁴ +3a² +4=0. No real solutions, let alone integer ones.Alternatively, maybe set a=1 and b= some number, but we saw that a=1 didn't work. Similarly, maybe set b=1. Let's try b=1. Then ab =n -1 =>a= n -1. Substitute into the equation:2(a*1 +2)^2 =a² +1=>2(a +2)^2 =a² +1=>2(a² +4a +4)=a² +1=>2a² +8a +8 =a² +1=>a² +8a +7=0. Solutions are a=(-8 ±√(64-28))/2= (-8 ±√36)/2=(-8 ±6)/2. So a=(-8+6)/2=-1 or a=(-8-6)/2=-7. Not positive. No solution.b=2. Then ab=n -1 =>a=(n -1)/2. Substitute into the equation:2(a*2 +2)^2 =a² +4=>2(2a +2)^2 =a² +4=>2(4a² +8a +4)=a² +4=>8a² +16a +8 =a² +4=>7a² +16a +4=0. Discriminant 256 -112=144. Solutions: (-16 ±12)/14. (-16 +12)/14=-4/14=-2/7; (-16 -12)/14=-28/14=-2. No positive solutions.b=3. ab=n -1 =>a=(n -1)/3. Substitute:2(a*3 +2)^2 =a² +9=>2(3a +2)^2 =a² +9=>2(9a² +12a +4)=a² +9=>18a² +24a +8 =a² +9=>17a² +24a -1=0. Discriminant 576 +68=644. Not a square.Hmm, no solutions here either. So seems like there are no solutions for a and b as positive integers, meaning that for n ≥2, there are no solutions? But wait, n=1 isn't captured here because when n=1, ab=0, which would require a=0 or b=0, but our parametrization requires a,b ≥1. Therefore, n=1 is the only solution.Alternatively, perhaps there is a larger n where A becomes a square again. Let's test n=4:Wait, already checked up to n=11, none worked. Maybe n=12:n=12:12⁴=20736, 4*12³=4*1728=6912, 5*144=720, 6*12=72.Total: 20736 +6912=27648; 27648 +720=28368; 28368 +72=28440. sqrt(28440)≈168.6, since 168²=28224, 169²=28561. Not a square.n=13:13⁴=28561, 4*13³=4*2197=8788, 5*169=845, 6*13=78.Total:28561 +8788=37349; 37349 +845=38194; 38194 +78=38272. sqrt(38272)≈195.6, 195²=38025, 196²=38416. Not a square.n=14:14⁴=38416, 4*14³=4*2744=10976, 5*196=980, 6*14=84.Total:38416 +10976=49392; 49392 +980=50372; 50372 +84=50456. sqrt(50456)≈224.6, 224²=50176, 225²=50625. Not a square.n=15:15⁴=50625, 4*15³=4*3375=13500, 5*225=1125, 6*15=90.Total:50625 +13500=64125; 64125 +1125=65250; 65250 +90=65340. sqrt(65340)=255.6, 255²=65025, 256²=65536. Not a square.n=16:16⁴=65536, 4*16³=4*4096=16384, 5*256=1280, 6*16=96.Total:65536 +16384=81920; 81920 +1280=83200; 83200 +96=83296. sqrt(83296)=~288.6, since 288²=82944, 289²=83521. Not a square.Hmm, seems like even as n increases, A is sandwiched between two consecutive squares but isn't a square itself. The difference between consecutive squares grows as n increases, but A seems to fall in between. So perhaps n=1 is the only solution.Alternatively, let's attempt to analyze A as a square. Suppose A = m². Then:n⁴ +4n³ +5n² +6n =m²Let me consider the expression n⁴ +4n³ +5n² +6n. Maybe complete the square in terms of n². Let's see:n⁴ +4n³ +5n² +6n = (n² +2n)^2 +n² +6n. As before.So (n² +2n)^2 +n² +6n = m²Let me set k =n² +2n. Then m² =k² +n² +6n. So we have m² =k² +n² +6n.But k =n² +2n, so substituting back:m² = (n² +2n)^2 +n² +6nBut perhaps express n² +6n in terms of k. Since k =n² +2n, then n² =k -2n. Therefore, n² +6n =k -2n +6n =k +4n. So:m² =k² +k +4nBut k =n² +2n, so substitute back:m² = (n² +2n)^2 + (n² +2n) +4n= n⁴ +4n³ +4n² +n² +2n +4n= n⁴ +4n³ +5n² +6nWhich brings us back. Not helpful.Alternatively, perhaps bound A between two consecutive squares. For n ≥2, can I find an m such that m² < A < (m+1)², hence A is not a square?Let's consider A =n⁴ +4n³ +5n² +6n. Let's compare this to (n² +2n)^2 =n⁴ +4n³ +4n². Then A = (n² +2n)^2 +n² +6n. So A is larger than (n² +2n)^2. Similarly, compare to (n² +2n +1)^2 =n⁴ +4n³ +6n² +4n +1. Then A =n⁴ +4n³ +5n² +6n = (n² +2n +1)^2 - (n² -2n -1). Wait, let's compute the difference:(n² +2n +1)^2 - A = [n⁴ +4n³ +6n² +4n +1] - [n⁴ +4n³ +5n² +6n] = (6n² -5n²) + (4n -6n) +1 =n² -2n +1=(n -1)^2So A = (n² +2n +1)^2 - (n -1)^2Therefore, A = [(n² +2n +1) - (n -1)][(n² +2n +1) + (n -1)] = (n² +n +2)(n² +3n)Which we had before. So, A is equal to (n² +3n)(n² +n +2). Let's see if we can bound this between two squares.Note that for n ≥2, n² +3n <n² +n +2? Let's check for n=2: 4 +6=10 vs 4 +2 +2=8. No, 10 >8. For n=3:9 +9=18 vs9 +3 +2=14. Still n² +3n >n² +n +2. So actually, the two factors are ordered as n² +3n >n² +n +2 for n ≥2. So we have A = (larger factor)(smaller factor). Let me denote F1 =n² +3n and F2 =n² +n +2. Then A =F1*F2.Let me see the geometric mean of F1 and F2. The geometric mean is sqrt(F1*F2)=sqrt(A). But if F1 and F2 are not equal, then sqrt(A) is less than (F1 +F2)/2. Wait, but since A is a product, and we are trying to see if it's a perfect square.Alternatively, if F1 and F2 are consecutive numbers, their product can be a square only if they are both squares. But F1 and F2 are not consecutive. Let's compute F1 - F2 = (n² +3n) - (n² +n +2) =2n -2. For n ≥2, this is ≥2.So F1 and F2 differ by 2n -2. For n ≥2, the difference is at least 2. Therefore, F1 and F2 are two distinct integers differing by at least 2. For their product to be a square, they must both be squares multiplied by some common factors. But since their gcd is?Earlier, we saw that gcd(F1, F2) divides 8. Let's compute gcd(n² +3n, n² +n +2). Using the Euclidean algorithm:gcd(n² +3n, n² +n +2) = gcd(n² +3n, (n² +3n) - (n² +n +2)) = gcd(n² +3n, 2n -2)Then, gcd(n² +3n, 2n -2). Let d divide n² +3n and 2n -2. Then d divides 2n -2 and n² +3n. Let me write n² +3n = q*(2n -2) + r. Let's perform polynomial division:Divide n² +3n by 2n -2:n² +3n = (2n -2)*(n/2 + 2) + (2n +4)Wait, but coefficients must be integers. Let me write it as:n² +3n = (2n -2)*( (n + 4)/2 ) - ( - (2n +4)/1 )Wait, maybe better to use the Euclidean algorithm steps:Let d divide 2n -2 and n² +3n. Then d divides 2*(n² +3n) -n*(2n -2) = 2n² +6n -2n² +2n =8n. And d divides 2n -2. Therefore, d divides gcd(8n, 2n -2). Let's compute gcd(8n, 2n -2).Let d divide 8n and 2n -2. Then d divides 8n and 2n -2. Compute gcd(8n, 2n -2):= gcd(2n -2, 8n mod (2n -2)).Compute 8n divided by (2n -2):8n =4*(2n -2) +8. So 8n mod (2n -2) =8.Therefore, gcd(2n -2,8). Thus, d divides gcd(2n -2,8). Therefore, gcd(F1, F2)=gcd(2n -2,8)/?Wait, backtracking:The gcd(n² +3n, n² +n +2)=gcd(2n -2,8). Wait, but gcd(n² +3n, n² +n +2)=gcd(2n -2,8). Therefore, gcd(F1, F2)=gcd(2n -2,8). Let me compute that.gcd(2n -2,8). Since 2n -2=2(n -1). So gcd(2(n -1),8). The gcd is 2*gcd(n -1,4). Therefore, gcd(F1, F2)=2*gcd(n -1,4).Therefore, the gcd of F1 and F2 is 2*gcd(n -1,4). So depending on n:If n-1 is divisible by 4, i.e., n ≡1 mod4, then gcd=2*4=8If n-1 is 2 mod4, then gcd=2*2=4If n-1 is odd, i.e., n is even, then gcd=2*1=2So, depending on n, gcd(F1,F2) is 8,4, or2.Therefore, F1 and F2 can be written as d * a and d * b, where d is gcd(F1,F2), and a and b are coprime. Then A = d² *a*b. For A to be a perfect square, a*b must be a perfect square. Since a and b are coprime, then a and b must each be perfect squares.Therefore, write F1 = d * a², F2 = d * b², where a and b are coprime. Then A =d² *a² *b²=(dab)^2. Therefore, A is a square if and only if F1 and F2 are each d times a square, with d being their gcd.Therefore, to have A be a square, we need both F1/d and F2/d must be squares. Let's explore this.Given that gcd(F1,F2)=d=2*gcd(n -1,4). Let's consider cases based on n mod4.Case1: n ≡1 mod4. Then gcd=8. So F1=8*a², F2=8*b². Then:F1 =n² +3n =8a²F2 =n² +n +2=8b²Subtracting these two equations: (n² +3n) - (n² +n +2)=2n -2=8a² -8b²Thus 2n -2=8(a² -b²) =>n -1=4(a² -b²)But since n ≡1 mod4, n -1≡0 mod4, so 4 divides n -1, which matches. Therefore, n -1=4(a² -b²). Let’s denote k =a² -b². Then n=4k +1.But also, from F1=8a²= n² +3n. Substitute n=4k +1:(4k +1)^2 +3*(4k +1)=16k² +8k +1 +12k +3=16k² +20k +4=8a²Simplify:16k² +20k +4=8a² =>Divide both sides by4:4k² +5k +1=2a²Similarly, from F2=8b²= n² +n +2. Substitute n=4k +1:(4k +1)^2 + (4k +1) +2=16k² +8k +1 +4k +1 +2=16k² +12k +4=8b²Divide by4:4k² +3k +1=2b²So now we have two equations:4k² +5k +1=2a² ...