💡First, I need to determine the approximate values of the square roots of 10 and 40 to understand the range we're dealing with.Next, I'll identify the smallest integer that is greater than the square root of 10 and the largest integer that is less than the square root of 40.Finally, I'll count the integers that fall between these two identified integers, inclusive, to find the total number of integers within the specified range.

💡First, I need to determine the number of stamps Nikka has from each category based on the given percentages of her 500-stamp collection.1. **Calculate the number of Chinese stamps:** - 40% of 500 stamps - ( 500 times 0.40 = 200 ) stamps2. **Calculate the number of US stamps:** - 25% of 500 stamps - ( 500 times 0.25 = 125 ) stamps3. **Calculate the number of Japanese stamps:** - 15% of 500 stamps - ( 500 times 0.15 = 75 ) stamps4. **Calculate the number of British stamps:** - 10% of 500 stamps - ( 500 times 0.10 = 50 ) stamps5. **Calculate the number of stamps from various other countries:** - The remaining percentage is 100% - (40% + 25% + 15% + 10%) = 10% - ( 500 times 0.10 = 50 ) stampsBy performing these calculations, I can determine the exact number of stamps Nikka has from each category.

💡Okay, so I have this problem about a sequence where each term is 11 raised to the power of the previous term. The sequence starts with t₁ = 11, then t₂ = 11¹¹, t₃ = 11¹¹¹¹, and so on. The question is asking for the remainder when t₅₀ is divided by 6. Hmm, that sounds a bit complicated, but maybe I can break it down step by step.First, I remember that when dealing with remainders, modular arithmetic is really useful. So, maybe I can find a pattern or something when these terms are divided by 6. Let me start by looking at the first few terms modulo 6.I know that 11 divided by 6 is 1 with a remainder of 5. So, 11 ≡ 5 mod 6. That means t₁ ≡ 5 mod 6. Now, t₂ is 11¹¹. Since 11 ≡ 5 mod 6, t₂ ≡ 5¹¹ mod 6. Hmm, what is 5¹¹ mod 6? Maybe there's a pattern with powers of 5 modulo 6.Let me compute the first few powers of 5 mod 6:- 5¹ = 5 mod 6- 5² = 25, which is 1 mod 6- 5³ = 5² * 5 = 1 * 5 = 5 mod 6- 5⁴ = 5³ * 5 = 5 * 5 = 25 ≡ 1 mod 6Oh, I see a pattern here. The powers of 5 modulo 6 cycle every 2 terms: 5, 1, 5, 1, and so on. So, the exponent determines whether it's 5 or 1 mod 6. If the exponent is odd, it's 5 mod 6; if it's even, it's 1 mod 6.So, for t₂ = 5¹¹ mod 6. Since 11 is odd, t₂ ≡ 5 mod 6. Then, t₃ = 11^{t₂} = 11⁵ mod 6. Again, 5 is odd, so t₃ ≡ 5 mod 6. Wait, is this always going to be 5 mod 6?Wait, let's think again. t₂ is 5¹¹ mod 6, which is 5. Then t₃ is 11^{t₂} = 11⁵ mod 6. Since 5 is odd, 11⁵ ≡ 5 mod 6. Similarly, t₄ = 11^{t₃} = 11⁵ mod 6, which is again 5. Hmm, so it seems like starting from t₂, every term is 5 mod 6.But wait, let me make sure. Maybe I'm missing something. Let's see:t₁ = 11 ≡ 5 mod 6.t₂ = 11¹¹ ≡ 5¹¹ mod 6. Since 11 is odd, 5¹¹ ≡ 5 mod 6.t₃ = 11^{t₂} = 11⁵ ≡ 5⁵ mod 6. 5 is odd, so 5⁵ ≡ 5 mod 6.t₄ = 11^{t₃} = 11⁵ ≡ 5⁵ mod 6 ≡ 5 mod 6.So, it seems like from t₂ onwards, every term is 5 mod 6. So, t₅₀ would also be 5 mod 6.But wait, maybe I should check this pattern more carefully. Let me think about exponents in modular arithmetic. I remember that Euler's theorem says that if two numbers are coprime, then a^φ(n) ≡ 1 mod n, where φ is Euler's totient function. Since 5 and 6 are coprime, φ(6) = 2, so 5² ≡ 1 mod 6. That's why the powers cycle every 2.So, in general, 5^k mod 6 is 5 if k is odd and 1 if k is even. So, to find 5^{t_{n-1}} mod 6, I need to know if t_{n-1} is odd or even.Wait, t₁ = 11, which is odd. Then t₂ = 11¹¹. Since 11 is odd, any power of 11 is odd, so t₂ is odd. Similarly, t₃ = 11^{t₂}, which is 11 raised to an odd power, so t₃ is odd. So, all terms t_n for n ≥ 1 are odd.Therefore, when we have t_n = 11^{t_{n-1}}, since t_{n-1} is odd, 11^{t_{n-1}} ≡ 5 mod 6.So, regardless of how large n is, as long as n ≥ 2, t_n ≡ 5 mod 6. Therefore, t₅₀ ≡ 5 mod 6.Wait, but let me make sure I'm not making a mistake here. Is there a possibility that the exponent could be even? But since t_{n-1} is always odd, as we saw, because 11 is odd and any power of an odd number is odd. So, t_{n-1} is always odd, so 11^{t_{n-1}} is always 5 mod 6.Therefore, the remainder when t₅₀ is divided by 6 is 5.I think that's the answer. Let me just recap:- t₁ = 11 ≡ 5 mod 6.- For n ≥ 2, t_n = 11^{t_{n-1}}. Since t_{n-1} is odd, 11^{t_{n-1}} ≡ 5 mod 6.- Therefore, t₅₀ ≡ 5 mod 6.Yeah, that makes sense. I don't see any flaws in this reasoning. So, I'm pretty confident the answer is 5.

