💡To determine the integer closest to the value of (sqrt[3]{11^3 + 2^3}), I will follow these steps:1. **Calculate each term separately:** - (11^3 = 1331) - (2^3 = 8)2. **Sum the cubes:** - (1331 + 8 = 1339)3. **Find the cubic root:** - I know that (11^3 = 1331) and (12^3 = 1728). Therefore, (sqrt[3]{1339}) is slightly greater than 11 but less than 12.4. **Determine the closest integer:** - Since (1339) is much closer to (1331) ((11^3)) than to (1728) ((12^3)), the closest integer to (sqrt[3]{1339}) is (11).

💡Okay, so I have this problem about a class where there are talkers and silent people. The rules are that every talker is friends with at least one silent person, and a talker will stay silent if there's an odd number of their silent friends in the room. The goal is to prove that the teacher can invite at least half the class to an extracurricular session in such a way that all talkers remain silent.Hmm, let me try to break this down. First, I need to understand the setup. There are two types of students: talkers and silent ones. Each talker has at least one silent friend. Now, if a talker is in a room and the number of their silent friends present is odd, they'll stay silent. Otherwise, they might talk. The teacher wants to invite as many students as possible, at least half, but ensure that all talkers stay silent.So, the key here is to select a subset of the class where for every talker in that subset, the number of their silent friends in the subset is odd. And this subset should be at least half the size of the entire class.Maybe I can model this as a graph problem. Let's think of the class as a graph where nodes represent students. Talkers are one set, and silent students are another set. Edges connect talkers to their silent friends. So, it's a bipartite graph between talkers and silent students.In this graph, each talker has at least one edge because every talker is friends with at least one silent person. Now, the problem is to find a subset of nodes (students) such that for every talker in this subset, the number of their silent friends in the subset is odd. And this subset should be as large as possible, at least half the class.This seems related to something called an "odd cover" in graph theory, where you want a subset of vertices such that every vertex in a certain set has an odd number of neighbors in the subset. In this case, the talkers need to have an odd number of silent friends in the subset.I remember that in bipartite graphs, especially with certain properties, you can use linear algebra over the field GF(2) to solve such problems. Maybe I can represent the adjacency matrix of the bipartite graph and look for a vector that satisfies certain parity conditions.Let me denote the set of talkers as T and silent students as S. The adjacency matrix A would have rows corresponding to talkers and columns to silent students. Each entry A_{i,j} is 1 if talker i is friends with silent student j, and 0 otherwise.We need to find a subset X of the entire class (T ∪ S) such that for every talker t in X, the number of silent students s in X that are friends with t is odd. Also, |X| ≥ (|T| + |S|)/2.Wait, actually, the problem says "at least half the class," so |X| ≥ (|T| + |S|)/2. Since the class is T ∪ S, the total number of students is |T| + |S|.But how do we ensure that the subset X includes enough students? Maybe we can think of selecting either all silent students or a combination of talkers and silent students such that the talkers in X have an odd number of silent friends in X.Alternatively, perhaps we can use some kind of greedy algorithm. Start by selecting silent students and then see how many talkers can be added without violating the odd condition.But I'm not sure if a greedy approach would work here because adding a talker might affect multiple silent students. It could create dependencies where selecting one talker requires selecting or deselecting others.Maybe another approach is to consider the problem as a system of linear equations over GF(2). For each talker, we have an equation that the sum of their silent friends in the subset X must be 1 mod 2. The variables are the silent students, indicating whether they are included in X or not.But wait, the subset X can include both talkers and silent students. So, actually, the variables would be all students, but the equations are only for the talkers in X. Hmm, this might complicate things because the equations depend on whether a talker is in X or not.Alternatively, perhaps we can fix the subset of silent students first and then determine which talkers can be included. If we select a subset of silent students, then any talker who has an odd number of friends in this subset can be included in X. The challenge is to choose the subset of silent students such that the number of talkers who can be included is maximized, and the total size is at least half the class.This seems like a combinatorial optimization problem. Maybe there's a theorem or a known result that can help here. I recall something about matchings and covers in bipartite graphs, but I'm not sure if that directly applies.Wait, another thought: if we can find a matching where each talker is matched to a silent friend, then perhaps we can use that to construct the subset X. But I don't see immediately how that would ensure the odd condition.Alternatively, maybe we can use induction. Suppose the statement is true for a class of size n, and then show it holds for n+1. But I need to figure out how to extend the subset X when adding a new student.Let me try to think of a small example to get some intuition. Suppose there are 3 students: two talkers and one silent student. Each talker is friends with the silent student. If the teacher invites the silent student and one talker, then the talker in the room has one silent friend, which is odd, so they stay silent. The