(1)4k² +3k +1=2b² ...(2)Subtract (2) from (1):(4k² +5k +1) - (4k² +3k +1)=2k=2a² -2b² =>k= a² -b²But from the earlier substitution, n=4k +1, and k= a² -b². Therefore, k must be a positive integer (since n≥1 and n=4k +1, so k≥0. But if k=0, n=1, which we already considered. Let's check k=0:If k=0:Equation (1):0 +0 +1=2a² =>1=2a², which is impossible. Therefore, k must be at least1.So k≥1. Let's look for integer solutions (k,a,b) with k≥1 and equations (1) and (2).But how can we solve these equations? Let’s subtract k= a² -b² from both equations.From equation (1):4k² +5k +1=2a²But a² =k +b². Substitute into equation (1):4k² +5k +1=2(k +b²)=2k +2b²From equation (2):4k² +3k +1=2b² =>2b²=4k² +3k +1. Substitute into the above:4k² +5k +1=2k + (4k² +3k +1)Simplify right-hand side:2k +4k² +3k +1=4k² +5k +1Which is equal to the left-hand side. Therefore, the equations are consistent given k= a² -b². But this doesn't help us find specific solutions. Therefore, we can only use trial and error.Start with k=1:From k=1, n=4*1 +1=5.Check equations:Equation (1):4*1 +5*1 +1=4+5+1=10=2a² =>a²=5. Not integer.Equation (2):4*1 +3*1 +1=8=2b² =>b²=4 =>b=2. But a²=5 isn't integer. Contradiction.k=2:n=4*2 +1=9.Equation(1):4*4 +5*2 +1=16 +10 +1=27=2a² =>a²=13.5. Not integer.Equation(2):4*4 +3*2 +1=16 +6 +1=23=2b² =>b²=11.5. Not integer.k=3:n=13.Equation(1):4*9 +5*3 +1=36 +15 +1=52=2a² =>a²=26. Not integer.Equation(2):4*9 +3*3 +1=36 +9 +1=46=2b² =>b²=23. Not integer.k=4:n=17.Equation(1):4*16 +5*4 +1=64 +20 +1=85=2a² =>a²=42.5. Nope.Equation(2):4*16 +3*4 +1=64 +12 +1=77=2b² =>b²=38.5. Nope.k=5:n=21.Equation(1):4*25 +5*5 +1=100 +25 +1=126=2a² =>a²=63. Not integer.Equation(2):4*25 +3*5 +1=100 +15 +1=116=2b² =>b²=58. Not integer.So up to k=5, no solutions. This suggests that for n ≡1 mod4, there are no solutions other than possibly n=1, which was already a solution but doesn't fit into this case (since k=0 is invalid here).Case2: n ≡2 mod4. Then gcd=4. So F1=4a², F2=4b². Thus:n² +3n=4a² ...(3)n² +n +2=4b² ...(4)Subtract (4) from (3):(n² +3n) - (n² +n +2)=2n -2=4a² -4b² =>n -1=2(a² -b²)Since n ≡2 mod4, n -1≡1 mod2. So n -1 is odd, which equals 2(a² -b²). But 2*(integer) is even, so LHS is odd and RHS even. Contradiction. Therefore, no solutions in this case.Case3: n ≡3 mod4. Then gcd=2. So F1=2a², F2=2b². Thus:n² +3n=2a² ...(5)n² +n +2=2b² ...(6)Subtract (6) from (5):2n -2=2a² -2b² =>n -1=a² -b²So n= a² -b² +1Now, substitute into equation (5):n² +3n=2a²But n= a² -b² +1. Let's substitute:(a² -b² +1)^2 +3(a² -b² +1)=2a²Expand the square:a⁴ -2a²b² +b⁴ +2a² -2b² +1 +3a² -3b² +3=2a²Combine like terms:a⁴ -2a²b² +b⁴ +2a² -2b² +1 +3a² -3b² +3 -2a²=0Simplify:a⁴ -2a²b² +b⁴ +3a² -5b² +4=0This equation seems complicated. Maybe trial and error with small a and b.Alternatively, since n= a² -b² +1, and n≡3 mod4, then a² -b² +1≡3 mod4 =>a² -b²≡2 mod4. Since squares mod4 are 0 or1. Therefore:If a is even: a²≡0 mod4; if a is odd: a²≡1 mod4. Same with b.Possible cases:Case a even, b even: a² -b²≡0 -0=0≡0 mod4. 0≡2 mod4? No.Case a even, b odd: a² -b²≡0 -1= -1≡3 mod4. 3≡2 mod4? No.Case a odd, b even: a² -b²≡1 -0=1≡1 mod4. 1≡2 mod4? No.Case a odd, b odd: a² -b²≡1 -1=0≡0 mod4. 0≡2 mod4? No.Therefore, no solutions exist in this case because a² -b² ≡2 mod4 is impossible. Hence, no solutions for n ≡3 mod4.Case4: n ≡0 mod4. Then gcd=2*gcd(n -1,4). Since n≡0 mod4, n -1≡3 mod4. Thus, gcd(n -1,4)=1, so gcd(F1,F2)=2*1=2. Similar to Case3.Therefore, F1=2a², F2=2b², leading to:n² +3n=2a² ...(7)n² +n +2=2b² ...(8)Subtract (8)-(7):-2n +2=2b² -2a² =>-n +1= b² -a² =>n= a² -b² +1Same as in Case3. Then similarly substitute into equation (7):(a² -b² +1)^2 +3(a² -b² +1)=2a²Which leads to the same complicated equation. But since n≡0 mod4, let's see the parity. Let n=4k. Then from n= a² -b² +1=4k.Thus, a² -b²=4k -1. Let's see if possible.But squares mod4 are 0 or1. So a² -b² mod4 can be:If a even, b even:0 -0=0a even, b odd:0 -1= -1≡3a odd, b even:1 -0=1a odd, b odd:1 -1=0Thus, a² -b² ≡0,1,3 mod4. But 4k -1≡3 mod4. Therefore, to have a² -b²≡3 mod4, we need a even and b odd.Therefore, a is even, b is odd. Let a=2p, b=2q +1, where p,q are integers. Then:a² -b²=4p² - (4q² +4q +1)=4(p² -q² -q) -1=4k -1Therefore, 4(p² -q² -q) -1=4k -1 =>p² -q² -q=kSo n=4k=4(p² -q² -q)But also, from equation (7):n² +3n=2a². Substitute n=4k, a=2p:(4k)^2 +3*4k=2*(4p²)=>16k² +12k=8p²Divide by4:4k² +3k=2p²Similarly, from equation (8):n² +n +2=2b². Substitute n=4k, b=2q +1:(16k²) +4k +2=2*(4q² +4q +1)=>16k² +4k +2=8q² +8q +2Subtract 2:16k² +4k=8q² +8qDivide by4:4k² +k=2q² +2qBut we have from k= p² -q² -q. Substitute into the above equation:4(p² -q² -q)^2 + (p² -q² -q)=2q² +2qThis seems very complex. Let's try small values.Let’s attempt small q and p. Let q=0:Then from k=p² -0 -0=p².From equation 4k² +k=0 +0=0. So 4k² +k=0. Only solution k=0. Then p=0. Then n=4k=0. But n=0 might not be considered a natural number. If allowed, n=0 gives A=0, which is a square. But assuming natural numbers start at 1, skip.q=1:From equation 4k² +k=2*1 +2*1=4.Also, k=p² -1 -1=p² -2.So substitute k= p² -2 into 4(p² -2)^2 + (p² -2)=4Expand:4(p⁴ -4p² +4) +p² -2=4p⁴ -16p² +16 +p² -2=4p⁴ -15p² +14=4Thus:4p⁴ -15p² +10=0Let x=p²:4x² -15x +10=0Discriminant=225 -160=65. Not a square. No integer solution.q=2:4k² +k=2*4 +2*2=8 +4=12Also, k=p² -4 -2=p² -6Substitute:4(p² -6)^2 + (p² -6)=12Expand:4(p⁴ -12p² +36) +p² -6=4p⁴ -48p² +144 +p² -6=4p⁴ -47p² +138=12Thus:4p⁴ -47p² +126=0Let x=p²:4x² -47x +126=0Discriminant=2209 -2016=193. Not a square. No solution.q=3:4k² +k=2*9 +2*3=18 +6=24k=p² -9 -3=p² -12Substitute:4(p² -12)^2 + (p² -12)=24Expand:4(p⁴ -24p² +144) +p² -12=4p⁴ -96p² +576 +p² -12=4p⁴ -95p² +564=24Thus:4p⁴ -95p² +540=0Let x=p²:4x² -95x +540=0Discriminant=9025 -8640=385. Not a square. No solution.q= -1: But q is a non-negative integer. Disregard.This approach isn't yielding solutions. Perhaps there are no solutions for n ≡0 mod4.Therefore, in all cases, except for n=1, there are no solutions. Thus, the only natural number n for which A is a perfect square is n=1.Final AnswerThe only natural number is boxed{1}.</think>To determine all natural numbers ( n ) for which the number ( A = n^4 + 4n^3 + 5n^2 + 6n ) is a perfect square, we start by evaluating ( A ) for small values of ( n ):- For ( n = 1 ): [ A = 1^4 + 4 cdot 1^3 + 5 cdot 1^2 + 6 cdot 1 = 1 + 4 + 5 + 6 = 16 = 4^2 ] This is a perfect square.- For ( n = 2, 3, 4, ldots, 10 ), the values of ( A ) are not perfect squares.Next, we factorize ( A ):[A = n(n + 3)(n^2 + n + 2)]We analyze the factors and their greatest common divisors (GCD). For ( A ) to be a perfect square, the product of the factors must be a square, implying each prime factor must have an even exponent. We use the Euclidean algorithm and properties of GCD to determine the possible values of ( n ).We express ( A ) as a difference of squares:[A = (n^2 + 2n + 1)^2 - (n - 1)^2]This factors into:[A = (n^2 + n + 2)(n^2 + 3n)]We check if this product can be a perfect square by considering different cases for ( n ) modulo 4. Through detailed analysis, we find that for ( n equiv 1 mod 4 ), ( n equiv 2 mod 4 ), ( n equiv 3 mod 4 ), and ( n equiv 0 mod 4 ), there are no solutions other than ( n = 1 ).After extensive trials and analysis, we conclude that the only natural number ( n ) for which ( A ) is a perfect square is ( n = 1 ).[boxed{1}]]

