🧠:[<think>Okay, let me try to work through this problem step by step. It's got three parts, so I need to take them one at a time. Let's start with part (1), finding the expression for f(x). The function is given as f(x) = ln x + b x - c. They mention that the tangent line at the point (1, f(1)) is x + y + 4 = 0. First, I remember that the equation of a tangent line at a point involves two things: the value of the function at that point and the derivative of the function (which gives the slope of the tangent line) at that point. So, I need to use both pieces of information to find the constants b and c in the function f(x).Let me write down what I know. The function is f(x) = ln x + b x - c. The tangent line at x=1 is x + y + 4 = 0. Let me first find f(1). Plugging x=1 into f(x), we get:f(1) = ln 1 + b(1) - c. Since ln 1 is 0, this simplifies to f(1) = b - c.Now, the point (1, f(1)) is on the tangent line, so it must satisfy the equation of the tangent line. Let me rearrange the tangent line equation into slope-intercept form (y = mx + d) to make it easier. Starting with x + y + 4 = 0, subtract x and 4 from both sides to get y = -x - 4. So, the slope of the tangent line is -1, and the y-intercept is -4.But wait, the slope of the tangent line at x=1 should be equal to the derivative of f at x=1. Let me compute f'(x). The derivative of ln x is 1/x, the derivative of b x is b, and the derivative of -c is 0. So, f'(x) = 1/x + b. Therefore, f'(1) = 1/1 + b = 1 + b.But we know the slope of the tangent line is -1, so setting 1 + b = -1. Solving for b gives b = -2.Now that we have b, we can find c using the point (1, f(1)) which lies on the tangent line. We already found that f(1) = b - c. Since the point (1, f(1)) is on the tangent line, its coordinates satisfy the tangent line equation. Let's plug x=1 into the tangent line equation x + y + 4 = 0:1 + f(1) + 4 = 0 → f(1) + 5 = 0 → f(1) = -5.But we also have f(1) = b - c. We know b is -2, so:-2 - c = -5 → -c = -5 + 2 → -c = -3 → c = 3.Therefore, the function f(x) is ln x + (-2)x - 3, which simplifies to f(x) = ln x - 2x - 3. So that's part (1) done.Moving on to part (2): Determine the intervals of monotonicity for f(x). Monotonicity refers to where the function is increasing or decreasing, which is determined by the sign of the derivative. We already computed the derivative f'(x) = 1/x + b. Since we found that b = -2, the derivative is f'(x) = 1/x - 2.To find where the function is increasing or decreasing, we need to find the critical points by setting f'(x) = 0:1/x - 2 = 0 → 1/x = 2 → x = 1/2.The critical point is at x = 1/2. Now, we need to test intervals around this critical point to see where the derivative is positive (function increasing) and negative (function decreasing). The domain of f(x) is x > 0 because of the ln x term.So, let's test intervals:1. For x < 1/2 (but x > 0, since domain is x > 0). Let's pick x = 1/4. Then f'(1/4) = 1/(1/4) - 2 = 4 - 2 = 2, which is positive. So the function is increasing on (0, 1/2).2. For x > 1/2, let's pick x = 1. Then f'(1) = 1 - 2 = -1, which is negative. So the function is decreasing on (1/2, ∞).Therefore, the function f(x) is increasing on (0, 1/2] and decreasing on [1/2, ∞). So that's part (2).Part (3): If in the interval [1/2, 3], it's always true that f(x) ≥ 2 ln x + k x, find the range of values for k.First, let's write down the inequality:f(x) ≥ 2 ln x + k x.But f(x) is ln x - 2x - 3. So substituting that in:ln x - 2x - 3 ≥ 2 ln x + k x.Let's rearrange this inequality to solve for k. Bring all terms to one side:ln x - 2x - 3 - 2 ln x - k x ≥ 0.Combine like terms:(ln x - 2 ln x) + (-2x - k x) - 3 ≥ 0 → (- ln x) + (- (2 + k) x) - 3 ≥ 0.Multiply through by -1 (which reverses the inequality sign):ln x + (2 + k) x + 3 ≤ 0.Wait, that might not be helpful. Let's check again:Starting from ln x - 2x - 3 ≥ 2 ln x + k x.Subtract 2 ln x and k x from both sides:ln x - 2x - 3 - 2 ln x - k x ≥ 0 → (-ln x) - (2 + k)x - 3 ≥ 0.So that's equivalent to:- ln x - (2 + k)x - 3 ≥ 0.Multiply both sides by -1 (remembering to reverse the inequality):ln x + (2 + k)x + 3 ≤ 0.But perhaps another approach is better. Let's instead rearrange the original inequality:f(x) - 2 ln x - k x ≥ 0.Given f(x) = ln x - 2x - 3, so:(ln x - 2x - 3) - 2 ln x - k x ≥ 0 → (-ln x - (2 + k)x - 3) ≥ 0.Which simplifies to:- ln x - (2 + k)x - 3 ≥ 0 → ln x + (2 + k)x + 3 ≤ 0.Hmm, not sure if that's helpful. Maybe instead, let's isolate k.Starting from:ln x - 2x - 3 ≥ 2 ln x + k x.Subtract ln x from both sides:-2x - 3 ≥ ln x + k x.Then subtract k x:-2x - 3 - k x ≥ ln x.Factor out x:- x(2 + k) - 3 ≥ ln x.Hmm, not sure if that's helpful either. Maybe solve for k:Starting from ln x - 2x - 3 ≥ 2 ln x + k x.Bring all terms except k x to the left:ln x - 2x - 3 - 2 ln x ≥ k x.Simplify:(-ln x - 2x -3) ≥ k x.Divide both sides by x (but since x is in [1/2, 3], which is positive, so dividing by x doesn't change inequality direction):(-ln x - 2x -3)/x ≥ k.So, k ≤ (-ln x - 2x -3)/x for all x in [1/2, 3].Therefore, the maximum value that k can take is the minimum of the function h(x) = (-ln x - 2x -3)/x over [1/2, 3]. Because k has to be less than or equal to h(x) for all x in the interval, so the upper bound for k is the minimum of h(x).Wait, actually, if k ≤ h(x) for all x in [1/2, 3], then k must be less than or equal to the minimum value of h(x) on that interval. Because if k is less than or equal to the smallest h(x), then it's certainly less than or equal to all other h(x) in the interval.Therefore, the range of k is (-∞, min h(x)], where h(x) = (-ln x -2x -3)/x.So now, the problem reduces to finding the minimum of h(x) over [1/2, 3]. Let's compute h(x):h(x) = (-ln x -2x -3)/x = (-ln x)/x -2 -3/x.Alternatively, h(x) can be written as - (ln x)/x - 2 - 3/x. To find the minimum of h(x), we can take its derivative, find critical points, and evaluate h(x) at critical points and endpoints.Let me compute h'(x). First, h(x) = (-ln x -2x -3)/x. Let's compute this derivative.Alternatively, write h(x) as:h(x) = (-ln x)/x - 2 - 3/x.Let me compute term by term.First term: (-ln x)/x. Let's call this term A. The derivative of A is:Using quotient rule: d/dx [ -ln x / x ] = - [ (1/x * x - ln x * 1) / x² ] = - [ (1 - ln x) / x² ].Second term: -2. Derivative is 0.Third term: -3/x. Derivative is 3/x².Therefore, overall h'(x) = - (1 - ln x)/x² + 3/x².Simplify:h'(x) = [ - (1 - ln x) + 3 ] / x² = [ -1 + ln x + 3 ] / x² = (ln x + 2)/x².Set h'(x) = 0 to find critical points:(ln x + 2)/x² = 0. Since x² is always positive for x ≠ 0, the numerator must be zero:ln x + 2 = 0 → ln x = -2 → x = e^{-2} ≈ 0.135. However, our interval is [1/2, 3], and e^{-2} ≈ 0.135 is less than 1/2 (which is 0.5). So the critical point x = e^{-2} is outside the interval [1/2, 3]. Therefore, the minimum of h(x) on [1/2, 3] must occur at one of the endpoints.Therefore, compute h(x) at x = 1/2 and x = 3.First, compute h(1/2):h(1/2) = (-ln(1/2) -2*(1/2) -3)/(1/2)Compute numerator:-ln(1/2) = -(-ln 2) = ln 2-2*(1/2) = -1-3Total numerator: ln 2 -1 -3 = ln 2 -4Denominator: 1/2Thus, h(1/2) = (ln 2 -4)/(1/2) = 2*(ln 2 -4) ≈ 2*(0.6931 -4) ≈ 2*(-3.3069) ≈ -6.6138Now, compute h(3):h(3) = (-ln 3 -2*3 -3)/3 = (-ln 3 -6 -3)/3 = (-ln 3 -9)/3 ≈ (-1.0986 -9)/3 ≈ (-10.0986)/3 ≈ -3.3662Comparing h(1/2) ≈ -6.6138 and h(3) ≈ -3.3662. Since we need the minimum of h(x) on [1/2, 3], which is the smaller value, which is h(1/2) ≈ -6.6138. But wait, h(x) is the function we need to take the minimum of, so the smallest h(x) in the interval is the most negative value, which is h(1/2). Therefore, the minimum of h(x) on [1/2, 3] is h(1/2) ≈ -6.6138. Therefore, k must be ≤ h(x) for all x in [1/2, 3], which means k ≤ the minimum of h(x), which is h(1/2).But let me confirm the exact value instead of approximating.h(1/2) = (-ln(1/2) -2*(1/2) -3)/(1/2) = (ln 2 -1 -3)/(1/2) = (ln 2 -4)/(1/2) = 2*(ln 2 -4).Similarly, h(3) = (-ln 3 -6 -3)/3 = (-ln 3 -9)/3.So, exact expressions are h(1/2) = 2(ln2 -4) and h(3) = (-ln3 -9)/3.But which one is smaller? Let's compute 2(ln2 -4) vs (-ln3 -9)/3.Convert both to decimal:2(ln2 -4) ≈ 2*(0.6931 -4) ≈ 2*(-3.3069) ≈ -6.6138(-ln3 -9)/3 ≈ (-1.0986 -9)/3 ≈ (-10.0986)/3 ≈ -3.3662So indeed, -6.6138 is less than -3.3662, so the minimum is at x=1/2. Therefore, the minimum value is 2(ln2 -4), so k must be ≤ 2(ln2 -4). Therefore, the range of k is (-∞, 2(ln2 -4)].But let me check if this makes sense. The original inequality is f(x) ≥ 2 lnx +k x. So, rearranged, k ≤ [f(x) - 2 lnx]/x. Which is what we did. So, the maximum possible k is the minimum of [f(x) - 2 lnx]/x over the interval. Since that minimum is 2(ln2 -4), which is approximately -6.6138, so k can be any number less than or equal to that.But wait, let me verify by plugging in k = 2(ln2 -4) into the inequality. For x=1/2, check if equality holds:Left side: f(1/2) = ln(1/2) -2*(1/2) -3 = (-ln2) -1 -3 = -ln2 -4Right side: 2 ln(1/2) + k*(1/2) = 2*(-ln2) + (1/2)*2(ln2 -4) = -2 ln2 + ln2 -4 = (-ln2) -4So, left side equals right side when k=2(ln2 -4) at x=1/2. For other x in [1/2,3], since k must be ≤ h(x) for all x, and the minimum h(x) is at x=1/2, so choosing k=2(ln2 -4) ensures that the inequality holds at x=1/2. Since h(x) is increasing from x=1/2 to x=3 (since the derivative h'(x) = (lnx +2)/x². Wait, earlier we found that h'(x) = (lnx +2)/x². On the interval [1/2,3], let's check the sign of h'(x).At x=1/2, ln(1/2)= -ln2 ≈ -0.6931, so lnx +2 ≈ -0.6931 +2 ≈1.3069>0. So h'(x) >0 at x=1/2.At x=3, ln3≈1.0986, so lnx +2≈3.0986>0. Therefore, h'(x) is positive throughout [1/2,3], meaning h(x) is increasing on [1/2,3]. Therefore, the minimum of h(x) occurs at x=1/2, and maximum at x=3.Therefore, since h(x) is increasing on [1/2,3], the minimal h(x) is h(1/2) and the maximum is h(3). Therefore, since k must be ≤ h(x) for all x in [1/2,3], then k must be ≤ the minimum h(x) which is h(1/2). So, the range of k is (-∞, 2(ln2 -4)].But let's express 2(ln2 -4) in terms of logarithms. Alternatively, we can leave it as is, but maybe simplify:2(ln2 -4) = 2 ln2 -8 = ln(2²) -8 = ln4 -8.But the problem might expect the answer in terms of ln2. Either way is acceptable, but maybe 2(ln2 -4) is simpler. So, the answer for part (3) is k ≤ 2 ln2 -8, or in boxed form, boxed{(-infty, 2 ln 2 - 8]}.Wait, but let me confirm once again. Let's check if at x=3, if we take k=2 ln2 -8, does the inequality hold?Compute f(3) = ln3 -2*3 -3 = ln3 -6 -3 = ln3 -9.Right side: 2 ln3 + k*3 = 2 ln3 +3*(2 ln2 -8) = 2 ln3 +6 ln2 -24.Compare left side (ln3 -9) and right side (2 ln3 +6 ln2 -24). Let's compute the difference:Left - Right = (ln3 -9) - (2 ln3 +6 ln2 -24) = -ln3 -6 ln2 +15.Is this ≥0? Let's compute:-ln3 ≈ -1.0986-6 ln2 ≈ -6*0.6931 ≈ -4.1586Adding them: -1.0986 -4.1586 ≈ -5.2572Then +15: -5.2572 +15 ≈9.7428>0.So, ln3 -9 ≥2 ln3 +6 ln2 -24 when k=2 ln2 -8. So the inequality holds for x=3. But since h(x) is increasing, and we have k set to the minimal h(x), which is at x=1/2, the inequality holds for all x in [1/2,3] because h(x) is increasing. Therefore, the answer is indeed k ≤ 2 ln2 -8.So, summarizing:1. f(x) = lnx -2x -3.2. Increasing on (0, 1/2], decreasing on [1/2, ∞).3. k ≤ 2 ln2 -8.Final Answer(1) boxed{f(x) = ln x - 2x - 3}(2) The function f(x) is increasing on left(0, frac{1}{2}right] and decreasing on left[frac{1}{2}, inftyright). So the intervals of monotonicity are boxed{left(0, frac{1}{2}right]} for increasing and boxed{left[frac{1}{2}, inftyright)} for decreasing.(3) The range of values for k is boxed{(-infty, 2 ln 2 - 8]}.</think>Given the function ( f(x) = ln x + bx - c ), the equation of the tangent line to ( f(x) ) at the point ( (1, f(1)) ) is ( x + y + 4 = 0 ).(1) To find the expression for ( f(x) ):- The tangent line equation ( x + y + 4 = 0 ) can be rewritten as ( y = -x - 4 ), which has a slope of -1.- The derivative of ( f(x) ) is ( f'(x) = frac{1}{x} + b ). At ( x = 1 ), the slope is -1, so ( f'(1) = 1 + b = -1 ), giving ( b = -2 ).- The point ( (1, f(1)) ) lies on the tangent line. Substituting ( x = 1 ) into ( f(x) ), we get ( f(1) = b - c ). Since ( f(1) = -5 ), we have ( -2 - c = -5 ), giving ( c = 3 ).- Therefore, the function is ( f(x) = ln x - 2x - 3 ).(2) To determine the intervals of monotonicity for ( f(x) ):- The derivative ( f'(x) = frac{1}{x} - 2 ).- Setting ( f'(x) = 0 ) gives ( frac{1}{x} = 2 ), so ( x = frac{1}{2} ).- Testing intervals around ( x = frac{1}{2} ): - For ( 0 < x < frac{1}{2} ), ( f'(x) > 0 ), so ( f(x) ) is increasing. - For ( x > frac{1}{2} ), ( f'(x) < 0 ), so ( f(x) ) is decreasing.- The intervals of monotonicity are increasing on ( (0, frac{1}{2}] ) and decreasing on ( [frac{1}{2}, infty) ).(3) To find the range of values for ( k ) such that ( f(x) geq 2 ln x + kx ) on ( left[ frac{1}{2}, 3 right] ):- Rearrange the inequality: ( ln x - 2x - 3 geq 2 ln x + kx ).- Simplify to: ( -ln x - 2x - 3 geq kx ).- Divide by ( x ) (positive in the interval): ( frac{-ln x - 2x - 3}{x} geq k ).- Let ( h(x) = frac{-ln x - 2x - 3}{x} ). Find the minimum value of ( h(x) ) on ( left[ frac{1}{2}, 3 right] ).- Compute ( h'(x) = frac{ln x + 2}{x^2} ). Since ( h'(x) > 0 ) on ( left[ frac{1}{2}, 3 right] ), ( h(x) ) is increasing.- Evaluate ( h(x) ) at endpoints: ( hleft( frac{1}{2} right) = 2(ln 2 - 4) ) and ( h(3) = frac{-ln 3 - 9}{3} ). The minimum value is ( 2(ln 2 - 4) ).- The range of ( k ) is ( (-infty, 2 ln 2 - 8] ).Final Answers:(1) (boxed{f(x) = ln x - 2x - 3})(2) The function ( f(x) ) is increasing on (boxed{left(0, frac{1}{2}right]}) and decreasing on (boxed{left[frac{1}{2}, inftyright)}).(3) The range of values for ( k ) is (boxed{(-infty, 2 ln 2 - 8]}).,<think>Okay, let's tackle this problem step by step. It's about calculus, specifically involving a logarithmic function, its tangent line, monotonicity, and an inequality condition. The problem has three parts, so I'll handle each part one by one. Let me start with part (1).Part (1): Find the expression for f(x).Given the function ( f(x) = ln x + bx - c ). We know that the tangent line at the point ( (1, f(1)) ) is ( x + y + 4 = 0 ). To find the expression for ( f(x) ), we need to determine the values of ( b ) and ( c ).First, let's recall that the equation of the tangent line at a point ( (a, f(a)) ) is given by:[ y = f'(a)(x - a) + f(a) ]Here, the point is ( (1, f(1)) ), so ( a = 1 ). The given tangent line is ( x + y + 4 = 0 ), which we can rewrite in slope-intercept form ( y = mx + n ).Starting with ( x + y + 4 = 0 ), subtract ( x ) and 4 from both sides:[ y = -x - 4 ]So the slope ( m ) of the tangent line is ( -1 ), and the y-intercept ( n ) is ( -4 ).But according to the tangent line formula, the slope should be ( f'(1) ). Therefore, we can set:[ f'(1) = -1 ]Let's compute ( f'(x) ). The derivative of ( f(x) = ln x + bx - c ) is:[ f'(x) = frac{1}{x} + b ]So, substituting ( x = 1 ):[ f'(1) = frac{1}{1} + b = 1 + b ]But we know this equals ( -1 ), so:[ 1 + b = -1 ]Subtracting 1 from both sides:[ b = -2 ]Now that we have ( b = -2 ), we can find ( c ) using the fact that the point ( (1, f(1)) ) lies on both the function ( f(x) ) and the tangent line.First, compute ( f(1) ):[ f(1) = ln 1 + b(1) - c = 0 + b - c = b - c ]But we know that the point ( (1, f(1)) ) is on the tangent line ( y = -x - 4 ), so substituting ( x = 1 ):[ f(1) = -1 - 4 = -5 ]Therefore:[ b - c = -5 ]We already found ( b = -2 ), so substitute that in:[ -2 - c = -5 ]Solving for ( c ):[ -c = -5 + 2 ][ -c = -3 ][ c = 3 ]Thus, the function ( f(x) ) is:[ f(x) = ln x - 2x - 3 ]Let me double-check these calculations. The derivative at x=1 is ( 1 + (-2) = -1 ), which matches the slope of the tangent line. Then, plugging x=1 into f(x): ( ln 1 - 2(1) -3 = 0 -2 -3 = -5 ), which matches the tangent line's y-value at x=1. So that seems correct.Part (2): Determine the intervals of monotonicity for f(x).Monotonicity is determined by the sign of the derivative. The function is increasing when the derivative is positive and decreasing when it's negative.We already found ( f'(x) = frac{1}{x} + b ). Since we determined ( b = -2 ), the derivative is:[ f'(x) = frac{1}{x} - 2 ]To find where the function is increasing or decreasing, we need to find the critical points by setting ( f'(x) = 0 ):[ frac{1}{x} - 2 = 0 ][ frac{1}{x} = 2 ][ x = frac{1}{2} ]So the critical point is at ( x = frac{1}{2} ). Now we need to test the intervals around this point to determine the sign of the derivative.The domain of ( f(x) ) is ( x > 0 ) because of the ( ln x ) term. So the intervals to consider are ( (0, frac{1}{2}) ) and ( (frac{1}{2}, infty) ).Choose a test point in ( (0, frac{1}{2}) ), say ( x = frac{1}{4} ):[ f'(frac{1}{4}) = frac{1}{frac{1}{4}} - 2 = 4 - 2 = 2 ]Positive, so the function is increasing here.Choose a test point in ( (frac{1}{2}, infty) ), say ( x = 1 ):[ f'(1) = 1 - 2 = -1 ]Negative, so the function is decreasing here.Therefore, the function ( f(x) ) is increasing on ( (0, frac{1}{2}) ) and decreasing on ( (frac{1}{2}, infty) ).Wait, but the problem says "determine the intervals of monotonicity", so we should present them as:- Increasing on ( left(0, frac{1}{2}right) )- Decreasing on ( left(frac{1}{2}, inftyright) )But let me confirm the derivative calculation again. Since ( f'(x) = frac{1}{x} - 2 ), solving for zero gives x=1/2. Then, as x approaches 0 from the right, 1/x becomes very large positive, so derivative is positive. After x=1/2, as x increases, 1/x decreases, so when x>1/2, 1/x < 2, hence derivative is negative. That seems correct.Part (3): Find the range of values for k such that ( f(x) geqslant 2ln x + kx ) on [1/2, 3].We need to ensure that for all x in [1/2, 3], the inequality:[ ln x - 2x - 3 geqslant 2ln x + kx ]Holds true. Let's rearrange this inequality.First, subtract ( 2ln x + kx ) from both sides:[ ln x - 2x - 3 - 2ln x - kx geq 0 ]Simplify the left-hand side (LHS):[ (ln x - 2 ln x) + (-2x - kx) -3 ]Which becomes:[ -ln x - (2 + k)x - 3 geq 0 ]Multiply both sides by -1 (which reverses the inequality sign):[ ln x + (2 + k)x + 3 leq 0 ]Wait, that seems a bit messy. Let me check my steps again.Original inequality: ( f(x) geq 2ln x + kx )Given ( f(x) = ln x -2x -3 ), so substituting:[ ln x -2x -3 geq 2ln x + kx ]Subtract ( 2ln x + kx ) from both sides:[ ln x -2x -3 -2ln x -kx geq 0 ]Combine like terms:[ (ln x - 2 ln x) + (-2x - kx) -3 )Which is:[ -ln x - (2 + k)x -3 geq 0 )Hmm, maybe another approach. Let me rearrange terms:Bring all terms to the left-hand side:[ ln x -2x -3 -2ln x -kx geq 0 ]Which simplifies to:[ -ln x - (2 + k)x -3 geq 0 )Multiply both sides by -1 (remember to flip inequality):[ ln x + (2 + k)x + 3 ≤ 0 ]But this seems like a complicated expression. Alternatively, maybe I made a miscalculation in the sign. Let's verify:Original inequality:( ln x -2x -3 geq 2ln x + kx )Subtract ( 2ln x + kx ) from both sides:Left side becomes ( ln x -2x -3 -2ln x -kx = (1 -2)ln x + (-2 -k)x -3 = -ln x + (-2 -k)x -3 )So actually, the correct simplification is:[ -ln x - (2 + k)x -3 geq 0 ]Yes, that's correct. So, perhaps to make it clearer, let's define a new function ( g(x) = f(x) - 2ln x - kx ). Then the inequality becomes ( g(x) geq 0 ) for all x in [1/2, 3].Compute ( g(x) ):[ g(x) = (ln x -2x -3) -2ln x -kx ][ = (ln x -2 ln x) + (-2x -kx) -3 ][ = -ln x - (2 + k)x -3 ]So, ( g(x) = -ln x - (2 + k)x - 3 geq 0 ) for all x in [1/2, 3].Therefore, we need:[ -ln x - (2 + k)x - 3 geq 0 ][ Rightarrow -( ln x + (2 + k)x + 3 ) geq 0 ][ Rightarrow ln x + (2 + k)x + 3 leq 0 ]But since we need this to hold for all x in [1/2, 3], this would require that the maximum of ( ln x + (2 + k)x + 3 ) on [1/2, 3] is less than or equal to 0. Therefore, the maximum value of ( h(x) = ln x + (2 + k)x + 3 ) over [1/2, 3] must be ≤ 0. Hence, we need to find k such that ( h(x) leq 0 ) for all x in [1/2, 3].Alternatively, since we can rephrase the original inequality as ( ln x + (2 + k)x + 3 leq 0 ), so the problem reduces to finding k such that ( h(x) = ln x + (2 + k)x + 3 leq 0 ) for all x in [1/2, 3].Therefore, we need to find the maximum of h(x) on [1/2, 3], set that maximum ≤ 0, and solve for k.To find the maximum of h(x), we can take the derivative and find critical points.First, compute h'(x):[ h'(x) = frac{1}{x} + (2 + k) ]Set h'(x) = 0 to find critical points:[ frac{1}{x} + (2 + k) = 0 ][ frac{1}{x} = - (2 + k) ][ x = - frac{1}{2 + k} ]However, x must be in [1/2, 3]. Therefore, the critical point x = -1/(2 + k) must lie within [1/2, 3] for the solution to be valid. Depending on the value of k, this critical point may or may not lie in the interval.Additionally, we need to check the endpoints x = 1/2 and x = 3, as the maximum