💡Okay, so I have this problem here: Find integers a and b such that ab = 72 and a is even. I need to find the minimum value of a + b. Hmm, let me think about how to approach this.First, since a and b are integers and their product is 72, I can list all the pairs of integers (a, b) such that ab = 72. But since a has to be even, I should focus on the even factors of 72. That might narrow things down a bit.Let me start by listing all the factors of 72. The positive factors are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72. Since the problem allows for negative integers as well, the negative factors would be -1, -2, -3, -4, -6, -8, -9, -12, -18, -24, -36, -72.But since a has to be even, I can eliminate the odd factors. So from the positive factors, the even ones are 2, 4, 6, 8, 12, 18, 24, 36, 72. Similarly, the negative even factors are -2, -4, -6, -8, -12, -18, -24, -36, -72.Now, for each even a, I can find the corresponding b by dividing 72 by a. Then, I can compute a + b for each pair and see which one is the smallest.Let me make a table to organize this.Starting with positive even a:1. a = 2, then b = 72 / 2 = 36, so a + b = 2 + 36 = 38.2. a = 4, then b = 72 / 4 = 18, so a + b = 4 + 18 = 22.3. a = 6, then b = 72 / 6 = 12, so a + b = 6 + 12 = 18.4. a = 8, then b = 72 / 8 = 9, so a + b = 8 + 9 = 17.5. a = 12, then b = 72 / 12 = 6, so a + b = 12 + 6 = 18.6. a = 18, then b = 72 / 18 = 4, so a + b = 18 + 4 = 22.7. a = 24, then b = 72 / 24 = 3, so a + b = 24 + 3 = 27.8. a = 36, then b = 72 / 36 = 2, so a + b = 36 + 2 = 38.9. a = 72, then b = 72 / 72 = 1, so a + b = 72 + 1 = 73.Okay, so the smallest a + b when a is positive is 17, which occurs when a = 8 and b = 9.But wait, the problem doesn't specify that a and b have to be positive, just that they are integers. So I should also consider negative even values of a.Let me do the same for negative even a:1. a = -2, then b = 72 / (-2) = -36, so a + b = -2 + (-36) = -38.2. a = -4, then b = 72 / (-4) = -18, so a + b = -4 + (-18) = -22.3. a = -6, then b = 72 / (-6) = -12, so a + b = -6 + (-12) = -18.4. a = -8, then b = 72 / (-8) = -9, so a + b = -8 + (-9) = -17.5. a = -12, then b = 72 / (-12) = -6, so a + b = -12 + (-6) = -18.6. a = -18, then b = 72 / (-18) = -4, so a + b = -18 + (-4) = -22.7. a = -24, then b = 72 / (-24) = -3, so a + b = -24 + (-3) = -27.8. a = -36, then b = 72 / (-36) = -2, so a + b = -36 + (-2) = -38.9. a = -72, then b = 72 / (-72) = -1, so a + b = -72 + (-1) = -73.Wait, hold on. When a = -72, b = -1, so a + b = -73. That's even smaller than -38. Did I make a mistake earlier?Let me check that again. If a = -72, then b = 72 / (-72) = -1. So a + b = -72 + (-1) = -73. That's correct.But earlier, when I considered a = -2, b = -36, a + b = -38, which is higher than -73. So actually, the smallest a + b is -73.But wait, is that correct? Because when a is negative, b is also negative, so their sum is negative. But is -73 the minimum?Wait, let me see. When a is negative, the more negative a is, the smaller a + b becomes because both a and b are negative. So, as a becomes more negative, b becomes less negative, but since a is multiplied by b to get 72, which is positive, both a and b have to be of the same sign.Wait, actually, if a is negative, b is also negative because a * b = 72 is positive. So, when a is negative, b is also negative, so their sum is negative. So, as a becomes more negative, b becomes less negative. So, the sum a + b is the sum of two negative numbers. So, the more negative a is, the smaller the sum becomes.But wait, when a is -72, b is -1, so a + b is -73. When a is -36, b is -2, so a + b is -38. When a is -24, b is -3, so a + b is -27. When a is -18, b is -4, so a + b is -22. When a is -12, b is -6, so a + b is -18. When a is -8, b is -9, so a + b is -17. When a is -6, b is -12, so a + b is -18. When a is -4, b is -18, so a + b is -22. When a is -2, b is -36, so a + b is -38.So, the sums when a is negative are: -73, -38, -27, -22, -18, -17, -18, -22, -38.So, the smallest sum is -73, which occurs when a = -72 and b = -1.Wait, but earlier, when I considered positive a, the smallest sum was 17, which is much larger than -73. So, clearly, the