🔑:To determine all natural numbers ( n ) for which the number ( A = n^4 + 4n^3 + 5n^2 + 6n ) is a perfect square, we start by factoring ( A ).1. Factorization of ( A ): [ A = n^4 + 4n^3 + 5n^2 + 6n = n(n+3)(n^2+n+2) ]2. Greatest Common Divisors (GCD) Analysis: - ( gcd(n, n+3) ) is either 1 or 3. - ( gcd(n, n^2+n+2) ) is either 1 or 2. - ( gcd(n+3, n^2+n+2) ) is either 1, 2, 4, or 8. This is because: [ n^2 + n + 2 = (n-2)(n+3) + 8 ] If ( p ) divides both ( n+3 ) and ( n^2+n+2 ), then ( p ) must divide 8.3. Case Analysis: - Case 1: All ( n, n+3, n^2+n+2 ) are square numbers. - If ( n ) and ( n+3 ) are square numbers, then ( n = 1 ) (since 1 is the only natural number such that both ( n ) and ( n+3 ) are squares). - For ( n = 1 ): [ n^2 + n + 2 = 1^2 + 1 + 2 = 4 ] Hence, ( n = 1 ) is a solution. - Case 2: ( n = 3k^2 Rightarrow n+3 = 3(k^2+1) ). - ( n+3 ) is never three times a square number. - Case 3: ( n = 2k^2 Rightarrow n+3 = 2k^2 + 3 ). - ( n+3 ) is never a square number because if ( 2k^2 + 3 = r^2 ) for integer ( r ), then: [ 2k^2 equiv r^2 pmod{3} ] This implies ( k equiv 0 pmod{3} ) and ( r equiv 0 pmod{3} ), leading to a contradiction. - Case 4: ( n+3 = 2k^2 Rightarrow n = 2k^2 - 3 ). - ( n ) is never a square number (same argument as in Case 3). - Case 5: ( n+3 = 4k^2 Rightarrow n = (2k)^2 - 3 ). - For ( n ) to be a square number, ( k = 1 ): [ n = 1 ] We have already seen that ( n = 1 ) is a solution. - Case 6: ( n+3 = 8k^2 = 2(2k)^2 ). - This is the same as Case 4.4. Conclusion: - From the above cases, ( A ) will be a perfect square if and only if ( n = 1 ).The final answer is ( boxed{ n = 1 } ).

🧠:[<think>Okay, so I need to figure out the minimum number of dice Vasya should use so that each ordered combination of three digits from 000 to 999 can be formed using three different dice, with each die showing one digit. Also, the digits 6 and 9 can't be rotated into each other. Hmm, this sounds like a problem related to combinatorics and possibly covering codes or something. Let me break it down.First, each die has 6 faces, and Vasya writes one digit on each face. The goal is to cover all possible 3-digit ordered combinations by selecting three different dice and arranging them in order. So, each die is responsible for one position in the 3-digit number, but since the order matters, we need to make sure that for any combination of digits in the hundreds, tens, and units place, there exists three dice (one for each place) that can show those digits when appropriately rotated. Also, since each die is assigned to a specific position (hundreds, tens, units), we might need to consider the digits needed for each place.Wait, but the problem says "selecting some three different dice and placing them with the appropriate sides up in the correct order." So, the three dice can be any three from the set, but they need to be different. So, perhaps the dice aren't pre-assigned to positions, but instead, for any three-digit number, you can pick three different dice and arrange them in order such that each die shows the respective digit when rotated. That complicates things a bit. So, the same die can be used in any position (hundreds, tens, or units) as long as it's part of a trio for that specific number. But each digit in the number must come from a different die. So, each die can contribute a digit to any position, but each position in the number must be covered by a different die.Therefore, the problem reduces to covering all 1000 three-digit combinations with trios of dice, where each trio can cover multiple combinations depending on the digits on their faces and the rotations. However, since each die has only 6 faces, each die can only represent 6 different digits. Also, since 6 and 9 can't be rotated into each other, we have to treat them as separate digits. So, effectively, each die can contribute up to 6 unique digits. But each digit from 0-9 must be covered in each of the three positions (hundreds, tens, units). Wait, actually, no. Because the problem is about ordered combinations, so each position (hundreds, tens, units) can have digits from 0-9, but each die can be used in any position. So, if a die has a certain set of digits, those digits can be used in any position. However, for a given three-digit number, say 123, we need three different dice, each contributing one of the digits 1, 2, 3. Similarly, for 456, we need three different dice contributing 4, 5, 6. But the same die can be part of different trios for different numbers. So, the problem is similar to a covering problem where each die is a set of digits (with 6 elements), and we need to cover all possible triples (a, b, c) where a, b, c are digits from 0-9, such that for each triple, there exists three different dice, one containing a, one containing b, and one containing c. The question is: what's the minimal number of dice needed so that the union of all trios of dice can cover all 1000 combinations.This seems related to the concept of covering codes or maybe the set cover problem. However, set cover is NP-hard, so perhaps there's a combinatorial design that applies here.First, let's note that each die can cover 6 digits. If we have n dice, each with 6 digits, then the total number of digits available across all dice is 6n. However, since each digit from 0-9 must be covered in each position, we need each digit to be present in multiple dice. Because for each digit d, there are 100 numbers where d is in the hundreds place, 100 where it's in the tens, and 100 where it's in the units. So, each digit needs to appear in at least three different dice (one for each position), but actually, since a die can be used in any position, perhaps the same die can cover a digit in multiple positions. Wait, but no. Because for a given number, each digit must come from a different die. Therefore, if a die has digit d on it, then d can be used in any position, but if you need two different numbers that have d in different positions, you need different dice for each occurrence. Wait, maybe not. Let me think.Suppose we have a die with digit d. Then, for any number where d is in the hundreds place, we can use that die as the hundreds die. Similarly, for a number where d is in the tens place, we can use that die as the tens die. However, if a number has d in both hundreds and tens place, we need two different dice with d. Therefore, each occurrence of a digit in a different position in a number must be covered by a different die. Therefore, for each digit d, how many times does it appear across all positions? In total, each digit appears 300 times (100 in each of the three positions). Each die with digit d can cover up to 100 numbers for each position. Wait, not exactly. Because if a die has digit d, it can be used in any position, but once it's used in a position for a number, it cannot be used for another digit in the same number.Wait, maybe this is getting too complicated. Let's step back.Each three-digit number requires three distinct dice, each contributing one digit. So, each die can contribute to multiple numbers, but each number requires three unique dice.If we model this as a hypergraph where each die is a vertex and each three-digit number is a hyperedge connecting three vertices (dice), then the problem reduces to covering all hyperedges with vertices such that each hyperedge is connected to three distinct vertices. However, constructing such a hypergraph where each die (vertex) can cover multiple hyperedges (numbers) by contributing one of its digits.But perhaps another approach is needed. Let's think about the digits required. Each die can have up to 6 digits. For each digit, we need it to be present on enough dice such that for every occurrence of that digit in any position (hundreds, tens, units), there is a die that can provide it. However, since the same die cannot be used for multiple digits in the same number, we need that for any three digits (with possible repeats), there are three different dice, each containing the respective digit. However, when digits are repeated in a number, like 111, you would need three different dice each containing the digit 1. Therefore, for each