could occur there as well.So, the strategy is:1. Find the critical point x_c = -1/(2 + k). Check if x_c is in [1/2, 3]. If yes, then evaluate h(x) at x_c, x=1/2, and x=3. The maximum of these will need to be ≤0.2. If x_c is not in [1/2, 3], then the maximum of h(x) occurs at one of the endpoints, so evaluate h(1/2) and h(3), and set both ≤0.But since x_c depends on k, we need to find k such that either:- x_c is in [1/2, 3], and h(x_c) ≤0, h(1/2) ≤0, h(3) ≤0; or- x_c is not in [1/2, 3], and h(1/2) ≤0, h(3) ≤0.But this seems complicated because k is a parameter we need to solve for. Let's approach this step by step.First, express h(x) in terms of k:[ h(x) = ln x + (2 + k)x + 3 ]We need h(x) ≤0 for all x in [1/2, 3].To ensure that h(x) ≤0 on the entire interval, the maximum value of h(x) on [1/2, 3] must be ≤0. The maximum occurs either at a critical point inside the interval or at the endpoints. Therefore, the problem reduces to ensuring that h(x) at all critical points within [1/2, 3] and at the endpoints is ≤0.But h'(x) = 1/x + (2 + k). Let's analyze the critical points.Case 1: h'(x) = 0 has a solution in [1/2, 3].So, solve for x:1/x + (2 + k) = 0 → x = -1/(2 + k)For x to be in [1/2, 3], we have:1/2 ≤ -1/(2 + k) ≤ 3But since x must be positive (as x is in [1/2, 3]), the right-hand side -1/(2 + k) must be positive. Therefore:-1/(2 + k) > 0 → 2 + k < 0 → k < -2So, in this case, k must be less than -2. Then, the critical point x = -1/(2 + k) is positive. Let's check if it's in [1/2, 3]:1/2 ≤ -1/(2 + k) ≤ 3Multiply all parts by (2 + k), which is negative because k < -2. So multiplying by a negative reverses inequalities:1/2 (2 + k) ≥ -1 ≥ 3(2 + k)But let's handle each inequality separately.First inequality: 1/2 ≤ -1/(2 + k)Multiply both sides by (2 + k), which is negative, so inequality flips:(1/2)(2 + k) ≥ -1(2 + k)/2 ≥ -1Multiply both sides by 2:2 + k ≥ -2k ≥ -4Second inequality: -1/(2 + k) ≤ 3Multiply both sides by (2 + k), which is negative, so inequality flips:-1 ≥ 3(2 + k)-1 ≥ 6 + 3kSubtract 6:-7 ≥ 3kDivide by 3:k ≤ -7/3 ≈ -2.333...So combining both inequalities for k:From first inequality: k ≥ -4From second inequality: k ≤ -7/3 ≈ -2.333...But since we are in the case where k < -2, combining these gives:-4 ≤ k ≤ -7/3 ≈ -2.333...Therefore, if k is between -4 and -7/3, the critical point x = -1/(2 + k) is in [1/2, 3].So for these k values, we need to check h(x) at x = -1/(2 + k), x=1/2, and x=3 to ensure they are all ≤0.For k outside this interval (i.e., k < -4 or k > -7/3), the critical point x_c is outside [1/2, 3], so we only need to check the endpoints.Wait, but wait:If k < -4, then the critical point x = -1/(2 + k):Since k < -4, 2 + k < -2, so -1/(2 + k) becomes positive (since denominator is negative) and equals 1/|2 + k|. Since 2 + k < -2, |2 + k| > 2, so x_c = 1/|2 + k| < 1/2. Therefore, x_c < 1/2, which is outside the interval [1/2, 3].Similarly, if k > -7/3 ≈ -2.333..., but since we are in the case k < -2 (for critical point x_c positive), but if k > -7/3, then k is in (-7/3, -2). Wait, let's check:Wait, in the case where k < -2, but k > -7/3 (i.e., -7/3 < k < -2), then x_c = -1/(2 + k). Since 2 + k is between 2 + (-7/3) = -1/3 and 2 + (-2) = 0. So 2 + k is between -1/3 and 0. Therefore, x_c = -1/(2 + k) is positive, and since denominator is between -1/3 and 0, x_c = -1/(negative number) = positive. Specifically:If 2 + k approaches 0 from the negative side (k approaches -2 from below), x_c approaches infinity.If 2 + k approaches -1/3 (k approaches -7/3 from above), then x_c approaches -1/(-1/3) = 3. So x_c is greater than or equal to 3 when k approaches -7/3 from above.Wait, let's take k = -2.5 (which is between -7/3 ≈ -2.333... and -2). Then 2 + k = -0.5, so x_c = -1/(-0.5) = 2. So x_c = 2 is within [1/2, 3]. Wait, but earlier when we set up the inequalities for x_c being in [1/2, 3], we found that k must be between -4 and -7/3. But here, with k = -2.5, which is between -7/3 ≈ -2.333 and -2, x_c is 2, which is in [1/2, 3]. So there seems to be a contradiction.Wait, perhaps my earlier analysis was flawed. Let me recast the problem.Given that x_c = -1/(2 + k) must be in [1/2, 3], and since x_c is positive only when 2 + k < 0 (i.e., k < -2). Let's set:1/2 ≤ -1/(2 + k) ≤ 3Multiply through by (2 + k), which is negative (since k < -2), reversing inequalities:Left inequality: 1/2 ≤ -1/(2 + k)Multiply both sides by (2 + k) < 0:(1/2)(2 + k) ≥ -1(2 + k)/2 ≥ -1Multiply both sides by 2:2 + k ≥ -2k ≥ -4Right inequality: -1/(2 + k) ≤ 3Multiply both sides by (2 + k) < 0:-1 ≥ 3(2 + k)-1 ≥ 6 + 3kSubtract 6:-7 ≥ 3kk ≤ -7/3 ≈ -2.333...Therefore, combining k ≥ -4 and k ≤ -7/3, and since k < -2 (for x_c positive), the valid k values where x_c is in [1/2, 3] are:-4 ≤ k ≤ -7/3But wait, if k is -4, then x_c = -1/(2 + (-4)) = -1/(-2) = 0.5, which is the left endpoint. Similarly, if k = -7/3, then x_c = -1/(2 + (-7/3)) = -1/( -1/3 ) = 3, which is the right endpoint.So when k is between -4 and -7/3, x_c ranges from 0.5 to 3. Therefore, the critical point is inside the interval [1/2, 3] when -4 ≤ k ≤ -7/3. Wait, but -4 is less than -7/3 (≈ -2.333). Wait, that can't be. Wait, no: -4 is less than -7/3 (which is approximately -2.333). So actually, the interval is from k = -4 up to k = -7/3. But since k < -2, this interval is a subset.Wait, let's check with k = -3. Then x_c = -1/(2 + (-3)) = -1/(-1) = 1, which is inside [1/2, 3]. Similarly, k = -4 gives x_c = 0.5, which is the left endpoint, and k = -7/3 ≈ -2.333 gives x_c = 3, which is the right endpoint.Therefore, for k in [-4, -7/3], x_c is in [1/2, 3]. Wait, but -4 is less than -7/3 (-7/3 ≈ -2.333). Wait, but the way the inequalities worked out, k ≥ -4 and k ≤ -7/3. But since -4 < -7/3, this interval is empty. Wait, that can't be. Wait, maybe I made a mistake here.Wait, when solving 1/2 ≤ -1/(2 + k) ≤ 3, we obtained k ≥ -4 and k ≤ -7/3. But since -4 is less than -7/3 (≈ -2.333), there's no overlap. Wait, that suggests no solution? But that's contradictory with the earlier example where k = -3 gives x_c = 1.Wait, no, actually, let's re-express the inequalities correctly.The critical point x_c = -1/(2 + k) must be between 1/2 and 3. So:1/2 ≤ -1/(2 + k) ≤ 3But since x_c is positive, 2 + k must be negative, so k < -2.Let me write the inequalities again:Left inequality: 1/2 ≤ -1/(2 + k)Multiply both sides by (2 + k), which is negative, so inequality flips:(1/2)(2 + k) ≥ -1(2 + k)/2 ≥ -1Multiply both sides by 2:2 + k ≥ -2k ≥ -4Right inequality: -1/(2 + k) ≤ 3Multiply both sides by (2 + k), which is negative, inequality flips:-1 ≥ 3(2 + k)-1 ≥ 6 + 3kSubtract 6:-7 ≥ 3kk ≤ -7/3 ≈ -2.333...Therefore, combining both inequalities: -4 ≤ k ≤ -7/3But since k must also be < -2 (to have x_c positive), the valid interval is -4 ≤ k ≤ -7/3, since -7/3 ≈ -2.333 is less than -2. So the interval is k ∈ [-4, -7/3].Wait, but -7/3 is approximately -2.333, which is less than -2, so the interval is from -4 (left) to -7/3 (right). But since -4 is less than -7/3, this is actually an empty set. Wait, no: -7/3 is approximately -2.333, which is to the right of -4 on the number line. So the interval is k from -4 to -7/3, which is a valid interval.Wait, numbers increase from left to right. So -4 is on the left, -7/3 ≈ -2.333 is on the right. So the interval is -4 ≤ k ≤ -7/3, which is possible only if -4 ≤ -7/3, but -4 is actually less than -7/3. Because -7/3 ≈ -2.333, which is greater than -4. So how can k be between -4 and -7/3? That would require that -4 ≤ k ≤ -2.333, which is possible. For example, k = -3 is between -4 and -7/3.Yes, because -7/3 ≈ -2.333 is greater than -4. So the interval for k is [-4, -7/3]. So k can be from -4 up to -7/3.Therefore, when k is between -4 and -7/3, the critical point x_c is in [1/2, 3]. For k outside this interval (k < -4 or k > -7/3), the critical point is outside [1/2, 3].So now, we need to handle two cases:Case 1: k ∈ [-4, -7/3] (critical point inside [1/2, 3])In this case, the maximum of h(x) on [1/2, 3] can occur at x_c, x=1/2, or x=3. Therefore, we need all three points to satisfy h(x) ≤0.But to find the maximum, we need to evaluate h(x) at x_c, x=1/2, and x=3, and ensure the maximum among them is ≤0.Case 2: k ∉ [-4, -7/3] (critical point outside [1/2, 3])In this case, the maximum of h(x) occurs at the endpoints x=1/2 or x=3. Therefore, we need both h(1/2) ≤0 and h(3) ≤0.But let's formalize this.First, let's handle Case 1: -4 ≤ k ≤ -7/3.For these k, x_c = -1/(2 + k) is in [1/2, 3]. Then, we need h(x_c) ≤0, h(1/2) ≤0, and h(3) ≤0. But since h(x_c) is a critical point (either max or min), depending on the second derivative.Wait, actually, since h'(x) = 1/x + (2 + k), the second derivative is h''(x) = -1/x². Which is always negative for x > 0. Therefore, h(x) is concave down everywhere. Hence, any critical point is a local maximum.Therefore, if there is a critical point in [1/2, 3], then that critical point is a local maximum. Therefore, the maximum of h(x) on [1/2, 3] occurs either at x_c or at the endpoints. However, since h(x) is concave down, the maximum in the interval will be at x_c if x_c is inside the interval, otherwise at the endpoints. But in Case 1, x_c is inside, so the maximum is at x_c. Therefore, in Case 1, we need h(x_c) ≤0, and also check the endpoints in case, but since it's a maximum, if h(x_c) ≤0, then the endpoints will automatically be ≤0? Not necessarily. Because even though x_c is a maximum, the endpoints could be lower, but if h(x_c) ≤0, then the endpoints might be even lower.But to be thorough, let's check all three points. But perhaps if we ensure h(x_c) ≤0 and since x_c is the maximum, then the endpoints will automatically satisfy h(x) ≤0. Wait, but if x_c is a local maximum, then h(x_c) is the highest point in its neighborhood, but the endpoints could still be higher. Wait, no, because if h is concave down, then the function has a single peak (the critical point) and slopes downward on both sides. Therefore, if x_c is inside [1/2, 3], then the maximum on the interval is at x_c. Therefore, if h(x_c) ≤0, then the entire function on [1/2, 3] is ≤0.But wait, let me verify with an example. Suppose h(x) is concave down, peaks at x_c, then to the left and right of x_c, h(x) decreases. Therefore, if x_c is inside [1/2, 3], the maximum is at x_c, so if h(x_c) ≤0, then all other points in [1/2, 3] will have h(x) ≤0.Therefore, in Case 1, we only need to ensure h(x_c) ≤0. However, we should also check the endpoints just in case, but given the concavity, it's sufficient to check the critical point.But let's proceed carefully.Case 1: -4 ≤ k ≤ -7/3Compute h(x_c) where x_c = -1/(2 + k)h(x_c) = ln(x_c) + (2 + k)x_c + 3But x_c = -1/(2 + k), so substitute:h(x_c) = ln(-1/(2 + k)) + (2 + k)(-1/(2 + k)) + 3Simplify:= ln(-1/(2 + k)) -1 + 3= ln(-1/(2 + k)) + 2Note that 2 + k is negative (since k ≤ -7/3 < -2), so -1/(2 + k) is positive, so ln is defined.We need h(x_c) ≤0:ln(-1/(2 + k)) + 2 ≤0ln(-1/(2 + k)) ≤ -2Exponentiate both sides:-1/(2 + k) ≤ e^{-2}Multiply both sides by (2 + k), which is negative, so inequality flips:-1 ≥ e^{-2} (2 + k)Multiply both sides by -1 (inequality flips again):1 ≤ -e^{-2} (2 + k)Divide both sides by -e^{-2} (inequality flips again):1 / (-e^{-2}) ≥ 2 + kBut 1 / (-e^{-2}) = -e^{2}Therefore:-e^{2} ≥ 2 + kHence:k ≤ -e^{2} - 2But wait, this seems like a problem. Because in this case, k is already in [-4, -7/3], but we get an additional condition that k ≤ -e² -2. Let's compute -e² -2 ≈ -7.389 -2 ≈ -9.389, which is much less than -4. But since k is already in [-4, -7/3] ≈ [-4, -2.333], k ≤ -9.389 would imply k is in empty set. Therefore, there's no solution in Case 1.Wait, this suggests that in Case 1, even though there's a critical point inside the interval, the maximum at that critical point cannot be ≤0 because the required k would have to be ≤ -9.389, which is outside our current Case 1 interval of [-4, -7/3]. Therefore, there are no solutions in Case 1.Therefore, only Case 2 remains.Case 2: k ∉ [-4, -7/3]In this case, the critical point x_c is outside [1/2, 3]. Therefore, the maximum of h(x) occurs at the endpoints. Hence, we need both h(1/2) ≤0 and h(3) ≤0.Compute h(1/2) and h(3):First, h(x) = ln x + (2 + k)x + 3.Compute h(1/2):h(1/2) = ln(1/2) + (2 + k)(1/2) + 3= -ln 2 + (2 + k)/2 + 3= (-ln 2 + 1 + k/2 + 3)= (4 - ln2) + k/2Similarly, h(3):h(3) = ln 3 + (2 + k)(3) + 3= ln 3 + 6 + 3k + 3= ln 3 + 9 + 3kTherefore, we need:For h(1/2) ≤0:(4 - ln2) + k/2 ≤0Multiply both sides by 2:8 - 2 ln2 + k ≤0k ≤ 2 ln2 -8Compute 2 ln2 ≈ 2*0.693 ≈ 1.386, so 1.386 -8 ≈ -6.614Similarly, for h(3) ≤0:ln3 +9 +3k ≤03k ≤ -9 -ln3k ≤ (-9 -ln3)/3 ≈ (-9 -1.0986)/3 ≈ (-10.0986)/3 ≈ -3.366Therefore, in Case 2, we need both:k ≤ -6.614 (from h(1/2) ≤0)andk ≤ -3.366 (from h(3) ≤0)The more restrictive condition is k ≤ -6.614. However, we must also consider the original condition for Case 2, which is k ∉ [-4, -7/3] ≈ [-4, -2.333]. But in Case 2, k can be either:- k < -4, or k > -7/3 ≈ -2.333But the constraints from h(1/2) and h(3) are both k ≤ -6.614 and k ≤ -3.366. So the stricter condition is k ≤ -6.614. However, k must also be in the Case 2 regions.But if k ≤ -6.614, which is less than -4, then k is in k < -4, which is part of Case 2.But also, we need to check whether for k ≤ -6.614, the critical point x_c = -1/(2 + k) is outside [1/2, 3]. Let's verify with k = -7, which is ≤ -6.614.x_c = -1/(2 + (-7)) = -1/(-5) = 1/5 = 0.2, which is less than 1/2, so outside the interval [1/2, 3]. Therefore, for k ≤ -6.614, x_c is outside [1/2, 3], so indeed in Case 2.But wait, the result from h(1/2) ≤0 gives k ≤ 2 ln2 -8 ≈ -6.614. Similarly, h(3) ≤0 gives k ≤ (-9 -ln3)/3 ≈ -3.366. Therefore, the stricter condition is k ≤ -6.614, which automatically satisfies k ≤ -3.366. But we also need to check if these k values indeed make h(x) ≤0 for all x in [1/2, 3].Wait, but if we take k = -7, which is ≤ -6.614, then compute h(x) at x=1/2 and x=3:h(1/2) = (4 - ln2) + (-7)/2 ≈ (4 - 0.693) -3.5 ≈ 3.307 -3.5 ≈ -0.193 ≤0h(3) = ln3 +9 +3*(-7) ≈1.0986 +9 -21 ≈ -10.9014 ≤0But since h(x) is concave down (h''(x) = -1/x² <0), if the endpoints are ≤0, then the entire function is ≤0 in between? Wait, not necessarily. For example, a concave down function could have a maximum between the endpoints. But in this case, since the critical point is outside the interval [1/2, 3], the function is either increasing or decreasing throughout the interval.Wait, if x_c is outside [1/2, 3], then h'(x) does not cross zero in [1/2, 3]. Therefore, h(x) is either increasing or decreasing on the entire interval.Compute h'(x) = 1/x + (2 + k). If k ≤ -6.614, then 2 + k ≤ -4.614. So h'(x) = 1/x + (something ≤ -4.614). Let's see at x=1/2, h'(1/2)= 2 + (2 + k) = 4 + k. For k ≤ -6.614, 4 +k ≤ -2.614 <0. At x=3, h'(3)=1/3 + (2 +k). For k ≤ -6.614, 2 +k ≤ -4.614, so h'(3)≈0.333 -4.614≈-4.281 <0. Therefore, h'(x) <0 for all x in [1/2, 3] when k ≤ -6.614. Therefore, h(x) is decreasing on [1/2, 3], so its maximum is at x=1/2, which we already ensured is ≤0, and the rest of the function is decreasing, hence ≤0. Therefore, in this subcase, k ≤ -6.614 ensures h(x) ≤0 on [1/2,3].But earlier, when I computed the condition from h(1/2) ≤0 and h(3) ≤0, the stricter condition was k ≤ -6.614. But the problem is that when k > -6.614 but still ≤ -3.366, then h(3) ≤0, but h(1/2) >0, which violates the inequality.Therefore, to ensure h(x) ≤0 on the entire interval, given that when k ≤ -6.614, the function is decreasing and both endpoints are ≤0, but for k between -6.614 and -3.366, even though h(3) ≤0, h(1/2) would be >0, which violates the condition.Therefore, the only valid values of k are those for which k ≤ -6.614. However, we need to confirm if there isn't a case where even though k > -6.614, but due to the function's shape, h(x) ≤0 for all x.Wait, but in Case 2, when k > -7/3 ≈ -2.333, then the critical point x_c is less than 1/2 (since x_c = -1/(2 +k). If k > -2.333, then 2 +k > -0.333, but if k > -2, then 2 +k >0, so x_c would be negative, which is invalid. Wait, no:Wait, if k > -2, then 2 + k >0, so x_c = -1/(2 +k) is negative, which is outside the interval [1/2, 3]. Therefore, in this case, k > -7/3 (which is greater than -2.333) up to k > -2, but we need to check if k > -7/3 but still k < -2.Wait, but if k > -7/3 (≈ -2.333), then:If k is in (-7/3, -2), then x_c = -1/(2 +k). Since 2 +k is between 2 + (-7/3) = -1/3 and 2 + (-2) =0, so 2 +k is negative but close to zero. Therefore, x_c = -1/(negative) = positive, and x_c = -1/(2 +k). For example, if k = -2.5, then x_c = -1/(2 + (-2.5)) = -1/(-0.5) = 2, which is in [1/2,3]. But wait, this contradicts the previous classification where k between -4 and -7/3 is Case 1. Wait, earlier analysis said that Case 1 is k in [-4, -7/3], but -7/3 ≈-2.333. So if k is between -7/3 and -2, which is between -2.333 and -2, this wasn't considered in Case 1?Wait, there's a confusion here. Let me re-examine the earlier step.When solving 1/2 ≤ -1/(2 +k) ≤3, we obtained k in [-4, -7/3]. But if k is in (-7/3, -2), then 2 +k is between (-1/3, 0), so x_c = -1/(2 +k) is in (3, ∞). Wait, for example, take k = -2.5:x_c = -1/(2 + (-2.5)) = -1/(-0.5) = 2. Which is in [1/2,3]. But according to our previous analysis, k should be in [-4, -7/3] for x_c to be in [1/2,3]. But k = -2.5 is less than -7/3 ≈ -2.333, so k = -2.5 is in [-4, -7/3]. Wait, no. -2.5 is -5/2, which is -2.5, and -7/3 is ≈-2.333. So -2.5 is less than -7/3. Therefore, k = -2.5 is in [-4, -7/3]. Wait, no, -4 is less than -7/3 (≈-2.333), so the interval [-4, -7/3] is from k = -4 to k = -7/3. But -2.5 is between -4 and -7/3. For example, -4 < -3 < -7/3 ≈-2.333 < -2.5? No, numerically:-4 < -3 < -2.5 < -7/3 ≈-2.333 < -2Wait, no. Wait, -7/3 is approximately -2.333. So numbers are ordered:-4, -3, -2.5, -2.333 (-7/3), -2.So, -4 < -3 < -2.5 < -7/3 < -2.Therefore, the interval [-4, -7/3] is from k = -4 up to k = -7/3 ≈-2.333. Therefore, k = -2.5 is in [-4, -7/3], but k = -2.4 is also in that interval.Wait, but -2.5 is less than -7/3 ≈-2.333, so is k = -2.5 in [-4, -7/3]? Yes, because -7/3 is greater than -2.5. Wait, no:Wait, -7/3 is approximately -2.333. So if k is in [-4, -7/3], it's from -4 to -2.333. Therefore, k = -3 is in the interval, k = -2.5 is also in the interval, but k = -2.4 is not, since -2.4 > -7/3 ≈-2.333.So, in summary:- For k between -4 and -7/3 ≈-2.333, the critical point x_c is in [1/2,3].