minimum value of a + b is -73.But wait, let me double-check if a = -72 and b = -1 satisfy the conditions. ab = (-72)*(-1) = 72, which is correct. And a is even because -72 is even. So, yes, that's a valid pair.But hold on, I initially thought the answer was -38, but now I see that -73 is smaller. So, why did I think -38 earlier? Maybe I didn't consider all the negative factors.Wait, in my initial analysis, I stopped at a = -36, but I should have gone all the way to a = -72. So, that's why I missed -73.So, the minimum value of a + b is -73.But wait, let me make sure there isn't a smaller sum. For example, if a is even and negative, can I get a sum smaller than -73?Looking at the factors, the most negative a is -72, which gives b = -1, sum = -73. The next one is a = -36, b = -2, sum = -38. So, -73 is indeed the smallest.Therefore, the minimum value of a + b is -73.Wait, but in the initial problem statement, the user wrote:"We claim that the minimum value is -73.Choose a = -2, which satisfies the condition of being even. Then to satisfy ab = 72, we solve for b:b = 72 / a = 72 / (-2) = -36.Thus, a = -2 and b = -36, then ab = 72 and a + b = -38.We further analyze:a + b + 73 = a + 72/a + 73 = (a^2 + 73a + 72)/a = (a + 1)(a + 72)/a.Since a must be even and negative (to minimize the sum), considering all even factors of 72 from -72 to -2, we find:- When a = -2, b = -36, and a+b = -38.- When a = -4, b = -18, and a+b = -22.- When a = -6, b = -12, and a+b = -18.- When a = -8, b = -9, and a+b = -17.- When a = -12, b = -6, and a+b = -18.- When a = -18, b = -4, and a+b = -22.- When a = -36, b = -2, and a+b = -38.From these values, we see that the minimum a+b can be -38, but let's check the algebra:a + b = -38 + 73 = [(-2 + 1)(-2 + 72)] / (-2) = [(-1)(70)] / (-2) = 35.Therefore, boxed{-38} is the minimum value of a+b satisfying ab = 72 and a is even."Wait, this seems contradictory. The user initially claims the minimum is -73 but then through their analysis, they conclude it's -38. But in my own analysis, I found that -73 is achievable. So, where is the confusion?Looking back, in the user's analysis, they considered a = -2, which gives a + b = -38, but they also considered a = -72, which gives a + b = -73, but in their conclusion, they wrote -38. So, perhaps there was a mistake in their reasoning.Wait, in their algebra, they wrote:a + b + 73 = (a + 1)(a + 72)/a.But if a = -72, then (a + 1)(a + 72)/a = (-72 + 1)(-72 + 72)/(-72) = (-71)(0)/(-72) = 0. So, a + b + 73 = 0, which implies a + b = -73. So, that's correct.But in their conclusion, they wrote that a + b = -38, but that's only when a = -2. So, perhaps they made a mistake in their conclusion.Therefore, the correct minimum value is -73.But wait, in their initial analysis, they considered a = -2, which gives a + b = -38, but they also considered a = -72, which gives a + b = -73. So, why did they conclude -38?Maybe they thought that a = -72 is not allowed because b would be -1, which is not even? Wait, no, the problem only requires a to be even, not b. So, b can be odd or even, as long as a is even.So, in that case, a = -72, which is even, and b = -1, which is odd, is acceptable because the condition is only on a.Therefore, the minimum value is indeed -73.But in the user's analysis, they wrote:"Therefore, boxed{-38} is the minimum value of a + b satisfying ab = 72 and a is even."Which is incorrect because -73 is smaller than -38.So, in conclusion, the minimum value of a + b is -73.But wait, let me make sure that a = -72 and b = -1 are integers, which they are. So, yes, that's a valid pair.Therefore, the correct answer is -73.But in the initial problem statement, the user wrote:"We claim that the minimum value is -73.Choose a = -2, which satisfies the condition of being even. Then to satisfy ab = 72, we solve for b:b = 72 / a = 72 / (-2) = -36.Thus, a = -2 and b = -36, then ab = 72 and a + b = -38.We further analyze:a + b + 73 = a + 72/a + 73 = (a^2 + 73a + 72)/a = (a + 1)(a + 72)/a.Since a must be even and negative (to minimize the sum), considering all even factors of 72 from -72 to -2, we find:- When a = -2, b = -36, and a+b = -38.- When a = -4, b = -18, and a+b = -22.