digit, the number of dice containing that digit must be at least equal to the maximum number of times that digit appears in any number. Since numbers can have up to three of the same digit, we need at least three dice containing each digit. Wait, but 111 is a valid number, so we need three different dice each with 1. Similarly for 222, etc. Therefore, for each digit d, we need at least three dice that have d. Because in the number ddd, we need three different dice each with d. So, for each digit, the number of dice containing d must be at least 3. Similarly, if a number has two of the same digit, like 112, we need two different dice with 1 and one die with 2.Therefore, each digit needs to be present on at least three different dice. Since there are 10 digits (0-9), each needing three dice, that would be 10*3 = 30 "digit slots". However, each die can hold 6 digits, so the minimal number of dice would be at least ceiling(30 / 6) = 5. But wait, that's assuming that we can perfectly distribute the digits so that each digit is on exactly three dice, and each die holds six different digits. However, this might not be possible due to overlapping requirements.But maybe my reasoning is off. Let's check. If we need three dice for each digit, and each die can contain six digits, then with five dice, we can have 5*6=30 digit slots, which is exactly 3 per digit (10 digits * 3). However, this requires that each digit is on exactly three dice, and each die has six distinct digits. But is such a design possible? This is similar to a combinatorial design where each element (digit) is in exactly three subsets (dice), and each subset has six elements. This is called a (10, 6, 3) design? Wait, no. More precisely, a Balanced Incomplete Block Design (BIBD) with parameters (v, k, λ) where v=10, k=6, and each pair occurs in λ blocks. But here, we want each element to be in exactly r blocks, so it's a (v, k, r) design where r=3. The necessary conditions for a BIBD are that vr = bk and λ(v-1) = r(k-1). But here, we are not considering pairs, so maybe it's a different design. However, even if it's not a BIBD, perhaps such a design exists where each of the 10 digits is in exactly 3 dice, each die has 6 digits. Let's check if 10*3 = 30 = 5*6, so yes, 5 dice. So, if such a design exists, then 5 dice would suffice. But does this design exist?But wait, in this design, we need that each digit is in exactly 3 dice, and each die has 6 digits. Let me try to construct such a design.Take the 10 digits: 0,1,2,3,4,5,6,7,8,9.We need to create 5 dice, each with 6 digits, such that each digit appears in exactly 3 dice.This is similar to a 3-regular hypergraph on 10 vertices with 5 hyperedges each of size 6. Hmm. Is this possible?Alternatively, perhaps using group theory or combinatorial methods. Let me see.Suppose we arrange the digits in some systematic way. For example, divide the 10 digits into 5 groups of 2, and each die includes 6 digits by combining three of these groups. Wait, maybe not. Alternatively, use a finite projective plane or something, but I think those usually have parameters where the number of points is a prime power. 10 is not a prime power.Alternatively, maybe arrange each die to have 6 digits, ensuring that each digit is in 3 dice. Let's see. Let me attempt to construct such a set.Let me list the 5 dice:Die 1: 0,1,2,3,4,5Die 2: 0,1,2,6,7,8Die 3: 0,3,4,6,7,9Die 4: 1,3,5,6,8,9Die 5: 2,4,5,7,8,9Now, let's check how many times each digit appears.0: Die 1, Die 2, Die 3 → 3 times1: Die 1, Die 2, Die 4 → 3 times2: Die 1, Die 2, Die 5 → 3 times3: Die 1, Die 3, Die 4 → 3 times4: Die 1, Die 3, Die 5 → 3 times5: Die 1, Die 4, Die 5 → 3 times6: Die 2, Die 3, Die 4 → 3 times7: Die 2, Die 3, Die 5 → 3 times8: Die 2, Die 4, Die 5 → 3 times9: Die 3, Die 4, Die 5 → 3 timesWow, that works! Each digit is in exactly 3 dice, each die has 6 digits, and there are 5 dice. So, such a design exists. Therefore, in theory, 5 dice could be possible. But wait, does this design actually satisfy the problem's requirements? Let's see.The problem states that any ordered combination of three digits can be obtained by selecting three different dice and placing them with the appropriate sides up. So, for any three digits a, b, c (with possible repeats), there must exist three different dice, one containing a, one containing b, and one containing c. However, in our design, each digit is on three different dice. So, if a, b, c are all distinct, then we need three different dice each containing one of a, b, c. Since each digit is on three dice, but we need to choose one die for each digit such that all three dice are different. But is that possible?Wait, for example, suppose we have digits a, b, c, each on three different dice. If the three sets of dice for a, b, c are such that there is at least one trio where all three dice are distinct, then it works. But if, for instance, the dice for a and b overlap in all their dice, then maybe not. Let's take an example from the constructed design above.Take digits 0,1,2. Each is in three dice:0: Die 1, Die 2, Die 31: Die 1, Die 2, Die 42: Die 1, Die 2, Die 5So, if we need to get the number 012, we need three different dice, each containing 0, 1, and 2. However, 0 is in dice 1,2,3; 1 is in 1,2,4; 2 is in 1,2,5. So, possible trios:Take Die 3 (0), Die 4 (1), Die 5 (2). That would work. So, 0 from Die3, 1 from Die4, 2 from Die5. So, the trio Die3, Die4, Die5 can produce 0,1,2. Similarly, other combinations.But what about a number with repeated digits, say 000. Each 0 needs to come from a different die. Since 0 is in Die1, Die2, Die3. So, we can take Die1, Die2, Die3, each showing 0. So that works. Similarly, for 111, we can take Die1, Die2, Die4 (since 1 is in those three dice). Wait, Die1, Die2, Die4 each have 1. So, 1 from Die1, Die2, Die4. But Die1 and Die2 are different dice. So, 111 can be formed by selecting Die1, Die2, Die4, each showing 1. Wait, but each die can only contribute one digit per number. So, for 111, we need three different dice, each with 1. In our design, 1 is on Die1, Die2, Die4. So, we can use those three dice. Similarly, 222 is on Die1, Die2, Die5. So, use Die1, Die2, Die5.But what about a number like 121? We need three different dice: one with 1, one with 2, one with 1. But since we need three different dice, we can use Die1 (which has 1 and 2), but we can't use the same die for both 1s. Wait, 1 is on Die1, Die2, Die4. 2 is on Die1, Die2, Die5. So, to get 1,2,1, we need two different dice for the two 1s and one die for the 2. So, possible: Die4 (1), Die5 (2), Die2 (1). That works: Die4 shows 1, Die5 shows 2, Die2 shows 1. But the problem states that the three dice must be different. So, Die4, Die5, Die2 are all different. So that's okay. So, even with repeated digits, as long as each digit is present on at least three different dice, we can cover all combinations.Therefore, if such a design exists where each digit is on three different dice, and the design allows for any three digits (with possible repeats) to have three distinct dice each containing one of the digits, then 5 dice would suffice. Since we constructed such a design above, 5 dice might be the answer.However, wait, the problem mentions that "the digits 6 and 9 do not transform into each other when rotated 180 degrees." So, we have to ensure that 6 and 9 are treated as separate digits and cannot be confused by rotation. Therefore, when we put 6 or 9 on a die, we have to make sure that they are distinct and cannot be rotated to become each other. So, in our design above, 6 and 9 are separate digits. In the constructed example, Die2 has 6, Die3 has 6, Die4 has 6; Die3 has 9, Die4 has 9, Die5 has 9. So, they are on different dice. Therefore, when constructing the numbers, if a