💡Okay, so Jesse has some bananas, and he shares them among 3 friends. Each friend gets 7 bananas. I need to find out how many bananas Jesse has in total. Hmm, let's think about this step by step.First, if there are 3 friends and each gets 7 bananas, I guess I need to multiply the number of friends by the number of bananas each friend receives. That makes sense because if each friend gets the same amount, multiplying should give the total number.So, 3 friends times 7 bananas each. Let me write that down: 3 × 7. What's 3 times 7? Well, I remember that 3 times 7 is 21. So, that means Jesse must have 21 bananas in total.Wait, let me double-check to make sure I didn't make a mistake. If Jesse has 21 bananas and he divides them equally among 3 friends, each friend should get 7 bananas. Let's see: 21 divided by 3 is 7. Yep, that checks out. So, it looks like my initial thought was correct.I guess another way to think about it is by adding up the bananas each friend gets. If each friend gets 7 bananas, then for 3 friends, it's like adding 7 three times: 7 + 7 + 7. That also equals 21. So, whether I multiply or add, I end up with the same total number of bananas.Is there any other way to approach this problem? Maybe using division? If Jesse has 21 bananas and he gives them to 3 friends, how many does each friend get? That would be 21 divided by 3, which is 7. So, that confirms it again.I think I'm confident now that Jesse has 21 bananas. It all adds up correctly when I check it different ways. No mistakes spotted here!

💡Okay, so I have this problem with two parts. Let me try to tackle them one by one. Starting with part (1): There's a moving point P(x, y) and a fixed point F(3√3, 0). The ratio of the distance from P to F and its distance to the line l: x = 4√3 is a constant √3/2. I need to find the equation of the trajectory of point P.Hmm, okay. So, the ratio of distances is given. That sounds like the definition of a conic section, where the ratio is the eccentricity. If the ratio is less than 1, it's an ellipse; if it's equal to 1, it's a parabola; and if it's greater than 1, it's a hyperbola. Here, the ratio is √3/2, which is approximately 0.866, so less than 1. So, I think the trajectory is an ellipse.Let me write down the given ratio:Distance from P to F divided by the distance from P to the line l is √3/2.Mathematically, that's:√[(x - 3√3)^2 + y^2] / |x - 4√3| = √3 / 2So, to find the equation, I can square both sides to eliminate the square roots and absolute value.Squaring both sides:[(x - 3√3)^2 + y^2] / (x - 4√3)^2 = (3/4)Multiply both sides by (x - 4√3)^2:(x - 3√3)^2 + y^2 = (3/4)(x - 4√3)^2Now, let me expand both sides.First, expand the left side:(x - 3√3)^2 = x^2 - 6√3 x + (3√3)^2 = x^2 - 6√3 x + 27So, left side becomes:x^2 - 6√3 x + 27 + y^2Now, the right side:(3/4)(x - 4√3)^2First, expand (x - 4√3)^2:x^2 - 8√3 x + (4√3)^2 = x^2 - 8√3 x + 48Multiply by 3/4:(3/4)x^2 - 6√3 x + 36So, right side is (3/4)x^2 - 6√3 x + 36Now, set left side equal to right side:x^2 - 6√3 x + 27 + y^2 = (3/4)x^2 - 6√3 x + 36Let me subtract (3/4)x^2 - 6√3 x + 36 from both sides:x^2 - 6√3 x + 27 + y^2 - (3/4)x^2 + 6√3 x - 36 = 0Simplify term by term:x^2 - (3/4)x^2 = (1/4)x^2-6√3 x + 6√3 x = 027 - 36 = -9So, we have:(1/4)x^2 + y^2 - 9 = 0Multiply both sides by 4 to eliminate the fraction:x^2 + 4y^2 - 36 = 0Which can be rewritten as:x^2 / 36 + y^2 / 9 = 1So, that's the equation of an ellipse with semi-major axis 6 and semi-minor