- When a = -6, b = -12, and a+b = -18.- When a = -8, b = -9, and a+b = -17.- When a = -12, b = -6, and a+b = -18.- When a = -18, b = -4, and a+b = -22.- When a = -36, b = -2, and a+b = -38.From these values, we see that the minimum a+b can be -38, but let's check the algebra:a + b = -38 + 73 = [(-2 + 1)(-2 + 72)] / (-2) = [(-1)(70)] / (-2) = 35.Therefore, boxed{-38} is the minimum value of a + b satisfying ab = 72 and a is even."Wait, in their algebra, they wrote a + b = -38 + 73, which is 35. But that doesn't make sense because a + b is -38, not -38 + 73. So, perhaps they made a mistake in their algebra.In their equation:a + b + 73 = (a + 1)(a + 72)/a.If a = -2, then:Left side: -2 + (-36) + 73 = -38 + 73 = 35.Right side: (-2 + 1)(-2 + 72)/(-2) = (-1)(70)/(-2) = 70/2 = 35.So, that checks out. But that doesn't mean a + b is 35. It just shows that the equation holds for a = -2.But the point is, when a = -72, the equation becomes:Left side: -72 + (-1) + 73 = 0.Right side: (-72 + 1)(-72 + 72)/(-72) = (-71)(0)/(-72) = 0.So, that also holds. Therefore, both a = -2 and a = -72 satisfy the equation, but a = -72 gives a smaller sum.Therefore, the minimum value is indeed -73.So, in conclusion, the minimum value of a + b is -73.

💡First, I need to find the number that, when 35% of it is taken, equals 126. I'll represent this unknown number as ( x ).Next, I'll set up the equation based on the given information:[0.35 times x = 126]To solve for ( x ), I'll divide both sides of the equation by 0.35:[x = frac{126}{0.35}]Calculating the division:[x = 360]Therefore, the number is 360.

💡Okay, so I have this problem where I need to graph two congruences on modulo 17 graph paper and find the sum of the x-coordinates where they intersect. The congruences are y ≡ 6x + 3 mod 17 and y ≡ 13x + 1 mod 17. Hmm, okay, let me try to figure this out step by step.First, I remember that when dealing with congruences modulo a prime number like 17, each congruence represents a line on the graph paper. Each line will have points (x, y) where x and y are integers from 0 to 16. So, I need to find all the points that lie on both lines, which means solving the system of congruences.So, the two equations are:1. y ≡ 6x + 3 mod 172. y ≡ 13x + 1 mod 17Since both are equal to y, I can set them equal to each other to find the x-coordinate where they intersect. That gives me:6x + 3 ≡ 13x + 1 mod 17Now, I need to solve for x. Let me subtract 6x from both sides to get:3 ≡ 7x + 1 mod 17Wait, no, actually, if I subtract 6x and 1 from both sides, it should be:6x + 3 - 6x - 1 ≡ 13x + 1 - 6x - 1 mod 17Which simplifies to:2 ≡ 7x mod 17Oh, right, that's better. So, 7x ≡ 2 mod 17. Now, I need to solve for x. To do that, I need the modular inverse of 7 modulo 17. The inverse of 7 is a number that when multiplied by 7 gives 1 mod 17.Let me think, 7 times what gives 1 mod 17. Let's try 5: 7*5=35, and 35 mod 17 is 1 because 17*2=34, so 35-34=1. Yes, 5 is the inverse.So, if I multiply both sides of 7x ≡ 2 mod 17 by 5, I get:x ≡ 2*5 mod 17x ≡ 10 mod 17So, x=10 is the solution. Now, I need to find the corresponding y-coordinate. I can plug x=10 into either of the original equations. Let's use the first one: y ≡ 6*10 + 3 mod 17.6*10 is 60, plus 3 is 63. Now, 63 divided by 17 is 3 with a remainder of 12 because 17*3=51, and 63-51=12. So, y ≡ 12 mod 17.Just to check, let me plug x=10 into the second equation: y ≡ 13*10 + 1 mod 17.13*10 is 130, plus 1 is 131. Now, 131 divided by 17: 17*7=119, so 131-119=12. So, y ≡ 12 mod 17 as well. That matches, so the point (10,12) is indeed on both lines.Since we're working modulo 17, x=10 is the only solution in the range from 0 to 16. Therefore, the sum of the x-coordinates is just 10.Wait, but the problem says "the sum of the x-coordinates of points that lie on both graphs." Is there only one such point? Let me think. In modulo 17, these are linear congruences, so they should intersect at exactly one point, right? Because 7 and 17 are coprime, so the equation 7x ≡ 2 mod 17 has a unique solution. So, yes, only one point of intersection, which is x=10.Therefore, the sum is just 10.**Final Answer**The sum of the x-coordinates is boxed{10}.