number requires a 6 or a 9, we need to ensure that the die used has that specific digit. Since 6 and 9 are separate, they can't be substituted via rotation, so they must be explicitly on the die. Since in our design, they are on separate dice, that should be okay.But let me check a specific example. Suppose we have the number 666. According to our design, 6 is on Die2, Die3, Die4. So, we can use Die2, Die3, Die4 each showing 6. Similarly, 999 is on Die3, Die4, Die5. So, Die3, Die4, Die5. For a number like 696, we need Die2 (6), Die3 or Die4 or Die5 (9), and Die2, Die3, Die4 (6). But wait, we need three different dice. So, 6 is on Die2, Die3, Die4; 9 is on Die3, Die4, Die5. So, for 696, first 6: Die2, second 9: Die5, third 6: Die3. So, Die2, Die5, Die3. All different. Each shows the required digit. That works.So, considering that 6 and 9 are distinct and cannot be rotated into each other, our design still works because they are on different dice. Therefore, the minimal number of dice is 5.But wait, I need to verify if this is indeed the case. Let me check if all numbers can be formed with this set of dice.Take a problematic number, say 000. As we saw, Die1, Die2, Die3 all have 0. So, that's covered. How about 012? Die1 has 0,1,2, but we need three different dice. So, Die1 (0), Die2 (1), Die5 (2). Wait, Die2 has 1, but Die5 has 2. Wait, Die1 has 0,1,2,3,4,5. Die2 has 0,1,2,6,7,8. Die3 has 0,3,4,6,7,9. Die4 has 1,3,5,6,8,9. Die5 has 2,4,5,7,8,9. So, to get 0,1,2:0 can be from Die1, Die2, Die3.1 can be from Die1, Die2, Die4.2 can be from Die1, Die2, Die5.So, we need three different dice. For example, 0 from Die3, 1 from Die4, 2 from Die5. All different. So, that works.Another number: 369. 3 is on Die1, Die3, Die4. 6 is on Die2, Die3, Die4. 9 is on Die3, Die4, Die5. So, 3 from Die1, 6 from Die2, 9 from Die5. Different dice. Works.How about 123? 1 from Die4, 2 from Die5, 3 from Die3. All different.What about 111? As before, Die1, Die2, Die4.How about 678? 6 from Die2, 7 from Die3, 8 from Die4. All different.It seems like this design works. So, 5 dice would be sufficient. But before I conclude, let me think if there are any numbers that cannot be formed. For example, 999: Die3, Die4, Die5. 606: Die2 (6), Die3 or Die4 or Die5 (0), Die2 (6). Wait, need three different dice. So, 6 from Die2, 0 from Die1, 6 from Die3. But Die3 doesn't have 6? Wait, Die3 has 0,3,4,6,7,9. Oh, Die3 does have 6. So, 6 from Die2, 0 from Die1, 6 from Die3. So, three different dice: Die1, Die2, Die3. Each showing 0,6,6. Wait, but 606 is the number. So, first digit 6, second 0, third 6. So, first die: 6 (from Die2), second die: 0 (from Die1), third die: 6 (from Die3). That's valid. All different dice.Another edge case: 696. 6 from Die2, Die3, Die4. 9 from Die3, Die4, Die5. So, first 6 from Die2, 9 from Die5, second 6 from Die3. So, Die2, Die5, Die3. All different. Good.So, based on this, it seems that 5 dice is sufficient. But why does the problem mention "the minimum number of dice"? Maybe I'm missing something here. Let me check.Wait, the problem says Vasya must write one digit on each face of several dice. So, each die has 6 digits, one on each face. So, each face has a digit, and when you rotate the die, you can show any of the six digits. However, when constructing a number, you need to select three different dice and place them with the appropriate side up. So, each die can show any of its six digits. So, the key is that each die can contribute any one of its six digits to a number, but each die can only contribute one digit per number. Therefore, for any three-digit number, you need three different dice, each of which has the required digit on one of their faces.Therefore, the problem reduces to: for each digit d (0-9), how many dice have d on one of their faces. Since, in the worst case, for a number like ddd, you need three different dice each containing d. Therefore, as we thought before, each digit must be on at least three dice. If each digit is on exactly three dice, and each die has six digits, then 5 dice would suffice (since 10 digits * 3 = 30, 5 dice * 6 = 30). And we constructed such a set. So, 5 dice should be the minimal number. But in the problem statement, the answer is likely 5, but I need to check if there's a mistake in my reasoning.Wait, but in the problem statement, it says "the digits 6 and 9 do not transform into each other when rotated 180 degrees". So, if a die has a 6, it cannot be rotated to show a 9, and vice versa. So, this means that 6 and 9 are distinct and must be treated as separate digits. In our constructed set, we have both 6 and 9 present on different dice. So, in Die2, Die3, Die4, we have 6; in Die3, Die4, Die5, we have 9. Therefore, they are on separate dice and cannot be confused. Therefore, the condition is satisfied.Therefore, 5 dice should be the minimal number. However, let me check existing literature or similar problems. Wait, I recall that in some covering problems, the minimal number is 10/2 = 5, but sometimes you need more due to overlapping constraints. However, since we have an explicit construction, that should be proof. Therefore, the answer is 5. But I need to make sure there isn't a mistake in the design.Wait, let's check another number: 555. 5 is on Die1, Die4, Die5. So, Die1, Die4, Die5. Each die can show 5. So, that works. How about 012: Die3 (0), Die4 (1), Die5 (2). Wait, Die3 has 0,3,4,6,7,9. So, Die3 has 0. Die4 has 1,3,5,6,8,9. So, Die4 has 1. Die5 has 2,4,5,7,8,9. So, Die5 has 2. Therefore, Die3, Die4, Die5 can show 0,1,2. Correct.Another number: 777. 7 is on Die2, Die3, Die5. So, Die2, Die3, Die5. Each can show 7. Perfect.Hmm, seems solid. So, unless there's a hidden constraint I'm missing, 5 dice should work. Maybe the answer is 5.But let me see. Wait, actually, in the problem statement, it says "Vasya must write one digit on each face of several dice". So, each face of a die has one digit. Each die can show any of its six digits. So, the die is a cube with six digits. So, the key is that when constructing a number, you pick three different dice, and for each die, you can choose which face to show. So, the die is not assigned to a position; you just pick three dice, arrange them in order, and rotate each to show the required digit. So, the three dice must contain the three digits (in order). So, as long as for any three digits a, b, c (allowing repeats), there exist three different dice such that die1 has a, die2 has b, die3 has c. That's exactly the same as our problem.Therefore, if the construction allows that for any a, b, c, there are three different dice with a, b, c, then 5 dice suffice. Since we have each digit on three different dice, and for any three digits, you can pick one die for each digit such that they are all different. Since each digit is on three dice, which are different from each other (as in the constructed example), then you can choose one die per digit without overlapping. For example, for digits a, b, c, each has three dice. The intersection of their dice sets might have overlaps, but since each has three options, you can choose non-overlapping ones. For example, if a is on dice X, Y, Z; b is on dice A, B, C; c is on dice D, E, F. If there's overlap, say a is on X, Y, Z; b is on X, B, C; c is on X, D, E. Then, choosing Y for a, B for b, D for c. So, non-overlapping.In our constructed example, each digit is on three dice, but the assignment is such that for any two digits, their sets of dice overlap in at most one die. Let's check:Take digits 0 and 1. 0 is on Die1, Die2, Die3. 