🧠:[<think>Okay, so I need to find the equation of the tangent line to the circle C: (x-2)² + (y-3)² = 25 that passes through the point P(-1, 7). Hmm, let me recall how to approach this. First, I remember that a tangent line to a circle is perpendicular to the radius at the point of contact. So, if I can find the point where the tangent line touches the circle, maybe I can use the slope of the radius to find the slope of the tangent line. Then, with the slope and the point P, I can write the equation. But wait, I don't know the point of contact yet. Hmm, maybe there's another way.Alternatively, I know that the equation of a line can be written in the form y = mx + b, where m is the slope and b is the y-intercept. Since the line passes through P(-1, 7), substituting x = -1 and y = 7 into the equation gives 7 = -m + b, so b = m + 7. So the equation of the line can be written as y = mx + (m + 7). But this line has to be tangent to the circle. The condition for a line to be tangent to a circle is that the distance from the center of the circle to the line must be equal to the radius. Let me use that. The circle C has center at (2, 3) and radius 5 (since 25 is 5 squared). The distance from the center (2, 3) to the line y = mx + (m + 7) should be 5. The formula for the distance from a point (x₀, y₀) to the line Ax + By + C = 0 is |Ax₀ + By₀ + C| / sqrt(A² + B²). Let me write the line equation in the standard form. Starting from y = mx + m + 7, subtract y: mx - y + m + 7 = 0. So, A = m, B = -1, C = m + 7. Therefore, the distance from (2, 3) to this line is |m*2 + (-1)*3 + m + 7| / sqrt(m² + (-1)²) = |2m - 3 + m + 7| / sqrt(m² + 1) = |3m + 4| / sqrt(m² + 1). This distance must equal the radius 5. So:|3m + 4| / sqrt(m² + 1) = 5Let me square both sides to eliminate the absolute value and the square root:(3m + 4)² / (m² + 1) = 25Multiply both sides by (m² + 1):(3m + 4)² = 25(m² + 1)Expand the left side:9m² + 24m + 16 = 25m² + 25Bring all terms to one side:9m² + 24m + 16 - 25m² - 25 = 0Combine like terms:-16m² + 24m - 9 = 0Multiply both sides by -1 to make it easier:16m² - 24m + 9 = 0This is a quadratic equation in terms of m. Let's solve for m using the quadratic formula. The quadratic is 16m² -24m +9=0. So, a=16, b=-24, c=9.Discriminant D = b² -4ac = (-24)² - 4*16*9 = 576 - 576 = 0Wait, discriminant is zero, so there's only one solution. But that can't be right. Wait, if the discriminant is zero, that means there's exactly one tangent line? But a point outside the circle should have two tangent lines. Wait, maybe the point is on the circle? Let me check.Wait, the point P(-1,7). Let's check if P is on the circle. Substitute x=-1 and y=7 into the left side of the circle equation:(-1 -2)^2 + (7 -3)^2 = (-3)^2 + (4)^2 = 9 + 16 = 25. Which equals the right side. So point P is actually ON the circle. Therefore, there's only one tangent line at point P. Wait, but the problem says "draw a tangent line to the circle passing through point P". But if P is on the circle, then the tangent line at P is the only one passing through P. So maybe that's why the quadratic equation gave discriminant zero.But let me double-check. If P is on the circle, then the tangent at P is the only tangent line through P. So, that seems right. Therefore, the equation we found is the tangent at P.Wait, but the problem is presented as if there might be two tangent lines, but since P is on the circle, there's only one. Let me confirm again. Let me calculate the distance from P to the center.Center is (2,3). Distance between (2,3) and (-1,7): sqrt[(2 - (-1))² + (3 -7)^2] = sqrt[3² + (-4)^2] = sqrt[9+16] = sqrt[25] = 5. Which is exactly the radius. Therefore, P is on the circle. Therefore, the tangent line at P is the only tangent line passing through P. Therefore, the equation we found is the tangent at P.But let's make sure. Let's compute m from the quadratic equation:From 16m² -24m +9=0.m = [24 ± sqrt(0)] / (2*16) = 24 /32 = 3/4.So m = 3/4. Therefore, the slope is 3/4, and the equation is y = (3/4)x + (3/4 +7) = (3/4)x + 31/4.But let me confirm if this is indeed the tangent line at point P(-1,7).Alternatively, we can compute the tangent line at point P(-1,7) on the circle (x-2)^2 + (y-3)^2 =25.The formula for the tangent line at point (x₁,y₁) on the circle (x-h)^2 + (y-k)^2 = r² is (x₁ - h)(x - h) + (y₁ -k)(y -k) = r². Wait, no, let me recall. Actually, the standard equation is (x₁ - h)(x - h) + (y₁ - k)(y - k) = r²? Wait, maybe not. Wait, the equation of the tangent line at (x₁, y₁) is (x₁ - h)(x - h) + (y₁ - k)(y - k) = r². Wait, let me check with an example. Wait, if the circle is centered at (h,k), then the radius vector is from (h,k) to (x₁,y₁). The tangent line is perpendicular to this radius. So the tangent line equation can be written as (x₁ - h)(x - h) + (y₁ -k)(y -k) = r²? Wait, let me test with a simple circle. Let's take the circle x² + y² = r². Then the tangent line at (x₁,y₁) is x x₁ + y y₁ = r². So for the general circle (x-h)^2 + (y -k)^2 = r², the tangent line at (x₁,y₁) is (x₁ - h)(x - h) + (y₁ - k)(y - k) = r². Wait, substituting (x₁,y₁) into the left-hand side gives (x₁ - h)^2 + (y₁ -k)^2 = r², which matches the circle equation. So that's correct.So, applying this formula, the tangent line at P(-1,7) is:(-1 -2)(x -2) + (7 -3)(y -3) = 25Simplify:(-3)(x -2) + 4(y -3) = 25Multiply out:-3x +6 +4y -12 =25Combine constants:6 -12 = -6, so:-3x +4y -6 =25Bring constants to the right:-3x +4y =31So, the equation is -3x +4y =31. Let's write this in slope-intercept form:4y =3x +31y = (3/4)x +31/4Which matches the earlier result. So, that's correct. Therefore, the equation of the tangent line is y = (3/4)x + 31/4. However, the problem asks for the general equation. The general equation is usually written in standard form, which would be 3x -4y +31=0, but let me check.Wait, the previous steps gave us -3x +4y =31. Multiplying both sides by -1 gives 3x -4y = -31. So, 3x -4y +31=0. Wait, no: -3x +4y -31=0. Wait, the equation was -3x +4y =31, so bringing all terms to the left: -3x +4y -31=0. But usually, we prefer the coefficient of x to be positive. So multiplying by -1: 3x -4y +31=0. Wait, no: original equation: -3x +4y =31. To get standard form, move all terms to the left: -3x +4y -31=0. If we multiply by -1: 3x -4y +31=0. Hmm, but 3x -4y +31=0 would be 3x -4y = -31. Let me verify with the slope-intercept form. If we write 3x -4y +31=0, then solving for y: -4y = -3x -31 => y = (3/4)x + 31/4. Which is correct. Alternatively, if we write -3x +4y -31=0, solving for y: 4y =3x +31, same result. So both forms are correct. Depending on the convention. If they want the general equation, maybe either is acceptable, but perhaps they want the form without negative coefficients. So 3x -4y +31=0. Let me check the problem statement. It says "the general equation of this tangent line". The general equation is usually Ax + By + C =0. So either -3x +4y -31=0 or 3x -4y +31=0. But likely, they prefer the coefficient of x to be positive. So 3x -4y +31=0. However, let me check with the original calculation. When we used the distance formula, we ended up with m=3/4 and the equation y = (3/4)x +31/4. Let me convert that to standard form. Multiply both sides by 4: 4y =3x +31. Rearranged: -3x +4y -31=0, which is equivalent to 3x -4y +31=0. So both forms are correct. But depending on how the question expects the answer, maybe either is acceptable. However, looking back at the circle equation, it's written as (x-2)^2 + (y-3)^2=25, so they may prefer integers without common factors and A positive. So 3x -4y +31=0. Let me check if this line actually passes through P(-1,7). Substitute x=-1, y=7 into 3x -4y +31: 3*(-1) -4*7 +31 = -3 -28 +31=0. Yes, so it passes through P. Also, distance from center (2,3) to the line 3x -4y +31=0 is |3*2 -4*3 +31| / sqrt(3² + (-4)^2) = |6 -12 +31| /5= |25| /5=5, which is equal to the radius. So correct. Therefore, the general equation is 3x -4y +31=0. Alternatively, written as -3x +4y -31=0, but since the problem didn't specify, either is correct. But to match standard conventions, I think 3x -4y +31=0 is preferred. Let me check the quadratic equation solution again. When we solved for m, we found m=3/4, then constructed y= (3/4)x +31/4. Converting to standard form: 3x -4y +31=0. So that's the answer. Wait, but let me check once more: if the point is on the circle, there's only one tangent line, which is the tangent at that point. So, since P(-1,7) is on the circle, the tangent line at P is the only tangent line passing through P, so this must be the answer. Therefore, the equation is 3x -4y +31=0. Alternatively, maybe the problem expects the answer in a different form? Let me see. If they want it in the form Ax + By + C =0, then 3x -4y +31=0. Alternatively, if written as -3x +4y -31=0, but perhaps they want the coefficients to be integers with no common factors and A positive. Since 3, -4, 31 have no common factors, 3x -4y +31=0 is the answer. Alternatively, maybe they want it in the form like y = mx + b, but the problem says "general equation", which usually refers to the standard form Ax + By + C =0.Therefore, the final answer is 3x -4y +31=0. So in boxed form: boxed{3x - 4y + 31 = 0}Final AnswerThe general equation of the tangent line is boxed{3x - 4y + 31 = 0}.</think>Given the circle ( C: (x-2)^2 + (y-3)^2 = 25 ) and the point ( P(-1, 7) ), we need to find the equation of the tangent line to the circle passing through point ( P ).First, we verify that point ( P ) lies on the circle:[(-1 - 2)^2 + (7 - 3)^2 = (-3)^2 + 4^2 = 9 + 16 = 25]Since the point ( P ) is on the circle, there is exactly one tangent line at ( P ).The general equation of the tangent line at point ( (x_1, y_1) ) on the circle ( (x-h)^2 + (y-k)^2 = r^2 ) is given by:[(x_1 - h)(x - h) + (y_1 - k)(y - k) = r^2]Substituting ( (x_1, y_1) = (-1, 7) ), ( (h, k) = (2, 3) ), and ( r^2 = 25 ):[(-1 - 2)(x - 2) + (7 - 3)(y - 3) = 25]Simplifying this equation:[-3(x - 2) + 4(y - 3) = 25][-3x + 6 + 4y - 12 = 25][-3x + 4y - 6 = 25][-3x + 4y = 31]Rewriting in standard form:[3x - 4y + 31 = 0]Thus, the general equation of the tangent line is (boxed{3x - 4y + 31 = 0}).,<think>Okay, let me try to figure out how to find the equation of the tangent line to the circle C: (x-2)² + (y-3)² = 25 that passes through the point P(-1, 7). Hmm, I remember that a tangent line to a circle is perpendicular to the radius at the point of contact. So maybe I can use that property here. Let's start by visualizing the problem. The circle has center at (2, 3) and radius 5 because 25 is 5 squared. The point P(-1, 7) is outside the circle, which makes sense because we can draw two tangent lines from an external point to a circle. Wait, but the problem just says "a tangent line," so maybe there are two possible answers, but they might want the general equation which includes both? Hmm, not sure yet. Let me check if P is indeed outside the circle. To verify, the distance from P to the center should be greater than the radius. The distance between (2,3) and (-1,7) is sqrt[(2 - (-1))² + (3 - 7)²] = sqrt[(3)² + (-4)²] = sqrt[9 + 16] = sqrt[25] = 5. Wait, that's exactly equal to the radius. So, hold on, if the distance is equal to the radius, that means the point P is on the circle. But if P is on the circle, then there's only one tangent line at point P. Wait, but let me recalculate that distance to be sure. From (-1,7) to (2,3): the x difference is 2 - (-1) = 3, the y difference is 3 - 7 = -4. So the distance is sqrt(3² + (-4)²) = sqrt(9 + 16) = sqrt(25) = 5. Which is exactly the radius. So the point P is actually on the circle. Therefore, the tangent line at point P is the line that touches the circle only at P and is perpendicular to the radius OP, where O is the center (2,3). Therefore, the tangent line at P(-1,7) is perpendicular to the radius OP. So the slope of OP is (7 - 3)/(-1 - 2) = (4)/(-3) = -4/3. Therefore, the slope of the tangent line is the negative reciprocal, which is 3/4. Then, using the point-slope form of the line equation: y - 7 = (3/4)(x + 1). Let's write that out: y = (3/4)x + (3/4) + 7. Simplify 3/4 + 7: 7 is 28/4, so 28/4 + 3/4 = 31/4. Therefore, the equation is y = (3/4)x + 31/4. To write this in standard form, multiply both sides by 4: 4y = 3x + 31, so -3x + 4y - 31 = 0, or 3x - 4y + 31 = 0. Wait, but maybe they want it in general form Ax + By + C = 0. Let me check. Alternatively, maybe I made a mistake here. Wait, if the point P is on the circle, then the tangent line is unique. But the problem says "draw a tangent line to the circle passing through point P." So if P is on the circle, there is only one tangent line. But let me verify if my calculation that P is on the circle is correct. Plugging P(-1,7) into the circle equation: (x-2)^2 + (y-3)^2 = (-1-2)^2 + (7-3)^2 = (-3)^2 + 4^2 = 9 + 16 = 25. Yes, that's exactly the radius squared. So P is indeed on the circle. Therefore, there is only one tangent line at P. Therefore, the equation is y = (3/4)x + 31/4, or rearranged as 3x - 4y + 31 = 0. Wait, but let me confirm this again. The slope of OP is (7-3)/(-1 -2) = 4/-3 = -4/3, so the slope of the tangent line should be 3/4. Then, using point P(-1,7), the equation is y -7 = (3/4)(x +1). Multiply both sides by 4: 4(y -7) = 3(x +1) → 4y -28 = 3x +3 → 3x -4y +31 =0. Yes, that seems correct. So the general equation is 3x -4y +31 =0. But let me check if this line is indeed tangent to the circle. The distance from the center (2,3) to the line 3x -4y +31 =0 should be equal to the radius 5. The formula for distance is |3*2 -4*3 +31| / sqrt(3² + (-4)^2) = |6 -12 +31| /5 = |25| /5 =5. Which is equal to the radius. So that's correct. Therefore, the equation is 3x -4y +31 =0. But wait, the problem says "the general equation of this tangent line". Since P is on the circle, there's only one tangent line. So maybe the answer is 3x -4y +31=0. But let me check again. Alternatively, maybe I messed up the direction of the slope. The slope of the radius is -4/3, so the tangent line slope is 3/4. So yes, that's correct. Then the equation is correct. Alternatively, maybe there's another approach. Let's see. Another way to find the tangent line from an external point to a circle is using the formula for tangent lines. But in this case, since the point is on the circle, this formula reduces to the equation we found. Alternatively, if the point were outside, we would have two tangent lines, but here since it's on the circle, only one. So I think the answer is 3x -4y +31=0. Wait, but let me check once more. Let me find another point on the tangent line. For example, if x =0, then 3*0 -4y +31=0 → -4y= -31 → y=31/4=7.75. So (0, 7.75) is on the line. Let me check the distance from this point to the center (2,3). sqrt[(0-2)^2 + (7.75-3)^2] = sqrt[4 + (4.75)^2] = sqrt[4 + 22.5625] = sqrt[26.5625] ≈5.15, which is greater than 5, so this point is outside the circle, which makes sense. But the line is tangent, so only touching at P(-1,7). Let me plug x=-1 into the line equation: 3*(-1) -4y +31=0 → -3 -4y +31=0 → -4y +28=0 → y=7. Correct. So when x=-1, y=7, which is point P. Therefore, the line passes through P and is tangent at P. Therefore, this is the correct equation. So I think the answer is 3x -4y +31=0.Final AnswerThe general equation of the tangent line is boxed{3x - 4y + 31 = 0}.</think>Given the circle ( C: (x-2)^2 + (y-3)^2 = 25 ) and the point ( P(-1, 7) ), we need to find the equation of the tangent line passing through ( P ).First, we verify that the point ( P ) lies on the circle. Substituting ( P(-1, 7) ) into the circle equation:[(-1-2)^2 + (7-3)^2 = (-3)^2 + 4^2 = 9 + 16 = 25]This confirms that ( P ) is on the circle.Next, we determine the slope of the radius ( OP ) where ( O ) is the center ((2, 3)). The slope of ( OP ) is:[frac{7 - 3}{-1 - 2} = frac{4}{-3} = -frac{4}{3}]The slope of the tangent line, being perpendicular to ( OP ), is the negative reciprocal:[frac{3}{4}]Using the point-slope form of the line equation with point ( P(-1, 7) ):[y - 7 = frac{3}{4}(x + 1)]Multiplying through by 4 to clear the fraction:[4(y - 7) = 3(x + 1)]Expanding and rearranging into standard form:[4y - 28 = 3x + 3 implies 3x - 4y + 31 = 0]Verification by calculating the distance from the center ((2, 3)) to the line ( 3x - 4y + 31 = 0 ) confirms it equals the radius 5:[frac{|3(2) - 4(3) + 31|}{sqrt{3^2 + (-4)^2}} = frac{|6 - 12 + 31|}{5} = frac{25}{5} = 5]Thus, the general equation of the tangent line is (boxed{3x - 4y + 31 = 0}).]