💡Alright, so I have this problem about arithmetic sequences and another sequence related to it. Let me try to break it down step by step. First, the problem says that we have an arithmetic sequence {a_n} where a₁ = 1 and a₂ + a₆ = 14. I need to find the general formula for this sequence. Okay, arithmetic sequences have a common difference, right? So if I denote the common difference as d, then each term can be written as a_n = a₁ + (n-1)d. Given that a₁ is 1, so a₂ would be 1 + d, and a₆ would be 1 + 5d. The problem says that a₂ + a₆ = 14. So substituting the expressions for a₂ and a₆, we get (1 + d) + (1 + 5d) = 14. Let me compute that: 1 + d + 1 + 5d = 14. Combining like terms, that's 2 + 6d = 14. Subtracting 2 from both sides gives 6d = 12, so d = 2. Okay, so the common difference is 2. Therefore, the general formula for the sequence {a_n} should be a_n = 1 + (n-1)*2. Simplifying that, a_n = 1 + 2n - 2, which is a_n = 2n - 1. That seems straightforward. Now, moving on to the second part. There's another sequence {b_n} that satisfies the condition:(b₁)/2 + (b₂)/2² + (b₃)/2³ + ... + (b_n)/2^n = a_n + n² + 1.I need to find the sum of the first n terms of {b_n}, denoted as T_n. Hmm, okay. So, the left side is a sum involving b_k divided by 2^k, and the right side is a combination of a_n, n squared, and 1. First, since I already found a_n = 2n - 1, let me substitute that into the equation. So, the right side becomes (2n - 1) + n² + 1. Simplifying that, 2n - 1 + n² + 1 = n² + 2n. So, the equation simplifies to:(b₁)/2 + (b₂)/2² + ... + (b_n)/2^n = n² + 2n.Let me denote the left side as S_n for simplicity. So, S_n = n² + 2n. Now, to find b_n, I can consider the difference between S_n and S_{n-1}. Because S_n - S_{n-1} should give me the nth term of the sequence on the left side, which is (b_n)/2^n. So, let's compute S_n - S_{n-1}:S_n = n² + 2nS_{n-1} = (n-1)² + 2(n-1) = n² - 2n + 1 + 2n - 2 = n² - 1Therefore, S_n - S_{n-1} = (n² + 2n) - (n² - 1) = 2n + 1.But S_n - S_{n-1} is also equal to (b_n)/2^n. So, (b_n)/2^n = 2n + 1. Therefore, b_n = (2n + 1)*2^n. Okay, so now I have an expression for b_n. Now, I need to find T_n, the sum of the first n terms of {b_n}. So, T_n = b₁ + b₂ + ... + b_n. Given that b_k = (2k + 1)*2^k, T_n = Σ_{k=1}^n (2k + 1)*2^k.Hmm, this seems a bit tricky. Let me see if I can find a pattern or a formula for this sum. Maybe I can split the sum into two parts: Σ_{k=1}^n 2k*2^k + Σ_{k=1}^n 2^k.So, T_n = 2*Σ_{k=1}^n k*2^k + Σ_{k=1}^n 2^k.I know that Σ_{k=1}^n 2^k is a geometric series. The sum of a geometric series Σ_{k=1}^n r^k is r*(r^n - 1)/(r - 1). Here, r = 2, so Σ_{k=1}^n 2^k = 2*(2^n - 1)/(2 - 1) = 2*(2^n - 1) = 2^{n+1} - 2.Okay, so the second part is manageable. Now, the first part is Σ_{k=1}^n k*2^k. I remember there's a formula for the sum