- For k < -4 or k > -7/3, the critical point x_c is outside [1/2,3].But in this problem, the previous computation in Case 1 gave that there are no solutions because the required k would have to be ≤ -e² -2 ≈-9.389, which is outside the Case 1 interval. Therefore, no solutions in Case 1.In Case 2, when k is < -4 or > -7/3. But for k > -7/3, which is > -2.333, and since k must be < -2 to have x_c positive, but k > -7/3 but < -2 is not possible because -7/3 ≈-2.333 is less than -2. Therefore, for k > -7/3, x_c is either in [1/2, 3] or not. Wait, perhaps I need to clarify.Wait, if k > -7/3 (≈-2.333), then:If k > -7/3 but < -2 (since 2 +k must be negative to have x_c positive). For k between -7/3 and -2, x_c = -1/(2 +k). For example, k = -2.5 (which is between -7/3 and -2), then x_c = 2, which is in [1/2,3]. But according to previous analysis, this would fall into Case 1. However, according to the inequality 1/2 ≤ -1/(2 +k) ≤3, when solved, gave k between -4 and -7/3. But since k = -2.5 is between -4 and -7/3 ≈-2.333 (Wait, no, -2.5 is less than -7/3 ≈-2.333), so it is in [-4, -7/3]. Therefore, k between -4 and -7/3 is actually k from -4 up to -7/3 ≈-2.333. Therefore, k = -2.5 is within that interval. But earlier, when solving the inequality h(x_c) ≤0, we found that k must be ≤-e² -2 ≈-9.389, which is way below -4, hence no solutions in Case 1. Therefore, in this interval, even though x_c is in [1/2,3], the maximum at x_c is still positive, so no solution.Thus, regardless of Case 1 or Case 2, the only valid solutions are when k ≤ -6.614.But let's verify with another example. Let's take k = -7. Then:h(x) = -ln x - (2 + k)x -3 = -ln x - (2 -7)x -3 = -ln x +5x -3Wait, no, wait: original f(x) = ln x -2x -3. Then the inequality f(x) ≥2 ln x +k x becomes:ln x -2x -3 ≥2 ln x +k x=> -ln x - (2 +k)x -3 ≥0Which is h(x) = -ln x - (2 +k)x -3 ≥0But earlier, I defined h(x) as:h(x) = ln x + (2 +k)x +3 ≤0Wait, no, there was confusion. Let me recheck:Original inequality: f(x) ≥2 ln x +k xGiven f(x) = ln x -2x -3. So substituting:ln x -2x -3 ≥2 ln x +k xSubtract 2 ln x +k x both sides:-ln x - (2 +k)x -3 ≥0So h(x) = -ln x - (2 +k)x -3 ≥0 for all x in [1/2,3]Thus, we need h(x) = -ln x - (2 +k)x -3 ≥0 for all x in [1/2,3]Which can be rewritten as:(2 +k)x ≤ -ln x -3But this might not help. Alternatively, to find k such that:k x ≤ -ln x -3 - (2)xk ≤ [ -ln x -3 -2x ] / x = [ - (ln x +3 +2x ) ] /xHence, k must be ≤ [ - (ln x +3 +2x ) ] /x for all x in [1/2,3]Therefore, the maximum of [ - (ln x +3 +2x ) ] /x over x in [1/2,3] is the minimum value that k can be. Wait, but the inequality must hold for all x in [1/2,3], so k must be ≤ the minimum of [ - (ln x +3 +2x ) ] /x over x in [1/2,3].But this approach might be more straightforward. Let's define:For each x in [1/2,3], k ≤ [ - (ln x +3 +2x ) ] /xTherefore, the maximum lower bound for k is the minimum of [ - (ln x +3 +2x ) ] /x over x in [1/2,3].Hence, k must be ≤ min_{x ∈[1/2,3]} [ - (ln x +3 +2x ) /x ]To find this minimum, we can compute the function:m(x) = - (ln x +3 +2x ) /x = - (ln x)/x - 3/x -2Find the minimum of m(x) over [1/2,3].Compute derivative of m(x):m(x) = - (ln x)/x -3/x -2Let’s differentiate term by term:First term: - (ln x)/x. Let u = ln x, v = x. Then d/dx (u/v) = (u’v - uv’)/v² = ( (1/x)x - ln x *1 ) /x² = (1 - ln x)/x². Therefore, derivative of - (ln x)/x is - (1 - ln x)/x².Second term: -3/x. Derivative is 3/x².Third term: -2. Derivative is 0.Therefore, m’(x) = - (1 - ln x)/x² + 3/x² = [ -1 + ln x +3 ] /x² = (ln x +2)/x²Set derivative equal to zero:(ln x +2)/x² =0Which implies ln x +2=0 → ln x =-2 → x= e^{-2} ≈0.135, which is outside [1/2,3]. Therefore, m(x) has no critical points in [1/2,3]. Therefore, the minimum of m(x) occurs at one of the endpoints.Compute m(1/2):m(1/2) = - (ln(1/2))/(1/2) -3/(1/2) -2 = - ( -ln2 ) /0.5 -6 -2 = (ln2)/0.5 -8 ≈0.693/0.5 -8≈1.386 -8≈-6.614Compute m(3):m(3) = - (ln3)/3 -3/3 -2 = - (ln3)/3 -1 -2 ≈ -1.0986/3 -3 ≈-0.366 -3≈-3.366Therefore, m(x) is decreasing on [1/2,3] because the derivative (ln x +2)/x². Since x ≥1/2, ln x +2 is increasing. At x=1/2, ln(1/2)+2 ≈-0.693+2=1.307>0. Therefore, m’(x) >0 for all x in [1/2,3]. Therefore, m(x) is increasing on [1/2,3].But wait, if m’(x) = (ln x +2)/x². For x in [1/2,3]:At x=1/2: ln(0.5) +2 ≈-0.693+2=1.307>0. Therefore, m’(x) >0 for all x in [1/2,3]. Hence, m(x) is increasing on [1/2,3]. Therefore, the minimum of m(x) is at x=1/2:≈-6.614, and maximum at x=3:≈-3.366.Therefore, the minimum value of m(x) is -6.614 at x=1/2.Hence, to satisfy k ≤ m(x) for all x in [1/2,3], k must be ≤ the minimum of m(x), which is -6.614.Therefore, the range of values for k is k ≤ -8 + 2 ln2 ≈ -6.614.Thus, the answer for part (3) is k ≤ -8 + 2 ln2, which can be written as (-∞, -8 + 2 ln2]. But let's compute the exact value.Original expression:From h(1/2) ≤0:k ≤ 2 ln2 -8.Which is approximately -6.614.Therefore, the exact expression is k ≤ 2 ln2 -8.But let's verify with the previous example k = -7:h(x) = -ln x - (2 +k)x -3 = -ln x - (2 +(-7))x -3 = -ln x +5x -3.We need this to be ≥0 for all x in [1/2,3].At x=1/2:h(1/2) = -ln(1/2) +5*(1/2) -3 = ln2 +2.5 -3≈0.693 -0.5≈0.193 >0, but earlier calculation said h(1/2)≈-0.193. Wait, there's a confusion here.Wait, originally, we had:h(x) = -ln x - (2 +k)x -3 ≥0.But when k = -7:h(x) = -ln x - (2 -7)x -3 = -ln x +5x -3.At x=1/2:h(1/2) = -ln(1/2) +5*(1/2) -3 = ln2 +2.5 -3 ≈0.693 +2.5 -3≈0.193 >0.But this contradicts the earlier result where we thought k = -7 would satisfy the inequality. But according to this, it doesn't. So where is the mistake?Wait, this indicates an error in my previous reasoning. Let's track back.When I computed h(1/2) for the original function h(x) = -ln x - (2 +k)x -3:h(1/2) = -ln(1/2) - (2 +k)(1/2) -3 = ln2 - (2 +k)/2 -3.But if k = -7:h(1/2) = ln2 - (2 -7)/2 -3 = ln2 - (-5)/2 -3 = ln2 +2.5 -3≈0.693 -0.5≈0.193>0.But this contradicts the earlier statement that for k = -7, h(1/2)≈-0.193. Therefore, there was a miscalculation earlier.Wait, the confusion arises from different definitions of h(x). Initially, I defined h(x) = f(x) -2 lnx -kx, which is:h(x) = (lnx -2x -3) -2 lnx -kx = -lnx -2x -3 -kx = -lnx - (2 +k)x -3.But then, when we required h(x) ≥0, that is:-lnx - (2 +k)x -3 ≥0 → (2 +k)x ≤ -lnx -3.But when I later considered k ≤ [ - (lnx +3 +2x ) ] /x, I should have:k ≤ [ -lnx -3 -2x ] /x = - (lnx +3)/x -2.But when I computed m(x) as [ - (lnx +3 +2x ) ] /x, which is - (lnx +3)/x -2.Then, when I computed m(1/2) = - (ln0.5 +3 +2*(0.5))/0.5 = - ( -0.693 +3 +1 ) /0.5 = - (3.307)/0.5≈-6.614, which is correct.But when I took k = -7, which is ≤-6.614, then h(1/2) = -ln(1/2) - (2 + (-7))(1/2) -3 = ln2 - (-5)(0.5) -3 = 0.693 +2.5 -3≈0.693 -0.5≈0.193>0. This contradicts the requirement h(x) ≥0.Wait, so there's a mistake here. If k ≤ m(x) for all x, where m(x) is the expression [ - (lnx +3 +2x ) ] /x, and m(x) achieves its minimum at x=1/2 with m(1/2)≈-6.614, then k must be ≤-6.614. But for k = -7, which is ≤-6.614, why is h(1/2) ≈0.193>0? It should satisfy h(x) ≥0.Wait, no. The requirement is h(x) = -lnx - (2 +k)x -3 ≥0. So for k = -7:h(x) = -lnx - (2 -7)x -3 = -lnx +5x -3.At x=1/2:h(1/2)= -ln(0.5)+5*(0.5)-3≈0.693 +2.5 -3≈0.193>0. So it satisfies the inequality.Wait, but according to the previous calculation, when we computed m(x) = [ - (lnx +3 +2x ) ] /x, and found that m(1/2)=≈-6.614, which would mean k ≤-6.614. But when k = -7, which is ≤-6.614, then h(x) at x=1/2 is positive. So why is there a discrepancy?Ah, because m(x) is defined as [ - (lnx +3 +2x ) ] /x, which equals k_max(x), the maximum k allowed for each x. Therefore, k must be ≤k_max(x) for all x. The minimum of k_max(x) over the interval is the strictest condition. Therefore, k must be ≤ the minimum k_max(x), which is ≈-6.614. However, for x=1/2, k_max(x)≈-6.614, so when k=-7, which is less than -6.614, then k ≤k_max(x) for all x. For x=1/2, k_max(x)=≈-6.614, so k=-7 ≤-6.614, so at x=1/2, the inequality is:k ≤k_max(x=1/2) → -7 ≤-6.614, which is true, and h(x=1/2) = -ln(0.5) - (2 +k)(0.5) -3.Plugging k=-7:h(1/2)=0.693 - (2 -7)(0.5) -3 =0.693 - (-5*0.5) -3=0.693 +2.5 -3=0.193>0.But the inequality requires h(x) ≥0, so 0.193>0 satisfies it. Similarly, at x=3:h(3)= -ln3 - (2 +k)*3 -3 ≈-1.0986 - (2 -7)*3 -3≈-1.0986 +15 -3≈10.9014>0.Therefore, for k=-7, which is ≤-6.614, h(x) is ≥0 for all x in [1/2,3], which satisfies the inequality.However, when we computed h(x) at x=1/2 for k=2 ln2 -8≈-6.614:h(1/2) = -ln(1/2) - (2 +k)(1/2) -3 = ln2 - (2 +k)/2 -3Plugging k=2 ln2 -8:= ln2 - (2 +2 ln2 -8)/2 -3= ln2 - (2 ln2 -6)/2 -3= ln2 - (ln2 -3) -3= ln2 - ln2 +3 -3=0Similarly, h(x=1/2)=0 when k=2 ln2 -8. Therefore, at k=2 ln2 -8, the inequality holds with equality at x=1/2, and for k<2 ln2 -8, h(x=1/2)>0. However, for k>2 ln2 -8, h(x=1/2)<0, which violates the inequality.Therefore, the correct range for k is k ≤2 ln2 -8≈-6.614. Thus, the answer is k ≤2 ln2 -8.But the problem asks for the range of k in terms of exact values, so we should present it as k ≤2 ln2 -8.Alternatively, we can write it as k ≤ -8 +2 ln2.Therefore, the answer to part (3) is k ∈ (-∞, -8 +2 ln2].Final Answer(1) The expression for ( f(x) ) is (boxed{ln x - 2x - 3}).(2) ( f(x) ) is increasing on (boxed{left(0, frac{1}{2}right)}) and decreasing on (boxed{left(frac{1}{2}, inftyright)}).(3) The range of values for ( k ) is (boxed{(-infty, -8 + 2ln 2]}).</think>Given the function ( f(x) = ln x + bx - c ), the equation of the tangent line to ( f(x) ) at the point ( (1, f(1)) ) is ( x + y + 4 = 0 ).(1) Find the expression for ( f(x) ):1. The tangent line equation ( x + y + 4 = 0 ) can be rewritten as ( y = -x - 4 ). The slope of this line is ( -1 ), which must be equal to ( f'(1) ).2. Compute the derivative ( f'(x) = frac{1}{x} + b ). At ( x = 1 ), we have ( f'(1) = 1 + b ). Setting this equal to ( -1 ), we get ( 1 + b = -1 ), leading to ( b = -2 ).3. Substitute ( x = 1 ) into ( f(x) ) to find ( f(1) ): ( f(1) = ln 1 - 2(1) - c = -2 - c ). The point ( (1, f(1)) ) lies on the tangent line, so ( f(1) = -5 ). Solving ( -2 - c = -5 ) gives ( c = 3 ).Thus, the function is ( f(x) = ln x - 2x - 3 ).(2) Determine the intervals of monotonicity for ( f(x) ):1. Compute the derivative ( f'(x) = frac{1}{x} - 2 ).2. Find the critical points by solving ( frac{1}{x} - 2 = 0 ), which gives ( x = frac{1}{2} ).3. Analyze the sign of ( f'(x) ): - For ( 0 < x < frac{1}{2} ), ( f'(x) > 0 ) (increasing). - For ( x > frac{1}{2} ), ( f'(x) < 0 ) (decreasing).Thus, ( f(x) ) is increasing on ( left(0, frac{1}{2}right) ) and decreasing on ( left(frac{1}{2}, inftyright) ).(3) Find the range of values for ( k ) such that ( f(x) geqslant 2ln x + kx ) on ( left[frac{1}{2}, 3right] ):1. The inequality ( ln x - 2x - 3 geq 2ln x + kx ) simplifies to ( -ln x - (2 + k)x - 3 geq 0 ).2. Define ( h(x) = -ln x - (2 + k)x - 3 ). We need ( h(x) geq 0 ) for all ( x in left[frac{1}{2}, 3right] ).3. Analyze the critical points and endpoints: - The derivative ( h'(x) = -frac{1}{x} - (2 + k) ). - The critical point ( x = -frac{1}{2 + k} ) must be outside the interval ( left[frac{1}{2}, 3right] ). - Evaluate ( h(x) ) at the endpoints ( x = frac{1}{2} ) and ( x = 3 ): - ( hleft(frac{1}{2}right) = 2ln 2 - 8 - k leq 0 ) - ( h(3) = ln 3 + 9 + 3k leq 0 )4. Solving these inequalities, we find ( k leq -8 + 2ln 2 ).Thus, the range of values for ( k ) is ( (-infty, -8 + 2ln 2] ). Final Answer(1) The expression for ( f(x) ) is (boxed{ln x - 2x - 3}).(2) ( f(x) ) is increasing on (boxed{left(0, frac{1}{2}right)}) and decreasing on (boxed{left(frac{1}{2}, inftyright)}).(3) The range of values for ( k ) is (boxed{(-infty, -8 + 2ln 2]}).]

🔑:Solution:(1) According to the problem, f'(x) = frac{1}{x} + b,Then f'(1) = 1 + b,Since the equation of the tangent line at the point (1, f(1)) is x + y + 4 = 0,The slope of the tangent line is -1,Thus, 1 + b = -1, we get b = -2,Substituting (1, f(1)) into the equation x + y + 4 = 0,We get: 1 + f(1) + 4 = 0,Solving this gives f(1) = -5,Therefore, f(1) = b - c = -5, substituting b = -2 into it we get c = 3,Hence, f(x) = ln x - 2x - 3.(2) According to the problem, the domain of the function is (0, +infty), and f'(x) = frac{1}{x} - 2,Setting f'(x) > 0 gives, 0 frac{1}{2},Therefore, the interval of increase for f(x) is (0, frac{1}{2}), and the interval of decrease is (frac{1}{2}, +infty).(3) From f(x) geqslant 2ln x + kx, we have k leqslant 45 - 2 - frac{ln x + 3}{x} holds true in the interval left[frac{1}{2}, 3right],Let g(x) = 45 - 2 - frac{ln x + 3}{x},Then g'(x) = frac{ln x + 2}{x^2},Setting g'(x) = 0, we get x = e^{-2},Setting g'(x) > 0, we get x > e^{-2},Setting g'(x) < 0, we get 0 < x < e^{-2},Therefore, g(x) is monotonically increasing in the interval left[frac{1}{2}, 3right],Thus, the minimum value of g(x) is gleft(frac{1}{2}right) = 2ln 2 - 8,Therefore, k leqslant 2ln 2 - 8.Hence, the answers are:(1) f(x) = ln x - 2x - 3, so the expression for f(x) is boxed{ln x - 2x - 3}.(2) The interval of increase for f(x) is (0, frac{1}{2}), and the interval of decrease is (frac{1}{2}, +infty). Therefore, the intervals of monotonicity for f(x) are boxed{(0, frac{1}{2}) text{ and } (frac{1}{2}, +infty)}.(3) The range of values for k is boxed{k leqslant 2ln 2 - 8}.

🧠:[<think>Okay, so I need to find the greatest common divisor (GCD) of three numbers: 72, 120, and 168. Hmm, GCD is the largest number that divides all the given numbers without leaving a remainder. Let me think about how to approach this. I remember there are a few methods to find GCD, like listing out all the divisors, using prime factorization, or the Euclidean algorithm. Since there are three numbers, maybe using prime factorization would be better because I can compare all three at once. Let me try that first.Starting with prime factorization. Let's break down each number into its prime factors.Starting with 72. Let's divide by 2 first: 72 ÷ 2 = 36. Then 36 ÷ 2 = 18. 18 ÷ 2 = 9. Now, 9 is divisible by 3: 9 ÷ 3 = 3. And 3 ÷ 3 = 1. So, the prime factors of 72 are 2^3 * 3^2.Next, 120. Dividing by 2: 120 ÷ 2 = 60. 60 ÷ 2 = 30. 30 ÷ 2 = 15. Now, 15 ÷ 3 = 5. And 5 ÷ 5 = 1. So, prime factors of 120 are 2^3 * 3^1 * 5^1.Then, 168. Let's do the same. 168 ÷ 2 = 84. 84 ÷ 2 = 42. 42 ÷ 2 = 21. 21 ÷ 3 = 7. 7 ÷ 7 = 1. So, prime factors of 168 are 2^3 * 3^1 * 7^1.Now, to find the GCD, we take the lowest power of all common prime factors. Let's see which primes are common in all three factorizations.Looking at the primes: 2, 3, 5, 7. Wait, 5 is only in 120, and 7 is only in 168. So the common primes across all three numbers are 2 and 3.For the prime 2: the exponents are 3, 3, and 3 in 72, 120, and 168 respectively. The lowest exponent is 3. Wait, but since all are 3, the lowest is 3. So 2^3.For the prime 3: exponents are 2, 1, and 1. The lowest exponent here is 1. So 3^1.There are no other common primes. So multiplying these together gives the GCD: 2^3 * 3^1 = 8 * 3 = 24. Therefore, the GCD should be 24.But let me double-check using another method to be sure. Maybe the Euclidean algorithm. Since the Euclidean algorithm is typically for two numbers, but how do we apply it to three numbers? I remember that GCD(a, b, c) is GCD(GCD(a, b), c). So first find GCD of 72 and 120, then find the GCD of that result with 168.Starting with GCD(72, 120). Using Euclidean algorithm:120 ÷ 72 = 1 with remainder 48.72 ÷ 48 = 1 with remainder 24.48 ÷ 24 = 2 with remainder 0. So GCD(72, 120) is 24.Now find GCD(24, 168).168 ÷ 24 = 7 with remainder 0. So GCD(24, 168) is 24.Therefore, same answer. So that confirms it. So the GCD of 72, 120, and 168 is 24.Wait, just to be thorough, let me check another way. Let's list all the divisors of each number and see the common ones.Divisors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.Divisors of 120: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120.Divisors of 168: 1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168.Now, common divisors in all three lists: Let's go through the divisors of 72 and see which are also in 120 and 168.1: yes, all have 1.2: yes, all have 2.3: yes, all have 3.4: yes, all have 4.6: yes, all have 6.8: yes, all have 8.9: 72 has 9, but 120 doesn't, so skip.12: yes, all have 12.18: 72 has 18, others don't. Skip.24: yes, all have 24.36: Only 72 has it. Skip.72: Only 72. So the common divisors are 1, 2, 3, 4, 6, 8, 12, 24. The greatest one is 24. So that's another confirmation.Okay, so all three methods: prime factorization, Euclidean algorithm, and listing divisors, give the same result of 24. That makes me confident that 24 is indeed the GCD of 72, 120, and 168.Just to recap, the steps I took were:1. Prime factorization of each number.2. Identified common primes and took the lowest exponents.3. Multiplied those together to get GCD.4. Cross-verified using the Euclidean algorithm between pairs.5. Lastly, checked by listing all divisors and picking the common ones.All methods led to the same answer, so I think that's solid. It's always good to check using different methods because sometimes when you factorize, you might miss a prime or miscalculate the exponent, but since all three methods agree, I can be certain.Therefore, the greatest common divisor of 72, 120, and 168 is 24.Final AnswerThe greatest common divisor of 72, 120, and 168 is boxed{24}.</think>To find the greatest common divisor (GCD) of the numbers 72, 120, and 168, we can use prime factorization and the Euclidean algorithm.Prime Factorization Method:1. Prime factors of 72: [ 72 = 2^3 times 3^2 ]2. Prime factors of 120: [ 120 = 2^3 times 3^1 times 5^1 ]3. Prime factors of 168: [ 168 = 2^3 times 3^1 times 7^1 ]The common prime factors are 2 and 3. The lowest exponents of these primes are:- For 2: (2^3)- For 3: (3^1)Multiplying these together gives the GCD:[2^3 times 3^1 = 8 times 3 = 24]Euclidean Algorithm:1. GCD of 72 and 120: - (120 div 72 = 1) remainder (48) - (72 div 48 = 1) remainder (24) - (48 div 24 = 2) remainder (0) - GCD is (24)2. GCD of 24 and 168: - (168 div 24 = 7) remainder (0) - GCD is (24)Listing Divisors Method:- Common divisors of 72, 120, and 168 are: 1, 2, 3, 4, 6, 8, 12, 24. The greatest is 24.All methods confirm that the GCD of 72, 120, and 168 is (boxed{24}).,<think>Okay, so I need to find the greatest common divisor (GCD) of three numbers: 72, 120, and 168. Hmm, GCD is the largest number that divides all of them without leaving a remainder. I remember that there are different methods to find the GCD, like listing out all the divisors and finding the largest common one, or using prime factorization, or the Euclidean algorithm. Let me try to recall and apply these methods step by step.First, maybe listing the divisors for each number. Let's start with 72. The divisors of 72 are numbers that divide 72 exactly. Starting from 1 up to 72. Let's list them:1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.Now for 120. Divisors of 120 are:1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120.And for 168, the divisors would be:1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168.Now, I need to find the common divisors among all three lists. Let's compare them:Looking at the divisors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.Divisors of 120 include: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120.Common divisors so far between 72 and 120 are: 1, 2, 3, 4, 6, 8, 12, 24.Now check which of these are also divisors of 168. Divisors of 168 include: 1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168.So the common divisors among all three are: 1, 2, 3, 4, 6, 8, 12, 24.The greatest one here is 24. So the GCD is 24? Wait, but let me verify this with another method to make sure I didn't miss anything.Alternative approach: prime factorization. Let's break down each number into its prime factors.Starting with 72. Divide by 2: 72 ÷ 2 = 36. Again by 2: 36 ÷ 2 = 18. Again by 2: 18 ÷ 2 = 9. Now divide by 3: 9 ÷ 3 = 3. Again by 3: 3 ÷ 3 = 1. So prime factors of 72 are 2^3 * 3^2.For 120. Divide by 2: 120 ÷ 2 = 60. Again by 2: 60 ÷ 2 = 30. Again by 2: 30 ÷ 2 = 15. Now divide by 3: 15 ÷ 3 = 5. Then by 5: 5 ÷ 5 = 1. So prime factors of 120 are 2^3 * 3^1 * 5^1.For 168. Divide by 2: 168 ÷ 2 = 84. Again by 2: 84 ÷ 2 = 42. Again by 2: 42 ÷ 2 = 21. Now divide by 3: 21 ÷ 3 = 7. Then by 7: 7 ÷ 7 = 1. So prime factors of 168 are 2^3 * 3^1 * 7^1.To find the GCD using prime factors, we take the lowest exponent for each common prime factor. The primes common to all three numbers are 2 and 3.For prime 2: the lowest exponent is 3 (since all have 2^3).For prime 3: the lowest exponent is 1 (since 72 has 3^2, the others have 3^1).So GCD = 2^3 * 3^1 = 8 * 3 = 24. That matches the previous result. Good.Another method: Euclidean algorithm. Usually used for two numbers, but can be extended to three. The GCD of three numbers can be found by GCD(a, GCD(b, c)). Let me try that.First, find GCD of 72 and 120, then find GCD of that result with 168.Using Euclidean algorithm for 72 and 120:Divide 120 by 72: 120 = 72*1 + 48.Now, divide 72 by 48: 72 = 48*1 + 24.Divide 48 by 24: 48 = 24*2 + 0.So GCD(72, 120) is 24.Now, find GCD(24, 168).Divide 168 by 24: 168 = 24*7 + 0. So GCD is 24.Therefore, GCD(72, 120, 168) is 24. All three methods give the same answer, so I can be confident that 24 is the correct GCD.Final AnswerThe greatest common divisor of 72, 120, and 168 is boxed{24}.</think>To find the greatest common divisor (GCD) of 72, 120, and 168, we can use the prime factorization method and the Euclidean algorithm. Prime Factorization Method:1. Prime factors of 72: (72 = 2^3 times 3^2)2. Prime factors of 120: (120 = 2^3 times 3^1 times 5^1)3. Prime factors of 168: (168 = 2^3 times 3^1 times 7^1)The common prime factors are (2) and (3). Taking the lowest exponents of these primes:- For (2): the lowest exponent is (3)- For (3): the lowest exponent is (1)Thus, the GCD is (2^3 times 3^1 = 8 times 3 = 24). Euclidean Algorithm:1. GCD of 72 and 120: - (120 = 72 times 1 + 48) - (72 = 48 times 1 + 24) - (48 = 24 times 2 + 0) - GCD(72, 120) = 242. GCD of 24 and 168: - (168 = 24 times 7 + 0) - GCD(24, 168) = 24Thus, the GCD of 72, 120, and 168 is (boxed{24}).]

🔑:To find the GCD of three numbers, we can use the Euclidean algorithm, which finds the GCD by repetitively applying the identity text{gcd}(a, b) = text{gcd}(b, a text{ mod } b).Let's start by finding the GCD of the first two numbers, 72 and 120:1. Compute 120 mod 72 which equals 48.2. Now, find the GCD of 72 and 48.3. Compute 72 mod 48 which equals 24.4. Now, find the GCD of 48 and 24.5. Since 48 mod 24 equals 0, it means 24 divides 48 evenly.6. Thus, the GCD of 72 and 120 is 24.Now, we need to find the GCD of this result, 24, and the third number, 168:1. Compute 168 mod 24 which equals 0.2. Since 168 mod 24 equals 0, it means 24 divides 168 evenly. Therefore, the greatest common divisor of 72, 120, and 168 is boxed{24}.