🔑:The center of circle C: (x-2)^2+(y-3)^2=25 is C(2, 3), and its radius r=5. Point P is on the circle. According to the problem, let's assume the equation is y-7=k(x+1), which simplifies to kx-y+7+k=0. Since line l is tangent to circle C: (x-2)^2+(y-3)^2=25, the distance from the center of the circle to line l equals the radius, i.e., d= frac {|2k-3+7+k|}{ sqrt {k^{2}+1}}=5. Solving this equation yields k= frac {3}{4}, Therefore, the equation of line l is y-7= frac {3}{4}(x+1), which simplifies to 3x-4y+31=0. Hence, the answer is: boxed{3x-4y+31=0}. Given the equation of circle C: (x-2)^2+(y-3)^2=25 with center C(2, 3) and radius r=5, and point P on the circle, we can set the equation of tangent line l. By using the distance from a point to a line formula and considering the tangent condition, we solve for k to find the equation of the required tangent line. This problem provides the equation of a circle and asks for the equation of a tangent line passing through a given point. It focuses on the distance from a point to a line formula, the standard equation of a circle, and the relationship between a line and a circle, making it a medium-level question.

🧠:[<think>Okay, so I need to calculate this big expression: (10 × 19 × 20 × 53 × 100 + 601) divided by 13. Hmm, that looks a bit intimidating at first glance because of all those numbers multiplied together. Let me try to break it down step by step. First, maybe I should handle the multiplication part before adding 601. Let me write that out: 10 × 19 × 20 × 53 × 100. That's a lot of numbers multiplied together. Maybe I can simplify this multiplication by grouping numbers that are easy to multiply first. Let me see... 10 and 20 are easy to multiply. 10 × 20 is 200. Then, 200 × 100 is 20,000. Wait, hold on, maybe I should rearrange the numbers to make the multiplication easier. Let me rearrange the multiplication in a different order. Let's do 10 × 20 × 100 first, since those are powers of 10 and might be easier. So 10 × 20 is 200, then 200 × 100 is 20,000. So that part is 20,000. Then, the remaining numbers are 19 and 53. So now I have 20,000 × 19 × 53. Hmm, okay, let's tackle 19 × 53 first. What's 19 × 53? Let me calculate that. 19 times 50 is 950, and 19 times 3 is 57, so adding those together, 950 + 57 = 1007. So 19 × 53 is 1007. Then, multiplying that by 20,000. So 20,000 × 1007. Let me do that step by step. 20,000 × 1000 is 20,000,000, and 20,000 × 7 is 140,000. So adding those together, 20,000,000 + 140,000 = 20,140,000. So the first part of the expression, 10 × 19 × 20 × 53 × 100, equals 20,140,000. Then, we add 601 to that. So 20,140,000 + 601 = 20,140,601. Now, the entire expression is 20,140,601 divided by 13. So we need to compute 20,140,601 ÷ 13. Alright, now how do I divide such a large number by 13? Maybe I can use long division here. Let me recall how long division works. Alternatively, perhaps I can break down the number into smaller parts that are divisible by 13 and then add the results. Let's try that approach.First, let's look at the number 20,140,601. Let me write it down: 20,140,601. Let me split this number into parts that I can handle more easily. Maybe start from the leftmost digits. Let's see:13 into 20 goes 1 time (since 13×1=13), subtract 13 from 20, we get 7. Bring down the next digit, which is 1, making it 71. 13 into 71 goes 5 times (13×5=65). Subtract 65 from 71, remainder 6. Bring down the next digit, which is 4, making it 64. 13 into 64 goes 4 times (13×4=52). Subtract 52 from 64, remainder 12. Bring down the next digit, which is 0, making it 120. 13 into 120 goes 9 times (13×9=117). Subtract 117 from 120, remainder 3. Bring down the next digit, which is 6, making it 36. 13 into 36 goes 2 times (13×2=26). Subtract 26 from 36, remainder 10. Bring down the next digit, which is 0, making it 100. 13 into 100 goes 7 times (13×7=91). Subtract 91 from 100, remainder 9. Bring down the next digit, which is 1, making it 91. 13 into 91 goes 7 times (13×7=91). Subtract 91, remainder 0.Wait, let me verify that. Let me write this long division step by step to make sure I didn't make a mistake.Dividing 20,140,601 by 13:13 | 20,140,601First, 13 into 20 is 1 (13×1=13), subtract 13 from 20, remainder 7. Bring down 1: 71.13 into 71 is 5 (13×5=65), subtract 65 from 71, remainder 6. Bring down 4: 64.13 into 64 is 4 (13×4=52), subtract 52 from 64, remainder 12. Bring down 0: 120.13 into 120 is 9 (13×9=117), subtract 117 from 120, remainder 3. Bring down 6: 36.13 into 36 is 2 (13×2=26), subtract 26 from 36, remainder 10. Bring down 0: 100.13 into 100 is 7 (13×7=91), subtract 91 from 100, remainder 9. Bring down 1: 91.13 into 91 is 7 (13×7=91), subtract 91, remainder 0.So the quotient is 1 (from 20), then 5 (from 71), then 4 (from 64), then 9 (from 120), then 2 (from 36), then 7 (from 100), then 7 (from 91). So putting that all together: 1,549,277.Therefore, 20,140,601 divided by 13 is 1,549,277. Let me check if that's correct. Let me multiply 1,549,277 by 13 to see if we get back 20,140,601.Calculating 1,549,277 × 13. Let's break it down:1,549,277 × 10 = 15,492,7701,549,277 × 3 = 4,647,831Adding those two together: 15,492,770 + 4,647,831.15,492,770 + 4,647,831:15,492,770 + 4,000,000 = 19,492,77019,492,770 + 647,831 = 20,140,601Yes! That adds up to 20,140,601. So the division was done correctly.Therefore, the final answer is 1,549,277.But let me just verify again step by step to make sure I didn't make a mistake in the multiplication earlier. Let me check the initial multiplication part again.First part: 10 × 19 × 20 × 53 × 100. Let me compute this step by step.10 × 19 = 190190 × 20 = 3,8003,800 × 53. Let's calculate that. 3,800 × 50 = 190,000; 3,800 × 3 = 11,400. Adding those gives 190,000 + 11,400 = 201,400.Then, 201,400 × 100 = 20,140,000. Correct. Then add 601: 20,140,000 + 601 = 20,140,601. That's correct.Divided by 13, we get 1,549,277. And we checked that 1,549,277 × 13 = 20,140,601. So that's all correct.Therefore, the answer is 1,549,277.Final Answerboxed{1549277}</think>To calculate ((10 times 19 times 20 times 53 times 100 + 601) div 13):1. First, compute the product (10 times 19 times 20 times 53 times 100): - Group the numbers to simplify multiplication: [ (10 times 20 times 100) times (19 times 53) ] - Calculate (10 times 20 times 100 = 20000) - Calculate (19 times 53): [ 19 times 53 = 1007 ] - Multiply the results: [ 20000 times 1007 = 20140000 ]2. Add 601 to the result: [ 20140000 + 601 = 20140601 ]3. Divide the result by 13 using long division: - Dividing 20,140,601 by 13 gives 1,549,277. Verification by multiplying back confirms the result: [ 1,549,277 times 13 = 20,140,601 ]Thus, the final answer is (boxed{1549277}).,<think>Okay, so I need to calculate this big expression: (10 × 19 × 20 × 53 × 100 + 601) divided by 13. Hmm, that's a lot of multiplication and then adding 601 before dividing by 13. Let me think about how to approach this step by step. Maybe breaking it down into smaller parts would help. First, let me tackle the multiplication part: 10 × 19 × 20 × 53 × 100. That seems like a lot, but maybe I can multiply them step by step. Let me start with the first two numbers: 10 × 19. 10 times 19 is straightforward, right? That's 190. Then I have to multiply that result by 20. So 190 × 20. Let me calculate that: 190 × 20. Well, 19 × 2 is 38, and then add the two zeros from 190 and 20, so that's 3800. Wait, let me check that again. 190 × 20: 100 × 20 is 2000, 90 × 20 is 1800, so 2000 + 1800 = 3800. Yeah, that's correct. Next, I need to multiply that 3800 by 53. Hmm, 3800 × 53. That might be a bit more involved. Maybe I can break it down into 3800 × 50 and 3800 × 3, then add them together. Let's see: 3800 × 50. Well, 3800 × 10 is 38,000, so 3800 × 50 is 38,000 × 5, which is 190,000. Then 3800 × 3 is 11,400. Adding those together: 190,000 + 11,400 = 201,400. So 3800 × 53 is 201,400. Now, multiply that by 100. So 201,400 × 100. That's just adding two zeros to the end, making it 20,140,000. Wait, no. Wait, 201,400 × 100. Let me check again. 201,400 × 10 is 2,014,000. Then ×10 again would be 20,140,000. Yes, correct. So the entire multiplication part is 20,140,000. Then we have to add 601 to that. So 20,140,000 + 601. That should be 20,140,601. Now, we have to divide this sum by 13. So 20,140,601 ÷ 13. Hmm, dividing such a big number by 13. Maybe there's a smarter way than just long division here. Let me see if I can simplify this. Alternatively, maybe I can check if 20,140,601 is divisible by 13 without a remainder. Let's try dividing step by step. Alternatively, maybe using modular arithmetic or some divisibility rules. Wait, the divisibility rule for 13 is a bit complicated, but let's recall. One method for checking divisibility by 13 is to take the last digit, multiply it by 4, and subtract it from the rest of the number. Repeat until you get a number that's easy to check. Let's try that. Starting with 20,140,601. Let's take the last digit, which is 1, multiply by 4: 1×4=4. Subtract that from the remaining number: 2,014,060 - 4 = 2,014,056. Now repeat the process with 2,014,056. Last digit is 6, multiply by 4: 24. Subtract from 201,405: 201,405 - 24 = 201,381. Again: last digit 1, multiply by 4: 4. Subtract from 20,138: 20,138 - 4 = 20,134. Continue: last digit 4, multiply by 4: 16. Subtract from 2,013: 2,013 - 16 = 1,997. Again: last digit 7, multiply by 4: 28. Subtract from 199: 199 - 28 = 171. Now, check if 171 is divisible by 13. 13 × 13 is 169, so 13 × 13 = 169, 171 - 169 = 2, remainder 2. So the original number divided by 13 would have a remainder of 2? Wait, but that contradicts if I thought maybe the entire expression was divisible by 13. But since we added 601, maybe not. Hmm, perhaps I made a mistake in the divisibility test steps. Let me double-check. Alternatively, maybe instead of using the divisibility rule, I can perform the division step by step. Let's try that. Divide 20,140,601 by 13. Let me write it out as long division. 13 | 20,140,601Starting from the leftmost digits. 13 goes into 20 once (1), 1×13=13. Subtract 13 from 20: 7. Bring down the next digit: 1, making it 71. 13 into 71: 5 times (5×13=65). Subtract 65 from 71: 6. Bring down the next digit: 4, making it 64. 13 into 64 is 4 times (4×13=52). Subtract 52 from 64: 12. Bring down the next digit: 0, making it 120. 13 into 120 is 9 times (9×13=117). Subtract 117 from 120: 3. Bring down the next digit: 6, making it 36. 13 into 36 is 2 times (2×13=26). Subtract 26 from 36: 10. Bring down the next digit: 0, making it 100. 13 into 100 is 7 times (7×13=91). Subtract 91 from 100: 9. Bring down the next digit: 1, making it 91. 13 into 91 is exactly 7 times (7×13=91). Subtract 91: 0. So putting that all together, the quotient is 1,549,277 with a remainder of 0. Wait, but according to this long division, there's no remainder. But earlier when I tried the divisibility rule, I ended up with a remainder of 2. Which one is correct? There must be a mistake in one of the methods. Let me check my long division steps again. Maybe I messed up a step. Let's go through it again step by step. Dividing 20,140,601 by 13:1. 13 into 20: 1. 1×13=13. Subtract 13 from 20: 7. Bring down 1: 71.2. 13 into 71: 5. 5×13=65. Subtract 65: 6. Bring down 4: 64.3. 13 into 64: 4. 4×13=52. Subtract 52: 12. Bring down 0: 120.4. 13 into 120: 9. 9×13=117. Subtract 117: 3. Bring down 6: 36.5. 13 into 36: 2. 2×13=26. Subtract 26: 10. Bring down 0: 100.6. 13 into 100: 7. 7×13=91. Subtract 91: 9. Bring down 1: 91.7. 13 into 91: 7. 7×13=91. Subtract 91: 0. Remainder 0.So the quotient is 1 (from first step), 5 (second), 4 (third), 9 (fourth), 2 (fifth), 7 (sixth), 7 (seventh). So 1,549,277. And the remainder is 0. But this contradicts the divisibility test which suggested a remainder of 2. Where did I go wrong there?Let me retrace the divisibility steps:Original number: 20,140,601Step 1: Last digit 1, 1×4=4. Subtract from 2,014,060.1 becomes 2,014,060 - 4 = 2,014,056.Wait, but actually, when we take the original number 20,140,601, the last digit is 1. The rest is 2,014,060. So 2,014,060 - (1×4) = 2,014,060 -4 = 2,014,056.Then next step: Take 2,014,056. Last digit 6, 6×4=24. Rest of the number is 201,405.6, so 201,405 -24=201,381.Next: 201,381. Last digit 1, 1×4=4. Rest is 20,138.1, so 20,138 -4=20,134.Next: 20,134. Last digit 4, 4×4=16. Rest is 2,013.4, so 2,013 -16=1,997.Next: 1,997. Last digit 7, 7×4=28. Rest is 199.7, so 199 -28=171.171 divided by 13: 13×13=169, 171-169=2. So remainder 2. But according to the long division, the remainder is 0. So which is correct?Hmm, maybe the divisibility rule is not accurate here? Wait, let me check the original number. According to long division, the remainder is 0. But according to the divisibility steps, it's 2. That suggests a mistake in the divisibility test steps.Wait, perhaps I made a mistake in subtracting. Let me check each step again:Original number: 20,140,601Step 1: Last digit 1, multiply by 4: 4. Subtract from 20,140,60 (the rest). Wait, 20,140,601 minus 1*4 would be 20,140,601 -4? No, no. Wait, the method is: Take the last digit, multiply by 4, subtract that from the rest of the number. So the rest of the number is 20,140,60, and subtract 4. So 20,140,60 - 4 = 20,140,56. So 20,140,56. Then the new number is 20,140,56.Wait, but perhaps I need to keep leading zeros? Wait, not sure. Maybe I messed up the number of digits when subtracting. Let me write it out properly.Original number: 20,140,601Step 1: Take last digit (1), multiply by 4: 4. Subtract from the remaining number: 2,014,060 - 4 = 2,014,056. So the new number is 2,014,056.Step 2: Take last digit of 2,014,056, which is 6, multiply by 4: 24. Subtract from remaining number: 201,405 -24 = 201,381.Step 3: Take last digit of 201,381, which is 1, multiply by 4: 4. Subtract from remaining number: 20,138 -4 = 20,134.Step 4: Take last digit of 20,134, which is 4, multiply by 4: 16. Subtract from remaining number: 2,013 -16 = 1,997.Step 5: Take last digit of 1,997, which is 7, multiply by 4: 28. Subtract from remaining number: 199 -28 = 171.Step 6: Take last digit of 171, which is 1, multiply by 4: 4. Subtract from remaining number: 17 -4 = 13.Now, 13 is divisible by 13, so the remainder is 0. Wait, but previously I stopped at 171 and said 171-169=2. But if we continue the steps:From 171, take last digit 1, multiply by 4=4. Subtract from 17: 17 -4=13. Since 13 is divisible by 13, the remainder is 0. So actually, the original number is divisible by 13. So why did I stop at 171? Because I thought 171 divided by 13 gives 13×13=169, remainder 2. But according to the divisibility rule, we need to continue the process until we get a small number. So continuing:171: Last digit 1, multiply by 4=4. Subtract from 17: 17 -4=13. 13 is divisible by 13, so the original number is divisible by 13. So the remainder is 0. So my mistake earlier was not continuing the divisibility test steps until reaching a manageable number. Therefore, the correct remainder is 0. Therefore, 20,140,601 ÷13 is 1,549,277 with remainder 0. Therefore, the answer is 1,549,277. But wait, let me confirm with actual multiplication. Let's multiply 13 × 1,549,277 and see if it equals 20,140,601.Calculating 1,549,277 ×13:1,549,277 ×10=15,492,7701,549,277 ×3=4,647,831Add them together: 15,492,770 +4,647,831.15,492,770 +4,647,831. Let's add:15,492,770+4,647,831=20,140,601. Yes, that's correct. So 13 ×1,549,277=20,140,601, which matches. So the division is exact, no remainder. Therefore, the entire expression (10×19×20×53×100+601) ÷13 equals 1,549,277.But let me check again each step because this is a lot of computation and easy to make a mistake.First, 10×19=190. Then 190×20=3,800. Then 3,800×53.Wait, 3,800 ×50=190,000 and 3,800×3=11,400. 190,000+11,400=201,400. Then 201,400×100=20,140,000. Then adding 601 gives 20,140,601. Divided by13 gives 1,549,277. Yes, seems correct. And verifying that 13×1,549,277=20,140,601. Correct. So the answer is 1,549,277. But to make sure I didn't make a mistake in the multiplication steps, let me recompute 10×19×20×53×100.10×19=190.190×20=3,800.3,800×53. Let's compute 3,800×50=190,000 and 3,800×3=11,400. 190,000+11,400=201,400.201,400×100=20,140,000. Then add 601: 20,140,000 +601=20,140,601.Divide by13: 20,140,601 ÷13=1,549,277. Correct. Therefore, the final answer is 1,549,277.Final Answerboxed{1549277}</think>To calculate ((10 times 19 times 20 times 53 times 100 + 601) div 13), we proceed as follows:1. Calculate the product step-by-step: - (10 times 19 = 190) - (190 times 20 = 3800) - (3800 times 53 = 201,400) (breaking down into (3800 times 50 + 3800 times 3)) - (201,400 times 100 = 20,140,000)2. Add 601 to the product: - (20,140,000 + 601 = 20,140,601)3. Divide the result by 13: - Performing long division of (20,140,601) by (13) gives (1,549,277) with a remainder of (0).Verification:- Multiplying (13 times 1,549,277) confirms the result is (20,140,601).Thus, the final answer is (boxed{1549277}).]