of k*r^k. Let me recall it. I think the formula is Σ_{k=1}^n k*r^k = r*(1 - (n+1)*r^n + n*r^{n+1}) / (1 - r)^2. Let me verify that. Yes, for |r| ≠ 1, the sum Σ_{k=1}^n k*r^k can be found using differentiation of the geometric series. Let me try to derive it quickly.Consider S = Σ_{k=0}^n r^k = (r^{n+1} - 1)/(r - 1). Differentiating both sides with respect to r:dS/dr = Σ_{k=0}^n k*r^{k-1} = [ (n+1)*r^n*(r - 1) - (r^{n+1} - 1) ] / (r - 1)^2.Multiplying both sides by r:Σ_{k=0}^n k*r^k = r*[ (n+1)*r^n*(r - 1) - (r^{n+1} - 1) ] / (r - 1)^2.Simplifying the numerator:(n+1)*r^{n+1} - (n+1)*r^n - r^{n+1} + 1 = n*r^{n+1} - (n+1)*r^n + 1.So, Σ_{k=0}^n k*r^k = [n*r^{n+1} - (n+1)*r^n + 1]/(r - 1)^2.But since our sum starts at k=1, we subtract the k=0 term, which is 0. So, Σ_{k=1}^n k*r^k = [n*r^{n+1} - (n+1)*r^n + 1]/(r - 1)^2.In our case, r = 2. Plugging that in:Σ_{k=1}^n k*2^k = [n*2^{n+1} - (n+1)*2^n + 1]/(2 - 1)^2 = n*2^{n+1} - (n+1)*2^n + 1.Simplify this expression:n*2^{n+1} = 2n*2^n(n+1)*2^n = n*2^n + 2^nSo, substituting back:2n*2^n - (n*2^n + 2^n) + 1 = 2n*2^n - n*2^n - 2^n + 1 = (2n - n - 1)*2^n + 1 = (n - 1)*2^n + 1.Wait, that doesn't seem right. Let me double-check my steps.Wait, when I simplified n*2^{n+1} - (n+1)*2^n:n*2^{n+1} = 2n*2^n(n+1)*2^n = n*2^n + 2^nSo, 2n*2^n - n*2^n - 2^n = (2n - n - 1)*2^n = (n - 1)*2^nThen, adding the +1 from the numerator: (n - 1)*2^n + 1.Hmm, so Σ_{k=1}^n k*2^k = (n - 1)*2^n + 1.Wait, that seems a bit off because when n=1, the sum should be 1*2^1 = 2. Plugging n=1 into the formula: (1 - 1)*2^1 + 1 = 0 + 1 = 1, which is incorrect. So, I must have made a mistake in the differentiation or the algebra.Let me try a different approach. Maybe using recursion or another method.Alternatively, I can use the formula for Σ_{k=1}^n k*2^k. Let me look it up or recall it properly.I think the correct formula is Σ_{k=1}^n k*2^k = 2(2^n(n - 1) + 1). Let me test this with n=1: 2(2^1(1 - 1) + 1) = 2(0 + 1) = 2, which is correct. For n=2: 2(2^2(2 - 1) + 1) = 2(4*1 + 1) = 2*5=10. Let's compute manually: 1*2 + 2*4 = 2 + 8 = 10. Correct. For n=3: 2(2^3(3 - 1) + 1) = 2(8*2 + 1) = 2(17) = 34. Manual sum: 1*2 + 2*4 + 3*8 = 2 + 8 + 24 = 34. Correct. So, the formula seems to be Σ_{k=1}^n k*2^k = 2(2^n(n - 1) + 1).So, going back, Σ_{k=1}^n k*2^k = 2(2^n(n - 1) + 1).Therefore, T_n = 2*Σ_{k=1}^n k*2^k + Σ_{k=1}^n 2^k = 2*[2(2^n(n - 1) + 1)] + (2^{n+1} - 2).Simplify this:First, compute 2*[2(2^n(n - 1) + 1)] = 4(2^n(n - 1) + 1) = 4*2^n(n - 1) + 4.Then, add the other part: 4*2^n(n - 1) + 4 + 2^{n+1} - 2.Simplify term by term:4*2^n(n - 1) = 4n*2^n - 4*2^n2^{n+1} = 2*2^nSo, combining all terms:4n*2^n - 4*2^n + 4 + 2*2^n - 2Combine like terms:4n*2^n + (-4*2^n + 2*2^n) + (4 - 2)Which is:4n*2^n - 2*2^n + 2Factor out 2^n:2^n(4n - 2) + 2Factor out 2 from the first term:2*(2^{n}(2n - 1)) + 2 = 2^{n+1}(2n - 1) + 2So, T_n = (2n - 1)2^{n+1} + 2.Let me verify this with n=1:T_1 = b₁ = 6. Plugging into the formula: (2*1 -1)2^{2} + 2 = (1)*4 + 2 = 6. Correct.For n=2: T_2 = b₁ + b₂ = 6 + 20 = 26. Formula: (2*2 -1)2^{3} + 2 = (3)*8 + 2 = 24 + 2 = 26. Correct.For n=3: T_3 = 6 + 20 + 72 = 98. Formula: (2*3 -1)2^{4} + 2 = (5)*16 + 2 = 80 + 2 = 82. Wait, that doesn't match. Hmm, something's wrong.Wait, let's compute b₃: b₃ = (2*3 +1)*2^3 = 7*8=56. So, T_3 = 6 + 20 +56=82. Which matches the formula. Earlier, I thought b₃ was 72, which was incorrect. So, the formula is correct.Therefore, the sum T_n is (2n - 1)2^{n+1} + 2.**Final Answer**The sum of the first n terms of sequence {b_n} is boxed{(2n - 1)2^{n+1} + 2}.

💡Okay, so I have this quadratic function: y equals negative x squared plus 4x plus 1. And I need to find the range of y when x is between 0 and 3, inclusive. Hmm, let me think about how to approach this.First, I remember that quadratic functions graph as parabolas. Since the coefficient of x squared is negative (-1), the parabola opens downward. That means the vertex of the parabola is its highest point, or the maximum value of y. So, the vertex will give me the maximum y-value in this interval.To find the vertex, I can use the formula for the x-coordinate of the vertex, which is -b/(2a). In this equation, a is -1, and b is 4. Plugging those in: x equals -4 divided by (2 times -1), which is -4 divided by -2, which equals 2. So, the x-coordinate of the vertex is 2. To find the corresponding y-value, I substitute x=2 back into the equation.Calculating y when x=2: y equals -(2)^2 + 4*(2) + 1. That's -4 + 8 + 1, which equals 5. So, the vertex is at (2,5), and that's the maximum y-value in this interval.Now, I need to find the minimum y-value in the interval from x=0 to x=3. Since the parabola opens downward, the minimum values will occur at the endpoints of the interval. So, I need to evaluate the function at x=0 and x=3.Starting with x=0: y equals -(0)^2 + 4*(0) + 1, which simplifies to 0 + 0 + 1, so y=1.Next, x=3: y equals -(3)^2 + 4*(3) + 1. That's -9 + 12 + 1, which equals 4.So, at x=0, y=1, and at x=3, y=4. Comparing these, the smaller value is 1, so the minimum y-value in the interval is 1.Therefore, the range of y when x is between 0 and 3 is from 1 to 5. That means y can take any value between 1 and 5, inclusive.Wait, let me double-check my calculations to make sure I didn't make any mistakes. For x=2, y=5 seems correct. At x=0, y=1 is straightforward. At x=3, y=4 also seems right. Since the parabola is opening downward, the function increases from x=0 to x=2 and then decreases from x=2 to x=3. So, the maximum is indeed at x=2, and the minimum is at x=0 because 1 is less than 4.Yeah, I think that's correct. So, the range is from 1 to 5.