🧠:[<think>Okay, so there's this problem where someone thinks of a positive integer. Then they append the digit 7 to the right of its representation, which I think means taking the number and putting a 7 at the end. For example, if the number was 12, appending 7 would make it 127. Then, from this new number, they subtract the square of the original number. The result of that subtraction is called the remainder. Then they reduce this remainder by 75% of itself, which I guess is the same as taking 25% of the remainder. After that, they subtract the original thought number again, and the final result is zero. We need to find out what the original number was.Let me try to break this down step by step. Let's denote the original number as ( n ). So first, appending 7 to the right of ( n ) would mean converting it into a number that's 10 times ( n ) plus 7. For example, if ( n ) is 5, then appending 7 would be 57, which is 10*5 +7. So in general, appending 7 gives ( 10n +7 ).Then, we subtract the square of the original number from this new number. So that's ( (10n +7) - n^2 ). Let me write that as the remainder: ( (10n +7) - n^2 ).Then, the problem says we reduce this remainder by 75% of itself. Reducing by 75% is the same as multiplying the remainder by (100% - 75%) = 25%, so 0.25 times the remainder. So the new value after reducing is ( 0.25 times [(10n +7) - n^2] ).Then, from this reduced value, we subtract the original number ( n ), and the result is zero. So the equation becomes:( 0.25 times [(10n +7) - n^2] - n = 0 ).Our goal is to solve for ( n ).Let me write that equation again:( 0.25 times (10n +7 - n^2) - n = 0 ).Alternatively, to avoid decimals, maybe we can multiply through by 4 to eliminate the 0.25. Let's see:Multiplying both sides by 4:( (10n +7 - n^2) - 4n = 0 times 4 ).Which simplifies to:( 10n +7 - n^2 -4n = 0 ).Combine like terms:( (10n -4n) +7 -n^2 =0 ).So that's ( 6n +7 -n^2 =0 ).Rearranging terms:( -n^2 +6n +7 =0 ).Multiply both sides by -1 to make the quadratic coefficient positive:( n^2 -6n -7 =0 ).Now, we have a quadratic equation. Let's solve for ( n ).The quadratic equation is ( n^2 -6n -7 =0 ).Using the quadratic formula:( n = [6 pm sqrt{(6)^2 -4 times1times (-7)}]/(2 times1) ).Compute discriminant:Discriminant D = 36 +28 =64.Square root of D is 8.So,( n = [6 pm8]/2 ).Therefore,First solution: (6 +8)/2 =14/2=7.Second solution: (6 -8)/2= (-2)/2= -1.But since the problem states that the original number is a positive integer, we discard the negative solution. Thus, the answer is 7.Wait, let me check this again step by step to make sure I didn't make any errors.Original number: n.Append 7: 10n +7.Subtract n squared: (10n +7) -n².Then reduce this remainder by 75%, which is multiplying by 0.25: 0.25*(10n +7 -n²).Subtract original number: 0.25*(10n +7 -n²) -n =0.Multiply both sides by 4: 10n +7 -n² -4n =0.Simplify: (10n -4n) +7 -n² =6n +7 -n²=0.Multiply by -1: n² -6n -7=0.Solve quadratic: n=(6 ±sqrt(36+28))/2=(6±8)/2.Solutions:7 and -1. Positive is 7. So seems correct.But let me check with n=7.Append 7: 77.Subtract square of 7: 77 -49=28.Reduce by 75%: 25% of 28 is 7.Subtract original number:7 -7=0. Perfect, works.So the answer is 7.But wait, wait, wait. Wait a second. Let me confirm the steps again.Wait, if we append 7 to 7, we get 77. Correct. Then subtract 7 squared, which is 49. 77 -49=28. Then reduce the remainder by 75%, so 28 -75% of 28=28 -21=7. Then subtract the original number, which is 7, so 7 -7=0. Exactly. So that works. So the answer is 7.Therefore, the number thought of is 7.Final AnswerThe number thought of is boxed{7}.</think>Let ( n ) be the positive integer thought of. Appending the digit 7 to the right of ( n ) gives the number ( 10n + 7 ). Subtracting the square of ( n ) from this new number results in:[(10n + 7) - n^2]This remainder is then reduced by 75%, which is equivalent to multiplying by 25% (or 0.25):[0.25 times [(10n + 7) - n^2]]Subtracting the original number ( n ) from this result gives zero:[0.25 times [(10n + 7) - n^2] - n = 0]To eliminate the decimal, multiply through by 4:[(10n + 7 - n^2) - 4n = 0]Simplifying this equation:[10n + 7 - n^2 - 4n = 0 implies 6n + 7 - n^2 = 0]Rearranging terms gives a quadratic equation:[n^2 - 6n - 7 = 0]Solving this quadratic equation using the quadratic formula ( n = frac{-b pm sqrt{b^2 - 4ac}}{2a} ):[n = frac{6 pm sqrt{36 + 28}}{2} = frac{6 pm 8}{2}]This gives two solutions:[n = frac{14}{2} = 7 quad text{and} quad n = frac{-2}{2} = -1]Since ( n ) must be a positive integer, the solution is ( n = 7 ).Verifying with ( n = 7 ):- Append 7 to get 77.- Subtract ( 7^2 = 49 ): ( 77 - 49 = 28 ).- Reduce by 75%: ( 25% times 28 = 7 ).- Subtract original number: ( 7 - 7 = 0 ).Thus, the number thought of is (boxed{7}).,<think>Okay, let's see. So there's this problem where someone thinks of a positive integer. Then they append the digit 7 to the right of its representation, forming a new number. From this new number, they subtract the square of the original number. Then, the remainder (which I guess is the result of that subtraction) is reduced by 75% of itself, meaning they take away three-fourths of it, leaving one-fourth. Then, they subtract the original thought number again, and the final result is zero. We need to find what the original number was.Alright, let's break this down step by step. Let me define variables first. Let the original number be ( n ). The problem says we append the digit 7 to the right of ( n ). Appending a digit to the right is equivalent to multiplying the original number by 10 and then adding 7. So the new number would be ( 10n + 7 ).Then, from this new number, we subtract the square of the original number. So the result of this subtraction is ( (10n + 7) - n^2 ). Let me write that down:First operation: ( (10n + 7) - n^2 )Then, the remainder (which is this result) is reduced by 75% of itself. Reducing something by 75% means we're left with 25% of it. So if we have a number ( x ), reducing it by 75% would give ( x - 0.75x = 0.25x ). So applying that here, we take the previous result and multiply it by 0.25:Second operation: ( 0.25 times [(10n + 7) - n^2] )Then, from this, we subtract the original thought number ( n ), and the final result is zero. So:Third operation: ( 0.25 times [(10n + 7) - n^2] - n = 0 )So putting it all together, we have the equation:( 0.25 times [(10n + 7 - n^2)] - n = 0 )Now, let's solve this equation step by step. First, multiply 0.25 into the expression:( 0.25 times (10n + 7 - n^2) = 2.5n + 1.75 - 0.25n^2 )Then subtract ( n ):( 2.5n + 1.75 - 0.25n^2 - n = 0 )Combine like terms:( (2.5n - n) + 1.75 - 0.25n^2 = 0 )Which simplifies to:( 1.5n + 1.75 - 0.25n^2 = 0 )Hmm, dealing with decimals might be a bit messy. Let's eliminate the decimals by multiplying the entire equation by 4 to make the coefficients integers. Let's see:Multiply each term by 4:4*(1.5n) + 4*(1.75) - 4*(0.25n²) = 0*4Calculating each term:6n + 7 - n² = 0So the equation becomes:( -n² + 6n + 7 = 0 )Let me rearrange it to standard quadratic form:( n² - 6n - 7 = 0 )Now, this is a quadratic equation in terms of ( n ). Let's solve for ( n ) using the quadratic formula. The quadratic formula states that for an equation ( ax² + bx + c = 0 ), the solutions are:( x = frac{-b pm sqrt{b² - 4ac}}{2a} )Here, ( a = 1 ), ( b = -6 ), and ( c = -7 ). Plugging these into the formula:( n = frac{-(-6) pm sqrt{(-6)² - 4*1*(-7)}}{2*1} )Simplify step by step:First, the numerator:- The first term is ( -(-6) = 6 )- The discriminant (the part under the square root) is ( (-6)² - 4*1*(-7) = 36 - (-28) = 36 + 28 = 64 )So the square root of 64 is 8.Therefore, the solutions are:( n = frac{6 pm 8}{2} )Calculating both possibilities:1. ( n = frac{6 + 8}{2} = frac{14}{2} = 7 )2. ( n = frac{6 - 8}{2} = frac{-2}{2} = -1 )But the problem states that the original number is a positive integer, so we discard the negative solution ( n = -1 ).Thus, the solution is ( n = 7 ).Wait a minute. Let me check if this works. Let's verify each step with ( n = 7 ).First, append 7 to the right: 7 becomes 77. Then subtract the square of the original number, which is 7² = 49. So 77 - 49 = 28.Then, reduce the remainder by 75%, which is 75% of 28 = 21. So 28 - 21 = 7. Then subtract the original number again: 7 - 7 = 0. That works out perfectly.Therefore, the original number is indeed 7.But wait, let me check if there might be another solution. The quadratic equation had two roots, but we discarded -1 because it's negative. But maybe there's a possibility of another positive solution? Let's check. The quadratic equation was ( n² - 6n - 7 = 0 ), which only has two roots, 7 and -1. Since -1 is not positive, 7 is the only solution.Alternatively, maybe I made a mistake in setting up the equation. Let me check the steps again.Original number: n.Append 7: 10n + 7.Subtract square: (10n + 7) - n².Reduce remainder by 75%: 25% of remainder is left, so 0.25*(10n +7 -n²).Then subtract n: 0.25*(10n +7 -n²) - n = 0.So 0.25*(10n +7 -n²) = n.Multiply both sides by 4: 10n +7 -n² = 4n.Bring all terms to one side: -n² +10n +7 -4n =0 → -n² +6n +7 =0 → n² -6n -7 =0.Yes, same quadratic equation. So steps are correct.Therefore, 7 is the correct answer.But let me check if there's another way this could have multiple solutions. For example, suppose the original number was a two-digit number. Let's test n=13. Let's see:Append 7: 137. Subtract n²=169: 137 - 169 = -32. Then reduce by 75%: -32*0.25 = -8. Then subtract n: -8 -13 = -21 ≠0. Not zero.Another example, n=5. Append 7:57. Subtract 25: 57-25=32. Reduce by 75%: 8. Subtract 5: 3 ≠0.n=8. Append 7:87. Subtract 64:23. Reduce by 75%:5.75. Subtract 8:5.75-8= -2.25≠0.n=7. As before, works. So 7 is the only solution.Alternatively, maybe I need to check if the original number is multi-digit? For example, suppose n=10. Append 7:107. Subtract 100:7. Reduce by 75%:1.75. Subtract 10: -8.25≠0. Doesn't work.Alternatively, n=14. Append 7:147. Subtract 196:147-196= -49. Reduce by75%: -49*0.25=-12.25. Subtract 14: -26.25≠0. Not zero.So no, seems like 7 is unique. Therefore, the answer is 7.Final AnswerThe number thought of is boxed{7}.</think>Let the original number be ( n ). Appending the digit 7 to the right of ( n ) gives the number ( 10n + 7 ). Subtracting the square of the original number from this new number results in:[(10n + 7) - n^2]Reducing this remainder by 75% leaves 25% of the remainder, which is:[0.25 times [(10n + 7 - n^2)]]Subtracting the original number ( n ) from this result gives zero:[0.25 times [(10n + 7 - n^2)] - n = 0]Multiplying through by 4 to clear the decimal:[10n + 7 - n^2 - 4n = 0]Simplifying:[-n^2 + 6n + 7 = 0]Rewriting in standard quadratic form:[n^2 - 6n - 7 = 0]Solving this quadratic equation using the quadratic formula ( n = frac{-b pm sqrt{b^2 - 4ac}}{2a} ) where ( a = 1 ), ( b = -6 ), and ( c = -7 ):[n = frac{6 pm sqrt{36 + 28}}{2} = frac{6 pm 8}{2}]This gives the solutions:[n = frac{14}{2} = 7 quad text{and} quad n = frac{-2}{2} = -1]Since ( n ) must be a positive integer, the solution is ( n = 7 ).Verifying with ( n = 7 ):- Appending 7 gives 77.- Subtracting ( 7^2 = 49 ) gives ( 77 - 49 = 28 ).- Reducing by 75% leaves ( 28 times 0.25 = 7 ).- Subtracting the original number gives ( 7 - 7 = 0 ).Thus, the number thought of is (boxed{7}).]

🔑:3.14 Let a number x be thought of. Then, following the text of the condition, we get the numbers10 x+7, 10 x+7-x^{2} and the remainder frac{25}{100}left(10 x+7-x^{2}right).Then frac{1}{4}left(10 x+7-x^{2}right)-x=0, or x^{2}-6 x-7=0. Only the value x=7 works.Answer: 7.

🧠:[<think>Okay, so I need to figure out whether the statement "x + y ≤ 1" is a sufficient, necessary, both, or neither condition for "x ≤ 1/2 and y ≤ 1/2". Let me start by understanding what each of these terms means in the context of logical conditions.First, a sufficient condition means that if the condition is met, then the statement must be true. But it might not be the only way for the statement to be true. On the other hand, a necessary condition means that for the statement to be true, the condition must be met, but meeting the condition doesn't guarantee the statement is true. So, if it's both sufficient and necessary, then they imply each other. If neither, then they don't have that relationship.Let me break down the problem. We have two real numbers, x and y. The first condition is x + y ≤ 1. The second condition is x ≤ 1/2 and y ≤ 1/2. The question is asking about the relationship between these two conditions. Specifically, is "x + y ≤ 1" a sufficient, necessary, both, or neither condition for "x ≤ 1/2 and y ≤ 1/2"?Let me consider each direction.First, suppose that "x ≤ 1/2 and y ≤ 1/2" is true. Then, adding these two inequalities, we get x + y ≤ 1/2 + 1/2 = 1. So, if both x and y are less than or equal to 1/2, then their sum is certainly less than or equal to 1. That means that the condition "x ≤ 1/2 and y ≤ 1/2" implies the condition "x + y ≤ 1". Therefore, "x + y ≤ 1" is a necessary condition for "x and y being ≤ 1/2". Because if "x and y ≤ 1/2" is true, then "x + y ≤ 1" must also be true. So, "x + y ≤ 1" is necessary for "x ≤ 1/2 and y ≤ 1/2".But is it sufficient? That is, does "x + y ≤ 1" guarantee that both x and y are ≤ 1/2? Let me test this with a counterexample. Suppose x = 0.6 and y = 0.4. Then, x + y = 1.0, which is ≤ 1. However, x = 0.6 is greater than 0.5. So here, even though x + y ≤ 1 is true, x ≤ 1/2 is false. Therefore, "x + y ≤ 1" does not guarantee that both x and y are ≤ 1/2. Therefore, "x + y ≤ 1" is not a sufficient condition for "x ≤ 1/2 and y ≤ 1/2".Therefore, putting this together: "x + y ≤ 1" is a necessary condition because the conjunction "x ≤ 1/2 and y ≤ 1/2" implies "x + y ≤ 1", but it's not sufficient because the converse isn't true. Therefore, the correct answer should be B: Necessary and insufficient condition.Wait, let me double-check. Let me ensure I didn't mix up the direction. The question says: "the statement 'x + y ≤ 1' is a ( ) condition for 'x ≤ 1/2 and y ≤ 1/2'". So, the structure is: [A] is a ( ) condition for [B]. So, the options are about A's relation to B. So, in logic terms, A is a condition for B. So, the question is: Is A a sufficient condition for B? (If A holds, then B holds). Or is A a necessary condition for B? (If B holds, then A holds). Or both, or neither.Wait, actually, when we say "A is a sufficient condition for B", it means A implies B. So, if you have A, then you have B. If "x + y ≤1" is a sufficient condition for "x ≤1/2 and y ≤1/2", that would mean that whenever x + y ≤1, then x and y are both ≤1/2. But as my counterexample showed, x + y can be 1 even if one of them is 0.6. So, A does not imply B. Therefore, it's not sufficient.On the other hand, if "x + y ≤1" is a necessary condition for "x ≤1/2 and y ≤1/2", that would mean that whenever "x ≤1/2 and y ≤1/2" is true, then "x + y ≤1" must be true. Which is the case, as we saw. So, B implies A. Therefore, A is a necessary condition for B.Therefore, since A is necessary but not sufficient for B, the correct answer is B: Necessary and insufficient condition.Yes, that seems right. Let me just think of another example to confirm. Suppose x = 0.5 and y = 0.5. Then, x + y = 1, which satisfies A. Here, both x and y are ≤0.5. Now, if x and y are both less than 0.5, say 0.4 and 0.4, then x + y = 0.8 ≤1. So, in that case, B is true, and A is true. But if I take x = 0.3 and y = 0.8, then x + y = 1.1, which does not satisfy A. So, in this case, B is not true (since y is not ≤0.5), and A is also not true. So, if B is false, A can be either true or false? Wait, no. Wait, B being false means that either x >0.5 or y >0.5. Then, in that case, A could still be true if, for example, x=0.6 and y=0.4, as before. So, even if B is false (because x is 0.6), A can still be true (0.6 +0.4=1). Therefore, the necessary condition is that B implies A, but not the other way around.Therefore, yes, A is necessary for B. So the answer is B. I think that's correct.Final Answerboxed{B}</think>To determine whether the statement (x + y leq 1) is a sufficient, necessary, both, or neither condition for (x leq frac{1}{2}) and (y leq frac{1}{2}), we analyze the logical relationship between these conditions.1. Necessary Condition: - If (x leq frac{1}{2}) and (y leq frac{1}{2}), then adding these inequalities gives (x + y leq frac{1}{2} + frac{1}{2} = 1). Therefore, (x + y leq 1) must hold if both (x) and (y) are at most (frac{1}{2}). This means (x + y leq 1) is a necessary condition for (x leq frac{1}{2}) and (y leq frac{1}{2}).2. Sufficient Condition: - To check if (x + y leq 1) is sufficient, we need to see if it guarantees both (x leq frac{1}{2}) and (y leq frac{1}{2}). Consider a counterexample: (x = 0.6) and (y = 0.4). Here, (x + y = 1 leq 1), but (x = 0.6) is not less than or equal to (frac{1}{2}). Thus, (x + y leq 1) does not imply both (x) and (y) are at most (frac{1}{2}), so it is not a sufficient condition.Therefore, (x + y leq 1) is a necessary but insufficient condition for (x leq frac{1}{2}) and (y leq frac{1}{2}).[boxed{B}],<think>Okay, so I have this problem here about conditions related to inequalities. Let me try to wrap my head around it step by step. The question is asking whether the statement "x + y ≤ 1" is a sufficient, necessary, both, or neither condition for "x ≤ 1/2 and y ≤ 1/2" where x and y are real numbers. The options are A, B, C, or D.First, let me recall what sufficient and necessary conditions mean. A sufficient condition means that if the condition is met, then the statement must be true. But it doesn't necessarily mean it's the only way the statement can be true. On the other hand, a necessary condition means that for the statement to be true, the condition must hold, but the condition alone might not be enough to guarantee the statement. If something is both sufficient and necessary, then they imply each other. If it's neither, then it doesn't work in either direction.So, the main statement here is "x + y ≤ 1" and the other statement is "x ≤ 1/2 and y ≤ 1/2". The question is asking about the relationship between these two.Let me break it down. Let's call the first condition P: x + y ≤ 1, and the second condition Q: x ≤ 1/2 and y ≤ 1/2. The problem is asking whether P is a sufficient, necessary, both, or neither condition for Q.Wait, hold on. Wait, actually, the phrasing is a bit confusing. Let me check again. It says: "the statement 'x + y ≤ 1' is a ( ) condition for 'x ≤ 1/2 and y ≤ 1/2'". So, actually, they're saying is P (x + y ≤1) a condition for Q (x ≤1/2 and y ≤1/2). So, in logical terms, does P imply Q (sufficient), does Q imply P (necessary), both, or neither?Wait, terminology here: If P is a sufficient condition for Q, that means P implies Q. If P is a necessary condition for Q, that means Q implies P. So, we have to see if P ⇒ Q, or Q ⇒ P, or both, or neither.So, let's analyze both directions.First, does P ⇒ Q? That is, if x + y ≤1, does that necessarily mean that x ≤1/2 and y ≤1/2?Hmm, let's see. Suppose x + y is less than or equal to 1. Does that force both x and y to be at most 1/2?Wait, let me think of a counterexample. Suppose x is 0.6 and y is 0.4. Then x + y = 1.0, which is ≤1. But x is 0.6, which is greater than 0.5. So, in this case, x + y ≤1 is true, but x ≤1/2 is false. Therefore, P does not imply Q. Therefore, P is not a sufficient condition for Q.Okay, so that rules out options A and C, which claim that it's sufficient.Now, check if P is a necessary condition for Q. That is, does Q imply P? In other words, if x ≤1/2 and y ≤1/2, then does x + y ≤1?Yes, because if both x and y are at most 0.5, then their sum is at most 0.5 + 0.5 = 1. Therefore, Q ⇒ P. So, P is a necessary condition for Q. But since P is not sufficient (as we saw in the previous counterexample), then P is a necessary and insufficient condition. So, that would be option B.Wait, let me confirm. So, necessary condition means that Q cannot be true without P being true. Since Q implies P, whenever Q is true, P must be true. Therefore, P is necessary for Q. However, just because P is true doesn't mean Q is true, as shown by the counterexample. Therefore, P is necessary but not sufficient, so the answer is B.I think that's the right reasoning.Final Answerboxed{B}</think>Given the problem, we need to determine whether the statement ( x + y leq 1 ) is a sufficient, necessary, both, or neither condition for ( x leq frac{1}{2} ) and ( y leq frac{1}{2} ).1. Sufficient Condition Check: We need to check if ( x + y leq 1 ) implies ( x leq frac{1}{2} ) and ( y leq frac{1}{2} ). A counterexample is provided where ( x = 0.6 ) and ( y = 0.4 ). Here, ( x + y = 1 leq 1 ), but ( x leq frac{1}{2} ) is false. Therefore, ( x + y leq 1 ) is not a sufficient condition.2. Necessary Condition Check: We need to check if ( x leq frac{1}{2} ) and ( y leq frac{1}{2} ) imply ( x + y leq 1 ). If both ( x ) and ( y ) are at most ( frac{1}{2} ), their sum is at most ( frac{1}{2} + frac{1}{2} = 1 ). Therefore, ( x leq frac{1}{2} ) and ( y leq frac{1}{2} ) imply ( x + y leq 1 ), making ( x + y leq 1 ) a necessary condition.Since ( x + y leq 1 ) is a necessary but not sufficient condition for ( x leq frac{1}{2} ) and ( y leq frac{1}{2} ), the answer is:[boxed{B}]]

🔑:If "x leq frac{1}{2} and y leq frac{1}{2}", then x + y leq frac{1}{2} + frac{1}{2} = 1 holds true, which means the necessity is valid.However, when x = 1 and y = 0, the condition x + y leq 1 is satisfied, but "x leq frac{1}{2} and y leq frac{1}{2}" does not hold, which means the sufficiency is not valid.Therefore, "x + y leq 1" is a necessary but not sufficient condition for "x leq frac{1}{2} and y leq frac{1}{2}".Hence, the answer is boxed{text{B}}.This problem primarily tests the ability to judge necessary and sufficient conditions, and understanding the relationship between inequalities is the key to solving this problem.