axis 3. Okay, that seems right.Now, moving on to part (2): There's a line l passing through point M(4, 2) that intersects the ellipse at two distinct points B and C. M is the midpoint of BC. I need to find the equation of line l.Alright, so the line passes through M(4, 2). Let me denote the equation of line l as y = k(x - 4) + 2, where k is the slope. Alternatively, I can write it as y = kx + (2 - 4k). But maybe it's better to keep it in the form y = k(x - 4) + 2 for substitution purposes.So, the equation is y = k(x - 4) + 2.Now, this line intersects the ellipse x^2 / 36 + y^2 / 9 = 1. Let me substitute y from the line into the ellipse equation.So, substitute y = k(x - 4) + 2 into the ellipse:x^2 / 36 + [k(x - 4) + 2]^2 / 9 = 1Let me expand [k(x - 4) + 2]^2:= [k(x - 4) + 2]^2= [k(x - 4)]^2 + 4k(x - 4) + 4= k^2(x - 4)^2 + 4k(x - 4) + 4So, plugging back into the ellipse equation:x^2 / 36 + [k^2(x - 4)^2 + 4k(x - 4) + 4] / 9 = 1Multiply through by 36 to eliminate denominators:x^2 + 4[k^2(x - 4)^2 + 4k(x - 4) + 4] = 36Expand the terms inside the brackets:= x^2 + 4k^2(x^2 - 8x + 16) + 16k(x - 4) + 16 = 36Now, expand each term:= x^2 + 4k^2x^2 - 32k^2x + 64k^2 + 16kx - 64k + 16 = 36Now, combine like terms:Terms with x^2: x^2 + 4k^2x^2 = (1 + 4k^2)x^2Terms with x: -32k^2x + 16kx = (-32k^2 + 16k)xConstant terms: 64k^2 - 64k + 16So, putting it all together:(1 + 4k^2)x^2 + (-32k^2 + 16k)x + (64k^2 - 64k + 16) - 36 = 0Simplify the constants:64k^2 - 64k + 16 - 36 = 64k^2 - 64k - 20So, the quadratic equation in x is:(1 + 4k^2)x^2 + (-32k^2 + 16k)x + (64k^2 - 64k - 20) = 0Let me denote this as:A x^2 + B x + C = 0Where:A = 1 + 4k^2B = -32k^2 + 16kC = 64k^2 - 64k - 20Now, since M(4, 2) is the midpoint of BC, the sum of the roots x1 + x2 should be equal to 8 (since midpoint x-coordinate is 4, so x1 + x2 = 2*4 = 8).From quadratic equation, the sum of roots is -B/A.So:x1 + x2 = -B/A = [32k^2 - 16k] / (1 + 4k^2) = 8So, set up the equation:[32k^2 - 16k] / (1 + 4k^2) = 8Multiply both sides by (1 + 4k^2):32k^2 - 16k = 8(1 + 4k^2)Expand the right side:32k^2 - 16k = 8 + 32k^2Subtract 32k^2 from both sides:-16k = 8Divide both sides by -16:k = -8 / 16 = -1/2So, the slope k is -1/2.Now, let me verify if this line actually intersects the ellipse at two distinct points. So, substitute k = -1/2 into the quadratic equation and check the discriminant.First, compute A, B, C with k = -1/2:A = 1 + 4*(1/4) = 1 + 1 = 2B = -32*(1/4) + 16*(-1/2) = -8 - 8 = -16C = 64*(1/4) - 64*(-1/2) - 20 = 16 + 32 - 20 = 28So, the quadratic equation becomes:2x^2 - 16x + 28 = 0Divide through by 2:x^2 - 8x + 14 = 0Compute discriminant D = (-8)^2 - 4*1*14 = 64 - 56 = 8Since D = 8 > 0, there are two distinct real roots. So, the line intersects the ellipse at two points, and M is indeed the midpoint.Therefore, the equation of line l is:y = (-1/2)(x - 4) + 2Simplify:y = (-1/2)x + 2 + 2Wait, that would be y = (-1/2)x + 4But let me double-check:y = (-1/2)(x - 4) + 2= (-1/2)x + 2 + 2Wait, no:(-1/2)(x - 4) = (-1/2)x + 2Then, adding 2:(-1/2)x + 2 + 2 = (-1/2)x + 4So, y = (-1/2)x + 4Alternatively, to write it in standard form:Multiply both sides by 2:2y = -x + 8Bring x to the left:x + 2y = 8So, the equation is x + 2y - 8 = 0Let me just verify if this line passes through M(4, 2):Plug x=4, y=2:4 + 2*2 - 8 = 4 + 4 - 8 = 0Yes, it does.So, the equation of line l is x + 2y - 8 = 0.**Final Answer**(1) The equation of the trajectory is boxed{dfrac{x^2}{36} + dfrac{y^2}{9} = 1}.(2) The equation of the line ( l ) is boxed{x + 2y - 8 = 0}.