1 is on Die1, Die2, Die4. Overlap on Die1 and Die2. Similarly, digits 0 and 2: Die1, Die2, Die3 and Die1, Die2, Die5. Overlap on Die1 and Die2. So, if I need to pick 0 and 1 on different dice, I can choose Die3 for 0, Die4 for 1, which are different. Similarly, for 0 and 2, Die3 and Die5.Therefore, even if two digits share two dice, you can pick the third die for one of them. Therefore, for any two digits, you can find a die for each that don't overlap, and similarly for three digits. Hence, such a design works.Therefore, the minimal number of dice required is 5. However, in some references, I recall that the problem of covering all triples with sets such that each element is in three sets, and each set has six elements, might require more. But given that we have an explicit construction, I think 5 is correct. Therefore, the answer is boxed{5}.Final Answerboxed{5}</think>Vasya needs to write one digit on each face of several dice such that any ordered combination of three digits from 000 to 999 can be obtained by selecting three different dice and placing them appropriately. The digits 6 and 9 cannot be transformed into each other by rotation.Key steps:1. Each die has 6 faces, and each digit (0-9) must be present on at least three different dice to cover all combinations, including repeated digits like 111.2. A combinatorial design is needed where each digit is on exactly three dice, and each die contains six different digits.3. A design with 5 dice was constructed, each containing 6 digits, ensuring each digit appears on exactly three dice: - Die 1: 0, 1, 2, 3, 4, 5 - Die 2: 0, 1, 2, 6, 7, 8 - Die 3: 0, 3, 4, 6, 7, 9 - Die 4: 1, 3, 5, 6, 8, 9 - Die 5: 2, 4, 5, 7, 8, 94. This design ensures that for any three digits (with possible repeats), there exist three different dice each containing one of the digits. The digits 6 and 9 are treated separately and do not overlap on the same die.Thus, the minimum number of dice required is (boxed{5}).,<think>Okay, so I have this problem where Vasya needs to write digits on dice such that any three-digit combination from 000 to 999 can be formed using three different dice. Also, the digits 6 and 9 can't be rotated to look like each other. The goal is to find the minimum number of dice required. Hmm, let me try to figure this out step by step.First, let me understand the problem again. Each die has a digit on each face, right? So each face of a die has a unique digit, I suppose, or can they repeat? The problem doesn't specify that the digits on a die must be unique, so maybe they can repeat. But Vasya can write "one digit on each face," which might just mean that each face has one digit, not necessarily unique. So, maybe digits can repeat on a die. But then again, if you have multiple copies of a digit on a die, does that help? Maybe, but maybe it's better to have unique digits to maximize coverage. Hmm.But the main point is that for any three-digit number, like 123 or 456, you should be able to pick three different dice, one for each digit, such that each die has the required digit on one of its faces. Also, the dice need to be arranged in order, so the first die represents the hundreds place, the second the tens, and the third the ones. So each die is assigned to a specific position. Wait, no, actually, the problem says "selecting some three different dice and placing them with the appropriate sides up in the correct order." So the three dice selected must be different, and then arranged in the correct order. So, the same die can't be used for multiple positions in the same combination. Therefore, each position (hundreds, tens, ones) must be covered by a different die.Therefore, Vasya needs a set of dice such that for each digit in each position (hundreds, tens, ones), there is a die that has that digit, and when forming a number, you use three different dice for the three digits. So, the key is that each digit (0-9) must be present on at least one die for each of the three positions. Wait, no, not necessarily. Wait, if a die can be assigned to any position, but each digit is on some die. Wait, maybe not. Let me think again.Suppose you have a set of dice. For any three-digit number, say ABC, you need to have three different dice: one die that has A on some face, another die that has B on some face, and another die that has C on some face. And these three dice must all be different. So, each die can be used in any position, but the three dice used for the three digits must all be different.Therefore, for each digit from 0 to 9, we need that digit to be present on at least one die. But since we can use the same die for different digits in different numbers, as long as in each particular number, the three dice are different. Wait, but actually, no. Wait, if you have a die that has, say, both 1 and 2 on its faces, then you can't use that die for both the hundreds and tens place in the same number. So if you need to write 112, you need three different dice: one die with 1 (for the hundreds place), another die with 1 (for the tens place), and a third die with 2 (for the ones place). Therefore, in this case, the die with 1 can't be used for both the hundreds and tens place in the same number. So, each digit in the three-digit number must be covered by a different die, even if the digits are the same.Therefore, this problem is similar to a covering problem where each die can "cover" digits, but for each occurrence of a digit in the three-digit number, a different die must be used. Therefore, if a number has repeated digits, like 111, we need three different dice each having a 1. Therefore, the number of dice required must be such that for each digit, the number of dice that have that digit is at least equal to the maximum number of times that digit can appear in a three-digit number. However, since the maximum number of times a digit can appear in a three-digit number is three (e.g., 111), but since we can't use the same die for multiple positions in the same number, each die can only supply one instance of a digit per number. Therefore, for each digit, the number of dice containing that digit must be at least three. Wait, but if we need to cover all possible numbers, even those with three of the same digit, then we need three different dice each containing that digit. Therefore, for each digit, the multiplicity (number of dice containing it) must be at least three.But that seems like we need 3 dice for each digit, but digits are 0-9, which would lead to 30 dice. But that's probably overkill. Wait, but maybe a single die can have multiple digits. For example, if a die has multiple digits, then it can cover multiple digits. But if a die is used in a number, it can only provide one digit. So, each die can be thought of as a set of digits, and when you use a die in a number, you choose one digit from its set. However, if a die is used in a number, it can't be used again in another position. Therefore, if a die has multiple digits, it can be used to cover different digits in different numbers, but not in the same number.But the problem requires that for each three-digit number, the three digits are each on different dice. Therefore, for each digit, the number of dice that contain that digit must be sufficient to cover the maximum number of times that digit can appear in a single number, which is three. Therefore, for each digit, we need at least three different dice that include that digit. So that even if a number has three of the same digit, like 777, we need three different dice each containing 7. Therefore, each digit must be present on at least three different dice. Therefore, the minimal number of dice would be such that every digit is covered by at least three dice, and the total number of dice is minimized. But each die can cover multiple digits. Therefore, we need a covering problem where each die is a subset of digits, and each digit is covered by at least three subsets (dice). The minimal number of subsets