🧠:[<think>Okay, so I need to find the number of 7-digit integers using the digits 0,1,2,3,4, where none of the blocks 22, 33, or 44 appear. Hmm, let's start by understanding the problem properly.First, the integers are 7-digit numbers, which means the first digit can't be 0. So even though we can use 0, it can't be in the first position. The digits allowed are 0,1,2,3,4, so that's five digits in total. But we have a restriction: the number shouldn't contain 22, 33, or 44 anywhere in it. So, consecutive same digits except for 0 and 1 are not allowed. Wait, actually, the forbidden blocks are 22, 33, 44. So, 00, 11 are allowed? Let me check the problem statement again. It says "none of the three blocks 22, 33 and 44". So only those three specific consecutive pairs are forbidden. So 00, 11, 222, etc., would be okay? Wait, the problem says "blocks", so maybe any occurrence of two 2s, two 3s, or two 4s in a row is forbidden. So, for example, 22 is forbidden, but 222 would contain two 22 blocks, right? So even 222 would be invalid because it has 22 twice. Similarly, 33 and 44 cannot be consecutive.So the problem is to count 7-digit numbers using digits 0-4, first digit not 0, and with no instances of 22, 33, or 44. Got it.This seems like a problem that can be solved with recurrence relations. Let me think about how to model this.Since the forbidden blocks are two consecutive identical digits for 2,3,4, the key is that after using a 2,3, or 4, the next digit must be different. However, 0 and 1 can be followed by any digit, including themselves. Wait, no, 0 and 1 can be followed by themselves? Because the forbidden blocks are only 22,33,44. So 00, 11 are allowed. So if I have a 0 or 1, the next digit can be 0,1,2,3,4, but if I have a 2,3, or 4, the next digit can't be the same as the previous one. So, for example, if the previous digit is 2, the next can be 0,1,3,4; if the previous is 3, next can be 0,1,2,4; if previous is 4, next can be 0,1,2,3; and if previous is 0 or 1, next can be anything. So maybe we can model this with states.Let me formalize this. Let's consider the problem as building the number digit by digit, keeping track of the previous digit. So the state would be the last digit used. Since the first digit can't be 0, the first digit has four possibilities: 1,2,3,4. Then, for each subsequent digit, depending on the previous digit, the number of choices varies.So let's define a recurrence where for each position i (from 1 to 7), and for each possible previous digit d (0,1,2,3,4), we keep track of the number of numbers of length i ending with digit d. However, since the first digit cannot be 0, we have to adjust the initial conditions accordingly.Let me formalize this. Let's denote:- Let a_i be the number of valid i-digit numbers ending with 0 or 1.- Let b_i be the number of valid i-digit numbers ending with 2.- Let c_i be the number of valid i-digit numbers ending with 3.- Let d_i be the number of valid i-digit numbers ending with 4.Wait, actually, grouping 0 and 1 together might simplify things since after 0 or 1, the next digit can be any of 0,1,2,3,4. Whereas after 2,3,4, the next digit can't be the same. So maybe:For each position i >= 1:- If the previous digit is 0 or 1 (so part of a_i), then the next digit can be 0,1,2,3,4. So the number of possibilities is 5 for each a_i.- If the previous digit is 2 (b_i), then the next digit can be 0,1,3,4 (4 choices).- Similarly, if the previous digit is 3 (c_i), next digit can be 0,1,2,4 (4 choices).- If the previous digit is 4 (d_i), next digit can be 0,1,2,3 (4 choices).So the total number of i-digit numbers would be a_i + b_i + c_i + d_i.But we need to set up the recurrence relations properly. Let's see.For i = 1 (the first digit):- Since it's a 1-digit number, and the first digit can't be 0. So:a_1: numbers ending with 0 or 1. But 0 is invalid for the first digit. So only numbers ending with 1. So a_1 = 1 (just the number 1).b_1: numbers ending with 2. So 1 possibility: 2. So b_1 = 1.c_1: numbers ending with 3: 1 possibility: 3. So c_1 = 1.d_1: numbers ending with 4: 1 possibility: 4. So d_1 = 1.Therefore, total 1-digit numbers: 1 + 1 + 1 + 1 = 4, which is correct (digits 1,2,3,4).For i = 2:a_2: numbers ending with 0 or 1. To get to 0 or 1 at position 2, the previous digit could have been anything (since after any digit you can put 0 or 1). However, since we are building valid numbers, we have to consider that if the previous digit was 0 or 1, we can still put 0 or 1. Wait, actually, the previous digit could be any digit (0,1,2,3,4), but for the second digit, if we want to end with 0 or 1, the previous digit could have been anything, but the previous digit's restrictions are already handled in the previous step. Hmm, maybe I need to think of a_i as the count of numbers of length i ending with 0 or 1, so:a_i = (number of numbers of length i-1 ending with any digit) * 2 (choices: 0 or 1). Wait, no, because if the previous digit was 0 or 1, you can still add 0 or 1. But actually, the transitions are:To end with 0 or 1 at position i, you can come from any number of length i-1 (regardless of what it ended with) and append 0 or 1. However, since the first digit can't be 0, but starting from the second digit, 0 is allowed. So for i >= 2:a_i = (total number of (i-1)-digit numbers) * 2.Similarly, b_i (ending with 2): to end with 2, the previous digit must not be 2. So you can come from numbers ending with 0,1,3,4. Similarly, c_i (ending with 3) comes from numbers not ending with 3, and d_i (ending with 4) comes from numbers not ending with 4.Wait, this might be a better way:At each step, the next digit depends on the previous digit:- If the previous digit was 0 or 1 (a_{i-1}), then you can choose 0,1,2,3,4. So from a_{i-1}, you can go to a_i (by choosing 0 or 1) with 2 options, or to b_i, c_i, d_i with 1 option each (2,3,4).Wait, perhaps breaking down:Let me redefine the states more precisely. Maybe instead of grouping 0 and 1, since they can be followed by any digit, and 2,3,4 have restrictions. Wait, perhaps it's better to have separate states for each previous digit. But that might complicate things since there are 5 digits. Alternatively, group into states based on whether the last digit is 0/1 or 2,3,4. Wait, let me think.Let me try defining two states:- State S: the last digit is either 0 or 1. From here, the next digit can be anything (0,1,2,3,4).- State T: the last digit is 2,3, or 4. From here, the next digit can be anything except the same as the previous digit. So if the last digit was 2, next can be 0,1,3,4; similarly for 3 and 4.But perhaps even more precise: Since after 2,3,4, you can't have the same digit, but the next digits can be 0,1, or the other two digits. Wait, maybe to make this manageable, we need to track the specific last digit. Because, for example, if the last digit is 2, the next can't be 2, but if the last digit is 3, the next can't be 3, etc. So the number of possibilities after a 2 is 4 (0,1,3,4), same as after 3 or 4. So maybe group them all together.Wait, here's another approach. Let's use separate recurrence relations for each possible last digit. Let me denote:For each position i (from 1 to 7):- Let f_i(d) be the number of i-digit numbers ending with digit d, where d is in {0,1,2,3,4}.But since the first digit can't be 0, we have to initialize accordingly.For i=1:f_1(0) = 0 (since first digit can't be 0)f_1(1) = 1f_1(2) = 1f_1(3) = 1f_1(4) = 1Total for i=1: 1+1+1+1 = 4.For i >= 2:To compute f_i(d), we need to consider all possible previous digits that can transition to d.If d is 0 or 1:Any previous digit can transition to d, except that if the previous digit was the same as d, but wait, no, the forbidden blocks are only 22,33,44. So even 00 or 11 is allowed. Therefore, for d in {0,1}, f_i(d) = sum over all previous digits (including d itself) of f_{i-1}(prev_digit). But wait, no. Wait, actually, since we can append d to any number of length i-1, regardless of what the previous digit was, because 00 and 11 are allowed. Wait, but if the previous digit was 2,3,4, we can still append 0 or 1. The forbidden blocks are only 22,33,44, so appending 0 or 1 after any digit is allowed.Therefore, for d in {0,1}:f_i(d) = (sum over all f_{i-1}(prev_d) for prev_d in {0,1,2,3,4}) * 1 (since we choose d specifically). Wait, no, more precisely:If we want to append a 0, then f_i(0) = sum of all f_{i-1}(prev_d) for prev_d in {0,1,2,3,4}, because we can append 0 to any number ending with any digit. Similarly, f_i(1) = sum of all f_{i-1}(prev_d). But this seems like f_i(0) + f_i(1) = 2 * sum of all f_{i-1}(prev_d). But since 0 and 1 are separate, if we consider each, f_i(0) = sum of f_{i-1}(prev_d), same for f_i(1). Wait, that would imply f_i(0) = f_i(1) = sum of f_{i-1}(prev_d). But that would mean if we have 5 choices for digits, but we are only appending 0 or 1. Wait, maybe this is incorrect.Wait, actually, if we are allowed to append 0 or 1 after any digit, then for each number of length i-1, we can create two numbers of length i: one ending with 0 and one ending with 1. But no, that's not correct. For each number of length i-1, you can append either 0 or 1, but you have to count them separately. So f_i(0) is equal to the total number of numbers of length i-1, since each can be extended by 0. Similarly, f_i(1) is equal to the total number of numbers of length i-1, since each can be extended by 1. Therefore:f_i(0) = total_{i-1}f_i(1) = total_{i-1}Where total_{i-1} = f_{i-1}(0) + f_{i-1}(1) + f_{i-1}(2) + f_{i-1}(3) + f_{i-1}(4)Similarly, for d in {2,3,4}:To append d, the previous digit cannot be d. So f_i(d) = total_{i-1} - f_{i-1}(d). Because total_{i-1} is all possible numbers, and subtracting those that end with d (since we can't have d after d). Therefore:f_i(2) = total_{i-1} - f_{i-1}(2)f_i(3) = total_{i-1} - f_{i-1}(3)f_i(4) = total_{i-1} - f_{i-1}(4)This seems like a good approach. Let's verify this with i=2.For i=2:total_1 = f_1(0) + f_1(1) + f_1(2) + f_1(3) + f_1(4) = 0 + 1 + 1 + 1 + 1 = 4Then:f_2(0) = total_1 = 4f_2(1) = total_1 = 4f_2(2) = total_1 - f_1(2) = 4 - 1 = 3f_2(3) = total_1 - f_1(3) = 4 - 1 = 3f_2(4) = total_1 - f_1(4) = 4 - 1 = 3Total_2 = 4 + 4 + 3 + 3 + 3 = 17Wait, let's check manually. For 2-digit numbers, first digit can't be 0. So first digit is 1,2,3,4. Second digit can be 0,1,2,3,4, but if the first digit is 2, the second can't be 2; if first digit is 3, second can't be 3; same for 4. Let's compute:First digit 1: second digit can be 0,1,2,3,4 → 5 numbers: 10,11,12,13,14First digit 2: second digit can be 0,1,3,4 → 4 numbers: 20,21,23,24First digit 3: second digit can be 0,1,2,4 → 4 numbers: 30,31,32,34First digit 4: second digit can be 0,1,2,3 → 4 numbers: 40,41,42,43Total 2-digit numbers: 5 + 4 + 4 + 4 = 17, which matches the recurrence's total_2 = 17. Good.Similarly, let's check f_2(0) = 4. The numbers ending with 0: 10,20,30,40. That's 4, which matches. f_2(1) = 4: 11,21,31,41. f_2(2) = 3: 12,32,42. f_2(3)=3:13,23,43. f_2(4)=3:14,24,34. Correct.So this recurrence seems to work. Therefore, we can proceed.So for each i from 1 to 7, we can compute f_i(0), f_i(1), f_i(2), f_i(3), f_i(4) based on the previous values.Given that, let's compute up to i=7.We already have i=1:f1: [0,1,1,1,1]total1 = 4i=2:f2(0) = total1 = 4f2(1) = total1 = 4f2(2) = total1 - f1(2) = 4 -1=3f2(3) = total1 -f1(3)=4-1=3f2(4)= total1 -f1(4)=4-1=3total2=4+4+3+3+3=17i=3:f3(0) = total2=17f3(1) = total2=17f3(2) = total2 - f2(2)=17-3=14f3(3)= total2 - f2(3)=17-3=14f3(4)= total2 - f2(4)=17-3=14total3=17+17+14+14+14= 17*2 +14*3=34 +42=76Check: total3=76i=4:f4(0) = total3=76f4(1) = total3=76f4(2)= total3 - f3(2)=76 -14=62f4(3)=76 -14=62f4(4)=76 -14=62total4=76+76+62+62+62=76*2 +62*3=152 +186=338i=5:f5(0)=total4=338f5(1)=338f5(2)=338 -62=276f5(3)=338 -62=276f5(4)=338 -62=276total5=338+338+276+276+276=338*2 +276*3=676 +828=1504i=6:f6(0)=total5=1504f6(1)=1504f6(2)=1504 -276=1228f6(3)=1504 -276=1228f6(4)=1504 -276=1228total6=1504+1504+1228+1228+1228=1504*2 +1228*3=3008 +3684=6692i=7:f7(0)=total6=6692f7(1)=6692f7(2)=6692 -1228=5464f7(3)=6692 -1228=5464f7(4)=6692 -1228=5464total7=6692+6692+5464+5464+5464=6692*2 +5464*3=13384 +16392=29776Wait, so according to this, total7=29776. But we need to be careful here. Because total7 includes all numbers of length 7, including those that might start with 0. Wait, no! Wait, the way we set up the recurrence, for i=1, f1(0)=0, which correctly accounts for the first digit not being 0. Then, for i>=2, when we compute f_i(0), it's allowed because starting from the second digit, 0 is permitted. Therefore, total7 counts all 7-digit numbers (with first digit not 0) that don't contain 22,33,44. Therefore, the answer should be total7=29776.But let me verify this with smaller i. For i=2, total2=17, which matches manual calculation. For i=3, total3=76. Let's verify with another method.Alternative approach: For 3-digit numbers, first digit can be 1,2,3,4. Let's consider each case.First digit 1:Second digit can be 0,1,2,3,4 (since after 1, any digit is allowed). Third digit:If second digit is 0 or 1: third digit can be anything (5 choices).If second digit is 2: third digit can't be 2 (4 choices).If second digit is 3: third digit can't be 3 (4 choices).If second digit is 4: third digit can't be 4 (4 choices).So number of numbers starting with 1:First digit 1. Second digit: 0,1,2,3,4. For each second digit:If second digit is 0: third digit can be 0,1,2,3,4 (5)Similarly for second digit 1: 5.Second digit 2: 4.Second digit 3: 4.Second digit 4: 4.Total for first digit 1: 5 +5 +4 +4 +4 = 22First digit 2:Second digit can be 0,1,3,4 (since after 2, can't be 2).For each second digit:If second digit is 0: third digit can be anything (5)If second digit is 1: third digit can be anything (5)If second digit is 3: third digit can't be 3 (4)If second digit is 4: third digit can't be 4 (4)So total for first digit 2: 5 +5 +4 +4=18Similarly, first digit 3:Second digit can be 0,1,2,4.Each of these:Second digit 0: 5Second digit 1:5Second digit 2:4Second digit 4:4Total:5+5+4+4=18Same as first digit 2.First digit 4:Same as first digit 2 and 3.Second digits:0,1,2,3Third digits:0:5,1:5,2:4,3:4Total:5+5+4+4=18Total 3-digit numbers:22 (first digit 1) +18+18+18=22 +54=76. Which matches total3=76. Good.Therefore, the recurrence seems to hold. Therefore, proceeding to i=7, total7=29776. Therefore, the answer is 29776.But wait, the problem says "formed by some or all of the five digits", but wait, the problem is about 7-digit integers. "formed by some or all of the five digits" probably means that each digit in the number must be among 0,1,2,3,4 (i.e., digits can be repeated, as long as they are in the set {0,1,2,3,4}), which is already considered in our approach. So our calculation is correct.Thus, the final answer is 29776, so boxed, boxed{29776}Final Answerboxed{29776}</think>To find the number of 7-digit integers formed by the digits 0, 1, 2, 3, and 4, such that these integers do not contain the blocks 22, 33, or 44, we use a recurrence relation approach. We define ( f_i(d) ) as the number of ( i )-digit numbers ending with digit ( d ) (where ( d ) is 0, 1, 2, 3, or 4) and ensure the first digit is not 0. The recurrence relations are derived based on the constraints that after using 2, 3, or 4, the next digit cannot be the same. Key Steps:1. Initialization for ( i = 1 ): - ( f_1(0) = 0 ) (since the first digit cannot be 0) - ( f_1(1) = 1 ) - ( f_1(2) = 1 ) - ( f_1(3) = 1 ) - ( f_1(4) = 1 ) - Total for ( i = 1 ): ( 0 + 1 + 1 + 1 + 1 = 4 )2. Recurrence Relations for ( i geq 2 ): - ( f_i(0) = text{total}_{i-1} ) - ( f_i(1) = text{total}_{i-1} ) - ( f_i(2) = text{total}_{i-1} - f_{i-1}(2) ) - ( f_i(3) = text{total}_{i-1} - f_{i-1}(3) ) - ( f_i(4) = text{total}_{i-1} - f_{i-1}(4) ) - ( text{total}_i = f_i(0) + f_i(1) + f_i(2) + f_i(3) + f_i(4) )3. Calculations for ( i = 2 ) to ( i = 7 ): - For ( i = 2 ): - ( f_2(0) = 4 ) - ( f_2(1) = 4 ) - ( f_2(2) = 3 ) - ( f_2(3) = 3 ) - ( f_2(4) = 3 ) - Total: ( 4 + 4 + 3 + 3 + 3 = 17 ) - For ( i = 3 ): - ( f_3(0) = 17 ) - ( f_3(1) = 17 ) - ( f_3(2) = 14 ) - ( f_3(3) = 14 ) - ( f_3(4) = 14 ) - Total: ( 17 + 17 + 14 + 14 + 14 = 76 ) - For ( i = 4 ): - ( f_4(0) = 76 ) - ( f_4(1) = 76 ) - ( f_4(2) = 62 ) - ( f_4(3) = 62 ) - ( f_4(4) = 62 ) - Total: ( 76 + 76 + 62 + 62 + 62 = 338 ) - For ( i = 5 ): - ( f_5(0) = 338 ) - ( f_5(1) = 338 ) - ( f_5(2) = 276 ) - ( f_5(3) = 276 ) - ( f_5(4) = 276 ) - Total: ( 338 + 338 + 276 + 276 + 276 = 1504 ) - For ( i = 6 ): - ( f_6(0) = 1504 ) - ( f_6(1) = 1504 ) - ( f_6(2) = 1228 ) - ( f_6(3) = 1228 ) - ( f_6(4) = 1228 ) - Total: ( 1504 + 1504 + 1228 + 1228 + 1228 = 6692 ) - For ( i = 7 ): - ( f_7(0) = 6692 ) - ( f_7(1) = 6692 ) - ( f_7(2) = 5464 ) - ( f_7(3) = 5464 ) - ( f_7(4) = 5464 ) - Total: ( 6692 + 6692 + 5464 + 5464 + 5464 = 29776 )Thus, the number of valid 7-digit integers is (boxed{29776}).,<think>Okay, so I need to find the number of 7-digit integers using the digits 0,1,2,3,4, but they can't contain the blocks 22, 33, or 44. Hmm, let me think. First, since it's a 7-digit integer, the first digit can't be 0. So that's something to remember. Also, the problem says "formed by some or all of the five digits," which I think means that each digit can be used multiple times, right? Because if we had to use each digit exactly once, we couldn't form a 7-digit number. So it's with repetition allowed, but avoiding the specific blocks 22, 33, and 44.So, this seems like a problem where I can use recursion or maybe dynamic programming. Let me recall how these kinds of problems are approached. When you need to count numbers without certain forbidden substrings, a common technique is to build up the number digit by digit, keeping track of the last digit to ensure that adding a new digit doesn't create one of the forbidden blocks.So, maybe I can model this with states. Each state represents the last digit used. Since the forbidden blocks are two of the same digits in a row (except 11 is allowed, since it's not forbidden), the states would need to track the previous digit to ensure that we don't place a 2 after a 2, 3 after a 3, or 4 after a 4. However, 0,1 can be followed by any digit, including themselves, right? Wait, the forbidden blocks are only 22, 33, 44. So, 00, 11 are allowed? The problem statement says "none of the three blocks 22,33 and 44". So yes, 00 and 11 are allowed. So, only consecutive 2s, 3s, or 4s are forbidden. So, the constraints are: after a 2, you can't place another 2; after a 3, you can't place another 3; after a 4, you can't place another 4. But after 0 or 1, you can place any digit, including another 0 or 1. And after 2, you can place 0,1,3,4; same with 3 and 4.Also, the first digit can't be 0, so we have to adjust for that. So, starting the number, the first digit has 4 choices (1,2,3,4). Then, each subsequent digit depends on the previous one.This seems like a standard recurrence problem. Let me define f(n, d) as the number of n-digit numbers ending with digit d, following the rules. Then, the total number of n-digit numbers would be the sum over d of f(n, d). But since the first digit can't be 0, we need to handle the first digit separately.Wait, maybe it's better to separate the first digit and then handle the rest. Let me structure it step by step.First digit: 1,2,3,4 (4 options). Then, for each subsequent digit (positions 2 to 7), we have to make sure that if the previous digit was 2, the current digit isn't 2; if previous was 3, current isn't 3; if previous was 4, current isn't 4. But if the previous digit was 0 or 1, then current can be any of 0,1,2,3,4. So, the transitions depend on the previous digit.Therefore, for each position from 2 to 7, the number of possibilities depends on the previous digit. So, this is a classic case for using a recurrence relation with states representing the last digit.Let me formalize this.Let’s denote:- For each position i from 1 to 7, and for each digit d in {0,1,2,3,4}, let’s define a function f(i, d) which is the number of i-digit numbers ending with digit d, following the rules.Then, the total number of 7-digit numbers is the sum over d in {1,2,3,4} of f(7, d), since the first digit cannot be 0.Wait, no. Wait, for i=1 (the first digit), f(1, 0) = 0 (since numbers can't start with 0), and f(1, d) = 1 for d=1,2,3,4. Then, for i >= 2, f(i, d) is computed based on the previous digits.But actually, for each i, the count for ending with d is the sum over all possible previous digits that can transition to d. So, for transitions:If the previous digit was 2, then the current digit cannot be 2. Similarly for 3 and 4. But if the previous digit was 0 or 1, then current can be anything.So, more formally, for i >= 2:- If d is 0, then the previous digit can be anything (0,1,2,3,4). So f(i, 0) = sum_{k=0}^4 f(i-1, k)- If d is 1, similarly, f(i, 1) = sum_{k=0}^4 f(i-1, k)- If d is 2, then the previous digit cannot be 2. So f(i, 2) = sum_{k in {0,1,3,4}} f(i-1, k)- Similarly, f(i, 3) = sum_{k in {0,1,2,4}} f(i-1, k)- f(i, 4) = sum_{k in {0,1,2,3}} f(i-1, k)But note that for the first digit, we have constraints. So, for i=1:- f(1, 0) = 0 (can't start with 0)- f(1, 1) = 1- f(1, 2) = 1- f(1, 3) = 1- f(1, 4) = 1Then, for each subsequent i from 2 to 7, we compute f(i, d) as above.But to compute this, maybe I can represent this as a matrix and perform dynamic programming.Alternatively, maybe find a pattern or a recursion formula.Alternatively, since the transitions are similar for digits 2,3,4 and different for 0,1, perhaps group them.Wait, let's see. Let me consider the total number of valid numbers of length i. Let’s denote T(i) = sum_{d=0}^4 f(i, d). However, for i=1, T(1) = 4 (digits 1,2,3,4). Then, for i >=2, T(i) would be sum_{d=0}^4 f(i, d). But perhaps we can relate T(i) to T(i-1). But maybe not directly, since the transitions have different constraints.Wait, but for digits 0 and 1, the next digit can be anything, but for digits 2,3,4, the next digit cannot be themselves. Hmm. Let's think about it.Alternatively, let's split the previous digits into two categories:- Category A: digits that can be followed by any digit (digits 0 and 1)- Category B: digits that cannot be followed by themselves (digits 2,3,4)So, for each i, let’s denote:- A(i) = number of i-digit numbers ending with 0 or 1- B(i) = number of i-digit numbers ending with 2,3, or 4Then, total T(i) = A(i) + B(i)But we need to find a recursion relation between A(i) and B(i). Let's see.For i=1:- A(1) = f(1,0) + f(1,1) = 0 + 1 = 1- B(1) = f(1,2) + f(1,3) + f(1,4) = 1 + 1 + 1 = 3- T(1) = 4For i >=2:- To compute A(i): the numbers ending with 0 or 1. These can be preceded by any digit. So, for each number ending with any digit (A(i-1) + B(i-1)), we can append 0 or 1. But wait, the previous digit could be anything, but the next digit is 0 or 1. So, A(i) = 2 * T(i-1)Wait, but if the previous digit is in A or B, you can append 0 or 1 regardless. So, yes, since appending 0 or 1 is allowed after any digit. So, A(i) = 2 * T(i-1)Similarly, for B(i): numbers ending with 2,3,4. However, after a digit in B (i.e., 2,3,4), we can't append the same digit. So, if the previous digit was a 2, we can't append 2, but we can append 3,4,0,1. Wait, but B(i) counts numbers ending with 2,3,4. So, to get B(i), we need to consider transitions from previous digits.Wait, maybe it's better to split B into individual digits. Wait, perhaps this approach is not straightforward. Let me try again.Alternatively, perhaps we can model A(i) and B(i) with the following recurrence:- To end with a digit in A (0 or 1), you can come from any digit (A or B) and append 0 or 1. So, A(i) = 2 * T(i-1)- To end with a digit in B (2,3,4), you can come from: - If the previous digit was in A (0 or 1), then you can append 2,3,4 (3 choices) - If the previous digit was in B (2,3,4), then you can append any of B except the same as previous. Wait, but since we don't track which specific digit in B it was, this complicates things. So, if we just have B(i-1), which is the count of numbers ending with 2,3,4, but each of these has different restrictions. For example, if the previous digit was 2, you can append 3,4,0,1 (but 0,1 are in A, so appending those would go to A). Wait, this seems messy. Maybe splitting B into individual digits would be better.Alternatively, maybe model each ending digit separately. Let's go back to the original definition with f(i, d).So, for each position i and digit d:- f(1, 0) = 0- f(1, 1) = 1- f(1, 2) = 1- f(1, 3) = 1- f(1, 4) = 1Then, for i >=2:- f(i, 0) = sum_{k=0}^4 f(i-1, k)- f(i, 1) = sum_{k=0}^4 f(i-1, k)- f(i, 2) = sum_{k=0}^4 f(i-1, k) - f(i-1, 2) [since you can't have 2 after 2]Similarly,- f(i, 3) = sum_{k=0}^4 f(i-1, k) - f(i-1, 3)- f(i, 4) = sum_{k=0}^4 f(i-1, k) - f(i-1, 4)Wait, that's an interesting way to think about it. For digits 2,3,4, the number of ways to end with them is equal to the total number of ways to have any digit before, minus the number of ways that ended with the same digit (to exclude the forbidden transition).But sum_{k=0}^4 f(i-1, k) is just T(i-1), the total number of (i-1)-digit numbers. So, f(i, 2) = T(i-1) - f(i-1, 2)Similarly for 3 and 4. For 0 and 1, it's T(i-1). But we also need to remember that for i >=2, the first digit was already handled (non-zero), but from the second digit onwards, 0 is allowed.Wait, actually, in this model, for i >=2, the numbers can have 0 in the second to seventh positions, but the first digit cannot be 0. So, perhaps the way we set up the recursion is correct.But let's test it with small cases.For i=1:f(1,0)=0, f(1,1)=1, f(1,2)=1, f(1,3)=1, f(1,4)=1. Total T(1)=4.For i=2:f(2,0) = T(1) =4f(2,1) = T(1) =4f(2,2) = T(1) - f(1,2) = 4 -1=3f(2,3) = T(1) - f(1,3)=4 -1=3f(2,4) = T(1) - f(1,4)=4 -1=3So, T(2) =4+4+3+3+3=17But let's check manually. For 2-digit numbers, first digit is 1,2,3,4. Second digit can be anything except if the first digit was 2,3,4, in which case second digit can't be the same. Let's compute:If first digit is 1: second digit can be 0,1,2,3,4 (5 options)If first digit is 2: second digit can be 0,1,3,4 (4 options)Similarly for first digits 3 and 4: each have 4 options.So total 5 + 4 +4 +4 =17. Which matches T(2)=17. Good.Similarly, for i=3:f(3,0)=T(2)=17f(3,1)=17f(3,2)=17 - f(2,2)=17 -3=14f(3,3)=17 -3=14f(3,4)=17 -3=14Thus, T(3)=17+17+14+14+14=76Let me verify manually for i=3. It might be tedious, but let's try.Total 3-digit numbers without 22,33,44. First digit: 1,2,3,4.For each first digit:- If first digit is 1: Second digit can be anything (0,1,2,3,4). For each second digit: - If second digit is 0,1: third digit can be anything (5 options each) - If second digit is 2: third digit can't be 2 (4 options) - If second digit is 3: third digit can't be 3 (4 options) - If second digit is 4: third digit can't be 4 (4 options) So, for first digit 1: Second digit 0: 5 options for third digit Second digit 1:5 Second digit 2:4 Second digit 3:4 Second digit 4:4 Total: 5+5+4+4+4 =22 But second digits: 0,1,2,3,4. So first digit 1 gives 5*5 - (forbidden transitions). Wait, maybe my way is not correct. Alternatively, since after first digit 1, second digit can be any, so 5 choices, then third digit depends on second digit. Let's compute: First digit 1, second digit 0: third digit can be 0,1,2,3,4 (5) Second digit 1: third digit 5 Second digit 2: third digit 4 (excluding 2) Second digit 3: third digit 4 (excluding 3) Second digit 4: third digit 4 (excluding 4) So total for first digit 1: 5 +5 +4 +4 +4 =22Similarly, first digit 2:Second digit can be 0,1,3,4 (4 options)For each second digit:- If second digit is 0: third digit can be anything (5)- If second digit is 1: third digit 5- If second digit is 3: third digit 4 (excluding 3)- If second digit is 4: third digit 4 (excluding 4)But wait, second digit 0,1,3,4. Let's count:Second digit 0: 5Second digit 1:5Second digit 3:4Second digit 4:4Total: 5+5+4+4=18Similarly, first digit 3:Same as first digit 2. Second digits: 0,1,2,4 (4 options)Third digits:If second digit 0:5Second digit 1:5Second digit 2:4 (can't be 2)Second digit 4:4 (can't be 4)Total:5+5+4+4=18Similarly, first digit 4:Same as above. Second digits:0,1,2,3. Third digits:0:5,1:5,2:4,3:4. Total:5+5+4+4=18Therefore, total 3-digit numbers:First digit 1:22First digits 2,3,4:18 eachTotal:22 +18*3=22 +54=76. Which matches T(3)=76. Good.So, the recursion seems to work.Therefore, the approach is:- For each digit position from 1 to 7: - If it's the first digit, set f(1,0)=0, f(1,1)=f(1,2)=f(1,3)=f(1,4)=1 - For positions i >=2: - f(i,0) = T(i-1) - f(i,1) = T(i-1) - f(i,2) = T(i-1) - f(i-1,2) - f(i,3) = T(i-1) - f(i-1,3) - f(i,4) = T(i-1) - f(i-1,4) - Then, T(i) = f(i,0)+f(i,1)+f(i,2)+f(i,3)+f(i,4)Since we need to compute up to 7 digits, we can compute this step by step.Let me compute the values step by step.Initialize:For i=1:f(1,0)=0f(1,1)=1f(1,2)=1f(1,3)=1f(1,4)=1T(1)=0+1+1+1+1=4i=2:f(2,0)=T(1)=4f(2,1)=4f(2,2)=4 - f(1,2)=4 -1=3f(2,3)=4 -1=3f(2,4)=4 -1=3T(2)=4+4+3+3+3=17i=3:f(3,0)=17f(3,1)=17f(3,2)=17 - f(2,2)=17 -3=14f(3,3)=17 -3=14f(3,4)=17 -3=14T(3)=17+17+14+14+14=76i=4:f(4,0)=76f(4,1)=76f(4,2)=76 - f(3,2)=76 -14=62f(4,3)=76 -14=62f(4,4)=76 -14=62T(4)=76+76+62+62+62= 76*2 +62*3=152 +186=338Wait, 76+76=152, 62*3=186, 152+186=338. Correct.i=5:f(5,0)=338f(5,1)=338f(5,2)=338 - f(4,2)=338 -62=276f(5,3)=338 -62=276f(5,4)=338 -62=276T(5)=338+338+276+276+276= 338*2 +276*3=676 +828=1504i=6:f(6,0)=1504f(6,1)=1504f(6,2)=1504 - f(5,2)=1504 -276=1228f(6,3)=1504 -276=1228f(6,4)=1504 -276=1228T(6)=1504+1504+1228+1228+1228=1504*2 +1228*3=3008 +3684=6692i=7:f(7,0)=6692f(7,1)=6692f(7,2)=6692 - f(6,2)=6692 -1228=5464f(7,3)=6692 -1228=5464f(7,4)=6692 -1228=5464T(7)=6692+6692+5464+5464+5464= let's compute:First, 6692*2=133845464*3=1639213384 +16392=29776But wait, this T(7) counts numbers of length 7 that can start with 0, which is invalid. Wait, but hold on. Wait, in our setup, f(i, d) for i=1 already accounts for the first digit not being 0. Because for i=1, f(1,0)=0. So, actually, all the numbers counted in T(i) are valid i-digit numbers (i.e., not starting with 0). Wait, but let's verify.Wait, for i=1, T(1)=4, which is correct: 1,2,3,4.For i=2, T(2)=17. Which includes numbers starting with 1,2,3,4 followed by any allowed digit. Since the first digit is already non-zero, the rest can include 0. So, the T(i) counts valid i-digit numbers. Therefore, T(7)=29776 is the total number of 7-digit numbers using digits 0-4, not starting with 0, and not containing 22,33,44.But wait, according to our calculations, T(7)=29776. Therefore, the answer is 29776. Let me check if that makes sense.Wait, but let me double-check the calculations step by step.Starting with i=1: T(1)=4i=2: T(2)=17 (correct as verified earlier)i=3: T(3)=76 (correct as verified earlier)i=4: T(4)=338. Let me check that again.For i=4:f(4,0)=T(3)=76f(4,1)=76f(4,2)=76 - f(3,2)=76 -14=62Similarly, f(4,3)=62, f(4,4)=62So sum:76+76=152; 62*3=186; 152+186=338. Correct.i=5: T(5)=338*2 + (338 -62)*3? Wait, no. Wait, for i=5:f(5,0)=T(4)=338f(5,1)=338f(5,2)=338 - f(4,2)=338 -62=276Similarly, f(5,3)=276, f(5,4)=276Thus, T(5)=338 +338 +276*3= 676 +828=1504. Correct.i=6: T(6)=1504*2 + (1504 -276)*3. Wait, 1504 -276=1228. So 1228*3=3684; 1504*2=3008. 3008 +3684=6692. Correct.i=7: T(7)=6692*2 + (6692 -1228)*3. 6692 -1228=5464. 5464*3=16392; 6692*2=13384. 13384 +16392=29776. Correct.Therefore, the total number of valid 7-digit numbers is 29,776.But let me just verify once again with a different approach to be safe.Alternative approach: Let's model this using linear recurrences.Looking at the recursion for T(i). Let's see if we can find a linear recurrence for T(i).From the previous calculations:T(1) =4T(2)=17T(3)=76T(4)=338T(5)=1504T(6)=6692T(7)=29776Let me see if there's a pattern here. Compute the differences or ratios:From 4 to17: multiply by ~4.2517 to76: multiply by ~4.4776 to338: ~4.447338 to1504: ~4.451504 to6692: ~4.456692 to29776: ~4.45So roughly multiplying by ~4.45 each time. That suggests a linear recurrence with a constant coefficient. Let's see.Assume that T(n) = a*T(n-1) + b*T(n-2) + ... Let's try to find a linear recurrence.Compute the terms:n | T(n)1 | 42 |173 |764 |3385 |15046 |66927 |29776Let's see:From T(1) to T(2): 17 = a*4 + ?But maybe it's a second-order linear recurrence. Let's try T(n) = x*T(n-1) + y*T(n-2)Set up equations:For n=3:76 =x*17 + y*4For n=4:338 =x*76 + y*17Solve:From first equation:76=17x +4ySecond equation:338=76x +17yMultiply first equation by 17: 1292=289x +68yMultiply second equation by4: 1352=304x +68ySubtract first multiplied equation from second multiplied:1352 -1292= (304x -289x) + (68y -68y)60=15x =>x=4Substitute x=4 into first equation:76=17*4 +4y =>76=68 +4y =>4y=8 =>y=2So, the recurrence is T(n)=4*T(n-1)+2*T(n-2)Check for n=4:4*76 +2*17=304 +34=338. Correct.For n=5:4*338 +2*76=1352 +152=1504. Correct.n=6:4*1504 +2*338=6016 +676=6692. Correct.n=7:4*6692 +2*1504=26768 +3008=29776. Correct.So, the recurrence relation is T(n)=4T(n-1)+2T(n-2). So, this is a homogeneous linear recurrence relation with characteristic equation r^2 -4r -2=0.But maybe we don't need the closed-form, since we already computed up to n=7. But interesting to see that the recurrence holds. So, given that, the answer is 29,776.Therefore, the number of 7-digit integers is boxed{29776}.Final Answerboxed{29776}</think>To find the number of 7-digit integers formed by some or all of the digits 0, 1, 2, 3, and 4, such that these integers do not contain the blocks 22, 33, or 44, we use a dynamic programming approach. We define ( f(i, d) ) as the number of ( i )-digit numbers ending with digit ( d ) that follow the rules. The key steps are as follows:1. Initialization for ( i = 1 ): - ( f(1, 0) = 0 ) (since the first digit cannot be 0) - ( f(1, 1) = 1 ) - ( f(1, 2) = 1 ) - ( f(1, 3) = 1 ) - ( f(1, 4) = 1 ) - Total ( T(1) = 4 )2. Recurrence relations for ( i geq 2 ): - ( f(i, 0) = T(i-1) ) - ( f(i, 1) = T(i-1) ) - ( f(i, 2) = T(i-1) - f(i-1, 2) ) - ( f(i, 3) = T(i-1) - f(i-1, 3) ) - ( f(i, 4) = T(i-1) - f(i-1, 4) ) - Total ( T(i) = f(i, 0) + f(i, 1) + f(i, 2) + f(i, 3) + f(i, 4) )3. Computing values step-by-step: - For ( i = 2 ): - ( T(2) = 17 ) - For ( i = 3 ): - ( T(3) = 76 ) - For ( i = 4 ): - ( T(4) = 338 ) - For ( i = 5 ): - ( T(5) = 1504 ) - For ( i = 6 ): - ( T(6) = 6692 ) - For ( i = 7 ): - ( T(7) = 29776 )Thus, the number of 7-digit integers that meet the given conditions is (boxed{29776}).]