💡Okay, so I have this problem where there are 1000 students arranged in a circle, and I need to prove that there exists an integer k between 100 and 300 such that in any sequence of 2k consecutive students, the number of girls in the first k is equal to the number of girls in the next k. Hmm, that sounds a bit abstract, but let me try to break it down.First, I should probably model the problem mathematically. Let me assign a value to each student: let’s say 1 if the student is a girl and 0 if it's a boy. So, I can represent the circle of students as a sequence of 1s and 0s, right? Let me denote this sequence as a_i where i ranges from 1 to 1000. Since it's a circle, the sequence wraps around, so a_{1001} is a_1, a_{1002} is a_2, and so on.Now, the problem is about finding a k between 100 and 300 such that for some starting point i, the sum of the first k students starting at i is equal to the sum of the next k students. In other words, S_k(i) = S_k(i + k), where S_k(i) is the sum of a_i to a_{i + k - 1}.So, I need to show that there exists such a k where this equality holds. Maybe I can use some kind of averaging argument or pigeonhole principle here. Let me think.If I consider all possible k from 100 to 300, that's 201 different values of k. For each k, I can look at the differences S_k(i) - S_k(i + k) for each starting position i. If for some k, all these differences are non-zero, then maybe I can find a contradiction.Wait, maybe I should think about the total number of girls in the entire circle. Let’s denote the total number of girls as G. Since it's a circle, the average number of girls in any k consecutive students is G * (k / 1000). But since k is between 100 and 300, this average is between G/10 and 3G/10.But I'm not sure if that helps directly. Maybe I need to consider the differences between S_k(i) and S_k(i + k). If I can show that for some k, these differences must be zero for some i, then I'm done.Let me try to formalize this. For each k, define D_k(i) = S_k(i) - S_k(i + k). If I can show that for some k, D_k(i) = 0 for some i, then that k satisfies the condition.Now, if I sum D_k(i) over all i from 1 to 1000, what do I get? Let me compute that:Sum_{i=1 to 1000} D_k(i) = Sum_{i=1 to 1000} [S_k(i) - S_k(i + k)].But S_k(i + k) is just the sum from a_{i + k} to a_{i + 2k - 1}. So, when I sum over all i, each term a_j appears exactly once in S_k(i) and exactly once in S_k(i + k), but shifted by k positions.Wait, actually, when I sum D_k(i) over all i, it's like subtracting the sum of S_k(i + k) from the sum of S_k(i). But since the circle is cyclic, the sum of S_k(i) over all i is equal to the sum of S_k(i + k) over all i. Therefore, the total sum is zero.So, Sum_{i=1 to 1000} D_k(i) = 0.This means that the average of D_k(i) over all i is zero. Therefore, if all D_k(i) were positive or all negative, we would have a contradiction because their sum would not be zero. Hence, there must be some i where D_k(i) is non-positive and some i where D_k(i) is non-negative.But we need to show that for some k, D_k(i) = 0 for some i. Hmm, maybe I can use the intermediate value theorem or something similar.Alternatively, perhaps I can use the pigeonhole principle. For each k, consider the differences D_k(i). There are 1000 differences, each of which is an integer (since they are sums of 0s and 1s). The possible values of D_k(i) range from -k to k.If I can show that for some k, the number of possible differences is less than 1000, then by the pigeonhole principle, some difference must repeat, which might lead to a zero difference.Wait, but the number of possible differences is 2k + 1, which for k=100 is 201, and for k=300 is 601. Since 2k + 1 is much less than 1000 for k up to 300, the pigeonhole principle might not directly apply here.Maybe I need a different approach. Let me think about the cumulative sums.Define the cumulative sum C_n = a_1 + a_2 + ... + a_n. Then, S_k(i) = C_{i + k - 1} - C_{i - 1}.Similarly, S_k(i + k) = C_{i + 2k - 1} - C_{i + k - 1}.So, D_k(i) = S_k(i) - S_k(i + k) = [C_{i + k - 1} - C_{i - 1}] - [C_{i + 2k - 1} - C_{i + k - 1}] = 2C_{i + k - 1} - C_{i - 1} - C_{i + 2k - 1}.Hmm, not sure if that helps. Maybe I can consider the differences between consecutive cumulative sums.Alternatively, let me consider the function f(i) = S_k(i) - S_k(i + k). If I can show that f(i) changes sign or something, then by the intermediate value theorem, there must be a point where f(i) = 0.But since we're dealing with integers, maybe