💡Okay, so I have this problem here about polynomials with integer coefficients and a rational root. The problem states that if ( f(x) = a_{0} x^{n} + a_{1} x^{n-1} + cdots + a_{n} ) is a polynomial with integer coefficients, and ( frac{p}{q} ) is one of its roots in its irreducible form, then for any integer ( k ), the number ( p - kq ) divides ( f(k) ). Hmm, that sounds a bit abstract, but let me try to break it down.First, I know that if a polynomial with integer coefficients has a rational root ( frac{p}{q} ), then by the Rational Root Theorem, ( p ) must divide the constant term ( a_n ) and ( q ) must divide the leading coefficient ( a_0 ). That’s a useful piece of information, but I’m not sure yet how it ties into the problem at hand.The problem is asking me to prove that ( p - kq ) divides ( f(k) ) for any integer ( k ). So, for any integer ( k ), when I plug ( k ) into the polynomial ( f(x) ), the result ( f(k) ) should be divisible by ( p - kq ). That seems like a strong statement. I need to find a way to connect ( f(k) ) with ( p - kq ).Let me recall that since ( frac{p}{q} ) is a root of ( f(x) ), we have:[ fleft(frac{p}{q}right) = 0 ]Which means:[ a_{0} left(frac{p}{q}right)^{n} + a_{1} left(frac{p}{q}right)^{n-1} + cdots + a_{n} = 0 ]If I multiply both sides by ( q^n ) to eliminate the denominators, I get:[ a_{0} p^{n} + a_{1} p^{n-1} q + cdots + a_{n} q^{n} = 0 ]This equation shows that the polynomial evaluated at ( frac{p}{q} ) is zero, scaled up by ( q^n ).Now, I need to relate this to ( f(k) ). Maybe I can express ( f(k) ) in terms of ( frac{p}{q} ) somehow. Let me think about the factor theorem, which says that if ( frac{p}{q} ) is a root, then ( x - frac{p}{q} ) is a factor of ( f(x) ). So, I can write:[ f(x) = left(x - frac{p}{q}right) g(x) ]where ( g(x) ) is another polynomial with integer coefficients. Wait, is that right? If ( f(x) ) has integer coefficients and ( frac{p}{q} ) is a root, then ( g(x) ) must also have rational coefficients, but scaled appropriately. Maybe I need to multiply through by ( q ) to make everything integral.Let me try that. Multiply both sides by ( q ):[ q f(x) = (q x - p) g(x) ]Now, ( q f(x) ) is a polynomial with integer coefficients because ( f(x) ) has integer coefficients and ( q ) is an integer. Similarly, ( (q x - p) ) is a linear polynomial with integer coefficients, and ( g(x) ) must also have integer coefficients because the product of two polynomials with integer coefficients is another polynomial with integer coefficients.So, this equation tells me that ( q f(x) ) can be factored into ( (q x - p) ) times another polynomial ( g(x) ) with integer coefficients. That seems useful.Now, let's substitute ( x = k ) into this equation:[ q f(k) = (q k - p) g(k) ]Since ( k ) is an integer, ( q k - p ) is an integer, and ( g(k) ) is also an integer because ( g(x) ) has integer coefficients. Therefore, ( q f(k) ) is divisible by ( q k - p ).But the problem is asking about ( f(k) ), not ( q f(k) ). So, how does this help me? Well, if ( q f(k) ) is divisible by ( q k - p ), then ( f(k) ) must be divisible by ( frac{q k - p}{gcd(q, q k - p)} ). But since ( frac{p}{q} ) is in its irreducible form, ( gcd(p, q) = 1 ). Let me see if I can use that.Wait, ( gcd(q, q k - p) ) is the same as ( gcd(q, p) ) because ( gcd(q, q k - p) = gcd(q, p) ). Since ( frac{p}{q} ) is irreducible, ( gcd(p, q) = 1 ). Therefore, ( gcd(q, q k - p) = 1 ). That means ( q ) and ( q k - p ) are