required such that each element (digit) is in at least three subsets.This is similar to a set cover problem but with multiplicity. The problem is to find the minimum number of subsets (dice) such that each element (digit) is covered at least three times. This is called the set multicover problem. In our case, the universe is the 10 digits (0-9), and each die is a subset of this universe. The goal is to cover each digit at least three times with the minimal number of subsets.But in the set multicover problem, the subsets can be of any size. However, in our problem, each die has six faces, so each subset (die) can have up to six digits. So each die can cover up to six different digits, each with multiplicity one. Therefore, the minimal number of dice needed would be such that each digit is covered at least three times, with each die contributing at most six distinct digits. Therefore, the minimal number of dice required would be the ceiling of (10 digits * 3 coverage) / 6 digits per die) = ceiling(30 / 6) = 5. But this is assuming that the digits can be perfectly partitioned into dice with no overlap. But since we need each digit to be in at least three dice, overlapping would require more dice.Wait, no. Wait, if we have 5 dice, each with 6 digits, that's 30 digit slots. Since we need each digit to be covered 3 times, 10 digits * 3 = 30. So, in theory, if we can arrange the digits such that each die has exactly 6 unique digits, and each digit is spread across exactly 3 dice, then 5 dice would be sufficient. However, this would require that each digit is present on exactly 3 dice, and each die contains 6 distinct digits. But 10 digits, each appearing 3 times, gives 30 total digit slots. 5 dice * 6 digits per die = 30, so that works arithmetically. However, we need to check if such a design is possible.But wait, this is similar to a combinatorial design problem, like a Block Design. Specifically, a (v, k, λ) design where v=10 (digits), k=6 (digits per die), and λ=3 (each digit appears in 3 dice). However, in Block Design terms, this would require that each pair of digits appears in λ blocks, but here we just need each digit to appear in 3 blocks (dice). So actually, it's a (v, k, r) design where r=3 (each element is in r blocks). The necessary conditions for such a design are that vr = bk, where b is the number of blocks (dice). Here, v=10, r=3, k=6, so 10*3 = b*6 => 30 = 6b => b=5. So yes, such a design is possible if the necessary conditions are met. But the necessary conditions also include that the pairwise intersections of blocks have certain properties, but perhaps in our case, since we don't have restrictions on pairs, just on each element appearing 3 times, maybe such a design exists.However, constructing such a design might be non-trivial, but perhaps it's possible. For example, if we can partition the digits into 5 groups, each group of 6 digits, such that each digit appears in exactly 3 groups. Wait, but 10 digits can't be divided into 5 groups of 6 without overlapping, since 5*6=30, which is three times the 10 digits, so each digit must be in exactly 3 groups. So yes, that's exactly the problem. So it's a question of whether such a design exists.But in reality, constructing such a design might be difficult, but let's assume for a moment that it is possible. Then, 5 dice would suffice. However, I need to check whether this is actually feasible. Let's try to construct such a design.We have 10 digits: 0,1,2,3,4,5,6,7,8,9. We need each digit to appear in exactly 3 dice, and each die contains 6 digits.Let's attempt to create 5 dice:Dice 1: 0,1,2,3,4,5Dice 2: 0,1,2,6,7,8Dice 3: 0,3,4,6,7,9Dice 4: 1,3,5,6,8,9Dice 5: 2,4,5,7,8,9Now, let's check how many times each digit appears:0: Dice 1, 2, 3 => 3 times1: Dice 1, 2, 4 => 3 times2: Dice 1, 2, 5 => 3 times3: Dice 1, 3, 4 => 3 times4: Dice 1, 3, 5 => 3 times5: Dice 1, 4, 5 => 3 times6: Dice 2, 3, 4 => 3 times7: Dice 2, 3, 5 => 3 times8: Dice 2, 4, 5 => 3 times9: Dice 3, 4, 5 => 3 timesWow, that actually works! Each digit is in exactly 3 dice, and each die has 6 digits. So this design satisfies all the conditions. Therefore, 5 dice would suffice.But wait, the problem has an additional constraint: the digits 6 and 9 cannot be rotated into each other. So, does this affect anything? Let me think. If a die contains a 6 or a 9, we need to ensure that when the die is placed with that digit up, it is not confused with the other. So, in other words, 6 and 9 must be distinct and cannot be on the same face or share a face through rotation. Therefore, when Vasya writes the digits on the dice, he must include both 6 and 9 as separate digits if needed, and not rely on rotating a 6 to get a 9 or vice versa.But in the design I created above, let's check if both 6 and 9 are included. Looking at the dice:Dice 1: 0,1,2,3,4,5 – no 6 or 9Dice 2: 0,1,2,6,7,8 – has 6Dice 3: 0,3,4,6,7,9 – has 6 and 9Dice 4: 1,3,5,6,8,9 – has 6 and 9Dice 5: 2,4,5,7,8,9 – has 9So, in this design, 6 appears on dice 2, 3, 4 and 9 appears on dice 3,4,5. Therefore, they are treated as separate digits, each appearing on three dice. Therefore, since they are explicitly included on different faces, there's no issue with rotation. So as long as 6 and 9 are distinct and not on the same face, this should be okay. In our design, 6 and 9 are on different dice (they are on the same dice 3 and 4, but on different faces). Wait, in dice 3 and 4, both 6 and 9 are present. But since they are different faces, that's okay. So each die that has both 6 and 9 just has them on separate faces. Therefore, when you need a 6, you can choose the face with 6, and when you need a 9, you choose the face with 9. Therefore, there is no conflict with the rotation restriction.Therefore, this design satisfies all the problem's conditions, and only uses 5 dice. Therefore, the minimal number of dice required is 5.But wait, wait a second. Let me check if there is a mistake here. The problem states that "the digits 6 and 9 do not transform into each other when rotated 180 degrees." So, when you put a die down, if it has a 6, you can't rotate it 180 degrees to make it a 9, and vice versa. Therefore, Vasya must ensure that if a die has a 6, it doesn't have a 9 on the opposite face (since rotating 180 degrees would flip it to the opposite face). Therefore, in our design, we need to ensure that 6 and 9 are not on opposite faces of the same die. However, in our current design, dice 3 and 4 have both 6 and 9. So, if Vasya writes both 6 and 9 on the same die, he needs to make sure they are not on opposite faces.But the problem doesn't say that 6 and 9 can't be on the same die, just that they can't be rotated into each other. Therefore, as long as 6 and 9 are on different faces (not opposite) on the same die, it's okay. But in reality, on a standard die, opposite faces add up to 7, but here we are not dealing with numbers but digits. However, the problem doesn't specify any constraints on the arrangement of digits on the die, except that 6 and 9 cannot be rotated into each other. Therefore, as long as 6 and 9 are not placed on faces that are opposite each other, since rotating 180 degrees would flip the die to the opposite face, which could turn a 6 into a 9 or vice versa if they are on opposite faces.Therefore, Vasya must ensure that if a die has both 6 and 9, they are not on opposite faces. However, the problem doesn't state that the dice need to follow any standard configuration (like opposite faces summing to 7), so Vasya can arrange the digits on the dice as he likes, as long as 6 and 9 are not on opposite faces.Therefore, in our design, even if a die has both 6 and 9, as long as they are placed on adjacent faces or non-opposite faces, it's acceptable. Therefore, the presence of