other talker is not invited, so it's okay. So, in this case, the teacher can invite two students, which is more than half of three.Another example: 4 students, two talkers and two silent. Each talker is friends with both silent students. If the teacher invites both silent students, then each talker has two silent friends, which is even, so they would not stay silent. That's bad. Alternatively, invite one silent student and one talker. Then the invited talker has one silent friend, which is odd, so they stay silent. The other talker is not invited, so it's okay. The total invited is two, which is half of four. So, that works.Wait, but in this case, inviting one silent and one talker gives exactly half. But the problem says "at least half," so that's acceptable.Another example: 5 students, three talkers and two silent. Each talker is friends with at least one silent. Suppose each talker is friends with both silent students. If the teacher invites both silent students and one talker, the invited talker has two silent friends, which is even, so they wouldn't stay silent. That's bad. Alternatively, invite one silent and two talkers. Each invited talker has one silent friend, which is odd, so they stay silent. The total invited is three, which is more than half of five. That works.Hmm, so in these small cases, it seems possible. Maybe the general idea is to pair talkers with silent students in such a way that each talker is "covered" by an odd number of silent students in the subset.Wait, another approach: think of the problem as choosing a subset of silent students such that the number of talkers who have an odd number of friends in this subset is at least half the class. But I'm not sure.Alternatively, maybe we can model this as a hypergraph where each talker is a hyperedge connecting their silent friends, and we need a hitting set that intersects each hyperedge in an odd number of vertices. But I don't know much about odd hitting sets.Wait, maybe linear algebra over GF(2) can help here. Let me formalize this.Let T be the set of talkers and S be the set of silent students. Let A be the adjacency matrix where rows are talkers and columns are silent students, with A_{t,s} = 1 if t is friends with s.We need to find a subset X ⊆ T ∪ S such that for every t ∈ X ∩ T, the number of s ∈ X ∩ S with A_{t,s} = 1 is odd. And |X| ≥ (|T| + |S|)/2.Let me denote x_s ∈ {0,1} indicating whether silent student s is in X, and y_t ∈ {0,1} indicating whether talker t is in X.Then, for each talker t in X, we have:∑_{s ∈ S} A_{t,s} x_s ≡ 1 mod 2.But this is only required if y_t = 1. If y_t = 0, there's no condition.This seems like a system of equations where the variables are x_s and y_t, but the equations are conditional on y_t.This complicates things because the system is not linear in the usual sense.Alternatively, perhaps we can fix y_t and then solve for x_s. But since y_t affects the equations, it's tricky.Wait, maybe another angle: consider that for each talker t, if we include t in X, we need an odd number of their silent friends in X. If we don't include t, we don't care about their silent friends.So, perhaps we can model this as a constraint satisfaction problem where for each talker t, if y_t = 1, then ∑ A_{t,s} x_s ≡ 1 mod 2.We need to maximize |X| = ∑ y_t + ∑ x_s, subject to these constraints.This is an integer linear programming problem, but maybe we can find a combinatorial solution.Alternatively, think about it probabilistically. If we randomly select silent students, what's the probability that a given talker has an odd number of friends selected? It's 1/2, assuming independence. So, on average, half the talkers would have an odd number of friends selected. But we need a deterministic solution.Wait, but maybe we can use some kind of parity argument. If we can find a subset of silent students such that the number of talkers with an odd number of friends in this subset is at least half the class.But how?Wait, another thought: since each talker has at least one silent friend, the adjacency matrix A has no all-zero rows. So, in linear algebra terms over GF(2), the rows are non-zero. Therefore, the row space has dimension at least 1.If we can find a vector x in GF(2)^{|S|} such that Ax has at least half the entries equal to 1, then we can set X to be the silent students corresponding to x and the talkers t where (Ax)_t = 1.But how do we ensure that Ax has at least half the entries equal to 1?Hmm, maybe using some properties of linear codes or something. I'm not sure.Alternatively, maybe we can use the probabilistic method. Consider choosing a random subset of silent students. For each talker, the probability that the number of their friends in the subset is odd is 1/2. So, the expected number of talkers with an odd number of friends is |T|/2. Therefore, there exists a subset of silent students such that at least |T|/2 talkers have an odd number of friends in it.But we need to include these talkers in X, so the total size would be |S'| + |T'|, where S' is the subset of silent students and T' is the subset of talkers with an odd number of friends in S'. We need |S'| + |T'| ≥ (|T| + |S|)/2.But I don't know if this approach can guarantee that |S'| + |T'| is at least half.Wait, maybe we can optimize the choice of S' to maximize |S'| + |T'|, where T' is the set of talkers with an odd number of friends in S'.This seems like a combinatorial optimization problem. Maybe we can use some kind of exchange argument or consider the problem as a matroid.Alternatively, think about it as a graph where we want to select a subset of silent students and as many talkers as possible who have an odd number of friends in the selected silent students.But I'm not sure.Wait, another idea: since each talker has at least one silent friend, we can pair each talker with one of their silent friends. Then, for each pair, we can decide whether