🔑:24. Ans: 29776Let an denote the number of such 1 -digit integers. Among these a_{n} integers, let b_{n} denote the number of those which end with 2 . By symmetry, the number of those which end with 3 (or 4 ) is also equal to b_{n}. Hencebegin{array}{l}a_{n}=underbrace{2 a_{n-1}}_{text {end with } 0 text { or } 1}+underbrace{3 b_{n}}_{text {end with } 2,3} text { or } 4 b_{n}=underbrace{2 a_{n-2}}_{text {end with } 02 text { or } 12}+underbrace{2 b_{n-1}}_{text {end with } 32 text { or } 42} end{array}Thinsbegin{aligned}a_{n}-2 a_{n-1} & =3 b_{n}=6 a_{n-2}+6 b_{n-1} & =6 a_{n-2}+2left(a_{n-1}-2 a_{n-2}right) a_{n} & =4 a_{n-1}+2 a_{n-2}end{aligned}We have a_{1}=4, a_{2}=4 times 5-3=17. By iterating we get a_{7}=29776.

🧠:[<think>Okay, let's try to tackle this problem. So, we have two natural numbers a and b. When we divide a² + b² by a + b, the quotient is q and the remainder is r. We need to find all pairs (a, b) such that q² + r = 1977. Hmm, okay, let me think step by step.First, let's recall how division with quotient and remainder works. For any integers m and n (with n > 0), there exist unique integers q and r such that m = n*q + r and 0 ≤ r < n. So in this case, m is a² + b² and n is a + b. Therefore, we can write:a² + b² = (a + b)*q + r, where 0 ≤ r < a + b.And we also have the condition that q² + r = 1977. So our goal is to find all pairs (a, b) such that these two equations hold.Let me write down the two equations:1. a² + b² = (a + b)*q + r2. q² + r = 1977Since r is the remainder when a² + b² is divided by a + b, r must be less than a + b. So r < a + b. Also, since a and b are natural numbers, a + b is at least 2 (if a and b are both 1). But let's not get ahead of ourselves.Let me try to eliminate r from the equations. From equation 2, we can express r as 1977 - q². Then substitute this into equation 1:a² + b² = (a + b)*q + (1977 - q²)Let's rearrange that:a² + b² = q*(a + b) + 1977 - q²Bring all terms to one side:a² + b² - q*(a + b) + q² - 1977 = 0Hmm, not sure if that helps. Maybe let's think about a² + b² in terms of (a + b)². Because (a + b)² = a² + 2ab + b², so a² + b² = (a + b)² - 2ab. Let's substitute that into equation 1:(a + b)² - 2ab = (a + b)*q + rSo:(a + b)² - 2ab = q*(a + b) + rBut from equation 2, r = 1977 - q². Substitute that in:(a + b)² - 2ab = q*(a + b) + 1977 - q²Hmm, maybe rearrange terms:(a + b)² - q*(a + b) + q² - 2ab - 1977 = 0Notice that (a + b)² - q*(a + b) + q² can be written as [(a + b) - q/2]^2 + (3q²)/4? Wait, maybe not. Alternatively, (a + b - q)^2 + 2q(a + b) - q²? Hmm, maybe not helpful. Let's see:Alternatively, perhaps consider variables. Let me denote s = a + b and p = ab. Then, since a² + b² = s² - 2p. So substituting into equation 1:s² - 2p = q*s + rBut r = 1977 - q², so:s² - 2p = q*s + 1977 - q²Rearranged:s² - q*s + q² - 1977 = 2pSo 2p = s² - q*s + q² - 1977Since p = ab, and s = a + b, we know that a and b are roots of the quadratic equation x² - s*x + p = 0. Which means that the discriminant must be non-negative for real roots (which they are, since a and b are natural numbers). So discriminant D = s² - 4p ≥ 0.But from 2p = s² - q*s + q² - 1977, so p = (s² - q*s + q² - 1977)/2. Therefore, D = s² - 4*(s² - q*s + q² - 1977)/2Let me compute D:D = s² - 2*(s² - q*s + q² - 1977)= s² - 2s² + 2q*s - 2q² + 3954= -s² + 2q*s - 2q² + 3954For the discriminant to be non-negative:-s² + 2q*s - 2q² + 3954 ≥ 0Hmm, this seems a bit complicated. Maybe there's another approach.Alternatively, since r = 1977 - q² and r must be less than a + b = s. So 1977 - q² < s. Also, since r ≥ 0, 1977 - q² ≥ 0 ⇒ q² ≤ 1977 ⇒ q ≤ sqrt(1977). Let's compute sqrt(1977). Since 44² = 1936 and 45² = 2025, so sqrt(1977) is between 44 and 45. So q ≤ 44. Therefore, possible values of q are integers from 0 up to 44. But since a and b are natural numbers, s = a + b ≥ 2. Also, when we divide a² + b² by s, the quotient q must satisfy q = floor((a² + b²)/s). Since a² + b² ≥ (s²)/2 (by Cauchy-Schwarz, since (a² + b²)(1 + 1) ≥ (a + b)^2), so a² + b² ≥ s²/2. Therefore, q ≥ (s²/2)/s = s/2. So q ≥ s/2.But q is an integer, so s/2 ≤ q < s. Wait, is that right? Let me think. If I divide a² + b² by s, then the quotient q is floor(a² + b² / s). So, a² + b² = q*s + r, with 0 ≤ r < s. So q = floor((a² + b²)/s). Since a² + b² ≥ s²/2, as per the inequality, (a² + b²)/s ≥ s/2. Therefore, q ≥ floor(s/2). So depending on whether s is even or odd, floor(s/2) is either s/2 or (s-1)/2. But q must be at least approximately s/2.But how does this help us? Hmm.Alternatively, perhaps let's think about possible values of q. Since q can be up to 44, maybe try to find q such that 1977 - q² is non-negative and less than s, where s is a + b. But since s is at least 2, and r = 1977 - q² must be less than s, then 1977 - q² < s. But s = a + b, which is related to a and b, which are natural numbers.Alternatively, perhaps express s in terms of q. Let me see. From equation 1:a² + b² = q*s + r = q*s + (1977 - q²)Therefore, a² + b² = q*s + 1977 - q²But also, a² + b² = (a + b)^2 - 2ab = s² - 2abSo:s² - 2ab = q*s + 1977 - q²Therefore:s² - q*s + q² - 1977 = 2abSo 2ab = s² - q*s + q² - 1977Since a and b are positive integers, ab is a positive integer, so the right-hand side must be positive. Therefore, s² - q*s + q² - 1977 > 0 ⇒ s² - q*s + q² > 1977Also, since 2ab is equal to that, and ab is positive, so the right-hand side must be even? Because 2ab is even. Therefore, s² - q*s + q² - 1977 must be even. So s² - q*s + q² must be odd (since 1977 is odd, odd minus odd is even). Wait, 1977 is odd, so s² - q*s + q² must be odd + even? Wait, let's check:s² - q*s + q² - 1977 must be even. Therefore, s² - q*s + q² must be even + 1977. Since 1977 is odd, even + odd is odd. Wait, no. Wait, the left-hand side is 2ab, which is even. Therefore, s² - q*s + q² - 1977 must be even. Therefore, s² - q*s + q² must be odd, because 1977 is odd. So s² - q*s + q² ≡ 1 mod 2.Now, s² and q² modulo 2: s² ≡ s mod 2, because squaring preserves parity. Similarly, q² ≡ q mod 2. Therefore:s² - q*s + q² ≡ s - q*s + q mod 2.Factor terms:s(1 - q) + q mod 2.Hmm, so s(1 - q) + q ≡ s - q*s + q mod 2.Let me factor:s(1 - q) + q = s - q*s + q = s*(1 - q) + q.If q is even or odd?Alternatively, perhaps check possible cases for parity of q and s.Case 1: q is even.If q is even, then q ≡ 0 mod 2.So s*(1 - 0) + 0 = s*1 + 0 = s mod 2.Therefore, the entire expression is s mod 2. Since s² - q*s + q² ≡ s mod 2. But we know s² - q*s + q² must be odd, so s must be odd.Case 2: q is odd.If q is odd, then q ≡ 1 mod 2.Then s*(1 - 1) + 1 = s*0 + 1 = 1 mod 2.Thus, the expression is 1 mod 2, which is odd. So in this case, regardless of s, if q is odd, then s² - q*s + q² is odd, which is required.Therefore, if q is even, s must be odd; if q is odd, s can be either even or odd, but since s*(1 - q) + q ≡ 1 mod 2, which is satisfied regardless of s. Wait, no, when q is odd, then regardless of s, the expression is 1 mod 2. So in that case, s can be any parity. But when q is even, then s must be odd for the expression to be odd. Okay, so that gives us some parity conditions.But maybe this is getting too detailed. Let's see if there's another approach.Given that r = 1977 - q² and r < s (since r is the remainder when divided by s). Therefore:1977 - q² < sBut s = a + b. Also, from equation 1:a² + b² = q*s + r = q*s + 1977 - q²But since a and b are positive integers, a² + b² ≥ (a + b)² / 2 = s² / 2 (by Cauchy-Schwarz inequality). Therefore:q*s + 1977 - q² ≥ s² / 2Let me rearrange this inequality:q*s + 1977 - q² - s² / 2 ≥ 0Multiply both sides by 2 to eliminate the fraction:2q*s + 2*1977 - 2q² - s² ≥ 0Rearranged:-s² + 2q*s - 2q² + 3954 ≥ 0Which is the same as the discriminant condition we had earlier. Hmm, interesting. So this inequality must hold. Let me write it as:s² - 2q*s + 2q² - 3954 ≤ 0Hmm, quadratic in s. Let's consider s as variable and q as parameter. Let's write the quadratic inequality:s² - 2q*s + (2q² - 3954) ≤ 0The quadratic equation s² - 2q*s + (2q² - 3954) = 0 has roots at s = [2q ± sqrt(4q² - 4*(2q² - 3954))]/2Simplify the discriminant:sqrt(4q² - 8q² + 15816) = sqrt(-4q² + 15816) = sqrt(15816 - 4q²)So the roots are:[2q ± sqrt(15816 - 4q²)] / 2 = q ± sqrt(15816 - 4q²)/2For real roots, 15816 - 4q² ≥ 0 ⇒ 4q² ≤ 15816 ⇒ q² ≤ 3954 ⇒ q ≤ sqrt(3954). Let's compute sqrt(3954). sqrt(3954) is approximately sqrt(4096)=64, so sqrt(3954) is about 62.89. So q ≤ 62. But we already know q ≤ 44 from before. So this is a weaker condition.Therefore, for each q from 0 to 44, we can compute the roots and find the range of s for which the inequality holds. The quadratic in s is ≤ 0 between its two roots. Therefore, s must be between q - sqrt(15816 - 4q²)/2 and q + sqrt(15816 - 4q²)/2. Since s must be a positive integer greater than or equal to 2, and s must also satisfy 1977 - q² < s (from r < s). So we have:s > 1977 - q²ands ∈ [ q - sqrt(15816 - 4q²)/2 , q + sqrt(15816 - 4q²)/2 ]Hmm, this seems quite involved. Maybe instead of dealing with quadratics, let's try to approach this differently.Let me think. We have:From equation 1 and 2:a² + b² = q*(a + b) + 1977 - q²Let me rearrange this:a² - q*a + b² - q*b + q² - 1977 = 0Hmm, perhaps complete the squares for a and b.a² - q*a = a² - q*a + (q²)/4 - (q²)/4 = (a - q/2)^2 - q²/4Similarly for b² - q*b:b² - q*b = (b - q/2)^2 - q²/4So substituting back:(a - q/2)^2 - q²/4 + (b - q/2)^2 - q²/4 + q² - 1977 = 0Simplify:(a - q/2)^2 + (b - q/2)^2 - q²/4 - q²/4 + q² - 1977 = 0Which simplifies to:(a - q/2)^2 + (b - q/2)^2 - q²/2 + q² - 1977 = 0Further:(a - q/2)^2 + (b - q/2)^2 + q²/2 - 1977 = 0Wait, that seems like:(a - q/2)^2 + (b - q/2)^2 = 1977 - q²/2But 1977 - q²/2 must be non-negative because the left side is a sum of squares. Therefore:1977 - q²/2 ≥ 0 ⇒ q² ≤ 3954 ⇒ q ≤ sqrt(3954) ≈ 62.89, which is consistent with earlier findings.But since q is an integer up to 44, this is okay.But we have (a - q/2)^2 + (b - q/2)^2 = 1977 - q²/2But a and b are natural numbers, so (a - q/2) and (b - q/2) can be either integers or half-integers, depending on whether q is even or odd. Let's see:If q is even, say q = 2k, then (a - k) and (b - k) must be integers, since a and b are integers. Then the equation becomes:(a - k)^2 + (b - k)^2 = 1977 - (2k)^2 / 2 = 1977 - 2k²Similarly, if q is odd, say q = 2k + 1, then (a - (2k + 1)/2) and (b - (2k + 1)/2) would be half-integers. Let me write them as (a - k - 0.5) and (b - k - 0.5). Then their squares would be (0.5)^2 = 0.25 each. So:(a - k - 0.5)^2 + (b - k - 0.5)^2 = 1977 - (2k + 1)^2 / 2But 1977 - (4k² + 4k + 1)/2 = 1977 - 2k² - 2k - 0.5 = 1976.5 - 2k² - 2kBut the left side is a sum of squares of half-integers, which would be multiples of 0.25. For example, if a - k - 0.5 = m + 0.5, where m is an integer, then (m + 0.5)^2 = m² + m + 0.25. Similarly for the other term. So adding two such terms would give an integer + 0.5. However, the right side is 1976.5 - 2k² - 2k, which is also a half-integer. Therefore, the equation can hold if the left side equals the right side. But this seems complicated because a and b must be integers. Let's see if we can avoid this by considering q even and odd separately.Case 1: q is even.Let q = 2k. Then:(a - k)^2 + (b - k)^2 = 1977 - 2k²Since a and b are natural numbers, (a - k) and (b - k) can be positive or negative integers. However, since a and b are at least 1, (a - k) ≥ 1 - k and (b - k) ≥ 1 - k. But since k is a positive integer (as q = 2k and q ≤ 44, so k ≤ 22), but actually k can be zero? Wait, q = 2k, q must be at least 0. But since a and b are natural numbers, s = a + b ≥ 2, and q ≥ floor((a² + b²)/s) ≥ floor(s²/2 / s) = floor(s/2) ≥ 1 (since s ≥ 2). So q must be at least 1, so k ≥ 1 when q is even. Wait, if q is even and q ≥ 1, then k ≥ 1. So k ranges from 1 to 22 (since q = 2k ≤ 44).So, for each k from 1 to 22, we can compute 1977 - 2k² and check if it can be expressed as the sum of two squares (since (a - k)^2 + (b - k)^2 must equal that number). Additionally, since a and b are natural numbers, (a - k) and (b - k) must be integers such that a = (a - k) + k ≥ 1, so (a - k) ≥ 1 - k. Similarly for b. So (a - k) and (b - k) can be negative integers as long as a and b are at least 1.But perhaps it's better to consider possible representations of 1977 - 2k² as a sum of two squares. For each k, check if 1977 - 2k² is non-negative and can be written as a sum of two squares, then find the corresponding a and b.Case 2: q is odd.q = 2k + 1. Then:(a - k - 0.5)^2 + (b - k - 0.5)^2 = 1977 - (2k + 1)^2 / 2Let me compute the right-hand side:1977 - (4k² + 4k + 1)/2 = 1977 - 2k² - 2k - 0.5 = 1976.5 - 2k² - 2kSo left-hand side is:(a - k - 0.5)^2 + (b - k - 0.5)^2 = 1976.5 - 2k² - 2kMultiply both sides by 4 to eliminate the fractions:[2(a - k - 0.5)]² + [2(b - k - 0.5)]² = 4*(1976.5 - 2k² - 2k) = 7906 - 8k² - 8kLet me set m = 2(a - k - 0.5) = 2a - 2k - 1 and n = 2(b - k - 0.5) = 2b - 2k - 1. Then m and n are integers because a and b are integers. So:m² + n² = 7906 - 8k² - 8kBut m = 2a - 2k -1 ⇒ a = (m + 2k + 1)/2Similarly, b = (n + 2k + 1)/2Since a and b are natural numbers, m + 2k + 1 and n + 2k + 1 must be even positive integers. Therefore, m and n must be odd integers. Because m = 2a - 2k -1, which is (even) - (even) -1 = odd. Similarly for n.Therefore, in this case, we need to find two odd integers m and n such that m² + n² = 7906 - 8k² - 8k. Then a and b can be derived as (m + 2k + 1)/2 and (n + 2k + 1)/2, which must be natural numbers.So, this seems complex, but maybe manageable. However, this approach requires checking each k for both even and odd cases, which could take time. But since the original problem is an Olympiad problem, perhaps there is a smarter way.Wait, let's note that 1977 is the value of q² + r, and r = 1977 - q². Since r must be less than s = a + b, and s must be at least 2, but also, from the division equation, a² + b² = q*s + r. So a² + b² = q*(a + b) + (1977 - q²). Therefore, rearranged:a² - q*a + b² - q*b = 1977 - q²Hmm. Let's consider this as a quadratic in a:a² - q*a + (b² - q*b - 1977 + q²) = 0Similarly, as a quadratic in a, discriminant would be q² - 4*(b² - q*b - 1977 + q²)But maybe this isn't helpful. Alternatively, consider symmetry between a and b.Alternatively, fix q and solve for s = a + b and p = ab. Recall earlier equations:From the two equations:1. s² - 2p = q*s + (1977 - q²)2. So 2p = s² - q*s + q² - 1977Additionally, since a and b are natural numbers, s and p must satisfy s ≥ 2, p ≥ 1, and the discriminant s² - 4p ≥ 0.So, substituting p from equation 2 into the discriminant inequality:s² - 4*((s² - q*s + q² - 1977)/2) ≥ 0 ⇒ s² - 2*(s² - q*s + q² - 1977) ≥ 0 ⇒ -s² + 2q*s - 2q² + 3954 ≥ 0 ⇒ s² - 2q*s + 2q² - 3954 ≤ 0Which is the same inequality as before. So for each q, we can solve this quadratic in s. The roots of the quadratic equation s² - 2q*s + 2q² - 3954 = 0 are:s = [2q ± sqrt(4q² - 4*(2q² - 3954))]/2 = [2q ± sqrt(4q² - 8q² + 15816)]/2 = [2q ± sqrt(-4q² + 15816)]/2 = [2q ± 2*sqrt(3954 - q²)]/2 = q ± sqrt(3954 - q²)Therefore, s must lie between q - sqrt(3954 - q²) and q + sqrt(3954 - q²). Since s must be a positive integer, and s must also satisfy r = 1977 - q² < s (since r < s). Therefore:s > 1977 - q²So, combining both conditions:q - sqrt(3954 - q²) < s < q + sqrt(3954 - q²)ands > 1977 - q²Therefore, for each q, we need to find integers s such that:max(q - sqrt(3954 - q²), 1977 - q² + 1) < s < q + sqrt(3954 - q²)But this is still complicated.Alternatively, note that since s must be greater than 1977 - q², and also s must be less than q + sqrt(3954 - q²). Therefore, 1977 - q² + 1 ≤ s ≤ floor(q + sqrt(3954 - q²))But 1977 - q² + 1 ≤ q + sqrt(3954 - q²)So:1978 - q² ≤ q + sqrt(3954 - q²)Rearranged:1978 - q² - q ≤ sqrt(3954 - q²)Square both sides (since both sides are non-negative):(1978 - q² - q)^2 ≤ 3954 - q²Expand the left side:(1978 - q(q + 1))^2 ≤ 3954 - q²This would lead to a quartic equation, which is very complicated. Maybe this is not the way to go.Perhaps instead of considering all possible q, we can find possible q such that 1977 - q² is positive and then check for possible s and a, b.Given that q ≤ 44 and 1977 - q² ≥ 0 ⇒ q ≤ 44 (as 44² = 1936 and 45² = 2025). So q ranges from 1 to 44.But this is a lot of cases. Maybe narrow down q by considering that r = 1977 - q² must be less than s = a + b. Also, from equation 1: a² + b² = q*s + r = q*s + (1977 - q²). So:a² + b² = q*(a + b) + 1977 - q²Let me rearrange this:a² - q*a + b² - q*b = 1977 - q²Let me consider this as:(a² - q*a) + (b² - q*b) = 1977 - q²This resembles two quadratic terms adding up to 1977 - q². Since a and b are natural numbers, maybe we can find values of q where 1977 - q² can be expressed as the sum of two such terms.Alternatively, maybe try specific values of q. Since 1977 is a specific number, let's see what q² is close to 1977. Since 44² = 1936, which is 1977 - 1936 = 41. So if q=44, then r=1977 - 44²=1977 - 1936=41. So r=41 < s. Therefore, s must be greater than 41. Also, s must satisfy the quadratic inequality for q=44.Let me check for q=44.Case q=44:r=1977 - 44²=1977 - 1936=41.So s must be greater than 41 (since r < s), and s must satisfy the quadratic inequality.From the quadratic in s:s² - 2*44*s + 2*44² - 3954 ≤ 0Compute:s² - 88s + 2*1936 - 3954 = s² -88s + 3872 -3954 = s² -88s -82 ≤0Solve s² -88s -82 ≤0Find roots:s = [88 ± sqrt(7744 + 328)]/2 = [88 ± sqrt(8072)]/2sqrt(8072) ≈ 89.85Therefore, roots are approximately (88 + 89.85)/2 ≈ 88.925 and (88 - 89.85)/2 ≈ -0.925So the inequality holds for s between approximately -0.925 and 88.925. Since s is positive integer greater than 41, s must be 42,43,...,88.But also, s must satisfy:From equation 2ab = s² -44s +44² -1977= s² -44s +1936 -1977= s² -44s -41Therefore, 2ab = s² -44s -41But ab must be a positive integer, so s² -44s -41 must be even and positive.Check evenness:s² -44s -41 ≡ s² -0*s -1 ≡ s² -1 mod 2s² is 0 or 1 mod 2, so s² -1 is -1 or 0 mod 2. Therefore, s² -1 is even if s is odd. So if s is odd, s² -1 is even; if s is even, s² -1 is odd. Therefore, s must be odd to make 2ab even. So s is odd and between 43 and 88 inclusive (since s>41 and s ≤88.925). So s can be 43,45,47,...,87.For each odd s from 43 to 87, compute 2ab =s² -44s -41. Then check if ab is integer and if a and b are positive integers such that a + b = s.For example, take s=43 (odd):2ab =43² -44*43 -41 =1849 -1892 -41= -1849 - no, wait, 43²=1849, 44*43=1892. Then 1849 -1892= -43, -43 -41= -84. So 2ab= -84. But ab can't be negative. So s=43 is invalid.Similarly, s=45:2ab=45² -44*45 -41=2025 -1980 -41=2025 -2021=4. So ab=2. So possible pairs (a,b) where a + b=45 and ab=2. The solutions are roots of x² -45x +2=0. The discriminant is 2025 -8=2017, which is not a perfect square. Therefore, no integer solutions.s=47:2ab=47² -44*47 -41=2209 -2068 -41=2209 -2109=100. So ab=50. So find a and b such that a + b=47 and ab=50. The quadratic is x² -47x +50=0. Discriminant=2209 -200=2009, not a square. No solutions.s=49:2ab=49² -44*49 -41=2401 -2156 -41=2401 -2197=204. ab=102. Find a + b=49, ab=102. Quadratic x² -49x +102=0. Discriminant=2401 -408=1993, not square.s=51:2ab=51² -44*51 -41=2601 -2244 -41=2601 -2285=316. ab=158. Find a + b=51, ab=158. Quadratic x² -51x +158=0. Discriminant=2601 -632=1969, not a square.s=53:2ab=53² -44*53 -41=2809 -2332 -41=2809 -2373=436. ab=218. a + b=53, ab=218. Quadratic: x² -53x +218=0. Discriminant=2809 -872=1937, not square.s=55:2ab=55² -44*55 -41=3025 -2420 -41=3025 -2461=564. ab=282. a + b=55, ab=282. Quadratic x² -55x +282=0. Discriminant=3025 -1128=1897, not square.s=57:2ab=57² -44*57 -41=3249 -2508 -41=3249 -2549=700. ab=350. a + b=57, ab=350. Quadratic x² -57x +350=0. Discriminant=3249 -1400=1849=43². Therefore, x=(57±43)/2. So x=(57+43)/2=50 or x=(57-43)/2=7. So a=50, b=7 or a=7, b=50. Both are natural numbers. Therefore, (50,7) and (7,50) are solutions. Let me check if these satisfy the original conditions.Let's check for a=50, b=7:Compute a² + b²=2500 + 49=2549. a + b=57. Divide 2549 by 57. 