🔑:begin{array}{l}(10 times 19 times 20 times 53 times 100+601) div 13=(2 times 19 times 53 times 10 times 10 times 100+601) div 13 =(2014 times 10000+601) div 13=(20140000+601) div 13=20140601 div 13=1549277end{array}The calculation is as follows:begin{array}{l}(10 times 19 times 20 times 53 times 100+601) div 13=(2 times 19 times 53 times 10 times 10 times 100+601) div 13 =(2014 times 10000+601) div 13=(20140000+601) div 13=20140601 div 13=1549277end{array}

🧠:[<think>Okay, so I need to figure out the functional relationship between the world population at the end of 2009, which is y billion, and the annual average growth rate x%. The starting point is the population at the end of 1999, which is 6 billion. Hmm, let's break this down step by step.First, the time period from the end of 1999 to the end of 2009 is 10 years. So we're looking at a span of 10 years. The growth rate is given as an annual average percentage, x%. I need to model the population growth over these 10 years with this growth rate.I remember that population growth can be modeled using exponential growth formulas. The general formula for exponential growth is:Final amount = Initial amount × (1 + growth rate)^number of periodsIn this case, the initial amount is 6 billion. The growth rate is x%, but I need to convert that percentage into a decimal for the formula. So x% as a decimal would be x/100. However, we have to be careful here—if the growth rate is x%, then the decimal equivalent is indeed x divided by 100. So, the growth factor per year is (1 + x/100).Since this growth happens every year for 10 years, we'll have to raise the growth factor to the power of 10. Therefore, the formula should be:y = 6 × (1 + x/100)^10Wait, let me verify this. Let's take an example. If the growth rate is 0%, then the population remains the same. Plugging x = 0 into the formula gives y = 6 × (1 + 0)^10 = 6 × 1 = 6. That makes sense. If the growth rate is 10%, which is quite high, then y = 6 × (1 + 0.10)^10. Calculating that, 1.1^10 is approximately 2.5937, so y ≈ 6 × 2.5937 ≈ 15.562 billion. That seems reasonable for a 10% growth rate over 10 years. Although in reality, such a high growth rate is unlikely, but mathematically, the formula works.Alternatively, sometimes continuous growth is modeled with e^(rt), but the problem mentions an annual average growth rate, which suggests it's compounded annually, not continuously. So the formula with (1 + x/100)^10 is appropriate here.Let me check if there's another way to interpret the problem. The annual average growth rate as x%—does that mean something different? Like, maybe geometric mean or something? But no, when they say annual average growth rate in the context of population, it's typically the compounded annual growth rate (CAGR), which is what this formula represents.Another way to think about it: if the population grows by x% each year, then each subsequent year's population is the previous year's multiplied by (1 + x/100). So after one year, it's 6*(1 + x/100). After two years, 6*(1 + x/100)^2, and so on until the tenth year, which would be 6*(1 + x/100)^10. So yes, that makes sense.Therefore, the functional relationship should be y = 6*(1 + x/100)^10. Alternatively, since x is a percentage, they might prefer expressing it as y = 6*(1 + 0.01x)^10. Both are equivalent; 1 + x/100 is the same as 1 + 0.01x. Depending on how they want it presented, either form is acceptable, but since the problem says "x%", writing it as (1 + x/100) might be clearer.Alternatively, maybe they want it in terms of decimal growth rate, but since x is given as a percentage, dividing by 100 is necessary. Let me see. For example, if x is 2, meaning 2%, then 1 + 2/100 = 1.02, which is correct.Yes, this seems right. So the final answer is y equals 6 times (1 plus x over 100) to the power of 10. So written in LaTeX, that would be y = 6 left(1 + frac{x}{100}right)^{10}.I think that's it. Let me just check once more. Start with 6 billion, grow by x percent each year for 10 years. Exponential growth formula. Yep, that's the standard way to compute it. So unless there's some nuance I'm missing here, like maybe the growth rate is simple interest instead of compound, but that's highly unlikely when talking about population growth. Populations grow exponentially, not linearly. A simple interest model would be linear growth: y = 6*(1 + 10*(x/100)) = 6 + 6*(x/10). But that's not realistic for population. So exponential is the way to go.Therefore, I think the correct function is y = 6*(1 + x/100)^10.Final AnswerThe functional relationship of ( y ) with respect to ( x ) is (boxed{6 left(1 + frac{x}{100}right)^{10}}).</think>The world population at the end of 1999 was 6 billion. We need to find the functional relationship of the population at the end of 2009, ( y ), with respect to an annual average growth rate of ( x% ).The time period from the end of 1999 to the end of 2009 is 10 years. Using the exponential growth formula, the population after 10 years can be modeled as:[y = 6 left(1 + frac{x}{100}right)^{10}]This formula accounts for the annual growth rate ( x% ), converted to a decimal by dividing by 100, and compounded annually over 10 years. Final AnswerThe functional relationship of ( y ) with respect to ( x ) is (boxed{6 left(1 + frac{x}{100}right)^{10}}).,<think>Alright, so I need to figure out the functional relationship between the world population y at the end of 2009 and the annual average growth rate x%. The starting population is 6 billion at the end of 1999. Hmm, okay. Let me think step by step.First, the time period between the end of 1999 and the end of 2009 is 10 years. So if the growth rate is x% per year, we can model the population growth using the formula for compound interest, right? Because population growth is typically exponential, similar to how money grows with compound interest.The compound interest formula is A = P(1 + r)^t, where:- A is the amount after t years,- P is the principal amount,- r is the annual growth rate in decimal,- t is the time in years.In this case, the principal P is 6 billion, the rate r is x% which needs to be converted to decimal (so x/100), and the time t is 10 years. So substituting these values into the formula, we should get y = 6*(1 + x/100)^10.Wait, let me double-check. The problem says "annual average growth rate of the population be x%", so yes, each year the population is multiplied by (1 + x/100). So after 10 years, it's 6*(1 + x/100)^10. That seems right.But hold on, is there another way to model population growth? Sometimes continuous growth is modeled with e^(rt), but the problem mentions an annual average growth rate, which suggests it's compounded annually, not continuously. So the formula with (1 + r)^t is correct here.Let me confirm with an example. Suppose the growth rate is 2% per year. Then after 10 years, the population should be 6*(1.02)^10. If I calculate that, 1.02^10 is approximately 1.21899, so 6*1.21899 ≈ 7.3139 billion. Does that make sense? Well, historical data shows that the population in 2009 was around 6.8 billion, so if the growth rate were 2%, the model would predict a bit higher, but maybe the actual growth rate was less. Anyway, the formula is correct regardless of specific numbers.Alternatively, if the growth rate were 0%, the population would stay at 6 billion, which the formula also gives because (1 + 0)^10 = 1, so y = 6*1 = 6. That checks out.Another test case: growth rate of 10%. Then it would be 6*(1.10)^10. 1.1^10 is approximately 2.5937, so 6*2.5937 ≈ 15.562 billion. That seems really high, but if the growth rate were 10% annually, which is extremely high, the population would indeed explode like that. So the formula works.Therefore, the functional relationship is y = 6*(1 + x/100)^10. But maybe we need to write it in terms of x as a percentage without converting? Wait, the problem says x%, so x is already a percentage, so converting it to decimal is necessary by dividing by 100.Alternatively, sometimes growth rates are expressed in formulas as decimals directly, but since the problem specifies x%, we have to use x/100 in the formula. So I think the answer is y = 6*(1 + x/100)^10. To write this more neatly, maybe as y = 6(1 + 0.01x)^10.Alternatively, using exponents, but I think both forms are equivalent. Let me check:x% as a decimal is x/100 = 0.01x. So 1 + 0.01x is the same as (1 + x/100). Therefore, 6*(1 + x/100)^10 is correct.So yes, the formula is y = 6*(1 + x/100)^10. Therefore, the functional relationship is y equals 6 multiplied by (1 plus x over 100) raised to the power of 10.Final AnswerThe functional relationship is boxed{y = 6left(1 + frac{x}{100}right)^{10}}.</think>The world population at the end of 1999 was 6 billion. We need to find the functional relationship between the population at the end of 2009 (y) and the annual average growth rate (x%).1. The time period between the end of 1999 and the end of 2009 is 10 years.2. The population growth can be modeled using the compound interest formula: ( A = P(1 + r)^t ), where: - ( A ) is the population after ( t ) years, - ( P ) is the initial population, - ( r ) is the annual growth rate in decimal, - ( t ) is the time in years.3. Substituting the given values: - ( P = 6 ) billion, - ( r = frac{x}{100} ) (converting x% to decimal), - ( t = 10 ) years.4. The formula becomes ( y = 6 left(1 + frac{x}{100}right)^{10} ).Thus, the functional relationship is (boxed{y = 6left(1 + frac{x}{100}right)^{10}}).]

🔑:(2) 60left(1+frac{x}{100}right)^{10}

🧠:[<think>Okay, so let's try to figure out this bonus distribution problem. Hmm, there are four employees, and they each voted for the distribution principle that benefits them the most, but it ended in a tie. So they decided to average the two distributions and round to the nearest 10 Ft. The youngest employee got 980 Ft. We need to find out what the others received. First, let me parse all the given information. The employees have different years of service: 14, 17, 21, and 31 years. Their salaries are 1500, 1600, 1800, and 2300 Ft. The total bonus amount is some number that was rounded to the nearest hundred. Then, after averaging the two distribution principles and rounding to the nearest ten, the youngest (which would be the one with 14 years of service?) received 980 Ft. So we need to figure out the total bonus first, maybe?Wait, the problem says "the vote did not result in a decision," which probably means it was a tie. Since there are four employees, maybe two voted for each method? But maybe each employee votes for the method that gives them more, so depending on the total amount and their proportions, their preferred method could differ. Then, the company decided to average the two distributions. So each employee gets the average of what they would have gotten under the salary-proportional method and the years-proportional method, then rounded to the nearest 10 Ft. The youngest one, with 14 years, got 980 Ft. So we need to find the total bonus amount first, perhaps?Since the total bonus was rounded to the nearest hundred, let's denote the exact total bonus before rounding as T. But actually, maybe the total bonus is a whole number of hundreds, since it's stated as "rounded to the nearest hundred." So the total bonus B is a multiple of 100. For example, if the actual amount was, say, 2450 Ft, they would round it to 2500 Ft. But maybe in this problem, they're just given that the total is a certain amount, already rounded. Wait, the problem says "a certain amount of money, rounded to the nearest hundred, is to be distributed..." So the total bonus is B, which is an exact multiple of 100. Hmm, maybe. Or maybe it's some amount that was rounded to the nearest hundred, so the actual total could be B ± 50. Wait, rounding to the nearest hundred could mean that the exact amount is within 50 of B. For example, if they say the total is 3400 Ft, the actual amount before rounding could be between 3350 and 3450. But does that matter here?But maybe the total bonus after averaging and rounding is such that each employee's amount is rounded to the nearest 10. So the final amounts are multiples of 10. The youngest received 980 Ft, which is a multiple of 10. So maybe each employee's average of the two distributions is then rounded to the nearest 10. Let me think. Let's denote:Total bonus amount: B (rounded to the nearest hundred). So B is an integer multiple of 100. But before rounding, the exact amount was somewhere between B - 50 and B + 50. Wait, but maybe they just take B as the total, already rounded. Hmm, the problem says "a certain amount of money, rounded to the nearest hundred, is to be distributed..." So maybe the exact amount is B, which is a multiple of 100. Or perhaps not. Wait, the problem might not specify whether the total is before or after rounding. This is a bit ambiguous.Wait, maybe the total amount is an exact number, but when they say "rounded to the nearest hundred," that's just for the purpose of informing the employees. So the actual total is a precise number, but it was presented as rounded to the nearest hundred. So the actual total could be, for example, 3450, which would be rounded to 3500. Hmm, but perhaps we can just consider B as the exact total, which is an integer. But maybe it's necessary to figure out the exact value based on the given information. Since when they average the two distributions and round to the nearest 10, the youngest gets 980, which is 980.0, so exact 980. So maybe the average for that employee was exactly 980, or maybe it was between 975 and 985, which would round to 980. But since the final answer is 980, perhaps it was exactly 980.5 or something, but maybe not. Since we have to calculate exact numbers, perhaps all the averages are exact multiples of 10, so that rounding doesn't change them. Hmm, but the problem says "rounded to the nearest 10 Ft", so they could round up or down. But the youngest received 980 Ft, so that was after rounding. So maybe the exact average was 975 to 984.999..., which rounds to 980. Wait, but 985 would round to 990. So the average must have been between 975 and 985. This seems complicated. Maybe we need to find the total bonus B first.Let me outline the steps:1. Let the total bonus be B (rounded to the nearest 100). So B is an integer multiple of 100.2. Under the first principle (salary proportion), each employee gets B * (salary_i / total_salary).3. Under the second principle (years proportion), each employee gets B * (years_i / total_years).4. Each employee then receives the average of these two amounts, which is (salary_portion + years_portion)/2, then rounded to the nearest 10 Ft.5. The youngest employee (14 years) received 980 Ft after rounding. So we can set up an equation for this employee.Let's denote the youngest employee as Employee 1 with salary S1 = 1500 Ft and years Y1 = 14.Total salary = 1500 + 1600 + 1800 + 2300 = let's calculate that.1500 + 1600 = 31003100 + 1800 = 49004900 + 2300 = 7200 Ft. Total salary is 7200 Ft.Total years = 14 + 17 + 21 + 31 = let's add:14 + 17 = 3131 + 21 = 5252 + 31 = 83 years. Total years is 83.So for Employee 1:Salary portion: B * (1500 / 7200) = B * (5/24) ≈ 0.208333BYears portion: B * (14 / 83) ≈ 0.168674BAverage of the two: (0.208333B + 0.168674B)/2 = (0.376007B)/2 ≈ 0.1880035BThis average, when rounded to the nearest 10 Ft, is 980. So:0.1880035B ≈ 980 ± 5 (since rounding to nearest 10)But maybe it's better to model the exact value.Let me denote the exact average as A1 = (B * (1500/7200) + B * (14/83)) / 2Which is B*( (1500/7200) + (14/83) ) / 2Compute the fractions:1500/7200 = 5/24 ≈ 0.208333...14/83 ≈ 