both 6 and 9 on the same die is allowed, as long as they are not opposite. So, our design is still valid.Therefore, with 5 dice, each containing 6 digits, and each digit appearing on 3 dice, and 6 and 9 handled appropriately, it seems possible. Therefore, 5 dice is the minimal number.But let me verify with an example. Suppose we need to form the number 699. Then, we need three different dice: one with 6, one with 9, and another with 9. But according to our design, 6 is on dice 2,3,4 and 9 is on dice 3,4,5. So, to form 699, we need one die with 6 (say dice 2), and two dice with 9 (dice 3 and 5). But can we use dice 3 for both the second and third 9? No, because we have to use three different dice. Therefore, for the two 9s, we need two different dice that have 9. But according to the design, 9 is on dice 3,4,5. Therefore, we can choose dice 3 and 4 for the two 9s, and dice 2 for the 6. That would work, using three different dice. Similarly, for three 9s, we need three different dice with 9, which are dice 3,4,5. Therefore, 999 can be formed by dice 3,4,5 each showing 9. So that works.Similarly, for a number like 666, we need three different dice with 6, which are dice 2,3,4. So yes, that works. Therefore, all numbers with repeated digits can be formed as long as each digit is on three different dice.Therefore, 5 dice should be sufficient.But wait, the problem says "Vasya must write one digit on each face of several dice". So each face has one digit, but multiple faces on a die can have the same digit? Wait, hold on. The problem says "one digit on each face", which I think means that each face has exactly one digit, but digits can be repeated on different faces. So, a die can have multiple faces with the same digit. But in our previous reasoning, we assumed that each die has distinct digits, but actually, the problem allows digits to be repeated on different faces of the same die.Wait, now this is a crucial point. If Vasya can write the same digit on multiple faces of a die, then perhaps we can have fewer dice. Because if a die can have multiple copies of the same digit, then even if a die is used in a number, it can provide multiple instances of that digit in different numbers, but not in the same number. Wait, no. Because when you form a three-digit number, you need to select three different dice. So, for the number 111, you need three different dice, each of which has a 1 on at least one face. Therefore, if a die has multiple 1s, it can still only contribute one 1 to a single number. Therefore, even if a die has multiple copies of a digit, it's still only counted once per number. Therefore, the key is that each digit must be present on at least three different dice, regardless of how many times it's present on each die.Therefore, even if you have a die with six 1s, it still only counts as one die that can provide a 1. Therefore, for the number 111, you need three different dice each with at least one 1. Therefore, having multiple copies on a die doesn't help with the coverage. Therefore, the minimal number of dice required is still determined by the need for each digit to be on three different dice, regardless of how many times it's on each die.Therefore, the initial reasoning holds: we need each digit to be on three different dice, and each die can hold up to six different digits. Therefore, the minimal number of dice is 5. However, the confusion arises if multiple copies of a digit on a single die count multiple times, but they don't because in any given number, a die can only be used once.Therefore, the answer is 5. But wait, let me check again.Wait, in the standard covering problem, if you have a set cover where each element needs to be covered three times, and each set can cover multiple elements, then the minimal number of sets required is ceil((number of elements * coverage)/set size). Here, 10 elements * 3 coverage = 30, divided by 6 elements per set gives 5. So yes, 5 sets. But the problem is that in set cover with multiplicity, overlapping elements in different sets can lead to needing more sets. However, in our case, since we can arrange the elements such that each element is covered exactly three times with 5 sets (as in the design above), then 5 is sufficient.Therefore, the answer should be 5. However, let me check if there are any other constraints. For example, the digits 0-9 are 10 digits. If each die can have up to 6 digits, and we need each digit to be on three dice, so 30 digit slots. 5 dice * 6 digits = 30. So it's a perfect fit. Therefore, if such a design exists, then 5 is possible. And as I constructed such a design above, it's possible. Therefore, 5 is the minimal number.But wait, let me check another angle. Suppose we try to use 4 dice. Each die can have 6 digits. 4 dice * 6 digits = 24 digit slots. But we need 10 digits * 3 = 30 digit slots. 24 < 30, so it's impossible. Therefore, 4 dice are insufficient. Therefore, 5 is indeed the minimal number.Therefore, the answer is boxed{5}.Final Answerboxed{5}</think>Vasya needs to write digits on dice such that any three-digit combination from 000 to 999 can be formed using three different dice. Each die can have up to six digits, and the digits 6 and 9 cannot be rotated into each other. To solve this, we need each digit (0-9) to appear on at least three different dice. This is because a three-digit number with repeated digits (e.g., 111) requires three different dice each containing that digit. The problem reduces to a set multicover problem where each digit must be covered at least three times. With 10 digits, each needing to appear three times, we require 30 digit slots. Each die can cover 6 digits, leading to a calculation of the minimum number of dice as follows:[text{Minimum number of dice} = leftlceil frac{10 times 3}{6} rightrceil = leftlceil frac{30}{6} rightrceil = 5]A combinatorial design confirms this: each digit appears exactly three times across five dice, each containing six digits. The constructed design ensures that 6 and 9 are on different faces of the same die if necessary, complying with the rotation constraint. Thus, the minimum number of dice required is (boxed{5}).]

🔑:Answer: 5.Solution. Since among the possible combinations there must be 000, 111, 222, ldots, 999, each digit must appear on at least three different faces (of different cubes), so there must be at least 30 digits on the faces in total, hence there must be at least 30: 6=5 cubes.On the other hand, we will show that if the digits are written such that each digit from 0 to 9 appears on the faces of three different cubes, then the requirement of the problem is satisfied. Consider any combination mathrm{ABC} of three digits. First, choose any cube that contains the digit A. The digit B appears on at least two other cubes besides, possibly, the first one, so we take any of them as the second. Finally, the digit C appears on at least one more cube besides the first and second cubes, so we take it as the third. By placing them in order with the corresponding faces up, we get the required combination ABC. Note that the reasoning does not depend on whether the digits A, B, and C are distinct or not.Let's provide a possible way to write the digits on the faces of five cubes such that each digit appears on the faces of three different cubes. Write all the digits from 0 to 9 in a row with each digit repeated three times: 000111222 . . .888999. The first five digits of this row will be written on the first faces of five different cubes, the second five digits will be written on the second faces of five different cubes, and so on.Grading criteria. Proof of the minimality of 5 cubes: 3 points. Providing a correct example for 5 cubes with justification: 4 points. Lack of justification: minus 2 points.