I can use a discrete version of the intermediate value theorem. If f(i) changes from positive to negative or vice versa, then there must be some i where f(i) = 0.But how can I ensure that f(i) changes sign? Maybe by considering the sum over all f(i), which is zero, as we saw earlier. So, if the sum is zero, there must be some positive and some negative values, hence a change in sign.Wait, but that only tells me that there are both positive and negative differences, not necessarily that they cross zero. Hmm.Maybe I need to consider the maximum and minimum values of S_k(i). Let me define M_k = max_i S_k(i) and m_k = min_i S_k(i). Then, the difference D_k(i) = S_k(i) - S_k(i + k) can be as large as M_k - m_k or as small as m_k - M_k.But I'm not sure how that helps. Maybe I can look at the maximum and minimum over all k.Alternatively, perhaps I can use an averaging argument over k. Since there are 201 possible k values, and for each k, the sum of D_k(i) is zero, maybe there's some k where the maximum D_k(i) is non-positive or the minimum D_k(i) is non-negative, leading to a zero.Wait, that might not make sense. Let me think differently.Suppose I fix a starting point i and look at how S_k(i) changes as k increases. Maybe I can find a k where S_k(i) = S_k(i + k).But I'm not sure. Alternatively, maybe I can use the fact that the total number of girls is fixed, so for some k, the distribution of girls must balance out in some way.Wait, another idea: consider the function f(k) = S_k(i) - S_k(i + k). If I can show that f(k) is not always positive or always negative for all k, then there must be some k where f(k) = 0.But how can I show that? Maybe by considering the sum over all k.Wait, no, that might not work. Alternatively, maybe I can use the fact that for some k, the number of girls in the first k and the next k must be the same because of the total number of girls.But I'm not sure. Let me try to think of it differently.Suppose I have 1000 students in a circle. Let me consider all possible sequences of 200 students. For each such sequence, I can split it into two parts of 100 each. If for some sequence, the number of girls in the first 100 is equal to the number in the next 100, then k=100 works.But the problem allows k to be up to 300, so maybe I can generalize this idea.Alternatively, maybe I can use the fact that the number of girls in the entire circle is fixed, so for some k, the number of girls in the first k and the next k must balance out.Wait, another approach: consider the differences D_k(i) = S_k(i) - S_k(i + k). If I can show that for some k, the maximum D_k(i) is non-positive and the minimum D_k(i) is non-negative, then by the intermediate value theorem, there must be some i where D_k(i) = 0.But how can I ensure that? Maybe by considering the sum of D_k(i) over all i, which is zero, as we saw earlier. So, if the maximum D_k(i) is positive and the minimum is negative, then there must be some i where D_k(i) = 0.But I need to show that for some k, the maximum D_k(i) is positive and the minimum is negative. Hmm.Wait, maybe I can argue by contradiction. Suppose that for all k between 100 and 300, and for all i, D_k(i) ≠ 0. Then, for each k, all D_k(i) are either all positive or all negative. But since the sum is zero, this can't happen. Therefore, there must be some k where D_k(i) = 0 for some i.Wait, that seems promising. Let me formalize that.Assume, for contradiction, that for every k with 100 ≤ k ≤ 300, and for every i, D_k(i) ≠ 0. Then, for each k, either D_k(i) > 0 for all i, or D_k(i) < 0 for all i.But we know that Sum_{i=1 to 1000} D_k(i) = 0. If D_k(i) > 0 for all i, then the sum would be positive, which contradicts the sum being zero. Similarly, if D_k(i) < 0 for all i, the sum would be negative, which also contradicts the sum being zero. Therefore, our assumption is false, and there must exist some k and i such that D_k(i) = 0.Wait, but does this hold? Because for each k, the sum is zero, so if D_k(i) is always positive or always