0.168674...Sum ≈ 0.376007...Divide by 2: ≈ 0.1880035...So A1 ≈ 0.1880035BAfter rounding to the nearest 10, it's 980. Therefore, the actual average A1 must be in [975, 985).Thus:975 ≤ A1 < 985But A1 = B*(5/24 + 14/83)/2Let me compute the exact value of the coefficient:First, 5/24 + 14/83. Let's compute this exactly.Find a common denominator for 24 and 83: 24*83 = 1992So 5/24 = (5*83)/1992 = 415/199214/83 = (14*24)/1992 = 336/1992Sum: 415 + 336 = 751/1992Divide by 2: 751/(1992*2) = 751/3984 ≈ 0.188004...So A1 = B * (751/3984)Therefore:975 ≤ B*(751/3984) < 985Multiply all parts by 3984/751:975 * (3984/751) ≤ B < 985 * (3984/751)Compute 975*(3984)/751 and 985*(3984)/751.First, compute 3984/751:3984 ÷ 751 ≈ 5.304... Let me compute:751 * 5 = 37553984 - 3755 = 229751 * 0.3 = 225.3229 - 225.3 = 3.7So 5.3 + 3.7/751 ≈ 5.3 + 0.0049 ≈ 5.3049So approximately 5.3049Therefore,975 * 5.3049 ≈ 975 * 5 + 975 * 0.3049= 4875 + 975 * 0.3 + 975 * 0.0049= 4875 + 292.5 + 4.7775 ≈ 4875 + 292.5 = 5167.5 + 4.7775 ≈ 5172.2775Similarly,985 * 5.3049 ≈ 985 * 5 + 985 * 0.3049= 4925 + 985 * 0.3 + 985 * 0.0049= 4925 + 295.5 + 4.8265 ≈ 4925 + 295.5 = 5220.5 + 4.8265 ≈ 5225.3265Therefore, B must be between approximately 5172.28 and 5225.33. But the total bonus B was rounded to the nearest hundred. So possible values for B are 5100, 5200, 5300, etc. But our calculated range is approximately 5172 to 5225, so the only multiple of 100 in this range is 5200. Because 5100 is too low (5172 is above 5100), 5200 is within the range, 5300 is above 5225. So B must be 5200.Wait, let me verify. If B is 5200, then:A1 = 5200 * (751/3984) ≈ 5200 * 0.1880035 ≈ 5200 * 0.188 ≈ 977.6Which is 977.6, which would round to 980. Correct.If B were 5100:A1 = 5100 * 0.1880035 ≈ 5100 * 0.188 ≈ 958.8, which would round to 960, which is not 980. So B must be 5200.Similarly, check 5300:A1 ≈ 5300 * 0.188 ≈ 996.4, which rounds to 1000. Not 980. So yes, B is 5200.So total bonus is 5200 Ft.Now, we need to compute for each employee their average of the two distributions and round to the nearest 10.First, let's compute the salary-based distribution and the years-based distribution for each employee.Total salary is 7200, total years is 83.Salary-based distribution:Employee 1 (1500 Ft salary): 5200 * (1500/7200) = 5200 * (5/24) ≈ 5200 * 0.208333 ≈ 1083.333 FtEmployee 2 (1600 Ft salary): 5200 * (1600/7200) = 5200 * (4/18) = 5200 * (2/9) ≈ 5200 * 0.222222 ≈ 1155.555 FtEmployee 3 (1800 Ft salary): 5200 * (1800/7200) = 5200 * (1/4) = 1300 FtEmployee 4 (2300 Ft salary): 5200 * (2300/7200) ≈ 5200 * 0.319444 ≈ 1661.111 FtYears-based distribution:Employee 1 (14 years): 5200 * (14/83) ≈ 5200 * 0.168674 ≈ 877.108 FtEmployee 2 (17 years): 5200 * (17/83) ≈ 5200 * 0.204819 ≈ 1065.060 FtEmployee 3 (21 years): 5200 * (21/83) ≈ 5200 * 0.253012 ≈ 1315.663 FtEmployee 4 (31 years): 5200 * (31/83) ≈ 5200 * 0.373494 ≈ 1942.168 FtNow, compute the average for each employee:Employee 1: (1083.333 + 877.108)/2 ≈ (1960.441)/2 ≈ 980.2205 FtRounded to nearest 10: 980 Ft (matches given info)Employee 2: (1155.555 + 1065.060)/2 ≈ (2220.615)/2 ≈ 1110.3075 Ft → Rounded to 1110 FtEmployee 3: (1300 + 1315.663)/2 ≈ (2615.663)/2 ≈ 1307.8315 Ft → Rounded to 1310 FtEmployee 4: (1661.111 + 1942.168)/2 ≈ (3603.279)/2 ≈ 1801.6395 Ft → Rounded to 1800 Ft or 1802? Wait, 1801.6395 is closer to 1800 or 1800? Wait, 1801.64 is 1.64 above 1800, and 0.36 below 1805? Wait, no, nearest 10. So 1801.64 is 1.64 away from 1800 and 8.36 away from 1810. So it should round to 1800. Wait, but 1801.64 is only 1.64 above 1800, which is less than 5, so it rounds down. So 1800.But wait, 1801.6395 is approximately 1801.64, which is 1801.6 when rounded to the nearest tenth, but we need to round to the nearest 10. So 1801.64 is between 1800 and 1810. The midpoint is 1805. Since 1801.64 is less than 1805, it rounds down to 1800. Therefore, 1800.But let's check these calculations again to make sure.First, Employee 1: average ≈ 980.22, rounds to 980. Correct.Employee 2:Salary portion: 5200*(1600/7200) = 5200*(2/9) ≈ 1155.555...Years portion: 5200*(17/83) ≈ 1065.060...Average: (1155.555 + 1065.060)/2 ≈ (2220.615)/2 ≈ 1110.3075. So 1110.3075 rounds to 1110. Correct.Employee 3:Salary portion: 5200*(1800/7200) = 5200*(1/4) = 1300Years portion: 5200*(21/83) ≈ 1315.663...Average: (1300 + 1315.663)/2 = 2615.663/2 ≈ 1307.8315. Rounded to nearest 10: 1310. Because 1307.8315 is 7.83 above 1300, which is closer to 1310? Wait, 1307.83 is 7.83 away from 1300 and 2.17 away from 1310. Wait, no. Wait, when rounding to the nearest 10, you look at the unit digit. 1307.83 is 1307.83, so the tens place is 0 (as in 1300). The next 10 is 1310. The value is 1307.83, which is 7.83 above 1300, so it's 7.83 away from 1300 and 2.17 away from 1310. Wait, no. Wait, 1307.83 is between 1300 and 1310. The halfway point is 1305. Since 1307.83 is above 1305, it rounds up to 1310. Ah, right. So 1307.83 rounds up to 1310. Correct.Employee 4:Salary portion: 5200*(2300/7200) ≈ 5200*(0.319444) ≈ 1661.111Years portion: 5200*(31/83) ≈ 1942.168Average: (1661.111 + 1942.168)/2 ≈ 3603.279/2 ≈ 1801.6395. As discussed, this is 1801.64, which is above 1800, but when rounding to the nearest 10, since the last digit before the tens place is 1 (from 1801.64), we look at the units digit which is 1, which is less than 5, so we round down? Wait, no. Wait, rounding to the nearest 10: if the number is 1801.64, the tens digit is 0 (in 1800), and the next digit is 1. Since 1 is less than 5, we round down to 1800. But wait, 1801.64 is 1801.64, which is 1.64 above 1800. So it's 1.64 away from 1800 and 8.36 away from 1810. Therefore, the nearest 10 is 1800. So it should round down to 1800. But some might argue if the 0.64 is considered, but no—when rounding to the nearest 10, you look at the units digit. If the units digit is less than 5, you round down. But here, the number is 1801.64, which has a units digit of 1 (in the ones place: 1801). So 1801.64 is between 1800 and 1810, and since the ones digit is 1, which is less than 5, you round down to 1800. Wait, but some might consider the decimal part. But when rounding to the nearest 10, typically, you look at the ones digit. For example, 1801.64 rounded to the nearest 10 is 1800 because the ones digit is 1. The decimal part doesn't affect rounding to the nearest 10. Wait, no—if you have 1801.64, the rule is to look at the digit in the ones place. If it's 5 or above, round up, otherwise round down. So 1801.64 has a 1 in the ones place, so you round down to 1800. Similarly, 1805.00 would round up to 1810. Therefore, 1801.64 rounds to 1800. However, depending on the rounding convention, sometimes people consider the first decimal digit. For example, if it's 1801.5, you might round to 1800 or 1810 depending on the convention. But in standard rounding rules, you look at the digit in the place you're rounding to. Wait, no—actually, when rounding to the nearest 10, you look at the digit in the ones place. If it's 5 or more, round up, otherwise round down. The decimal fractions are irrelevant. So 1801.64 would be 1800 when rounded to the nearest 10, because the ones digit is 1. But that seems counterintuitive because 1801.64 is closer to 1800 than 1810? Wait, 1801.64 is 1.64 away from 1800 and 8.36 away from 1810, so actually, it's closer to 1800. Therefore, it should round to 1800. So maybe the answer is 1800. But wait, let me confirm with the exact calculation. For the fourth employee:Average ≈ 1801.6395 Ft. When rounding to the nearest 10, we can write this as 1801.6395 ≈ 1801.6. The nearest multiple of 10 is 1800, since 1801.6 is only 1.6 above 1800, which is less than 5. So yes, 1800. So employee 4 gets 1800.But wait, the problem says "rounded to the nearest 10 Ft". So, if the amount is 1801.64, that would round to 1800, because 1801.64 is closer to 1800 (difference of 1.64) than to 1810 (difference of 8.36). So yes, 1800.Therefore, the amounts are:Employee 1: 980 FtEmployee 2: 1110 FtEmployee 3: 1310 FtEmployee 4: 1800 FtBut let me check once again to ensure there are no calculation errors.Total bonus B = 5200Employee 1 salary portion: 5200*(1500/7200) = 5200*(5/24) = (5200/24)*5 = 216.666...*5 = 1083.333...Employee 1 years portion: 5200*(14/83) ≈ (5200/83)*14 ≈ 62.6506*14 ≈ 877.108Average: (1083.333 + 877.108)/2 ≈ 1960.441/2 ≈ 980.2205, rounds to 980. Correct.Employee 2:Salary: 1600/7200 = 2/9 ≈ 0.222222, 5200*(2/9) ≈ 1155.555...Years: 17/83 ≈ 0.204819, 5200*0.204819 ≈ 1065.06Average: (1155.555 + 1065.06)/2 ≈ 2220.615/2 ≈ 1110.3075, rounds to 1110. Correct.Employee 3:Salary: 1800/7200 = 1/4, so 5200/4 = 1300Years: 21/83 ≈ 0.253012, 5200*0.253012 ≈ 1315.663Average: (1300 + 1315.663)/2 ≈ 2615.663/2 ≈ 1307.8315, rounds to 1310. Correct.Employee 4:Salary: 2300/7200 ≈ 0.319444, 5200*0.319444 ≈ 1661.111Years: 31/83 ≈ 0.373494, 5200*0.373494 ≈ 1942.168Average: (1661.111 + 1942.168)/2 ≈ 3603.279/2 ≈ 1801.6395, rounds to 1800. Correct.Therefore, the amounts received by the others (since the youngest got 980) are 1110, 1310, and 1800 Ft.So the answer should be:Employee 2: 1110 FtEmployee 3: 1310 FtEmployee 4: 1800 FtThus, the others received 1110, 1310, and 1800 Ft respectively.Final AnswerThe others received boxed{1110}, boxed{1310}, and boxed{1800} Ft respectively.</think>Given a total bonus amount ( B ) rounded to the nearest hundred, we need to determine the distribution among four employees based on the average of two principles: proportional to their salaries and proportional to their years of service. The youngest employee received 980 Ft after rounding to the nearest 10 Ft. 1. Total Salary and Total Years Calculation: - Total salary: ( 1500 + 1600 + 1800 + 2300 = 7200 ) Ft - Total years: ( 14 + 17 + 21 + 31 = 83 ) years2. Determining Total Bonus ( B ): - The youngest employee (14 years, 1500 Ft salary) received 980 Ft after averaging and rounding. - Salary portion: ( B times frac{1500}{7200} ) - Years portion: ( B times frac{14}{83} ) - Average of the two portions: ( frac{B}{2} left( frac{1500}{7200} + frac{14}{83} right) ) - Solving for ( B ) gives ( B = 5200 ) Ft (rounded to the nearest hundred).3. Calculating Each Employee's Share: - Salary-based distribution: - Employee 1: ( 5200 times frac{1500}{7200} = 1083.33 ) Ft - Employee 2: ( 5200 times frac{1600}{7200} = 1155.56 ) Ft - Employee 3: ( 5200 times frac{1800}{7200} = 1300 ) Ft - Employee 4: ( 5200 times frac{2300}{7200} = 1661.11 ) Ft - Years-based distribution: - Employee 1: ( 5200 times frac{14}{83} = 877.11 ) Ft - Employee 2: ( 5200 times frac{17}{83} = 1065.06 ) Ft - Employee 3: ( 5200 times frac{21}{83} = 1315.66 ) Ft - Employee 4: ( 5200 times frac{31}{83} = 1942.17 ) Ft4. Averaging and Rounding: - Employee 1 (youngest): ( frac{1083.33 + 877.11}{2} approx 980.22 ) → 980 Ft - Employee 2: ( frac{1155.56 + 1065.06}{2} approx 1110.31 ) → 1110 Ft - Employee 3: ( frac{1300 + 1315.66}{2} approx 1307.83 ) → 1310 Ft - Employee 4: ( frac{1661.11 + 1942.17}{2} approx 1801.64 ) → 1800 FtThe amounts received by the others are (boxed{1110}), (boxed{1310}), and (boxed{1800}) Ft respectively.,<think>Okay, let's see. There's a company that's giving out bonuses to four employees. The total amount of money is rounded to the nearest hundred. They had to choose between two distribution principles: one based on their monthly salaries and the other based on the years they've worked. But they couldn't decide because each employee voted for the principle that benefited them more. So, they ended up averaging the two bonuses for each person, rounding to the nearest 10 Ft. The youngest employee got 980 Ft. We need to find out how much the others received. The years worked are 14, 17, 21, 31, and the salaries are 1500, 1600, 1800, 2300 Ft respectively.First, let's parse the problem step by step. The key points are:1. Total bonus amount is rounded to the nearest hundred. Let's denote this total as T. Since it's rounded to the nearest hundred, T could be, for example, 1000, 1100, 1200, etc., but not something like 1050 or 1130.2. Two distribution principles: - First: Proportional to their monthly salaries. - Second: Proportional to the years they've worked.3. Each employee voted for the principle that was more favorable to them. The vote didn't result in a decision, so they compromised by taking the average of the two principles for each employee, then rounding to the nearest 10 Ft.4. The youngest employee (let's assume this is the one with the least years worked, which is 14 years) received 980 Ft after averaging and rounding.5. We need to find out what the other three employees received.So, the challenge here is to figure out the total bonus amount T, then compute the two possible distributions (salary-based and years-based), average them for each employee, round to the nearest 10 Ft, and see which total T results in the youngest employee (14 years) getting exactly 980 Ft.Since T is rounded to the nearest hundred, we might need to test possible values of T and see which one fits the given condition.First, let's note the employees:Employee A: 14 years, 1500 Ft salaryEmployee B: 17 years, 1600 Ft salaryEmployee C: 21 years, 2300 Ft salaryEmployee D: 31 years, 1800 Ft salaryWait, wait, hold on. The salaries are listed as 1500, 1600, 1800, 2300. But the years are 14,17,21,31. Wait, the problem says: "they have been working for 14, 17, 21, and 31 years, respectively, and their current salaries are 1500 Ft, 1600 Ft, 1800 Ft, and 2300 Ft, respectively."So, respectively means:Employee 1: 14 years, 1500 FtEmployee 2: 17 years, 1600 FtEmployee 3: 21 years, 1800 FtEmployee 4: 31 years, 2300 FtWait, no. Wait, the problem states: "they have been working for 14, 17, 21, and 31 years, respectively, and their current salaries are 1500 Ft, 1600 Ft, 1800 Ft, and 2300 Ft, respectively." So, the order is the same for both. So, the first employee has 14 years and 1500 salary, second 17 years and 1600 salary, third 21 years and 1800 salary, fourth 31 years and 2300 salary.So, employees in order:1. 14 years, 1500 Ft2. 17 years, 1600 Ft3. 21 years, 1800 Ft4. 31 years, 2300 FtThe youngest employee is the one with 14 years, so that's employee 1, who received 980 Ft.So, the problem is to figure out the total bonus T, then compute for each employee the average of the salary-based bonus and the years-based bonus, round to the nearest 10, and see which T makes employee 1's amount 980 Ft. Then compute the others accordingly.So, let's denote:For each employee i (i=1 to 4):Let S_i be their salary, Y_i be their years of service.Total salary sum: S_total = 1500 + 1600 + 1800 + 2300 = let's compute that.1500 + 1600 = 31003100 + 1800 = 49004900 + 2300 = 7200 Ft. So S_total = 7200 Ft.Total years sum: Y_total = 14 + 17 + 21 + 31 = let's compute.14 + 17 = 3131 + 21 = 5252 + 31 = 83 years. So Y_total = 83 years.Therefore, if using salary-based distribution, each employee's bonus would be (S_i / S_total) * TSimilarly, years-based distribution would be (Y_i / Y_total) * TThen, the average of these two is [(S_i / S_total) + (Y_i / Y_total)] / 2 * TThen, this average is rounded to the nearest 10 Ft.So, for employee 1 (14 years, 1500 Ft):Average bonus before rounding = [(1500/7200) + (14/83)] / 2 * TCompute this value, round to nearest 10, result is 980.So, we need to find T such that when we compute [(1500/7200 + 14/83)/2] * T and round to nearest 10, we get 980.Then, once T is found, compute the same for other employees.First, let's compute the coefficient for employee 1:Compute (1500/7200) = 0.208333...14/83 ≈ 0.1686747...Average of these two: (0.208333... + 0.1686747)/2 ≈ (0.377008) / 2 ≈ 0.188504So, the coefficient is approximately 0.188504Therefore, the average bonus before rounding is 0.188504 * TThis amount, when rounded to the nearest 10, is 980.So, 0.188504 * T must be in the interval [980 - 5, 980 + 5), i.e., [975, 985)Therefore,975 ≤ 0.188504 * T < 985Thus,975 / 0.188504 ≤ T < 985 / 0.188504Compute these values.First, compute 975 / 0.188504 ≈ 975 / 0.1885 ≈ Let's compute 975 / 0.1885.Multiply numerator and denominator by 1000: 975000 / 188.5 ≈Let's compute 975000 / 188.5:Divide 975000 by 188.5:First, 188.5 * 5000 = 188.5 * 5000 = 94,250So, 188.5 * 5000 = 94,250Subtract from 975,000: 975,000 - 942,500 = 32,500Wait, 975,000 divided by 188.5:Wait, 188.5 * 5000 = 942,500975,000 - 942,500 = 32,500Now, 32,500 / 188.5 ≈ 32,500 / 188.5 ≈ 172.41So total is 5000 + 172.41 ≈ 5172.41Similarly, 985 / 0.188504 ≈ 985 / 0.1885 ≈ 985000 / 188.5 ≈Same way:188.5 * 5000 = 942,500985,000 - 942,500 = 42,50042,500 / 188.5 ≈ 42,500 / 188.5 ≈ 225.46Thus, total ≈ 5000 + 225.46 ≈ 5225.46Therefore, T must be between approximately 5172.41 and 5225.46.But T is the total bonus amount, which is rounded to the nearest hundred. So possible T values are 5100, 5200, 5300.Wait, but according to the above, T must be approximately between 5172 and 5225. So the nearest hundred would be 5200. Because 5172 is closer to 5200 than 5100? 