57*44=2508, 2549 -2508=41. So quotient q=44, remainder r=41. Then q² + r=44² +41=1936 +41=1977. Yes, that works. Similarly for (7,50). So these are valid solutions.Continuing with higher s:s=59:2ab=59² -44*59 -41=3481 -2596 -41=3481 -2637=844. ab=422. a + b=59, ab=422. Quadratic x² -59x +422=0. Discriminant=3481 -1688=1793, not a square.s=61:2ab=61² -44*61 -41=3721 -2684 -41=3721 -2725=996. ab=498. a + b=61, ab=498. Quadratic x² -61x +498=0. Discriminant=3721 -1992=1729, not a square.s=63:2ab=63² -44*63 -41=3969 -2772 -41=3969 -2813=1156. ab=578. a + b=63, ab=578. Quadratic x² -63x +578=0. Discriminant=3969 -2312=1657, not a square.s=65:2ab=65² -44*65 -41=4225 -2860 -41=4225 -2901=1324. ab=662. a + b=65, ab=662. Quadratic x² -65x +662=0. Discriminant=4225 -2648=1577, not a square.s=67:2ab=67² -44*67 -41=4489 -2948 -41=4489 -2989=1500. ab=750. a + b=67, ab=750. Quadratic x² -67x +750=0. Discriminant=4489 -3000=1489, not a square.s=69:2ab=69² -44*69 -41=4761 -3036 -41=4761 -3077=1684. ab=842. a + b=69, ab=842. Quadratic x² -69x +842=0. Discriminant=4761 -3368=1393, not a square.s=71:2ab=71² -44*71 -41=5041 -3124 -41=5041 -3165=1876. ab=938. a + b=71, ab=938. Quadratic x² -71x +938=0. Discriminant=5041 -3752=1289, not a square.s=73:2ab=73² -44*73 -41=5329 -3212 -41=5329 -3253=2076. ab=1038. a + b=73, ab=1038. Quadratic x² -73x +1038=0. Discriminant=5329 -4152=1177, not a square.s=75:2ab=75² -44*75 -41=5625 -3300 -41=5625 -3341=2284. ab=1142. a + b=75, ab=1142. Quadratic x² -75x +1142=0. Discriminant=5625 -4568=1057, not a square.s=77:2ab=77² -44*77 -41=5929 -3388 -41=5929 -3429=2500. ab=1250. a + b=77, ab=1250. Quadratic x² -77x +1250=0. Discriminant=5929 -5000=929, not a square.s=79:2ab=79² -44*79 -41=6241 -3476 -41=6241 -3517=2724. ab=1362. a + b=79, ab=1362. Quadratic x² -79x +1362=0. Discriminant=6241 -5448=793, not a square.s=81:2ab=81² -44*81 -41=6561 -3564 -41=6561 -3605=2956. ab=1478. a + b=81, ab=1478. Quadratic x² -81x +1478=0. Discriminant=6561 -5912=649, not a square.s=83:2ab=83² -44*83 -41=6889 -3652 -41=6889 -3693=3196. ab=1598. a + b=83, ab=1598. Quadratic x² -83x +1598=0. Discriminant=6889 -6392=497, not a square.s=85:2ab=85² -44*85 -41=7225 -3740 -41=7225 -3781=3444. ab=1722. a + b=85, ab=1722. Quadratic x² -85x +1722=0. Discriminant=7225 -6888=337, not a square.s=87:2ab=87² -44*87 -41=7569 -3828 -41=7569 -3869=3700. ab=1850. a + b=87, ab=1850. Quadratic x² -87x +1850=0. Discriminant=7569 -7400=169=13². So x=(87±13)/2. x=(87+13)/2=50, x=(87-13)/2=37. So a=50, b=37 or a=37, b=50. Check if these satisfy the original condition.Check a=50, b=37:a² + b²=2500 +1369=3869. a + b=87. Divide 3869 by 87. 87*44=3828, 3869 -3828=41. So quotient q=44, remainder r=41. Then q² + r=44² +41=1936 +41=1977. Yes, works. Similarly for (37,50).So for q=44, we found two pairs: (50,7) and (7,50) when s=57, and (50,37) and (37,50) when s=87.Wait, but s=57 and s=87 both resulted in solutions. Let's check these:For s=57:We had a=50, b=7. Then s=57=50+7. Correct.For s=87:a=50, b=37. s=50+37=87. Correct.So both cases give valid solutions.Now, let's check if there are other q's that could give solutions.Let's try q=43:r=1977 -43²=1977 -1849=128. So r=128 < s. So s >128. But s must be greater than 128, and from the quadratic inequality:s² -2*43*s +2*43² -3954 ≤0s² -86s +2*1849 -3954= s² -86s +3698 -3954= s² -86s -256 ≤0Solving s² -86s -256 ≤0Roots are [86 ± sqrt(7396 +1024)]/2= [86 ± sqrt(8420)]/2. sqrt(8420)=91.85. So roots ≈(86 +91.85)/2≈88.9 and (86 -91.85)/2≈-2.9. Therefore, s must be ≤88.9. But s>128, which contradicts. Therefore, no solution for q=43.Similarly, q=42:r=1977 -42²=1977 -1764=213. So s>213. Quadratic inequality:s² -84s +2*1764 -3954= s² -84s +3528 -3954= s² -84s -426 ≤0Roots are [84 ± sqrt(7056 +1704)]/2= [84 ± sqrt(8760)]/2. sqrt(8760)=93.6. So roots≈(84 +93.6)/2≈88.8 and (84 -93.6)/2≈-4.8. So s must be ≤88.8, but s>213, impossible. No solutions.Similarly for lower q, s required would be even larger, but the quadratic inequality restricts s to be small. So likely only q=44 gives solutions. But let's check q=43,42,… maybe s somehow?Wait, but for q=44, we found two solutions. What about other q's?Wait, perhaps check q=29:But why q=29? Maybe if we note that 1977=44² +41=1936+41, but maybe there's another combination. For example, if q=29, then r=1977-841=1136. Then s must be greater than 1136. But s must be at least 1137. Then a and b would be huge numbers, but let's see:From equation:a² + b² =29*s +1136But s=a + b. So:a² + b²=29(a + b) +1136But a and b would be around s/2 each, so a² + b²≈ (s²)/2. So approximate equation:(s²)/2 ≈29s +1136 ⇒ s² -58s -2272≈0 ⇒ roots≈(58 ± sqrt(3364 +9088))/2=(58 ± sqrt(12452))/2≈(58 ± 111.6)/2. Positive root≈(58+111.6)/2≈84.8. So s≈85. But s must be >1136, which contradicts. So no solution.Alternatively, higher q's require s to be large, but the quadratic inequality restricts s to be low, so likely only q=44 gives solutions.Alternatively, check q=31:r=1977-31²=1977-961=1016. s>1016. But quadratic inequality:s² -62s +2*961 -3954= s² -62s +1922 -3954= s² -62s -2032 ≤0Roots≈[62 ± sqrt(3844 +8128)]/2=[62 ± sqrt(11972)]/2≈[62 ± 109.4]/2. Positive root≈(62+109.4)/2≈85.7. So s≤85.7, but s>1016, impossible.Similarly, other q's.Thus, the only possible solutions are for q=44, which gives us the pairs (50,7), (7,50), (50,37), and (37,50). Wait, but when s=57, we had (50,7) and (7,50), and when s=87, we had (50,37) and (37,50). Are there any other pairs for q=44?Wait, let's check s=99:But s must be less than q + sqrt(3954 - q²)=44 + sqrt(3954 -1936)=44 + sqrt(2018)≈44 +44.93≈88.93. So s must be less than 89. Therefore, s=87 is the maximum possible s for q=44.Thus, the only possible pairs are (50,7), (7,50), (50,37), (37,50).Wait, but 50 +37=87, which is s=87. And 50 +7=57. But 57 and 87 are both less than 89.So these are all possible pairs for q=44.But wait, could there be other pairs for other q's? Let's check another q, say q=43. Wait, earlier we saw q=43 requires s>128 but the quadratic restricts s≤88, so no. Let's check q=20. Maybe lower q.q=20:r=1977 -400=1577. s>1577. Quadratic inequality:s² -40s +2*400 -3954= s² -40s +800 -3954= s² -40s -3154 ≤0Roots≈[40 ± sqrt(1600 +12616)]/2=[40 ± sqrt(14216)]/2≈[40 ± 119.2]/2. Positive root≈(40+119.2)/2≈79.6. So s must be ≤79.6, but s>1577, impossible.Thus, no solutions.Alternatively, check q= even number like q=22:r=1977 -22²=1977 -484=1493. s>1493. Quadratic inequality:s² -44s +2*484 -3954= s² -44s +968 -3954= s² -44s -2986 ≤0Roots≈[44 ± sqrt(1936 +11944)]/2=[44 ± sqrt(13880)]/2≈[44 ± 117.8]/2. Positive root≈(44+117.8)/2≈80.9. So s≤80.9, conflicting with s>1493. Impossible.Therefore, no solutions for q=22.Thus, the only solutions are for q=44, giving four pairs: (50,7), (7,50), (50,37), (37,50). Wait, but when we checked s=57 and s=87, each gives two pairs. So total four pairs.But let me verify if these are distinct pairs. (50,7) and (7,50) are distinct, as well as (50,37) and (37,50). So total four pairs.But wait, the problem says "pairs of numbers (a, b)", so order matters? Or does it consider (a, b) and (b, a) as the same pair? The problem statement says "pairs of numbers", which in mathematics usually considers ordered pairs, but sometimes unordered. Let's check the original problem statement.The original problem is from the 19th IMO, problem 1, which asks for "all pairs of numbers (a, b)" such that q² + r = 1977. Looking at the original problem's solution, it's likely that ordered pairs are considered, so (a, b) and (b, a) are distinct. However, the problem might accept both as solutions. However, in the example given in the problem statement, the user mentions "pairs of numbers (a, b)", so likely ordered pairs. However, in Olympiad problems, sometimes unordered pairs are requested. Since the original problem is in Chinese, the exact phrasing might matter. But since we don't have the Chinese original, we can assume both ordered pairs are acceptable. However, in the answer, we need to put all possible ordered pairs.But according to our solution, for q=44, we have four ordered pairs: (50,7), (7,50), (50,37), (37,50). However, let's confirm these:For s=57, a=50, b=7 and a=7, b=50.For s=87, a=50, b=37 and a=37, b=50.So four pairs. But wait, why are there two different s values giving different pairs?Yes, because for q=44, different s values can satisfy the equations. Specifically, s=57 and s=87 both lead to valid pairs. Therefore, these four pairs are all valid.But let's check if there are other s values for q=44. For example, s=69. We tried s=69 and found no solution. Similarly, other s values didn't yield solutions. Only s=57 and s=87 gave valid solutions.Therefore, the answer should be these four pairs. However, let's check if there's another possible q.Wait, let me check q=31, but earlier it didn't work. Alternatively, what about q=21?r=1977 -21²=1977 -441=1536. s>1536. Quadratic inequality:s² -42s +2*441 -3954= s² -42s +882 -3954= s² -42s -3072 ≤0Roots≈[42 ± sqrt(1764 +12288)]/2=[42 ± sqrt(14052)]/2≈[42 ± 118.5]/2. Positive root≈(42+118.5)/2≈80.25. So s must be ≤80.25, contradicting s>1536. Impossible.Therefore, only q=44 yields solutions. Hence, the four pairs are the solutions.However, let me check if there's another pair with q=44 and s= something else. For example, s= 85. We saw that s=85 gave ab=1722, which didn't result in integer solutions. So only s=57 and s=87 worked.Therefore, the final answer is the four pairs: (50,7), (7,50), (50,37), (37,50). However, let me verify once again.For (50,37):a=50, b=37. Sum=87. a² + b²=2500+1369=3869. Divided by 87: 87*44=3828. 3869-3828=41. So q=44, r=41. Then q² + r=1936+41=1977. Correct.For (37,50): same result.For (50,7): sum=57. a² + b²=2500+49=2549. 57*44=2508. 2549-2508=41. q=44, r=41. q² + r=1977. Correct.For (7,50): same.Therefore, all four pairs are valid. Hence, these are the solutions.Final AnswerThe pairs are boxed{(50, 7)}, boxed{(7, 50)}, boxed{(50, 37)}, and boxed{(37, 50)}.</think>Given two natural numbers (a) and (b), we need to find all pairs ((a, b)) such that when (a^2 + b^2) is divided by (a + b), the quotient (q) and remainder (r) satisfy (q^2 + r = 1977).1. Division Relationship: [ a^2 + b^2 = (a + b)q + r ] where (0 leq r < a + b).2. Given Condition: [ q^2 + r = 1977 ] Substituting (r = 1977 - q^2) into the division relationship: [ a^2 + b^2 = (a + b)q + (1977 - q^2) ]3. Rearranging the Equation: [ a^2 + b^2 - q(a + b) + q^2 - 1977 = 0 ]4. Expressing in Terms of (s) and (p): Let (s = a + b) and (p = ab). Then: [ s^2 - 2p = qs + 1977 - q^2 ] Rearranging gives: [ 2p = s^2 - qs + q^2 - 1977 ]5. Discriminant Condition: For (a) and (b) to be natural numbers, the discriminant of the quadratic equation (x^2 - sx + p = 0) must be non-negative: [ s^2 - 4p geq 0 ] Substituting (p): [ s^2 - 2(s^2 - qs + q^2 - 1977) geq 0 ] Simplifying: [ -s^2 + 2qs - 2q^2 + 3954 geq 0 ]6. Checking Possible Values of (q): - (q) must be an integer such that (q^2 leq 1977), giving (q leq 44). - For each (q), solve for (s) and check if (r = 1977 - q^2 < s).7. Key Case (q = 44): - (r = 1977 - 44^2 = 41) - Solve (s^2 - 44s - 82 leq 0), leading to valid (s) values between approximately (q - sqrt{3954 - q^2}) and (q + sqrt{3954 - q^2}).8. Valid Solutions: - For (s = 57) and (s = 87), solving the quadratic equations for (a) and (b) yields pairs: - (s = 57): (a = 50, b = 7) and (a = 7, b = 50) - (s = 87): (a = 50, b = 37) and (a = 37, b = 50)Thus, the pairs ((a, b)) that satisfy the given conditions are:[boxed{(50, 7)}, boxed{(7, 50)}, boxed{(50, 37)}, boxed{(37, 50)}],<think>Okay, so I have this problem here where I need to find all pairs of natural numbers (a, b) such that when a² + b² is divided by a + b, the quotient is q and the remainder is r. And it's given that q² + r = 1977. Hmm, let's try to unpack this step by step.First, let me recall that when you divide a number by another number, say, dividing m by n, you get m = n*q + r, where 0 ≤ r < n. So in this case, a² + b² = (a + b)*q + r, with 0 ≤ r < a + b. Then, the problem states that q² + r = 1977. So I need to find all pairs (a, b) such that these conditions hold.Let me write down the equations:1. a² + b² = (a + b) * q + r2. q² + r = 19773. 0 ≤ r < a + bSince r is a remainder when dividing by a + b, it must be less than a + b, which is condition 3.Also, since a and b are natural numbers, a + b is at least 2 (since the smallest natural number is 1, so 1 + 1 = 2). Therefore, the divisor a + b is at least 2.Given that q² + r = 1977, and r is non-negative (as a remainder), so q² ≤ 1977. Therefore, q must be less than or equal to sqrt(1977). Let me compute sqrt(1977):Well, 44² = 1936 and 45² = 2025. So sqrt(1977) is between 44 and 45, so q can be at most 44. Therefore, possible values of q are from 0 up to 44. However, since q is the quotient when a² + b² is divided by a + b, which are both positive integers, the quotient q should also be a positive integer. Wait, but could q be zero? Let's see.If q were zero, then a² + b² = r. But since r < a + b, that would mean a² + b² < a + b. However, for natural numbers a and b, a² + b² is at least 1 + 1 = 2, and a + b is at least 2. So equality could hold if a = b = 1, then a² + b² = 2 and a + b = 2, so r = 0. Then q would be 1, because 2 divided by 2 is 1 with remainder 0. Wait, so if a = b = 1, then q = 1, r = 0, so q² + r = 1 + 0 = 1, which is much less than 1977. Therefore, q cannot be zero here because that would require a² + b² < a + b, which only occurs when a and b are 1 each, but even then the quotient is 1, not 0. So q must be at least 1.Therefore, q ranges from 1 to 44. So possible q values are integers from 1 up to 44.Given that, we can express r from the second equation: r = 1977 - q². Since r must be non-negative and less than a + b, we have:0 ≤ 1977 - q² < a + bSo 1977 - q² must be a non-negative integer, which implies that q² ≤ 1977, as we already have. Also, since r = 1977 - q², and a + b > r, so a + b > 1977 - q².But from the first equation, a² + b² = (a + b) * q + r. Since r = 1977 - q², substitute that into the first equation:a² + b² = (a + b) * q + (1977 - q²)So rearranged:a² + b² - (a + b) * q = 1977 - q²Hmm. Let me think if there's a way to express a² + b² in terms of (a + b) and maybe ab.Recall that a² + b² = (a + b)² - 2ab. Let's use that:(a + b)² - 2ab - (a + b) * q = 1977 - q²Let me rearrange this equation:(a + b)² - (a + b)*q - 2ab = 1977 - q²Hmm, not sure if that's helpful. Maybe I can bring all terms to one side:(a + b)² - (a + b)*q - 2ab - 1977 + q² = 0But this seems complicated. Maybe there's a better approach.Alternatively, since we have a² + b² divided by (a + b) gives quotient q and remainder r, so:a² + b² = q*(a + b) + rBut r = 1977 - q², so:a² + b² = q*(a + b) + 1977 - q²Let me write this as:a² + b² - q*(a + b) = 1977 - q²So maybe we can consider this as a quadratic in terms of a and b. But maybe instead, think of a and b as variables and try to find relationships.Alternatively, let's denote s = a + b and p = ab. Then, since a² + b² = s² - 2p.So substituting into the equation:s² - 2p = q*s + 1977 - q²So rearranged:s² - q*s - 2p = 1977 - q²But this seems like introducing more variables. However, maybe we can express p from here:2p = s² - q*s - (1977 - q²)So:p = (s² - q*s - 1977 + q²)/2But p must be a positive integer because a and b are natural numbers, so p = ab is a natural number.Therefore, s² - q*s - 1977 + q² must be even.Moreover, since s = a + b and p = ab, we can relate s and p to the roots of the quadratic equation x² - s*x + p = 0, whose roots are a and b. So for a and b to be natural numbers, the discriminant s² - 4p must be a perfect square (since the roots are integers). Let's note that.So discriminant D = s² - 4p must be a perfect square.From earlier, we have:p = (s² - q*s - 1977 + q²)/2Therefore, substitute p into D:D = s² - 4*(s² - q*s - 1977 + q²)/2Simplify:D = s² - 2*(s² - q*s - 1977 + q²)= s² - 2s² + 2q*s + 3954 - 2q²= -s² + 2q*s + 3954 - 2q²Therefore, D = -s² + 2q*s + 3954 - 2q² must be a perfect square.This seems complicated, but maybe we can find s in terms of q.Wait, from the equation:s² - q*s - 2p = 1977 - q²But we also know that r = 1977 - q², and r < s (since r < a + b = s). Therefore:1977 - q² < sBut s is a + b, which is at least 2, but given that r must be non-negative, we have:0 ≤ 1977 - q² < sSo s > 1977 - q²But from the first equation, s² - q*s - 2p = 1977 - q²But since p is positive (since a and b are natural numbers), then s² - q*s must be greater than 1977 - q². Therefore:s² - q*s > 1977 - q²Which can be written as:s² - q*s + q² > 1977Wait, that's s² - q*s + q² = (s - q/2)^2 + (3q²)/4, which is always positive. Not sure if that helps.Alternatively, maybe we can write s in terms of q. Let's see.We have:From p = (s² - q*s - 1977 + q²)/2But p = ab must be a positive integer. So s² - q*s - 1977 + q² must be even and positive.So:s² - q*s - 1977 + q² > 0So:s² - q*s + q² > 1977Hmm, similar to before.Alternatively, s² - q*s + q² > 1977But I'm not sure if that's helpful. Maybe trying specific values of q?Given that q can be from 1 to 44, and for each q, compute r = 1977 - q², then since r < s, we have s > r. But s = a + b, so s is at least something. Also, since a and b are natural numbers, s is at least 2.But s must be greater than 1977 - q². So for each q, we need s > 1977 - q². However, 1977 - q² must be positive, so q² < 1977. So q up to 44, as 44²=1936, which is less than 1977, and 45²=2025, which is more. So q can be 1 to 44.But let's think: if q is large, say q=44, then r=1977 - 44²=1977 - 1936=41. So r=41, and s>41. So s must be at least 42. Then, we have from the first equation:a² + b² = (a + b)*44 + 41But s = a + b ≥ 42, so let's set s = a + b, then a² + b² = 44s + 41.But a² + b² = (a + b)^2 - 2ab = s² - 2ab. Therefore:s² - 2ab = 44s + 41So 2ab = s² - 44s - 41Therefore, ab = (s² -44s -41)/2Since ab must be a positive integer, s² -44s -41 must be even and positive.But s is at least 42 (since s > r=41). Let's check s=42:ab=(42² -44*42 -41)/2=(1764 -1848 -41)/2=(-125)/2=-62.5. Not possible, since ab must be positive.s=43:ab=(43² -44*43 -41)/2=(1849 -1892 -41)/2=(-84)/2=-42. Negative. Still not possible.s=44:ab=(44² -44*44 -41)/2=(1936 -1936 -41)/2=(-41)/2=-20.5. Not possible.s=45:ab=(45² -44*45 -41)/2=(2025 -1980 -41)/2=(4)/2=2. So ab=2.So for s=45, ab=2. Since s=a + b=45 and ab=2. So we need to solve for a and b such that a + b=45 and ab=2. The solutions are the roots of x² -45x +2=0. The discriminant is 2025 -8=2017, which is not a perfect square. Therefore, no integer solutions. So no solution here.s=46:ab=(46² -44*46 -41)/2=(2116 -2024 -41)/2=(51)/2=25.5. Not integer.s=47:ab=(47² -44*47 -41)/2=(2209 -2068 -41)/2=(100)/2=50. So ab=50. Then, a + b=47, ab=50. So solving x² -47x +50=0. Discriminant=2209 -200=2009, which is not a perfect square. Therefore, no integer solutions.s=48:ab=(48² -44*48 -41)/2=(2304 -2112 -41)/2=(151)/2=75.5. Not integer.s=49:ab=(49² -44*49 -41)/2=(2401 -2156 -41)/2=(204)/2=102. So ab=102, s=49. Therefore, solving x² -49x +102=0. Discriminant=2401 -408=1993. Not a perfect square. No solution.s=50:ab=(50² -44*50 -41)/2=(2500 -2200 -41)/2=(259)/2=129.5. Not integer.s=51:ab=(51² -44*51 -41)/2=(2601 -2244 -41)/2=(316)/2=158. So ab=158, s=51. Then solving x² -51x +158=0. Discriminant=2601 -632=1969. Not a square.s=52:ab=(52² -44*52 -41)/2=(2704 -2288 -41)/2=(375)/2=187.5. Not integer.s=53:ab=(53² -44*53 -41)/2=(2809 -2332 -41)/2=(436)/2=218. So ab=218, s=53. So discriminant=2809 -872=1937. Not a square.s=54:ab=(54² -44*54 -41)/2=(2916 -2376 -41)/2=(499)/2=249.5. Not integer.s=55:ab= (55² -44*55 -41)/2= (3025 -2420 -41)/2= (564)/2=282. So ab=282, s=55. Solving x² -55x +282=0. Discriminant=3025 -1128=1897. Not a square.Hmm, this is getting tedious. Maybe there's a pattern here. For q=44, it seems like even when s increases, the discriminant isn't a perfect square. Maybe q=44 doesn't yield any solution. Let's try a smaller q.Let me try q=44, but maybe I made a mistake here. Wait, for q=44, r=41. Then s must be greater than 41, but when we tried s=45, 46, etc., the ab was non-integer or discriminant not a square. So perhaps q=44 does not yield any solutions. Let me check another q.Take q=43. Then r=1977 -43²=1977-1849=128. So r=128. Then s must be greater than 128.So s >=129.Then, similar approach: a² + b²=43*s +128But a² + b²= s² -2ab, so s² -2ab=43s +128Then 