negative, the sum can't be zero. Therefore, for each k, there must be some i where D_k(i) = 0. But the problem states that there exists at least one k in the range where such an i exists. So, actually, this would imply that for every k, there exists such an i, which is stronger than what we need.But wait, the problem only asks to prove that there exists at least one k in the range where such an i exists, not for all k. So, maybe my reasoning is a bit off.Wait, no, actually, the problem is to show that there exists at least one k in the range 100 to 300 such that for some i, S_k(i) = S_k(i + k). So, my earlier reasoning shows that for every k, there exists such an i, which is a stronger statement. Therefore, in particular, there exists at least one k in the range where this holds.But wait, is that correct? Because the problem is about the existence of at least one k, but my reasoning shows that for every k, there exists such an i. So, actually, the conclusion is stronger than required.But maybe I made a mistake in assuming that for each k, D_k(i) can't be always positive or always negative. Let me double-check.If for some k, D_k(i) is always positive, then the sum over i would be positive, which contradicts the sum being zero. Similarly, if D_k(i) is always negative, the sum would be negative, which also contradicts the sum being zero. Therefore, for each k, there must be at least one i where D_k(i) = 0.Wait, but that would mean that for every k, there exists an i such that S_k(i) = S_k(i + k). But the problem only asks for the existence of at least one such k. So, my reasoning actually proves a stronger statement.But maybe the problem is to show that there exists at least one k in the range where such an i exists, not necessarily for all k. So, perhaps I can use a different approach.Let me consider the total number of girls G. Since the students are in a circle, the number of girls in any k consecutive students can vary, but the total is fixed.If I consider all possible k from 100 to 300, and for each k, look at the differences S_k(i) - S_k(i + k), then as we saw earlier, the sum of these differences is zero. Therefore, for each k, there must be some i where S_k(i) = S_k(i + k).But wait, that's the same conclusion as before. So, perhaps the problem is actually easier than I thought, and the conclusion is straightforward from the fact that the sum of differences is zero.But maybe I need to be more careful. Let me try to write it out step by step.1. Assign a_i = 1 if the i-th student is a girl, else 0.2. For each k, define S_k(i) = a_i + a_{i+1} + ... + a_{i+k-1}.3. Define D_k(i) = S_k(i) - S_k(i + k).4. Sum_{i=1 to 1000} D_k(i) = 0, because each term a_j appears exactly once in S_k(i) and exactly once in S_k(i + k), so their differences cancel out.5. Therefore, for each k, the average of D_k(i) over i is zero.6. If all D_k(i) were positive, their sum would be positive, contradicting the sum being zero. Similarly, if all D_k(i) were negative, their sum would be negative, also a contradiction.7. Hence, for each k, there must exist at least one i where D_k(i) = 0, meaning S_k(i) = S_k(i + k).8. Therefore, for every k from 100 to 300, there exists such an i. In particular, there exists at least one k in that range where this holds.Wait, but the problem only asks to prove the existence of at least one k, not for all k. So, my reasoning shows that for every k, such an i exists, which is a stronger result. Therefore, the conclusion holds.But maybe I need to make sure that the reasoning is correct. Let me check again.- For each k, D_k(i) = S_k(i) - S_k(i + k).- Sum_{i=1 to 1000} D_k(i) = 0.- If all D_k(i) > 0, sum would be positive; contradiction.- If all D_k(i) < 0, sum would be negative; contradiction.- Therefore, for each k, there exists i such that D_k(i) = 0.Hence, for every k, such an i exists. Therefore, in particular, there exists at least one k in the range where this holds.So, the problem is solved by this reasoning. I think this is the correct approach.