5172 - 5100 = 72, 5200 - 5172 = 28, so 5172 is closer to 5200. Similarly, 5225 is 25 away from 5200 and 75 away from 5300. So 5225 would round to 5200. So T must be 5200.Wait, but let's confirm.If T is 5200, then the exact average bonus for employee 1 would be:0.188504 * 5200 ≈ 0.188504 * 5200Calculate 0.188504 * 5200:First, 0.1 * 5200 = 5200.08 * 5200 = 4160.008504 * 5200 ≈ 0.008 * 5200 = 41.6, and 0.000504*5200≈2.6208So total ≈ 520 + 416 + 41.6 + 2.6208 ≈ 520 + 416 = 936; 936 + 41.6 = 977.6; 977.6 + 2.6208 ≈ 980.2208So approximately 980.22 Ft, which would round to 980 Ft, as given. Perfect.Therefore, T is 5200 Ft.But let's check if T was rounded to the nearest hundred. The original amount before rounding must have been between 5150 and 5250 to round to 5200. But according to the problem, the total amount is "rounded to the nearest hundred". So the actual total amount could have been, say, 5172, but when rounded to the nearest hundred, it becomes 5200. Therefore, the exact total is 5200.But wait, perhaps we need to confirm whether T is 5200. Let's check.Alternatively, maybe the initial total was not 5200, but when they distributed via two principles and averaged, but the problem states that the total amount is rounded to the nearest hundred. Wait, the problem says: "a certain amount of money, rounded to the nearest hundred, is to be distributed..." So the total amount T is a multiple of 100. Therefore, T is exactly 5200.Therefore, T = 5200.So, the total bonus is 5200 Ft.Now, we need to compute for each employee the average of the salary-based and years-based distributions, then round to nearest 10 Ft.Let's compute for each employee.First, for employee 1 (14 years, 1500 Ft):Salary-based: (1500 / 7200) * 5200 = (1500/7200)*52001500/7200 = 5/24 ≈ 0.2083330.208333 * 5200 ≈ 0.208333 * 5200 ≈ 1083.333 FtYears-based: (14 / 83) * 5200 ≈ (0.1686747) * 5200 ≈ 0.1686747 * 5200 ≈ 877.10 FtAverage of these two: (1083.333 + 877.10)/2 ≈ (1960.433)/2 ≈ 980.2165 FtRounded to nearest 10 Ft: 980 Ft (since 980.2165 is closer to 980 than 990). So that checks out.Now, let's compute for the others.Employee 2: 17 years, 1600 FtSalary-based: (1600 / 7200) * 5200 = (1600/7200)*5200 = (2/9) * 5200 ≈ 1155.555 FtYears-based: (17 / 83) * 5200 ≈ 0.204819 * 5200 ≈ 1065.06 FtAverage: (1155.555 + 1065.06)/2 ≈ (2220.615)/2 ≈ 1110.3075 FtRounded to nearest 10: 1110 FtEmployee 3: 21 years, 1800 FtSalary-based: (1800 / 7200)*5200 = (1/4)*5200 = 1300 FtYears-based: (21/83)*5200 ≈ 0.253012 * 5200 ≈ 1315.66 FtAverage: (1300 + 1315.66)/2 ≈ 2615.66 / 2 ≈ 1307.83 FtRounded to nearest 10: 1310 FtEmployee 4: 31 years, 2300 FtSalary-based: (2300 / 7200)*5200 ≈ (0.319444)*5200 ≈ 0.319444*5200 ≈ 1661.11 FtYears-based: (31 / 83)*5200 ≈ 0.373494 * 5200 ≈ 1942.16 FtAverage: (1661.11 + 1942.16)/2 ≈ (3603.27)/2 ≈ 1801.635 FtRounded to nearest 10: 1800 Ft (since 1801.635 is closer to 1800 than 1810. Wait, 1801.635 is 1.635 above 1800, so it should round to 1800? Wait, no. Wait, rounding to the nearest 10: 1801.635 is 1.635 above 1800, so 1800 is the nearest multiple of 10. But actually, 1801.635 is 1801.64, which is 1.64 above 1800, and 8.36 below 1810. So it's closer to 1800. So yes, rounds to 1800.Wait, hold on. Wait, 1801.635 is between 1800 and 1810. The difference from 1800 is 1.635, and from 1810 is 8.365. Since 1.635 is less than 5, it rounds down. So yes, 1800.Therefore, the bonuses are:Employee 1: 980 FtEmployee 2: 1110 FtEmployee 3: 1310 FtEmployee 4: 1800 FtBut let's check the calculations again for each employee to make sure.Starting with Employee 2:Salary-based: 1600 / 7200 = 0.222222...0.222222 * 5200 = 1155.555... FtYears-based: 17 / 83 ≈ 0.2048190.204819 * 5200 ≈ 0.204819 * 5000 = 1024.095, plus 0.204819*200=40.9638, total ≈ 1065.059 FtAverage: (1155.555 + 1065.059)/2 ≈ 2220.614 / 2 ≈ 1110.307, which rounds to 1110. Correct.Employee 3:Salary-based: 1800 / 7200 = 0.250.25 * 5200 = 1300 FtYears-based: 21 / 83 ≈ 0.2530120.253012 * 5200 ≈ 0.25 * 5200 = 1300, 0.003012*5200≈15.6624, so total ≈ 1315.6624Average: (1300 + 1315.6624)/2 ≈ 2615.6624 / 2 ≈ 1307.8312, rounds to 1310. Correct.Employee 4:Salary-based: 2300 / 7200 ≈ 0.3194440.319444 * 5200 ≈ 0.319444 * 5000 = 1597.22, 0.319444*200=63.8888, total ≈ 1597.22 + 63.8888 ≈ 1661.1088Years-based: 31 / 83 ≈ 0.3734940.373494 * 5200 ≈ 0.373494 * 5000 = 1867.47, 0.373494*200=74.6988, total ≈ 1867.47 + 74.6988 ≈ 1942.1688Average: (1661.1088 + 1942.1688)/2 ≈ 3603.2776 / 2 ≈ 1801.6388Rounded to nearest 10: 1801.6388 is approximately 1801.64, which is 1.64 over 1800, so rounds to 1800 as per the nearest 10. Correct.Therefore, the final amounts are:Employee 1: 980 FtEmployee 2: 1110 FtEmployee 3: 1310 FtEmployee 4: 1800 FtBut let's confirm that the total distributed amount after averaging and rounding is still 5200. Let's check:980 + 1110 + 1310 + 1800 = 980 + 1110 = 2090; 2090 + 1310 = 3400; 3400 + 1800 = 5200. Perfect. So the total adds up to 5200, which matches the total T. So everything checks out.Therefore, the answer is that the others received 1110 Ft, 1310 Ft, and 1800 Ft respectively.Final AnswerThe other employees received boxed{1110}, boxed{1310}, and boxed{1800} Ft respectively.</think>Given a company's bonus distribution problem, we need to determine the amounts received by four employees based on averaging two distribution principles (proportional to salaries and years worked) and rounding to the nearest 10 Ft. The youngest employee received 980 Ft. Key steps:1. Total Bonus Calculation: - The total bonus amount ( T ) is rounded to the nearest hundred. Through calculations, we determined ( T = 5200 ) Ft.2. Salary and Years Proportions: - Total salary sum: ( 1500 + 1600 + 1800 + 2300 = 7200 ) Ft. - Total years sum: ( 14 + 17 + 21 + 31 = 83 ) years.3. Employee Calculations: - For each employee, calculate the average of the salary-based and years-based distributions, then round to the nearest 10 Ft.Employee 1 (14 years, 1500 Ft):- Salary-based: ( frac{1500}{7200} times 5200 approx 1083.33 ) Ft- Years-based: ( frac{14}{83} times 5200 approx 877.10 ) Ft- Average: ( frac{1083.33 + 877.10}{2} approx 980.22 ) Ft (rounded to 980 Ft)Employee 2 (17 years, 1600 Ft):- Salary-based: ( frac{1600}{7200} times 5200 approx 1155.56 ) Ft- Years-based: ( frac{17}{83} times 5200 approx 1065.06 ) Ft- Average: ( frac{1155.56 + 1065.06}{2} approx 1110.31 ) Ft (rounded to 1110 Ft)Employee 3 (21 years, 1800 Ft):- Salary-based: ( frac{1800}{7200} times 5200 = 1300 ) Ft- Years-based: ( frac{21}{83} times 5200 approx 1315.66 ) Ft- Average: ( frac{1300 + 1315.66}{2} approx 1307.83 ) Ft (rounded to 1310 Ft)Employee 4 (31 years, 2300 Ft):- Salary-based: ( frac{2300}{7200} times 5200 approx 1661.11 ) Ft- Years-based: ( frac{31}{83} times 5200 approx 1942.17 ) Ft- Average: ( frac{1661.11 + 1942.17}{2} approx 1801.64 ) Ft (rounded to 1800 Ft)Final amounts received by the employees:- Employee 1: 980 Ft- Employee 2: 1110 Ft- Employee 3: 1310 Ft- Employee 4: 1800 Ft[boxed{1110}, boxed{1310}, boxed{1800}]]

🔑:If the reward amount were to be divided in proportion to the payments, then the individual workers would receive 15/72, 16/72, 18/72, and 23/72 of the total amount, respectively. In the case of distribution in proportion to the number of years worked, they would receive 14/83, 17/83, 21/83, and 31/83 of the total amount, respectively. The average of these two fractional parts isfrac{1}{2}left(frac{15}{72}+frac{14}{83}right)=frac{2253}{11952}, quad frac{2552}{11952}, quad frac{3006}{11952}, quad frac{4141}{11952}The rewards paid to the workers according to the agreement are obtained by multiplying these averages by the total amount x to be distributed, and rounding the products to the nearest 10 Ft.The reward for the youngest worker, calculated by the said multiplication, had to fall between 975 and 985 Ft, because according to the announced rounding rule, 980 Ft results from these values. Therefore,975 leq frac{2253}{11952} x leq 985(Multiplying by the reciprocal of the coefficient of x on both sides, we get)5172 frac{684}{2253} leq x leq 5225 frac{795}{2253}(Since the multiplier used is positive).Among the limits obtained for x, there is only one round hundred, so x=5200 mathrm{Ft}.Using this value of x, the rewards for the other three workers, according to the above calculation principle, are 1110 Ft, 1310 Ft, and 1800 Ft, respectively. The adjustments made during rounding, both up and down, balance each other out, and the total amount of the rewards paid is 5200 Ft.

🧠:[<think>Okay, so I need to solve the equation x = f_n(x) where f_n(x) is the nth composition of the function f(x) = 1 + 2/x. Hmm, let's see. Let me start by understanding the problem step by step.First, f_1(x) is just f(x) = 1 + 2/x. So, solving x = f_1(x) would be solving x = 1 + 2/x. Let me check that. Multiply both sides by x: x² = x + 2, which gives x² - x - 2 = 0. Factoring: (x - 2)(x + 1) = 0, so solutions are x = 2 and x = -1. That makes sense.Now, for f_2(x), it's f(f_1(x)), so f applied to f_1(x). Since f_1(x) = 1 + 2/x, then f_2(x) = f(1 + 2/x) = 1 + 2/(1 + 2/x). Let me simplify that. The denominator is 1 + 2/x, which is (x + 2)/x. So 2 divided by that is 2 * x/(x + 2). So f_2(x) = 1 + (2x)/(x + 2). Combine the terms: [ (x + 2) + 2x ] / (x + 2) = (3x + 2)/(x + 2). Therefore, f_2(x) = (3x + 2)/(x + 2). So solving x = f_2(x) would be x = (3x + 2)/(x + 2). Multiply both sides by (x + 2): x(x + 2) = 3x + 2. Expand left side: x² + 2x = 3x + 2. Bring all terms to left: x² + 2x - 3x - 2 = 0 ⇒ x² - x - 2 = 0. Wait, that's the same equation as before! So solutions are again x = 2 and x = -1. Interesting. So even though f_2(x) is a different function, the fixed points are the same as f_1(x). Is that a coincidence?Wait, maybe this pattern continues? Let me check f_3(x). Let's compute f_3(x) = f(f_2(x)) = 1 + 2 / f_2(x). Since f_2(x) = (3x + 2)/(x + 2), so f_3(x) = 1 + 2 / [(3x + 2)/(x + 2)] = 1 + 2*(x + 2)/(3x + 2) = 1 + (2x + 4)/(3x + 2). Combine terms: [ (3x + 2) + 2x + 4 ] / (3x + 2 ) = (5x + 6)/(3x + 2). Therefore, f_3(x) = (5x + 6)/(3x + 2). Now set x = f_3(x): x = (5x + 6)/(3x + 2). Multiply both sides by (3x + 2): x(3x + 2) = 5x + 6. Expand left side: 3x² + 2x = 5x + 6. Bring all terms to left: 3x² + 2x - 5x - 6 = 0 ⇒ 3x² - 3x - 6 = 0. Divide by 3: x² - x - 2 = 0. Again! The same quadratic equation. So solutions x = 2 and x = -1. So no matter how many times we compose f, the fixed points remain x = 2 and x = -1? That's unexpected. Wait, maybe it's a general result. Let me try to see.Suppose that for all n ≥ 1, the equation x = f_n(x) reduces to x² - x - 2 = 0, so solutions are always 2 and -1. Then the answer would just be x = 2 and x = -1 for any n. But is that true?Wait, let's test with n=4. Compute f_4(x) = f(f_3(x)) = 1 + 2 / f_3(x). Since f_3(x) = (5x +6)/(3x +2), then 2 / f_3(x) = 2*(3x +2)/(5x +6). So f_4(x) = 1 + [6x +4]/[5x +6] = [5x +6 +6x +4]/[5x +6] = (11x +10)/(5x +6). Then set x = f_4(x): x = (11x +10)/(5x +6). Multiply both sides by (5x +6): 5x² +6x =11x +10. Bring all terms to left:5x² +6x -11x -10=0 ⇒5x² -5x -10=0. Divide by 5: x² -x -2=0. Same equation again! So same solutions x=2 and x=-1. Wow, so this seems to hold. So maybe regardless of n, composing f any number of times, the fixed points are always 2 and -1. That's fascinating.But why does this happen? Let me try to think of a reason. Maybe the function f has these fixed points, and applying f again doesn't introduce new fixed points. But in general, composing functions can introduce new fixed points. For example, if you have f(x) = -x, then f(f(x)) = x, so f_2(x) has all real numbers as fixed points, whereas f_1(x) only has x=0. So composition can add fixed points. But in our case, it seems that composition is not adding fixed points but keeping the same ones. So why?Perhaps because f is a Möbius transformation (a linear fractional function), and compositions of Möbius transformations have properties related to their fixed points. Let me recall. Möbius transformations are functions of the form (az + b)/(cz + d). The composition of two Möbius transformations is another Möbius transformation. The fixed points of a Möbius transformation can be found by solving z = (az + b)/(cz + d). For our function f(x) = 1 + 2/x, which can be written as (x + 2)/x, so a=1, b=2, c=1, d=0. The fixed points of f are indeed x=2 and x=-1. Now, when we compose f with itself multiple times, the resulting function f_n(x) is another Möbius transformation. Let me see if the fixed points remain the same.Suppose f_n(x) is a Möbius transformation (a_n x + b_n)/(c_n x + d_n). Then solving x = (a_n x + b_n)/(c_n x + d_n) leads to the quadratic equation c_n x² + (d_n - a_n)x - b_n = 0. If for each n, this equation is a multiple of x² - x -2 =0, then the solutions will always be 2 and -1. So perhaps c_n : (d_n - a_n) : (-b_n) = 1 : -1 : -2 for all n. That is, the coefficients are proportional to 1, -1, -2. If that's the case, then indeed, the quadratic equation will be the same, leading to the same roots.So maybe through induction, we can show that for each n, the composition f_n(x) has the form (p_n x + q_n)/(r_n x + s_n) where the coefficients satisfy p_n - s_n = r_n + q_n, or something like that, leading to the quadratic equation x² -x -2=0. Let's attempt an inductive proof.Base case: n=1. f_1(x) = (x + 2)/x. The equation x = (x + 2)/x gives x² =x +2 ⇒ x² -x -2=0. So holds.Assume that for some k ≥1, f_k(x) = (a x + b)/(c x + d) and solving x = f_k(x) leads to x² -x -2=0. Then f_{k+1}(x) = f(f_k(x)) = 1 + 2/f_k(x) = 1 + 2(c x + d)/(a x + b) = [ (a x + b) + 2(c x + d) ] / (a x + b ) = ( (a + 2c)x + (b + 2d) ) / (a x + b ).So f_{k+1}(x) = (A x + B)/(C x + D) where A = a + 2c, B = b + 2d, C = a, D = b.Now, solving x = f_{k+1}(x) gives x(C x + D) = A x + B ⇒ C x² + D x = A x + B ⇒ C x² + (D - A)x - B =0.By the induction hypothesis, for f_k(x), solving x = f_k(x) gives x² -x -2=0. So for f_k(x) = (a x + b)/(c x + d), we have c x² + (d -a)x -b =0 must be equivalent to x² -x -2=0. Therefore, c=1, d -a = -1, -b = -2. Therefore, c=1, b=2, d -a = -1.So from the induction hypothesis, we have for f_k(x): a, b, c, d with c=1, b=2, d = a -1.Now, let's compute coefficients for f_{k+1}(x):A = a + 2c = a + 2*1 = a + 2B = b + 2d = 2 + 2*(a -1) = 2 + 2a -2 = 2aC = aD = b = 2Therefore, f_{k+1}(x) = ( (a + 2)x + 2a ) / (a x + 2 )Now, let's form the quadratic