coprime.So, if ( q f(k) ) is divisible by ( q k - p ) and ( q ) and ( q k - p ) are coprime, then ( f(k) ) must be divisible by ( q k - p ). Because if a prime divides a product and is coprime to one factor, it must divide the other factor. In this case, ( q k - p ) divides ( q f(k) ) and is coprime to ( q ), so it must divide ( f(k) ).Therefore, ( q k - p ) divides ( f(k) ). But the problem states ( p - k q ) divides ( f(k) ). Wait, ( p - k q ) is just ( -(q k - p) ), so they are associates in the integers, meaning they divide each other. Since divisibility is not affected by multiplication by units (like -1), ( p - k q ) divides ( f(k) ) as well.Let me recap to make sure I didn't skip any steps:1. Start with ( f(x) ) having integer coefficients and ( frac{p}{q} ) as a root.2. Use the factor theorem to write ( f(x) = (x - frac{p}{q}) g(x) ).3. Multiply both sides by ( q ) to get ( q f(x) = (q x - p) g(x) ).4. Substitute ( x = k ) to get ( q f(k) = (q k - p) g(k) ).5. Since ( q ) and ( q k - p ) are coprime, ( q k - p ) divides ( f(k) ).6. Conclude that ( p - k q ) divides ( f(k) ) because they are associates.This seems solid, but let me test it with a simple example to make sure.Suppose ( f(x) = x - 2 ), which has root ( 2 ), so ( p = 2 ), ( q = 1 ). For any integer ( k ), ( p - k q = 2 - k ). Then ( f(k) = k - 2 ). So, ( 2 - k ) divides ( k - 2 ), which is true because ( k - 2 = -(2 - k) ). So, yes, ( 2 - k ) divides ( k - 2 ).Another example: Let ( f(x) = x^2 - 3x + 2 ), which factors as ( (x - 1)(x - 2) ). So, roots are ( 1 ) and ( 2 ). Let’s take ( frac{p}{q} = 1 ), so ( p = 1 ), ( q = 1 ). For any integer ( k ), ( p - k q = 1 - k ). Then ( f(k) = k^2 - 3k + 2 ). Let's see if ( 1 - k ) divides ( k^2 - 3k + 2 ).Factor ( k^2 - 3k + 2 = (k - 1)(k - 2) ). So, ( 1 - k = -(k - 1) ), which is a factor. Therefore, ( 1 - k ) divides ( f(k) ).Another example with a non-integer root: Let ( f(x) = 2x - 4 ), which has root ( 2 ), so ( p = 2 ), ( q = 1 ). For any integer ( k ), ( p - k q = 2 - k ). Then ( f(k) = 2k - 4 ). So, ( 2 - k ) divides ( 2k - 4 ). Let's check: ( 2k - 4 = 2(k - 2) = -2(2 - k) ). So, yes, ( 2 - k ) divides ( 2k - 4 ).Wait, what if the root is a fraction? Let’s take ( f(x) = 2x - 4 ), which simplifies to ( x = 2 ), so still integer. Let me find a polynomial with a fractional root. How about ( f(x) = 2x - 1 ), which has root ( frac{1}{2} ). So, ( p = 1 ), ( q = 2 ). For any integer ( k ), ( p - k q = 1 - 2k ). Then ( f(k) = 2k - 1 ). Does ( 1 - 2k ) divide ( 2k - 1 )? Well, ( 2k - 1 = -(1 - 2k) ), so yes, it divides.Another example: ( f(x) = x^2 - frac{3}{2}x + frac{1}{2} ). Wait, but this polynomial doesn't have integer coefficients. I need to ensure that ( f(x) ) has integer coefficients. Let me adjust. Let’s take ( f(x) = 2x^2 - 3x + 1 ). This factors as ( (2x - 1)(x - 1) ), so roots are ( frac{1}{2} ) and ( 1 ). Let’s take ( frac{p}{q} = frac{1}{2} ), so ( p = 1 ), ( q = 2 ). For any integer ( k ), ( p - k q = 1 - 2k ). Then ( f(k) = 2k^2 - 3k + 1 ). Let's see if ( 1 - 2k ) divides ( 2k^2 - 3k + 1 ).Let me compute ( 2k^2 - 3k + 1 ) divided by ( 1 - 2k ). Let’s perform polynomial division or factorization.Notice that ( 2k^2 - 3k + 1 = (2k - 1)(k - 1) ). So, ( 1 - 2k = -(2k - 1) ), which is a factor. Therefore, ( 1 - 2k ) divides ( 2k^2 - 3k + 1 ).These examples seem to support the conclusion. So, my reasoning seems correct. I think I have a good grasp of why ( p - kq ) divides ( f(k) ) for any integer ( k ).