2ab= s² -43s -128So ab=(s² -43s -128)/2We need s>128, so s>=129. Let's try s=129:ab=(129² -43*129 -128)/2=(16641 -5547 -128)/2=(10966)/2=5483. Then check discriminant: s² -4p=129² -4*5483=16641 -21932= -5291. Negative, so no solution.Wait, discriminant can't be negative because a and b are real numbers. Wait, but if the discriminant is negative, that would mean no real solutions, but a and b are natural numbers. Therefore, this case is impossible. So s=129 is invalid. Next s=130:ab=(130² -43*130 -128)/2=(16900 -5590 -128)/2=(16900 -5718)/2=11182/2=5591. Discriminant=130² -4*5591=16900 -22364= -5464. Still negative.Same issue. Wait, perhaps this approach is not working. Let me check if the discriminant can ever be positive here.Wait, discriminant D = s² -4p. From above, p=(s² -43s -128)/2. Therefore:D = s² -4*(s² -43s -128)/2 = s² -2*(s² -43s -128) = s² -2s² +86s +256 = -s² +86s +256.So D = -s² +86s +256. For D to be non-negative, -s² +86s +256 >=0.So s² -86s -256 <=0Solving s² -86s -256 <=0. The roots of the equation s² -86s -256=0 are s=(86 ± sqrt(86² +4*256))/2=(86 ± sqrt(7396 +1024))/2=(86 ± sqrt(8420))/2. sqrt(8420) is approx 91.76. So roots are (86 +91.76)/2≈177.76/2≈88.88 and (86 -91.76)/2≈-5.76/2≈-2.88. Therefore, inequality holds for s between -2.88 and 88.88. Since s is positive integer >=129, this inequality is not satisfied. Hence, discriminant D is negative for all s>=129. Therefore, no solutions for q=43.Hmm, that's a problem. So for q=43, there are no solutions because the discriminant is always negative. So maybe q=43 is invalid. Let's try q=42.Then r=1977 -42²=1977 -1764=213. So r=213. Therefore, s>213.So s>=214.Similarly, a² + b²=42*s +213Expressed in terms of s and ab:s² -2ab=42s +213Thus, 2ab= s² -42s -213ab=(s² -42s -213)/2Again, discriminant D = s² -4ab= s² -4*(s² -42s -213)/2= s² -2*(s² -42s -213)= s² -2s² +84s +426= -s² +84s +426For discriminant D to be non-negative:-s² +84s +426 >=0s² -84s -426 <=0Solving s² -84s -426=0. Roots: (84 ± sqrt(84² +4*426))/2=(84 ± sqrt(7056 +1704))/2=(84 ± sqrt(8760))/2. sqrt(8760)≈93.6. So roots≈(84 +93.6)/2≈177.6/2≈88.8 and (84 -93.6)/2≈-9.6/2≈-4.8. So inequality holds for s between -4.8 and 88.8. But s>=214, so again D is negative. Therefore, no solutions for q=42 either.This seems like a pattern. For large q, s needs to be large, but the discriminant becomes negative. So perhaps I need to consider smaller q where r=1977 - q² is smaller, so s can be smaller, which might make the discriminant positive.Let me try q=20. Then r=1977 -400=1577. Then s>1577. That's way too big. Wait, but maybe even with smaller q, s needs to be larger than r, which can be large.Wait, maybe the problem is that for larger q, r=1977 - q² is smaller, so s needs to be just larger than r. For example, take q=44, r=41, s>41. But in that case, the discriminant becomes negative. Wait, but maybe there's a sweet spot where s is just slightly larger than r, and the discriminant is a perfect square.Alternatively, perhaps q is small enough that r=1977 - q² is large, so s can be of the order of r, but then the equations might balance.Wait, maybe another approach. Let's note that q is the quotient when a² + b² is divided by a + b. Let's recall that when you divide a² + b² by a + b, the quotient is q = a - b if a > b, but only if a + b divides a² + b² exactly. But that's not generally the case. Wait, actually, let's perform polynomial division.Divide a² + b² by a + b. Let me treat a as a variable and b as a constant. Then:a² + b² divided by a + b. The first term is a, multiply (a + b) by a gives a² + a*b. Subtract from a² + b²: (a² + b²) - (a² + a*b) = -a*b + b² = b² -a*b. Now, divide this by a + b. The next term is -b, multiply (a + b)*(-b) gives -a*b -b². Subtracting from b² -a*b: (b² -a*b) - (-a*b -b²) = b² -a*b +a*b +b² = 2b². So the remainder is 2b². Wait, but this seems like the division of a² + b² by a + b gives quotient a - b and remainder 2b². But this is only if we fix b and treat a as the variable. But actually, since a and b are symmetric, maybe we can write the division as:a² + b² = (a + b)(a - b) + 2b². But if we continue dividing, then 2b² divided by a + b would be another term. Wait, perhaps the division is different.Wait, actually, if you divide a² + b² by a + b:The division process is as follows:Divide a² by a + b. The first term is a. Multiply (a + b) by a: a² + a*b. Subtract this from a² + b²: (a² + b²) - (a² + a*b) = -a*b + b². Then, divide -a*b by a + b. The next term is -b. Multiply (a + b)*(-b) = -a*b - b². Subtract this from -a*b + b²: (-a*b + b²) - (-a*b - b²) = 2b². So the quotient is a - b and the remainder is 2b². But this is only if we stop here. However, in integer division, we need the remainder to be less than the divisor. So 2b² must be less than a + b. But 2b² < a + b. Since a and b are natural numbers, this is a restrictive condition.Alternatively, maybe the division is different. Wait, perhaps there's a different way to express the division. For example, a² + b² divided by a + b. Let's suppose that the quotient q and remainder r satisfy a² + b² = q*(a + b) + r, with 0 ≤ r < a + b. Then, we can write:a² + b² - q*(a + b) = rBut since r < a + b, and r is non-negative, we can think of this as:a² + b² ≡ r mod (a + b), where 0 ≤ r < a + b.But I'm not sure if that helps. Alternatively, maybe express a² + b² in terms of (a + b). For example:a² + b² = (a + b)^2 - 2ab. So:(a + b)^2 - 2ab = q*(a + b) + rLet s = a + b, then:s² - 2ab = q*s + rBut r = 1977 - q², so:s² - 2ab = q*s + 1977 - q²Rearranged:s² - q*s - 2ab = 1977 - q²But I still have two variables here: s and ab. Unless I can relate them somehow. Since s = a + b and ab = product, maybe we can parameterize a and b in terms of s and ab.Alternatively, we can express ab in terms of s:ab = (s² - q*s - (1977 - q²))/2Then, since a and b are roots of the equation x² - s*x + ab = 0, discriminant must be a perfect square.So discriminant D = s² - 4ab = s² - 4*(s² - q*s -1977 + q²)/2 = s² - 2*(s² - q*s -1977 + q²) = s² - 2s² + 2q*s + 3954 - 2q² = -s² + 2q*s + 3954 - 2q²So D = -s² + 2q*s + (3954 - 2q²)This discriminant must be a perfect square. Let's denote D = k² for some integer k.Therefore, we have:-s² + 2q*s + (3954 - 2q²) = k²Rearranged:s² - 2q*s + (k² -3954 + 2q²) = 0This is a quadratic in s:s² - 2q*s + (k² -3954 + 2q²) = 0Solving for s using quadratic formula:s = [2q ± sqrt(4q² - 4*(k² -3954 + 2q²))]/2Simplify the discriminant under the square root:sqrt(4q² -4*(k² -3954 + 2q²)) = sqrt(4q² -4k² +15816 -8q²) = sqrt(-4k² -4q² +15816)Factor out -4:sqrt(-4(k² + q² -3954)) = 2*sqrt(- (k² + q² -3954))But sqrt of a negative number is not real. Wait, this suggests that my approach here is flawed, because we end up with a square root of a negative number unless the expression inside is non-positive. But k² is a square, so non-negative. Therefore, - (k² + q² -3954) must be non-negative. So:k² + q² -3954 ≤0 → k² ≤ 3954 - q²But since k² is non-negative, 3954 - q² must be non-negative. Therefore:q² ≤3954 → q ≤ sqrt(3954) ≈ 62.89. But q is at most 44, so that's okay.But how can sqrt(-4(k² + q² -3954)) be real? Unless k² + q² -3954 is negative or zero, but then the sqrt would be of a non-negative number. However, we have:sqrt(-4(k² + q² -3954)) = sqrt(4(3954 -k² -q²)) = 2*sqrt(3954 -k² -q²)Therefore, the expression simplifies to:s = [2q ± 2*sqrt(3954 -k² -q²)]/2 = q ± sqrt(3954 -k² -q²)Since s must be a positive integer, sqrt(3954 -k² -q²) must be an integer. Let me denote m = sqrt(3954 -k² -q²). Therefore:m² = 3954 -k² -q²Then, s = q ± mBut s = a + b must be positive, and since q is positive (as established earlier), s = q + m or s = q - m. However, since s must be positive, if we take s = q - m, then q - m >0, so m < q.Also, m must be a non-negative integer because it's a square root of an integer.So, combining these:m² +k² +q² =3954and s = q ± mBut s must also satisfy s > r =1977 - q²Recall that r =1977 - q², and since s > r, we have:q ± m >1977 - q²But s can be either q + m or q - m. So two cases:Case 1: s = q + mThen, q + m >1977 - q² → m >1977 - q² - qBut m is non-negative, so this inequality requires 1977 - q² - q < mBut m = sqrt(3954 -k² -q²). This seems complicated.Case 2: s = q - mThen, q - m >1977 - q² → -m >1977 - q² - q → m < q + q² -1977But m is non-negative, so q + q² -1977 must be positive. Therefore:q² + q -1977 >0Solving q² + q -1977 =0:q = [-1 ± sqrt(1 +4*1977)]/2 = [-1 ± sqrt(7909)]/2. sqrt(7909) ≈ 89. So positive root≈ (-1 +89)/2≈44. So when q≥44, q² + q -1977 ≥44² +44 -1977=1936 +44 -1977=1980 -1977=3>0. So for q≥44, q² +q -1977>0, but q≤44, so q=44. For q=44, q² +q -1977=1936 +44 -1977=1980 -1977=3>0. Therefore, for q=44, m<3. Since m is non-negative integer, m=0,1,2.But let's check for q=44, Case 2: s=q -m. So s=44 -m. But s must be > r=1977 -44²=41. So 44 -m>41→m<3. So m=0,1,2.But m² =3954 -k² -q²=3954 -k² -1936=2018 -k². So m²=2018 -k². Therefore:For m=0: 0=2018 -k²→k²=2018. Not a square.For m=1:1=2018 -k²→k²=2017. Not a square.For m=2:4=2018 -k²→k²=2014. Not a square.Hence, no solutions for q=44 in Case 2.For Case1: s=q +m. Then s=44 +m. Then s must be >41, which is always true since m is non-negative. So m can be any integer such that m²=3954 -k² -44²=3954 -k² -1936=2018 -k².So m² +k²=2018. So we need integer solutions (m,k) to m² +k²=2018.2018 factors: 2×1009. 1009 is a prime. Now, 1009 mod4=1, so it can be expressed as sum of two squares. Let's try to find integers m,k such that m² +k²=2018.Let me check sqrt(2018)≈44.9. So m and k can be up to 44.Check 44²=1936. 2018-1936=82. 82 is not a square.43²=1849. 2018-1849=169=13². So 43² +13²=1849 +169=2018. So (m,k)=(43,13) or (13,43). Also, since negative counterparts, but since m and k are non-negative, we have these.Similarly, other combinations: let's see 40²=1600. 2018-1600=418, not a square. 37²=1369. 2018-1369=649, which is not a square. 35²=1225. 2018-1225=793. Not a square. 32²=1024. 2018-1024=994. Not a square. 25²=625. 2018-625=1393. Not a square. So the only solutions are (43,13) and (13,43).Therefore, for q=44, m=43 or 13.But m = sqrt(3954 -k² -q²)=sqrt(2018 -k²). Wait, but earlier, we have m² +k²=2018. So for (m,k)=(43,13): m=43, k=13. Then s=q +m=44 +43=87. Then s=87. Then, from s=q +m=87, and s= a + b=87. Also, D=k²=13²=169. So discriminant D=169= (a - b)^2. Therefore, a - b=±13.Since a + b=87 and a - b=13 or -13. Then:If a - b=13, then solving:a + b=87a - b=13Adding: 2a=100→a=50→b=87 -50=37.If a - b=-13, then:a + b=87a - b=-13Adding: 2a=74→a=37→b=87 -37=50.So pairs (50,37) and (37,50). Let's check if these satisfy the original conditions.First, take a=50, b=37. Compute a² + b²=2500 +1369=3869. Divide by a + b=87. 3869 ÷87. 87×44=3828. 3869 -3828=41. So quotient q=44, remainder r=41. Then q² +r=44² +41=1936 +41=1977. Perfect. Similarly, a=37, b=50 gives the same result. So these are valid solutions.Similarly, check the other solution with m=13. Then s=44 +13=57. Then discriminant D=k²=43²=1849. So D=1849=(a - b)^2. Then a - b=±43. Since a + b=57.If a - b=43:a + b=57a - b=43Adding: 2a=100→a=50→b=57 -50=7.If a - b=-43:a + b=57a - b=-43Adding: 2a=14→a=7→b=57 -7=50.So pairs (50,7) and (7,50). Check these:a=50, b=7. a² +b²=2500 +49=2549. Divide by a +b=57. 57×44=2508. 2549 -2508=41. So quotient q=44, remainder r=41. q² +r=1936 +41=1977. Correct.Similarly, a=7, b=50 gives the same. So these are also solutions.Therefore, for q=44, we have four solutions: (50,37),(37,50),(50,7),(7,50).But wait, earlier when we tried s=45, we didn't find solutions. But here, s=87 and s=57 give us valid solutions. So maybe I missed these when incrementally checking s=45,46,... because these s values are smaller than the required s for larger q. Wait, but for q=44, s must be greater than r=41. So s can be 87 or 57, both greater than 41. Therefore, these solutions are valid.So, so far, we have four solutions from q=44. Let's check if there are solutions for other q.Take q=43. Wait, earlier attempts didn't find solutions, but let's try using this discriminant method.For q=43, r=1977 -43²=1977 -1849=128. So r=128. So s>128.But using the discriminant approach:We have D = -s² +2q*s +3954 -2q²= -s² +86s +3954 -3698= -s² +86s +256. Wait, for q=43, D=-s² +86s +256.Set D=k². Then:-s² +86s +256=k²Rearranged:s² -86s -256 +k²=0Same as before, solve for s:s=(86 ± sqrt(86² +4*(256 -k²)))/2=(86 ± sqrt(7396 +1024 -4k²))/2=(86 ± sqrt(8420 -4k²))/2To have integer s, sqrt(8420 -4k²) must be integer. Let’s denote m=sqrt(8420 -4k²). Then m must be even, since 8420 -4k² is divisible by 4. Let m=2n. Then:4n²=8420 -4k² → n² +k²=2105.So looking for integer solutions (n,k) such that n² +k²=2105.2105 factors into 5×421. 421 is a prime. Check if 421 is congruent 1 mod4: 421 divided by4 is 105.25, so 421≡1 mod4. Hence, 421 can be expressed as sum of two squares. Let's find such pairs.First, sqrt(2105)≈45.89. So n and k can be up to 45.Check for n=44: 44²=1936. 2105-1936=169=13². So 44² +13²=1936 +169=2105. So (n,k)=(44,13),(13,44),(44,-13), etc. Considering positive integers, (44,13) and (13,44).Therefore, m=2n=88 or 26.Then sqrt(8420 -4k²)=m=88 or 26.So:Case1: m=88. Then s=(86 +88)/2=174/2=87 or s=(86 -88)/2=-2/2=-1. Disregard negative s.Case2: m=26. Then s=(86 +26)/2=112/2=56 or s=(86 -26)/2=60/2=30.So possible s=87,56,30.Check s=87. Then D=k²=13²=169. So discriminant D=169=(a - b)^2.So a + b=87, a - b=±13. Then:If a - b=13:a + b=87a - b=13 → a=50, b=37If a - b=-13:a=37, b=50So pairs (50,37),(37,50). But wait, these are the same as before for q=44. But for q=43, would this be the case?Wait, s=87, which would correspond to q=43. Wait, but in our discriminant approach, we have s=87 for q=43. Let's check:For a=50, b=37: a + b=87. Then a² +b²=50² +37²=2500 +1369=3869. Divided by s=87. 87×44=3828, remainder 41. So quotient is 44, remainder 41. But we're considering q=43 here, which contradicts. Therefore, something is wrong.Ah, here's the confusion. The value of q in this case is 43, but when we computed the division, we get q=44. Therefore, there's inconsistency. Which means that my discriminant approach might be mixing up q values.Wait, this seems like a problem. When we set q=43, but the solutions obtained actually result in q=44. So likely, this method is overlapping q values. Therefore, this approach might not be reliable.Alternatively, the equations for different q can overlap, leading to same pairs (a,b) but different q, but since q is uniquely determined by (a,b), this suggests a mistake in the method.Therefore, perhaps I need to ensure that when solving for a particular q, the resulting quotient when dividing a² +b² by a +b is indeed q. So even if algebraically the equations allow for s and ab, we need to verify that q is indeed the quotient.For example, when we considered q=43 and got s=87, which gives a=50, b=37. But when we actually divide 3869 by87, we get q=44, not 43. Hence, this solution is invalid for q=43. Therefore, our discriminant approach might give mathematical solutions, but they don't satisfy the original problem's condition that q is the quotient. Hence, we need to filter out such cases.Therefore, this method needs an additional check: after finding a and b, compute q and r by dividing a² +b² by a +b, and verify that q is indeed the one we started with.In the previous example, for q=43, s=87 gives a=50, b=37, but the actual quotient is 44, so this should be rejected. Hence, solutions for q=43 do not exist, even though the discriminant method suggests so.Therefore, the only valid solutions are those found for q=44. Wait, but when we considered q=44, we got s=87 and s=57. Then for s=87, the quotient is 44, which matches q=44. Similarly, for s=57, let's check:Take a=50, b=7. Then a +b=57. a² +b²=2500 +49=2549. Divide by57. 57×44=2508. 2549-2508=41. So quotient q=44, remainder r=41. So correct.Similarly, a=7, b=50 same.Therefore, these solutions are valid for q=44. Hence, the only valid solutions are when q=44.Therefore, the pairs are (50,37),(37,50),(50,7),(7,50).Wait, but let's check other possible q's. Let me take q=29.Wait, let's think. If q is small, then r=1977 -q² is large, so s must be greater than r, which is large. For example, q=10, r=1977 -100=1877. Then s>1877. Then, a +b>1877, which would make a and b very large. But likely, the discriminant would be negative. Let me check with q=29.q=29, r=1977 -841=1136. So s>1136. Then:D = -s² +2*29*s +3954 -2*841= -s² +58s +3954 -1682= -s² +58s +2272Set this equal to k². Then:-s² +58s +2272 =k² → s² -58s -2272 +k²=0Quadratic in s: s=(58 ± sqrt(3364 +4*(2272 -k²)))/2=(58 ± sqrt(3364 +9088 -4k²))/2=(58 ± sqrt(12452 -4k²))/2.sqrt(12452 -4k²) must be integer. Let m=sqrt(12452 -4k²). Then m²=12452 -4k². So m must be even: m=2n. Then:4n²=12452 -4k² →n² +k²=31133113 is a prime? Let me check. Divided by 3: 3+1+1+3=8, not divisible by3. 5: last digit 3, no. 7: 3113/7=444.7, no. 11: 3113/11=283, which is 11*283=3113. 283 is a prime. 283≡3 mod4. Therefore, 3113=11*283. Since 283≡3 mod4, it cannot be expressed as sum of two squares. Therefore, n² +k²=3113 has no solutions. Therefore, q=29 has no solutions.Similarly, this suggests that only q=44 gives solutions. Let me check another q= floor(sqrt(1977))=44. So maybe only q=44 yields solutions. Alternatively, maybe q=43, but as we saw, the solutions actually have q=44.Wait, but let me check q=21. Then r=1977 -441=1536. s>1536. Then, D=-s² +42s +3954 -2*441= -s² +42s +3954 -882= -s² +42s +3072. Set equal to k²:-s² +42s +3072=k² →s² -42s -3072 +k²=0. Solve s=(42 ± sqrt(1764 +4*(3072 -k²)))/2=(42 ± sqrt(1764 +12288 -4k²))/2=(42 ± sqrt(14052 -4k²))/2. sqrt(14052 -4k²) must be integer. Let m=sqrt(14052 -4k²), so m even, m=2n. Then n² +k²=3513. Check if 3513 is expressible as sum of two squares. 3513 factors: 3×1171. 1171 is prime. 1171≡3 mod4, so cannot be expressed as sum of two squares. Hence, no solutions.Therefore, perhaps q=44 is the only one that works.Thus, the only solutions are the pairs we found earlier: (50,37),(37,50),(50,7),(7,50). Let's check another pair in these solutions.Take a=50, b=7. a + b=57. a² +b²=2500 +49=2549. 2549 divided by57 is 44*57=2508, remainder=2549-2508=41. q=44, r=41. q² +r=44² +41=1936 +41=1977. Correct.Similarly, a=37, b=50 gives same.Therefore, these four pairs are the solutions.Final AnswerThe pairs are boxed{(50, 7)}, boxed{(7, 50)}, boxed{(37, 50)}, and boxed{(50, 37)}.</think>Given (a, b in mathbb{N}), we need to find all pairs ((a, b)) such that when (a^2 + b^2) is divided by (a + b), the quotient (q) and remainder (r) satisfy (q^2 + r = 1977).1. Equations and Constraints: - (a^2 + b^2 = (a + b)q + r) - (q^2 + r = 1977) - (0 leq r < a + b)2. Expressing (r): - From (q^2 + r = 1977), we get (r = 1977 - q^2). - Since (r < a + b), we have (1977 - q^2 < a + b).3. Quadratic Formulation: - Using (a^2 + b^2 = (a + b)q + r), substitute (r): [ a^2 + b^2 = (a + b)q + (1977 - q^2) ] - Rewrite using (s = a + b) and (ab): [ s^2 - 2ab = qs + 1977 - q^2 ] - Solving for (ab): [ ab = frac{s^2 - qs - 1977 + q^2}{2} ]4. Discriminant Check: - The discriminant of the quadratic equation (x^2 - sx + ab = 0) must be a perfect square: [ s^2 - 4ab = -s^2 + 2qs + 3954 - 2q^2 ] - This discriminant must be a perfect square for (a) and (b) to be integers.5. Solving for (q): - For (q = 44), (r = 1977 - 44^2 = 41). - (s = a + b) must be greater than 41. - Solving for (s) and checking the discriminant, we find valid pairs ((a, b)) when (s = 87) and (s = 57).6. Valid Solutions: - For (s = 87), solving (a + b = 87) and (a - b = pm 13) gives pairs ((50, 37)) and ((37, 50)). - For (s = 57), solving (a + b = 57) and (a - b = pm 43) gives pairs ((50, 7)) and ((7, 50)).7. Verification: - Check that these pairs satisfy (q = 44) and (r = 41) when divided by (a + b), confirming (q^2 + r = 1977).The pairs ((a, b)) that satisfy the given conditions are:[boxed{(50, 7)}, boxed{(7, 50)}, boxed{(37, 50)}, boxed{(50, 37)}]]

🔑:Given a^{2}+b^{2}=q(a+b)+r,q^{2}+r=1977 quad(0 leqslant r < q^2),if q geq 45, then q^2 geq 2025 > 1977, a contradiction. If q leq 43, then r > 128, also a contradiction. Hence q=44, r=41, and a^{2}+b^{2}=44(a+b)+41.This simplifies to (a-22)^{2}+(b-22)^{2}=1009. The solutions are(a, b)=(50,37),(50,7),(37,50),(7,50).