equation for x = f_{k+1}(x):a x² + 2 x = (a + 2)x + 2aBring all terms to left:a x² + 2x - (a + 2)x -2a =0 ⇒ a x² - (a + 0)x -2a =0 ⇒ a x² -a x -2a =0Factor out a:a(x² -x -2)=0Since a is part of the Möbius transformation, and Möbius transformations have non-zero determinant (ad - bc ≠0). For f_k(x) = (a x + b)/(c x + d), determinant is a d - b c. From the induction hypothesis, c=1, d = a -1, b=2. So determinant is a*(a -1) - 2*1 = a² -a -2. For the transformation to be invertible, determinant ≠0. But since x² -x -2=0 has roots 2 and -1, if a is 2 or -1, determinant would be zero. Wait, but in our case, starting from f_1(x), which is (x + 2)/x, so a=1, b=2, c=1, d=0. Determinant is 1*0 -2*1 = -2 ≠0. Then f_2(x) is (3x + 2)/(x + 2). Here, a=3, b=2, c=1, d=2. Determinant is 3*2 -2*1=6-2=4≠0. Similarly, f_3(x) is (5x +6)/(3x +2), determinant 5*2 -6*3=10-18=-8≠0. So as long as a is not 2 or -1, determinant is non-zero. But in the inductive step, we have:When forming f_{k+1}(x) from f_k(x), we have coefficients A = a + 2, B = 2a, C = a, D=2. Then determinant for f_{k+1}(x) is A D - B C = (a + 2)*2 - 2a *a = 2a +4 -2a². For this to be non-zero, 2a +4 -2a² ≠0 ⇒ -2a² +2a +4 ≠0 ⇒ 2a² -2a -4 ≠0 ⇒ a² -a -2 ≠0 ⇒ (a -2)(a +1) ≠0. So as long as a ≠2 and a ≠ -1, determinant is non-zero. But in our previous steps, starting from a=1 (for f_1), then a=3 (f_2), then a=5 (f_3), then a=11 (Wait, no, for f_4(x) = (11x +10)/(5x +6), so a=11, but how does that follow? Wait, actually, the coefficients are following a pattern. Let's see:For f_1: a=1f_2: A = a + 2 =1 +2=3f_3: A =3 +2=5f_4: A=5 +2=7? Wait, but previously, f_3(x) is (5x +6)/(3x +2). Wait, the numerator coefficient for f_3 is 5, and then f_4's numerator coefficient would be 5 + 2*1=7? Wait, but when we computed f_4(x) earlier, the numerator was 11x +10, so A=11. Wait, perhaps my inductive step is missing something.Wait, let's recast. When we computed f_{k+1}(x) = ( (a + 2c)x + (b + 2d) ) / (a x + b ). But in the induction hypothesis, we had c=1, d = a -1, b=2. So substituting:A = a + 2*1 = a +2B = 2 + 2*(a -1) = 2 + 2a -2 =2aC = aD =2Therefore, f_{k+1}(x) = ( (a +2)x + 2a ) / (a x +2 )So the next a_{k+1} = a +2, but wait, in reality, when we computed f_2, starting from f_1 where a=1, then a_{2}=1 +2*1=3, which matches. Then f_3 would have a_{3}=3 +2*1=5, but when we computed f_3(x) earlier, we had (5x +6)/(3x +2). Wait, here a_{3}=5, which is a_{2} +2*1=3 +2=5. Then f_4 would be a_{4}=5 +2*1=7, but when we computed f_4(x), we had (11x +10)/(5x +6). Wait, here a_{4}=11? That's conflicting. Wait, perhaps my earlier calculation was wrong.Wait, let me recheck f_3(x):f_2(x) = (3x +2)/(x + 2). Then f_3(x)=f(f_2(x))=1 + 2/f_2(x)=1 + 2/[(3x +2)/(x +2)] =1 + 2*(x +2)/(3x +2)= [ (3x +2) + 2x +4 ] / (3x +2) = (5x +6)/(3x +2). So here, numerator is 5x +6, denominator is 3x +2. So a_{3}=5, c_{3}=3. But according to the induction step, in f_{k+1}(x) = ( (a +2)x + 2a ) / (a x +2 ). So if a_{k}=3, then a_{k+1}=3 +2=5, which is correct, and denominator is 3x +2. But in the denominator, for f_3(x), it's 3x +2, which is a_{k}x + 2, yes. So f_3(x) = (5x +6)/(3x +2). Then f_4(x) = f(f_3(x)) =1 +2/f_3(x)=1 +2*(3x +2)/(5x +6)= [5x +6 +6x +4]/(5x +6)= (11x +10)/(5x +6). So here, a_{4}=11, which is a_{3} +2* c_{3}=5 +2*3=11. Wait, but in the induction step above, we had f_{k+1}(x) as ( (a +2c)x + ... ). Wait, maybe the previous induction step was incorrect. Wait, in the induction step, when I computed A = a +2c, but c=1 for all k. Wait, no, in the induction hypothesis, c=1 for all k because in the induction step we set c=1. Wait, but in reality, when we computed f_3(x), c=3, but according to the previous induction step, c_{k+1}=a_{k}. Let me check.Wait, in the induction step, f_{k+1}(x) = (A x + B)/(C x + D ), where C = a_{k}. So in f_1(x): c=1. For f_2(x), C = a_{1}=1. But in reality, f_2(x) is (3x +2)/(x +2), so denominator is x +2, which is c_{2}=1. Wait, but when I computed f_3(x), the denominator is 3x +2, which would mean c_{3}=3. But according to the induction step, C_{k+1}=a_{k}. For f_3(x), C = a_{2}=3. So yes, denominator is 3x +2, where 3 is a_{2}, and 2 is b_{1} or something. Wait, maybe the induction step is correct. Let me track the coefficients properly.Let me formalize this. Let f_k(x) = (a_k x + b_k)/(c_k x + d_k ). Then f_{k+1}(x) = f(f_k(x)) =1 + 2/f_k(x) =1 + 2(c_k x + d_k )/(a_k x + b_k ) = [ (a_k x + b_k ) + 2(c_k x + d_k ) ] / (a_k x + b_k ) = [ (a_k + 2c_k )x + (b_k + 2d_k ) ] / (a_k x + b_k ).Therefore, the coefficients for f_{k+1}(x) are:a_{k+1} = a_k + 2c_kb_{k+1} = b_k + 2d_kc_{k+1} = a_kd_{k+1} = b_kSo this is the recursion relation. Starting from f_1(x) = (1x +2)/(1x +0), so a_1=1, b_1=2, c_1=1, d_1=0.Let's compute the coefficients step by step:For k=1:a_2 = a_1 + 2c_1 =1 +2*1=3b_2 = b_1 +2d_1=2 +2*0=2c_2 =a_1=1d_2 =b_1=2So f_2(x) = (3x +2)/(1x +2)Similarly, k=2:a_3 = a_2 + 2c_2=3 +2*1=5b_3 =b_2 +2d_2=2 +2*2=6c_3 =a_2=3d_3 =b_2=2Thus, f_3(x)=(5x +6)/(3x +2)k=3:a_4 =a_3 +2c_3=5 +2*3=11b_4 =b_3 +2d_3=6 +2*2=10c_4 =a_3=5d_4 =b_3=6Thus, f_4(x)=(11x +10)/(5x +6)k=4:a_5 =a_4 +2c_4=11 +2*5=21b_5 =b_4 +2d_4=10 +2*6=22c_5 =a_4=11d_5 =b_4=10So f_5(x)=(21x +22)/(11x +10)Hmm, so each time, the coefficients are following these recursions. Now, the quadratic equation for f_k(x) is c_k x² + (d_k -a_k )x -b_k =0. Let's compute this for each k:For k=1: c_1=1, d_1 -a_1=0 -1=-1, -b_1=-2. So equation: x² -x -2=0k=2: c_2=1, d_2 -a_2=2 -3=-1, -b_2=-2. So same equation: x² -x -2=0k=3: c_3=3, d_3 -a_3=2 -5=-3, -b_3=-6. So equation:3x² -3x -6=0. Divide by 3: x² -x -2=0k=4: c_4=5, d_4 -a_4=6 -11=-5, -b_4=-10. Equation:5x² -5x -10=0. Divide by 5: x² -x -2=0k=5: c_5=11, d_5 -a_5=10 -21=-11, -b_5=-22. Equation:11x² -11x -22=0. Divide by 11: x² -x -2=0So indeed, regardless of k, the quadratic equation reduces to x² -x -2=0, so solutions x=2 and x=-1.Therefore, even though the coefficients of f_n(x) get more complicated as n increases, the quadratic equation derived from x = f_n(x) is always a multiple of x² -x -2=0, hence the solutions are always x=2 and x=-1.This seems to hold for all n>0. Therefore, the solutions are always 2 and -1, regardless of n.But just to confirm, maybe check with n=5. Let's write f_5(x)=(21x +22)/(11x +10). Set x = (21x +22)/(11x +10). Multiply both sides by (11x +10): 11x² +10x =21x +22. Bring all terms to left:11x² +10x -21x -22=0⇒11x² -11x -22=0. Divide by 11: x² -x -2=0. Yep, same equation. So same solutions. Therefore, the conclusion is solid.So, the answer is that for any n >0, the solutions to x = f_n(x) are x=2 and x=-1.Final AnswerThe solutions are boxed{2} and boxed{-1}.</think>To solve the equation ( x = f_n(x) ) where ( f_n(x) ) is the nth composition of the function ( f(x) = 1 + frac{2}{x} ), we start by examining the fixed points of the function and its compositions.1. Base Case (n = 1): - The function ( f_1(x) = 1 + frac{2}{x} ). - Solving ( x = 1 + frac{2}{x} ) leads to the quadratic equation ( x^2 - x - 2 = 0 ). - Solutions: ( x = 2 ) and ( x = -1 ).2. Case n = 2: - The function ( f_2(x) = f(f_1(x)) = 1 + frac{2}{1 + frac{2}{x}} ). - Simplifying, we get ( f_2(x) = frac{3x + 2}{x + 2} ). - Solving ( x = frac{3x + 2}{x + 2} ) again leads to ( x^2 - x - 2 = 0 ). - Solutions: ( x = 2 ) and ( x = -1 ).3. Case n = 3: - The function ( f_3(x) = f(f_2(x)) = 1 + frac{2}{frac{3x + 2}{x + 2}} ). - Simplifying, we get ( f_3(x) = frac{5x + 6}{3x + 2} ). - Solving ( x = frac{5x + 6}{3x + 2} ) again leads to ( x^2 - x - 2 = 0 ). - Solutions: ( x = 2 ) and ( x = -1 ).4. General Case: - By induction, assume ( f_k(x) ) leads to the quadratic equation ( x^2 - x - 2 = 0 ). - For ( f_{k+1}(x) ), the function is a Möbius transformation. Solving ( x = f_{k+1}(x) ) results in a quadratic equation that is a multiple of ( x^2 - x - 2 = 0 ). - Coefficients of the Möbius transformation for ( f_{k+1}(x) ) lead to the same quadratic equation.Through induction and verification for higher compositions, we observe that the quadratic equation ( x^2 - x - 2 = 0 ) consistently arises, yielding solutions ( x = 2 ) and ( x = -1 ) for any ( n > 0 ). Final AnswerThe solutions are boxed{2} and boxed{-1}.,<think>Okay, so I need to solve this problem where we have a function f(x) = 1 + 2/x, and then we iterate it n times, called f_n(x), and find all solutions to the equation x = f_n(x) for any positive integer n. Hmm, interesting. Let me try to break this down step by step.First, let's make sure I understand the notation correctly. The problem defines f_1(x) as f(x), f_2(x) as f(f_1(x)), which is f(f(x)), and so on. So each subsequent f_n(x) is just applying f to the previous function. Therefore, f_n(x) is the function f composed with itself n times. So, for example, f_2(x) = f(f(x)) = 1 + 2/(1 + 2/x). And so on. Then, the problem asks to find all solutions x to the equation x = f_n(x). That is, find all fixed points of the function f after n iterations.Fixed points of a function are points where applying the function doesn't change their value. So, in this case, fixed points of f_n would be the x's that satisfy x = f_n(x). Since f is a rational function (a ratio of polynomials), each iteration f_n(x) will also be a rational function, albeit of higher degree each time. Therefore, the equation x = f_n(x) should be equivalent to a polynomial equation of degree n+1, right? Wait, let's check for n=1 first.For n=1, the equation is x = f(x) = 1 + 2/x. Multiplying both sides by x gives x^2 = x + 2, which rearranges to x^2 - x - 2 = 0. That's a quadratic equation, degree 2. So solutions are x = [1 ± sqrt(1 + 8)] / 2 = [1 ± 3]/2, so x=2 or x=-1. Okay, that's straightforward.For n=2, f_2(x) = f(f(x)) = 1 + 2/(1 + 2/x). Let me compute that step by step. Let's first compute f(x) = 1 + 2/x, then f(f(x)) = 1 + 2/(1 + 2/x). Let me simplify that:Denominator inside is 1 + 2/x, so 2 divided by that is 2 / (1 + 2/x). Let me write this as 2 / [(x + 2)/x] = 2x/(x + 2). Therefore, f_2(x) = 1 + 2x/(x + 2). Combining the terms, that's (x + 2 + 2x)/(x + 2) = (3x + 2)/(x + 2). So f_2(x) = (3x + 2)/(x + 2). Then the equation x = f_2(x) becomes x = (3x + 2)/(x + 2). Multiply both sides by (x + 2): x(x + 2) = 3x + 2. Expanding left side: x^2 + 2x = 3x + 2. Bring everything to left: x^2 + 2x - 3x - 2 = x^2 - x - 2 = 0. Wait, that's the same quadratic equation as for n=1! So the solutions are again x=2 and x=-1. That's unexpected. So for n=2, the fixed points are the same as for n=1?Wait, but that can't be, unless the function f is designed in such a way that its iterations stabilize quickly. Let me verify my calculation for f_2(x).Starting with f(x) = 1 + 2/x. Then f(f(x)) = 1 + 2/(1 + 2/x). Let's compute denominator: 1 + 2/x = (x + 2)/x. Then 2 divided by that is 2x/(x + 2). So f(f(x)) = 1 + 2x/(x + 2). Combine terms: (x + 2)/ (x + 2) + 2x/(x + 2) = [x + 2 + 2x]/(x + 2) = (3x + 2)/(x + 2). So that's correct. Then x = (3x + 2)/(x + 2). Multiply both sides: x(x + 2) = 3x + 2. x^2 + 2x = 3x + 2. x^2 - x - 2 = 0. So yes, same equation. So same solutions. So x=2 and x=-1.So for n=2, the fixed points are same as n=1. That's interesting. So maybe for any n, the fixed points are always x=2 and x=-1? But wait, the problem says "for n > 0". So perhaps regardless of how many times we iterate f, the fixed points remain the same? But that seems counterintuitive. Usually, composing functions can introduce more fixed points. For example, if we have f(x) = -x, then f_1(x) has fixed point 0, but f_2(x) = f(f(x)) = -(-x) = x, so every x is a fixed point. So in that case, fixed points increase. But here, it seems that even after composing, the fixed points remain the same.Wait, let me check for n=3. Let's compute f_3(x) = f(f_2(x)) = f((3x + 2)/(x + 2)) = 1 + 2 / [(3x + 2)/(x + 2)] = 1 + 2*(x + 2)/(3x + 2) = 1 + [2(x + 2)]/(3x + 2). Combine terms: [3x + 2 + 2x + 4]/(3x + 2) = (5x + 6)/(3x + 2). So f_3(x) = (5x + 6)/(3x + 2). Then setting x = (5x + 6)/(3x + 2). Multiply both sides by (3x + 2): x*(3x + 2) = 5x + 6. Left side: 3x^2 + 2x. Bring all terms to left: 3x^2 + 2x -5x -6 = 3x^2 -3x -6 = 0. Divide by 3: x^2 -x -2 =0. Same quadratic! So solutions x=2 and x=-1 again. So even for n=3, same fixed points.Wait, so this seems to hold. Maybe for any n, the equation x = f_n(x) reduces to x^2 -x -2=0? That is, no matter how many times we iterate f starting from x, the fixed points are always 2 and -1. That's fascinating. So perhaps regardless of n, the solutions are always x=2 and x=-1. Then why does the problem specify "for n > 0"? Maybe to ensure n is at least 1? Because n=0 would be trivial?But let's check n=4 just to confirm. Compute f_4(x) = f(f_3(x)) = f((5x + 6)/(3x + 2)) = 1 + 2 / [(5x + 6)/(3x + 2)] = 1 + 2*(3x + 2)/(5x + 6) = [5x + 6 + 6x + 4]/(5x + 6) = (11x + 10)/(5x + 6). Then set x = (11x +10)/(5x +6). Multiply both sides: x*(5x +6)=11x +10. Left side:5x^2 +6x. Bring all terms to left:5x^2 +6x -11x -10=5x^2 -5x -10=0. Divide by 5: x^2 -x -2=0. Again same equation. So x=2 and x=-1. So same solutions.So this pattern seems to hold. Each time, when we compute f_n(x), the equation x = f_n(x) simplifies to x^2 -x -2=0. Therefore, the solutions are always x=2 and x=-1, regardless of n. Hence, for any n >0, the solutions are 2 and -1. Therefore, the answer should be x=2 and x=-1.But wait, let me think again. Is there a possibility that higher iterations could introduce more fixed points? For example, maybe when solving x = f_n(x), the equation could factor into (x^2 -x -2) times another polynomial, leading to more roots? But in the examples above, n=1,2,3,4, each time the equation reduced to x^2 -x -2=0, which is degree 2, but the original equation is of degree n+1? Wait, for n=1, degree 2; n=2, equation is quadratic? Wait, but composing rational functions... Let me check the degree of f_n(x).Wait, the function f(x) =1 + 2/x is a Möbius transformation (a linear fractional transformation), except with a coefficient. Wait, Möbius transformations are functions of the form (ax + b)/(cx + d). Composing Möbius transformations corresponds to multiplying their matrices. So, composing f with itself multiple times would result in another Möbius transformation. Hence, each f_n(x) is also a Möbius transformation, i.e., a rational function of degree 1 (meaning numerator and denominator are both linear). Therefore, the equation x = f_n(x) will always result in a quadratic equation, as we saw. Therefore, regardless of n, the equation x = f_n(x) is quadratic, leading to two solutions. Hence, x=2 and x=-1 are the only solutions for any n.But how is that possible? Let me think. Since each iteration is a Möbius transformation, the composition is also a Möbius transformation, which can have at most two fixed points (unless it's the identity function, which has all points fixed). Therefore, unless the composed function is the identity function, which would mean every x is a solution, but in our case, when we computed up to n=4, we still got a quadratic equation, not an identity. So, unless after some n, the function f_n(x) becomes the identity function, which would require that f composed with itself n times is x. But in our calculations, even for n=4, f_4(x) = (11x +10)/(5x +6), which is not the identity function. Therefore, perhaps for all n, f_n(x) is a Möbius transformation different from the identity, hence having exactly two fixed points, which are the same as the original function f(x). That would mean that x=2 and x=-1 are fixed points for all iterations, and there are no other fixed points. Therefore, the answer is x=2 and x=-1 for any n>0.But let me verify with another approach. Let's consider the fixed points of f(x). The solutions to x =1 +2/x are x=2 and x=-1. Suppose we have a fixed point x of f(x), then x is also a fixed point of f_n(x), because if f(x)=x, then f(f(x))=f(x)=x, and so on. Therefore, fixed points of f are fixed points of all f_n. But could there be other fixed points for f_n that are not fixed points of f? For example, periodic points of period dividing n. But in our case, since we are solving x = f_n(x), these are periodic points of period dividing n. But in our calculations above, even for n=2,3,4, there are no new solutions. So, perhaps in this case, all periodic points are actually fixed points. That is, the function f is such that its iterations don't have any periodic points other than the fixed points. That's possible if the function is of finite order, meaning that after some iterations, it becomes the identity function. But in our case, f(x) =1 +2/x. Let's check if f(f(x)) is equal to x. Earlier, we saw f_2(x) = (3x +2)/(x +2). If we set that equal to x, we get x^2 -x -2=0, so x=2 and x=-1. So f_2(x) is not the identity. Similarly, f_3(x) is (5x +6)/(3x +2), which if set to x gives x^2 -x -2=0 again. So even after three iterations, not the identity. So perhaps this function has infinite order, meaning that no number of iterations will make it the identity function. Therefore, the only fixed points for any f_n are the original fixed points of f. Therefore, the solutions are always x=2 and x=-1.Alternatively, maybe there's a deeper reason here. Let's think about the functional equation. Suppose that x is a solution to x = f_n(x). Then, applying f to both sides, we get f(x) = f(f_n(x)) = f_{n+1}(x). But if x is a fixed point of f_n, then f(x) = f_{n+1}(x) = f_n(f(x)). So, unless f(x) is also a fixed point of f_n, this doesn't necessarily lead anywhere. But if x is a fixed point of f, then it's a fixed point of all f_n. So, the fixed points of f are fixed points of all f_n. But can there be fixed points of f_n that are not fixed points of f? In our calculations above, up to n=4, there are none, but is this always the case?Alternatively, let's suppose that x is a fixed point of f_n but not of f. Then x ≠ f(x), but x = f_n(x). That would mean that x is a periodic point of period dividing n. However, in our case, with f being a Möbius transformation, the dynamics can be more predictable. Möbius transformations have simple dynamics on the extended complex plane, classified by their trace, fixed points, etc. However, perhaps in this real line case, the behavior is such that the only fixed points for any iterate are the original fixed points.Alternatively, let's think in terms of functions. Let me suppose that for any n, the equation x = f_n(x) is equivalent to x^2 -x -2=0. Then, the solutions are always 2 and -1. Is there a mathematical theorem that could justify this?Alternatively, perhaps induction can be used here. Suppose that for some k ≥1, the equation x = f_k(x) reduces to x^2 -x -2=0. Then, for k+1, f_{k+1}(x) = f(f_k(x)). If f_k(x) = (ax + b)/(cx + d), then f(f_k(x)) = 1 + 2/[(ax + b)/(cx + d)] = 1 + 2(cx + d)/(ax + b) = (ax + b + 2cx + 2d)/(ax + b) = [(a + 2c)x + (b + 2d)]/(ax + b). Then, setting x = f_{k+1}(x) gives x = [(a + 2c)x + (b + 2d)]/(ax + b). Multiply both sides by (ax + b): ax^2 + bx = (a + 2c)x + (b + 2d). Rearranged: ax^2 + bx - (a + 2c)x - (b + 2d) = 0 ⇒ ax^2 + (b - a - 2c)x - (b + 2d) =0. Now, if we assume that the previous equation x = f_k(x) led to x^2 -x -2=0, then for f_k(x), we had the equation x^2 -x -2=0. Let's see if this can help. If f_k(x) satisfies x = f_k(x) ⇒ x^2 -x -2=0, then perhaps the coefficients of f_k(x) satisfy certain relations. Alternatively, perhaps the recursion for the coefficients a, b, c, d when composing f can be set up, leading to a pattern where the quadratic equation remains the same.Alternatively, let's note that if x is a solution to x = f(x), then it is a solution to x = f_n(x) for all n. Conversely, suppose x is a solution to x = f_n(x). Then x = f_n(x) = f(f_{n-1}(x)). If x is a fixed point of f_n, then applying f to both sides, f(x) = f_{n+1}(x). So f(x) is a fixed point of f_n as well. If the only fixed points of f_n are x=2 and x=-1, then f(x) must be one of these. Since f(2) =1 + 2/2=2, and f(-1)=1 + 2/(-1)=1 -2=-1. Therefore, both fixed points are mapped to themselves. Therefore, if we have a fixed point x of f_n, then f(x) is also a fixed point of f_n. So if x is a fixed point of f_n, then f(x) is also a fixed point. If we assume that the only fixed points are 2 and -1, which are fixed by f, then there are no other fixed points. Otherwise, if there were another fixed point x ≠2,-1, then f(x) would also have to be a fixed point of f_n, leading to a cycle. But since f is a Möbius transformation, which can have at most two fixed points, unless it's the identity, which has all fixed points. Therefore, unless f_n is the identity function, which would require that all x satisfy x = f_n(x), but in our case, f_n is not the identity function (as seen in examples), hence it can only have two fixed points, which must be the same as those of f. Therefore, the only solutions are x=2 and x=-1.Therefore, regardless of n, the solutions are x=2 and x=-1. Hence, the answer is that the solutions are x=2 and x=-1 for any n>0.Final AnswerThe solutions are boxed{2} and boxed{-1}.</think>To solve the problem of finding the solutions to ( x = f_n(x) ) for ( n > 0 ) where ( f(x) = 1 + frac{2}{x} ) and ( f_n(x) ) is the n-th iterate of ( f ), we start by examining the fixed points of the function ( f ) and its iterates.1. Fixed Points for ( n = 1 ): The equation ( x = f(x) ) becomes: [ x = 1 + frac{2}{x} ] Multiplying both sides by ( x ) gives: [ x^2 = x + 2 implies x^2 - x - 2 = 0 ] Solving this quadratic equation, we get the solutions: [ x = frac{1 pm sqrt{1 + 8}}{2} = frac{1 pm 3}{2} implies x = 2 text{ or } x = -1 ]2. Fixed Points for ( n = 2 ): The second iterate ( f_2(x) = f(f(x)) ) is calculated as: [ f_2(x) = 1 + frac{2}{1 + frac{2}{x}} = 1 + frac{2x}{x + 2} = frac{3x + 2}{x + 2} ] Setting ( x = f_2(x) ): [ x = frac{3x + 2}{x + 2} ] Multiplying both sides by ( x + 2 ) gives: [ x(x + 2) = 3x + 2 implies x^2 - x - 2 = 0 ] This is the same quadratic equation as before, leading to the solutions ( x = 2 ) and ( x = -1 ).3. Fixed Points for ( n = 3 ): The third iterate ( f_3(x) = f(f_2(x)) ) is: [ f_3(x) = 1 + frac{2}{frac{3x + 2}{x + 2}} = 1 + frac{2(x + 2)}{3x + 2} = frac{5x + 6}{3x + 2} ] Setting ( x = f_3(x) ): [ x = frac{5x + 6}{3x + 2} ] Multiplying both sides by ( 3x + 2 ) gives: [ x(3x + 2) = 5x + 6 implies 3x^2 - 3x - 6 = 0 implies x^2 - x - 2 = 0 ] Again, the solutions are ( x = 2 ) and ( x = -1 ).4. General Case for ( n ): By induction and the properties of Möbius transformations, each iterate ( f_n(x) ) remains a linear fractional transformation (Möbius transformation). Solving ( x = f_n(x) ) for any ( n ) results in a quadratic equation, which simplifies to ( x^2 - x - 2 = 0 ). This consistency across iterations confirms that the fixed points do not change.Thus, the solutions to ( x = f_n(x) ) for any ( n > 0 ) are always:[boxed{2} text{ and } boxed{-1}]]

🔑:x = -1 or 2. Solution f(x) = x iff 0 = x 2 - x - 2 = (x+1)(x-2). Now it is clear by a trivial induction that f n (x) = linear/linear. So f n (x) = x is a quadratic and so has at most two roots. But x = -1 and 2 are obviously roots, so they are the only roots. 6th Swedish 1966 © John Scholes [email protected] 30 December 2003 Last corrected/updated 30 Dec 03

🧠:[<think>Okay, so I need to find the smallest positive integer k such that there's exactly one prime number of the form kx + 60 where x is an integer between 0 and 10, inclusive. Hmm, let's break this down. First, the problem is asking for the smallest k where, when you plug in x values from 0 to 10 into the expression kx + 60, only one of those results is a prime number. The rest should be composite, I guess, or maybe 1? But primes are numbers greater than 1, so maybe 1 isn't considered prime here. Yeah, primes are numbers greater than 1 with exactly two distinct divisors. So, 1 isn't prime. So, for each k, I have to check the numbers 60 (when x=0), k + 60 (x=1), 2k + 60, ..., up to 10k + 60. And among these 11 numbers, exactly one has to be prime. My goal is to find the smallest such k. So, the plan is probably to check k starting from 1 upwards, and for each k, check how many primes are in that list. The first k where there's exactly one prime is the answer. But this might take a while, but maybe there's a smarter way. Let's see.First, let's note that when x=0, the number is 60. 60 is definitely composite (divisible by 2, 3, 5). So, the prime has to be in the other positions, x=1 to x=10. So, for each k, we can ignore x=0 since 60 is composite. Then, check kx + 60 for x from 1 to 10, and count how many primes there are. Find the smallest k where that count is exactly 1.So, starting with k=1:k=1: Check 1*1 +60=61 (prime), 1*2 +60=62 (composite), 1*3 +60=63 (composite), ..., up to 1*10 +60=70. So, numbers are 61,62,...,70. Let's count primes here. 61 is prime, 67 is prime, 71 is prime but 70 is 70, so wait, 10*1 +60=70. So 61,62,63,64,65,66,67,68,69,70. Primes here: 61, 67, and 71? Wait, 70 is not 71. Wait, x=10 would be 10*1 +60=70. So primes in that list are 61, 67. So two primes. So k=1 gives two primes. Not good, we need exactly one. So k=1 is out.k=2: Then numbers are 2*1 +60=62, 2*2 +60=64, ..., up to 2*10 +60=80. So numbers: 62,64,66,68,70,72,74,76,78,80. All even numbers, so none of these can be prime except 2. But all numbers here are even and greater than 2, so all composite. Wait, so for k=2, all numbers from x=1 to x=10 are even, so composite. So the count is zero primes. But the problem wants exactly one prime. So k=2 is also invalid.k=3: Let's check. 3*1 +60=63, 3*2 +60=66, 3*3 +60=69, ..., up to 3*10 +60=90. All numbers: 63,66,69,72,75,78,81,84,87,90. These are multiples of 3. 63=7*9, 66=6*11, etc. All of these numbers are divisible by 3. 63=3*21, 66=3*22, etc. So since all numbers are multiples of 3 and greater than 3, they are composite. So count of primes is zero. Not good. k=3 is out.k=4: Numbers: 4*1 +60=64, 4*2 +60=68, ..., up to 4*10 +60=100. Numbers: 64,68,72,76,80,84,88,92,96,100. All even numbers, so composite. Count of primes: zero. So k=4 is out.k=5: Numbers: 5*1 +60=65, 5*2 +60=70, 5*3 +60=75, ..., up to 5*10 +60=110. Numbers: 65,70,75,80,85,90,95,100,105,110. These are multiples of 5. All numbers are divisible by 5 and greater than 5, hence composite. So count is zero. k=5 is out.k=6: Numbers: 6*1 +60=66, 6*2 +60=72, ..., up to 6*10 +60=120. All numbers even and divisible by 6, hence composite. Count zero. k=6 out.k=7: Let's check. 7*1 +60=67, 7*2 +60=74, 7*3 +60=81, 7*4 +60=88, 7*5 +60=95, 7*6 +60=102, 7*7 +60=109, 7*8 +60=116, 7*9 +60=123, 7*10 +60=130.Now, check which are primes. 67 is prime. 74 is even, composite. 81=9*9, composite. 88 even, composite. 95=5*19, composite. 102 even, composite. 109 is prime. 116 even, composite. 123=3*41, composite. 130 even, composite. So primes here are 67 and 109. Two primes. So k=7 gives two primes. Not good. Need exactly one. So k=7 is out.k=8: Numbers: 8*1 +60=68, 8*2 +60=76, ..., up to 8*10 +60=140. All numbers: 68,76,84,92,100,108,116,124,132,140. All even, so composite. Count zero. k=8 is out.k=9: Numbers: 9*1 +60=69, 9*2 +60=78, 9*3 +60=87, 9*4 +60=96, 9*5 +60=105, 9*6 +60=114, 9*7 +60=123, 9*8 +60=132, 9*9 +60=141, 9*10 +60=150. Let's check primes here. 69=3*23, composite. 78 even, composite. 87=3*29, composite. 96 even, composite. 105=5*21, composite. 114 even, composite. 123=3*41, composite. 132 even, composite. 141=3*47, composite. 150 even, composite. All composite. Count zero. k=9 is out.k=10: Numbers: 10*1 +60=70, 10*2 +60=80, ..., up to 10*10 +60=160. All numbers even, composite. Count zero. k=10 is out.k=11: Let's check. 11*1 +60=71, 11*2 +60=82, 11*3 +60=93, 11*4 +60=104, 11*5 +60=115, 11*6 +60=126, 11*7 +60=137, 11*8 +60=148, 11*9 +60=159, 11*10 +60=170.Check primes: 71 is prime. 82 even. 93=3*31. 104 even. 115=5*23. 126 even. 137 is prime. 148 even. 159=3*53. 170 even. So primes: 71 and 137. Two primes. So k=11 is out.k=12: Numbers: 12*1 +60=72, 12*2 +60=84, ..., up to 12*10 +60=180. All even numbers, composite. Count zero.k=13: Let's compute. 13*1 +60=73, 13*2 +60=86, 13*3 +60=99, 13*4 +60=112, 13*5 +60=125, 13*6 +60=138, 13*7 +60=151, 13*8 +60=164, 13*9 +60=177, 13*10 +60=190.Check primes: 73 is prime. 86 even. 99=9*11. 112 even. 125=5^3. 138 even. 151 is prime. 164 even. 177=3*59. 190 even. So primes: 73 and 151. Two primes. k=13 out.k=14: Numbers: 14*1 +60=74, 14*2 +60=88, ..., up to 14*10 +60=200. All even, composite. Count zero.k=15: Numbers: 15*1 +60=75, 15*2 +60=90, ..., up to 15*10 +60=210. All multiples of 15, hence composite. Count zero.k=16: 16*1 +60=76, 16*2 +60=92, ..., up to 16*10 +60=220. All even, composite. Count zero.k=17: Let's see. 17*1 +60=77, 17*2 +60=94, 17*3 +60=111, 17*4 +60=128, 17*5 +60=145, 17*6 +60=162, 17*7 +60=179, 17*8 +60=196, 17*9 +60=213, 17*10 +60=230.Check primes: 77=7*11, composite. 94 even. 111=3*37. 128 even. 145=5*29. 162 even. 179 is prime. 196 even. 213=3*71. 230 even. So primes: 179. Only one prime here. So k=17 gives exactly one prime (179 at x=7). Wait, that seems to satisfy the condition. But let's verify if this is the smallest k. Wait, we went through k=1 to k=16 and they were all out. So k=17 is the first one that works. But let me double-check. Maybe I missed something. Let's check k=17 again.For k=17:x=0: 17*0 +60=60 (composite)x=1: 17+60=77 (7*11) compositex=2: 34+60=94 (even) compositex=3: 51+60=111 (3*37) compositex=4: 68+60=128 (even) compositex=5: 85+60=145 (5*29) compositex=6: 102+60=162 (even) compositex=7: 119+60=179 (prime)x=8: 136+60=196 (even) compositex=9: 153+60=213 (3*71) compositex=10:170+60=230 (even) compositeYes, only x=7 gives 179, which is prime. So k=17 has exactly one prime in that range. So is k=17 the answer? Wait, but let me check if between k=11 and k=17, maybe there's a smaller k that I might have miscalculated.Wait, let's check k=12 to k=16 again to confirm.k=12: All even numbers, composite. So no.k=13: Gave two primes. k=14: All even. k=15: Multiples of 15. k=16: All even. So yes, k=17 is indeed the next candidate. So maybe 17 is the answer.But wait, perhaps I made a mistake in k=7. Wait, when k=7, x=1 gives 67, prime. x=7 gives 7*7 +60=109, prime. So two primes. Correct. So k=7 is out.Similarly, k=5: all multiples of 5. k=7: two primes. k=11: two primes. k=13: two primes. So, 17 is the first k where exactly one prime exists in the list.But let me check k=19 just to see if perhaps there's a smaller k. Wait, no, k=17 is smaller than 19. Wait, but maybe there's a k between 1 and 17 that I missed? Let me check again.Wait, k=1: primes at x=1 (61) and x=6 (67). Wait, 61 and 67. Wait, x=10 gives 70, which is composite. So two primes. So k=1 has two primes.k=2: zero primes.k=3: zero.k=4: zero.k=5: zero.k=6: zero.k=7: two primes.k=8: zero.k=9: zero.k=10: zero.k=11: two primes.k=12: zero.k=13: two primes.k=14: zero.k=15: zero.k=16: zero.k=17: one prime.So, indeed, k=17 is the first k where there's exactly one prime. Therefore, the answer should be 17. But let me check k=19 for thoroughness.k=19: Numbers 19*1 +60=79 (prime), 19*2 +60=98 (composite), 19*3 +60=117 (composite), 19*4 +60=136 (composite), 19*5 +60=155 (composite), 19*6 +60=174 (composite), 19*7 +60=193 (prime), 19*8 +60=212 (composite), 19*9 +60=231 (composite), 19*10 +60=250 (composite). So primes at x=1 (79) and x=7 (193). Two primes. So k=19 is out.Similarly, k=18: even numbers. All composite. Zero primes.k=20: even numbers, composite. Zero.k=21: Let's check. 21*1 +60=81 (composite), 21*2 +60=102 (composite), ..., up to 21*10 +60=270. All numbers divisible by 3 and 21. So composite. Zero primes.So, yeah, k=17 is the first one. Wait, but let's check k=7 again. Wait, the problem states "for the integers 0 ≤ x ≤ 10". So x=0 to x=10 inclusive. For k=17, x=7 gives 179. So the problem counts x from 0 to 10, which gives 11 numbers. But for x=0, as mentioned, 60 is composite. So in all these k's, x=0 is not contributing a prime. So the primes come from x=1 to x=10. For k=17, x=7 is the only one. But is there a smaller k that I might have missed? Let me check k=7 again. For k=7, x=1: 67 (prime), x=7:109 (prime). So two primes. So no. k=17 is the first with one prime.Wait, perhaps k=19 is not needed. So the answer is 17. Therefore, boxed{17}.Final Answerboxed{17}</think>To find the smallest positive integer ( k ) such that there is exactly one prime number of the form ( kx + 60 ) for integers ( 0 le x le 10 ), we need to check each ( k ) starting from 1 and determine how many primes are formed by ( kx + 60 ) for ( x ) in the given range.1. For ( k = 1 ) to ( k = 16 ): - All values of ( k ) from 1 to 16 result in either zero, two, or more primes when checked for ( x ) from 0 to 10. None of these values satisfy the condition of exactly one prime.2. For ( k = 17 ): - Checking each ( x ) from 0 to 10: - ( x = 0 ): ( 17 cdot 0 + 60 = 60 ) (composite) - ( x = 1 ): ( 17 cdot 1 + 60 = 77 ) (composite) - ( x = 2 ): ( 17 cdot 2 + 60 = 94 ) (composite) - ( x = 3 ): ( 17 cdot 3 + 60 = 111 ) (composite) - ( x = 4 ): ( 17 cdot 4 + 60 = 128 ) (composite) - ( x = 5 ): ( 17 cdot 5 + 60 = 145 ) (composite) - ( x = 6 ): ( 17 cdot 6 + 60 = 162 ) (composite) - ( x = 7 ): ( 17 cdot 7 + 60 = 179 ) (prime) - ( x = 8 ): ( 17 cdot 8 + 60 = 196 ) (composite) - ( x = 9 ): ( 17 cdot 9 + 60 = 213 ) (composite) - ( x = 10 ): ( 17 cdot 10 + 60 = 230 ) (composite) - Only ( x = 7 ) results in a prime number (179).Thus, the smallest positive integer ( k ) such that there is exactly one prime number of the form ( kx + 60 ) for ( 0 le x le 10 ) is (boxed{17}).,<think>Okay, so I need to find the smallest positive integer k such that there's exactly one prime number in the sequence kx + 60 where x ranges from 0 to 10. That means for each x = 0,1,2,...,10, I plug it into the equation and check if the result is prime. But only one of those 11 numbers should be prime. And I need the smallest k possible. Alright, let's start thinking.First, when x=0, the number is k*0 + 60 = 60. So 60 is part of the sequence. But 60 is not a prime number. So the prime we're looking for has to be among x=1 to x=10. But wait, hold on, x can be 0, so the number 60 is part of the set. But 60 is composite, so we don't count that. So the prime must be in the numbers when x is from 1 to 10. So we have to have exactly one prime in those 10 numbers.So our task is to find the smallest k such that when you compute k*1 + 60, k*2 + 60, ..., k*10 + 60, exactly one of those is prime. And the rest are composite. Let's see.First, maybe start with k=1. Let's check how many primes are in 1 to 10 added to 60. So 61, 62, ..., 70. Let's list primes here: 61 is prime, 62 is even, 63 divisible by 3, 64 even, 65 divisible by 5, 66 even, 67 prime, 68 even, 69 divisible by 3, 70 even. So primes here are 61, 67. So two primes. So k=1 gives two primes, which is more than one. So not good.k=2: numbers are 2*1 +60=62, 2*2+60=64, 2*3+60=66, 2*4+60=68, 2*5+60=70, 2*6+60=72, 2*7+60=74, 2*8+60=76, 2*9+60=78, 2*10+60=80. All these numbers are even and greater than 2, so none are primes. So all composite. So how many primes? Zero. Not good, we need exactly one.k=3: Let's compute numbers:x=1: 3+60=63 (divisible by 3, composite)x=2:6+60=66 (divisible by 6, composite)x=3:9+60=69 (divisible by 3)x=4:12+60=72 (composite)x=5:15+60=75 (divisible by 5)x=6:18+60=78 (composite)x=7:21+60=81 (divisible by 3)x=8:24+60=84 (composite)x=9:27+60=87 (divisible by 3)x=10:30+60=90 (composite)So all these numbers are multiples of 3 or 5, so composite. So again, zero primes. Not good.k=4: let's check numbers:x=1:4+60=64 (even)x=2:8+60=68 (even)x=3:12+60=72 (even)x=4:16+60=76 (even)x=5:20+60=80 (even)x=6:24+60=84 (even)x=7:28+60=88 (even)x=8:32+60=92 (even)x=9:36+60=96 (even)x=10:40+60=100 (even)All even numbers, so composite. Zero primes. Not good.k=5: numbers:x=1:5+60=65 (divisible by 5)x=2:10+60=70 (divisible by 10)x=3:15+60=75 (divisible by 5)x=4:20+60=80 (divisible by 5)x=5:25+60=85 (divisible by 5)x=6:30+60=90 (divisible by 10)x=7:35+60=95 (divisible by 5)x=8:40+60=100 (divisible by 5)x=9:45+60=105 (divisible by 5)x=10:50+60=110 (divisible by 10)All multiples of 5 or 10, so composite. Zero primes. Not good.k=6: similar approach.x=1:6+60=66 (even)x=2:12+60=72 (even)... all the way up to x=10:60+60=120 (even). All even. Zero primes.k=7: Let's check.x=1:7+60=67 (prime)x=2:14+60=74 (even)x=3:21+60=81 (divisible by 3)x=4:28+60=88 (even)x=5:35+60=95 (divisible by 5)x=6:42+60=102 (even)x=7:49+60=109 (prime)x=8:56+60=116 (even)x=9:63+60=123 (divisible by 3)x=10:70+60=130 (even)So primes here are 67 and 109. So two primes. Not good.k=8: Let's check.x=1:8+60=68 (even)x=2:16+60=76 (even)x=3:24+60=84 (even)x=4:32+60=92 (even)x=5:40+60=100 (even)x=6:48+60=108 (even)x=7:56+60=116 (even)x=8:64+60=124 (even)x=9:72+60=132 (even)x=10:80+60=140 (even)All even. Zero primes. Not good.k=9:x=1:9+60=69 (divisible by 3)x=2:18+60=78 (divisible by 6)x=3:27+60=87 (divisible by 3)x=4:36+60=96 (even)x=5:45+60=105 (divisible by 5)x=6:54+60=114 (even)x=7:63+60=123 (divisible by 3)x=8:72+60=132 (even)x=9:81+60=141 (divisible by 3)x=10:90+60=150 (divisible by 10)All composite. Zero primes. Not good.k=10:x=1:10+60=70 (even)x=2:20+60=80 (even)...x=10:100+60=160 (even)All even. Zero primes. Not good.k=11:Let's check this one. Maybe here we have a prime.x=1:11+60=71 (prime)x=2:22+60=82 (even)x=3:33+60=93 (divisible by 3)x=4:44+60=104 (even)x=5:55+60=115 (divisible by 5)x=6:66+60=126 (even)x=7:77+60=137 (prime)x=8:88+60=148 (even)x=9:99+60=159 (divisible by 3)x=10:110+60=170 (even)So primes here are 71 and 137. Two primes. Not good.k=12:x=1:12+60=72 (even)x=2:24+60=84 (even)...All even. Zero primes.k=13:Check this.x=1:13+60=73 (prime)x=2:26+60=86 (even)x=3:39+60=99 (divisible by 3)x=4:52+60=112 (even)x=5:65+60=125 (divisible by 5)x=6:78+60=138 (even)x=7:91+60=151 (prime)x=8:104+60=164 (even)x=9:117+60=177 (divisible by 3)x=10:130+60=190 (even)So primes here are 73 and 151. Two primes. Not good.k=14:x=1:14+60=74 (even)x=2:28+60=88 (even)...All even. Zero primes.k=15:x=1:15+60=75 (divisible by 5)x=2:30+60=90 (divisible by 10)...All composite. Zero primes.k=16:x=1:16+60=76 (even)x=2:32+60=92 (even)...All even. Zero primes.k=17:Check this.x=1:17+60=77 (divisible by 7 and 11)x=2:34+60=94 (even)x=3:51+60=111 (divisible by 3)x=4:68+60=128 (even)x=5:85+60=145 (divisible by 5)x=6:102+60=162 (even)x=7:119+60=179 (prime)x=8:136+60=196 (even)x=9:153+60=213 (divisible by 3)x=10:170+60=230 (even)So here, x=7 gives 179, which is prime. Let's check if others are composite:x=1:77 is 7*11, composite.x=2:94 is even.x=3:111=3*37, composite.x=4:128 even.x=5:145=5*29, composite.x=6:162 even.x=7:179 prime.x=8:196 even.x=9:213=3*71, composite.x=10:230 even.So only one prime here, which is 179. So k=17 gives exactly one prime. But wait, is there a smaller k that I missed? Let's check k=7, 11, 13, etc., but in previous checks, k=7 had two primes, k=11 had two primes, k=13 had two primes, so 17 is the first one where only one prime is present.Wait, but let's check k=19:x=1:19+60=79 (prime)x=2:38+60=98 (even)x=3:57+60=117 (divisible by 3)x=4:76+60=136 (even)x=5:95+60=155 (divisible by 5)x=6:114+60=174 (even)x=7:133+60=193 (prime)x=8:152+60=212 (even)x=9:171+60=231 (divisible by 3)x=10:190+60=250 (even)So primes here are 79 and 193. Two primes. Not good.k=20:All even. So all composite. Zero primes.k=21:x=1:21+60=81 (divisible by 3)x=2:42+60=102 (even)...All composite. Zero primes.k=22:x=1:22+60=82 (even)x=2:44+60=104 (even)...All even. Zero primes.k=23:x=1:23+60=83 (prime)x=2:46+60=106 (even)x=3:69+60=129 (divisible by 3)x=4:92+60=152 (even)x=5:115+60=175 (divisible by 5)x=6:138+60=198 (even)x=7:161+60=221 (13*17=221? Let's check. 13*17 is 221. Yes. So composite)x=8:184+60=244 (even)x=9:207+60=267 (divisible by 3)x=10:230+60=290 (even)So primes here are 83. Only one prime. So k=23 gives exactly one prime. But 17 is smaller than 23. Wait, but we have k=17 with one prime. So k=17 is smaller.Wait, but maybe there's a k between 1 and 17 that gives one prime. Let's check k=19, which had two primes. What about k=15, k=16, which we already checked. How about k=12? All even. So no.Wait, maybe I skipped k=5 or something. Let's check.Wait, k=7: two primes, k=11: two primes, k=13: two primes, k=17: one prime, k=19: two primes, k=23: one prime. So 17 and 23 are the candidates. Since 17 is smaller, so 17 is the answer?Wait, but before jumping to conclusion, let me check k=17 again.For k=17, x=1 to 10:17*1 +60=77 (7*11, composite)17*2 +60=34+60=94 (even)17*3 +60=51+60=111 (3*37, composite)17*4 +60=68+60=128 (even)17*5 +60=85+60=145 (5*29, composite)17*6 +60=102+60=162 (even)17*7 +60=119+60=179 (prime)17*8 +60=136+60=196 (even)17*9 +60=153+60=213 (3*71, composite)17*10 +60=170+60=230 (even)So only x=7 gives a prime. So exactly one prime. So k=17 works. Now, check if there's a smaller k that also works. Let's check k=14,15,16. k=14: all even. k=15: composite. k=16: even. So no.k=12: all even.k=10: even.k=9: composites.k=8: even.k=7: two primes. So no.Wait, let's check k=18. Wait, k=18:x=1:18+60=78 (even)x=2:36+60=96 (even)...All even. Zero primes.k=19: two primes.k=20: even.k=21: composite.k=22: even.So next possible candidate is k=17. So is k=17 the smallest? Let's check k=5,6, etc. But earlier, for k=1 to 16, all had either zero or two primes. Wait, is that the case? Let me check k=10 again. For k=10, numbers from 70 to 160 in steps of 10. All even. So zero primes.Wait, wait, wait. Let me check k=7 again. For k=7, x=1:67 (prime), x=7:109 (prime). So two primes. So k=7 is out.Wait, how about k=17. Let's confirm once more. For k=17, x=7 gives 179, which is prime. All others composite. So that's exactly one. So 17 is valid. But let's check k=13 again. For k=13, primes at x=1 (73) and x=7 (151). Two primes. So k=13 is out.How about k=19. As above, two primes.k=23, one prime. So k=17 and k=23 work, but 17 is smaller.But wait, perhaps there's a smaller k that I missed. Let's check k=7, which gives two primes, k=11 gives two primes, k=13 two primes. k=17 one prime. So the next possible candidates are 17 and 23. So 17 is the answer.Wait, but let me check k=19 again. No, two primes. So okay.Wait, but perhaps k=5, k=7, etc. How about k=7, which gives 67 and 109. So two primes. So no.Wait, let me check k=7 again:k=7:x=1:7+60=67 (prime)x=2:14+60=74 (even)x=3:21+60=81 (composite)x=4:28+60=88 (even)x=5:35+60=95 (composite)x=6:42+60=102 (even)x=7:49+60=109 (prime)x=8:56+60=116 (even)x=9:63+60=123 (composite)x=10:70+60=130 (even)So two primes. So no.Similarly, k=11:x=1:71 (prime)x=7:137 (prime)Two primes.Similarly, k=13:x=1:73 (prime)x=7:151 (prime)Two primes.So that leaves k=17 as the first one with one prime.Wait, let me check k=25. For completeness, but 25 is larger than 17.x=1:25+60=85 (composite)x=2:50+60=110 (composite)x=3:75+60=135 (composite)x=4:100+60=160 (composite)x=5:125+60=185 (divisible by 5)x=6:150+60=210 (composite)x=7:175+60=235 (composite)x=8:200+60=260 (composite)x=9:225+60=285 (composite)x=10:250+60=310 (composite)Wait, but maybe x=7? 25*7=175+60=235. 235 is divisible by 5 (5*47). So composite. So zero primes. Not good.So k=17 is the first one where only one prime exists. Therefore, the answer is 17.Final AnswerThe smallest positive integer ( k ) is boxed{17}.</think>To find the smallest positive integer ( k ) such that there is exactly one prime number of the form ( kx + 60 ) for integers ( 0 le x le 10 ), we need to check each ( k ) starting from 1 and verify the number of primes in the sequence ( kx + 60 ) for ( x ) from 1 to 10 (since ( x = 0 ) gives 60, which is not prime).1. For ( k = 1 ): The numbers are 61, 62, ..., 70. Primes: 61, 67 (two primes).2. For ( k = 2 ): All numbers are even. No primes.3. For ( k = 3 ): All numbers are multiples of 3 or 5. No primes.4. For ( k = 4 ): All numbers are even. No primes.5. For ( k = 5 ): All numbers are multiples of 5. No primes.6. For ( k = 6 ): All numbers are even. No primes.7. For ( k = 7 ): The numbers are 67, 74, ..., 130. Primes: 67, 109 (two primes).8. For ( k = 8 ): All numbers are even. No primes.9. For ( k = 9 ): All numbers are multiples of 3. No primes.10. For ( k = 10 ): All numbers are even. No primes.11. For ( k = 11 ): The numbers are 71, 82, ..., 170. Primes: 71, 137 (two primes).12. For ( k = 12 ): All numbers are even. No primes.13. For ( k = 13 ): The numbers are 73, 86, ..., 190. Primes: 73, 151 (two primes).14. For ( k = 14 ): All numbers are even. No primes.15. For ( k = 15 ): All numbers are multiples of 5. No primes.16. For ( k = 16 ): All numbers are even. No primes.17. For ( k = 17 ): The numbers are 77, 94, ..., 230. Prime: 179 (one prime).After checking ( k ) values from 1 to 17, we find that ( k = 17 ) is the smallest value where exactly one prime number (179) exists in the sequence ( kx + 60 ) for ( 0 le x le 10 ).Thus, the smallest positive integer ( k ) is (boxed{17}).]

🔑:To find the smallest positive integer ( k ) such that there is exactly one prime number of the form ( kx + 60 ) for the integers ( 0 le x le 10 ), we need to analyze the expression ( kx + 60 ) for each ( x ) in the given range.1. Identify the form of the expression: [ kx + 60 ] We need to check for each ( x ) from 0 to 10 whether ( kx + 60 ) is a prime number.2. Check the values of ( x ): [ begin{aligned} &text{For } x = 0, quad k cdot 0 + 60 = 60 quad (text{not prime}) &text{For } x = 1, quad k cdot 1 + 60 = k + 60 &text{For } x = 2, quad k cdot 2 + 60 = 2k + 60 &text{For } x = 3, quad k cdot 3 + 60 = 3k + 60 &text{For } x = 4, quad k cdot 4 + 60 = 4k + 60 &text{For } x = 5, quad k cdot 5 + 60 = 5k + 60 &text{For } x = 6, quad k cdot 6 + 60 = 6k + 60 &text{For } x = 7, quad k cdot 7 + 60 = 7k + 60 &text{For } x = 8, quad k cdot 8 + 60 = 8k + 60 &text{For } x = 9, quad k cdot 9 + 60 = 9k + 60 &text{For } x = 10, quad k cdot 10 + 60 = 10k + 60 end{aligned} ]3. Determine the conditions for primality: We need exactly one of these expressions to be prime. We will test values of ( k ) that are relatively prime to 60, as these are more likely to yield prime numbers.4. Test values of ( k ): - For ( k = 1 ): [ begin{aligned} &k + 60 = 61 quad (text{prime}) &7k + 60 = 67 quad (text{prime}) end{aligned} ] Both are prime, so ( k = 1 ) does not work. - For ( k = 7 ): [ begin{aligned} &k + 60 = 67 quad (text{prime}) &7k + 60 = 109 quad (text{prime}) end{aligned} ] Both are prime, so ( k = 7 ) does not work. - For ( k = 11 ): [ begin{aligned} &k + 60 = 71 quad (text{prime}) &7k + 60 = 137 quad (text{prime}) end{aligned} ] Both are prime, so ( k = 11 ) does not work. - For ( k = 13 ): [ begin{aligned} &k + 60 = 73 quad (text{prime}) &7k + 60 = 151 quad (text{prime}) end{aligned} ] Both are prime, so ( k = 13 ) does not work. - For ( k = 17 ): [ begin{aligned} &k + 60 = 77 quad (text{not prime}) &7k + 60 = 179 quad (text{prime}) end{aligned} ] Only ( 7k + 60 ) is prime, so ( k = 17 ) works.Therefore, the smallest positive integer ( k ) such that there is exactly one prime number of the form ( kx + 60 ) for ( 0 le x le 10 ) is ( boxed{17} ).