to include the talker and their silent friend in X such that the talker has an odd number of silent friends in X.But pairing might not cover all cases, especially if talkers have multiple silent friends.Alternatively, maybe we can use induction. Suppose the statement is true for classes of size n, and then show it holds for n+1.But I need to figure out how to extend the subset X when adding a new student.Wait, let's try induction. Let's assume that for any class of size n, the teacher can invite at least n/2 students such that all talkers are silent. Now, consider a class of size n+1. Remove one student, apply the inductive hypothesis, and then see how to add the student back.But I'm not sure if this will work because adding a student might disrupt the conditions.Alternatively, maybe consider the case where there's only one silent student. Then, every talker is friends with that silent student. So, if we include the silent student and any subset of talkers, each included talker has one silent friend, which is odd, so they stay silent. Therefore, we can include all talkers and the silent student, which is the entire class, so certainly at least half.But if there are multiple silent students, it's more complicated.Wait, maybe if there are multiple silent students, we can partition them into groups and apply the inductive hypothesis.Alternatively, think about the problem as a hypergraph where each hyperedge connects a talker to their silent friends. We need a subset of vertices that intersects each hyperedge in an odd number of vertices.But I don't know much about hypergraphs.Wait, another approach: since each talker has at least one silent friend, the set of silent students forms a covering of the talkers. So, maybe we can use some covering properties.Alternatively, think about the problem in terms of linear algebra. Let me consider the adjacency matrix A as a matrix over GF(2). Each row corresponds to a talker, each column to a silent student. We need to find a vector x in GF(2)^{|S|} such that Ax has as many 1s as possible, and then set X to be the support of x plus the support of Ax.But how do we ensure that the total size is at least half?Wait, maybe the key is that the number of possible vectors x is 2^{|S|}, and for each x, the number of 1s in Ax can vary. By some averaging argument, there must exist an x such that the number of 1s in Ax is at least |T|/2. Then, the total size would be |S'| + |T'|, where S' is the support of x and T' is the support of Ax.But we need |S'| + |T'| ≥ (|T| + |S|)/2.Hmm, not sure.Wait, maybe another way: consider that for any subset S' of silent students, the number of talkers with an odd number of friends in S' is equal to the number of talkers with even number of friends in S' complemented by something.But I don't know.Wait, perhaps think about it as a system where we can choose S' to maximize |S'| + |T'|, where T' is the set of talkers with an odd number of friends in S'.This is similar to a maximum coverage problem, but with parity constraints.Alternatively, maybe we can model this as a flow network and find a maximum flow that satisfies the parity conditions.But I'm not sure.Wait, another idea: since each talker has at least one silent friend, we can construct a bipartite graph where talkers are connected to their silent friends. Then, we need to select a subset of silent students such that the number of talkers with an odd number of connections to this subset is as large as possible.This seems related to the concept of influence maximization, but with parity constraints.Alternatively, think about it as a binary matrix and look for a column vector x such that Ax has many 1s.But I don't know.Wait, maybe use the fact that in GF(2), the number of solutions to Ax = b is either 0 or 2^{n - rank(A)}.But I don't see how that helps here.Wait, another thought: if we can find a subset S' of silent students such that the number of talkers with an odd number of friends in S' is at least |T|/2, then we can set X = S' ∪ T', where T' is the set of talkers with an odd number of friends in S'. Then, |X| = |S'| + |T'|. We need |S'| + |T'| ≥ (|T| + |S|)/2.But how do we ensure that |S'| + |T'| is at least half?Wait, maybe consider that for any S', |T'| is the number of talkers with an odd number of friends in S'. So, |T'| can vary depending on S'.If we can choose S' such that |T'| is as large as possible, then maybe |S'| + |T'| will be large enough.But I don't know how to bound this.Wait, another approach: use the probabilistic method. Choose each silent student independently with probability 1/2. For each talker, the probability that the number of their friends in S' is odd is 1/2. So, the expected number of talkers with an odd number of friends is |T|/2. Therefore, there exists a subset S' such that |T'| ≥ |T|/2. Then, |X| = |S'| + |T'|. The expected |S'| is |S|/2. So, the expected |X| is |S|/2 + |T|/2 = (|S| + |T|)/2. Therefore, there exists a subset S' such that |X| ≥ (|S| + |T|)/2.Wait, that seems promising. So, by linearity of expectation, the expected size of X is exactly half the class. Therefore, there must exist a subset S' such that |X| ≥ half the class.But does this work? Because we're dealing with probabilities, but we need a deterministic subset.Yes, the probabilistic method shows that there exists such a subset, but it doesn't construct it. However, the problem only asks to prove that such a subset exists, not to construct it.So, maybe this is the way to go. Use the probabilistic method to show that there exists a subset S' of silent students such that the number of talkers with an odd number of friends in S' is at least |T|/2, and then the total size |S'| + |T'| is at least (|T| + |S|)/2.But wait, the expected |S'| is |S|/2, and the expected |T'| is |T|/2, so the expected |X| is (|T| + |S|)/2. Therefore, there must exist a subset S' where |X| ≥ (|T| + |S|)/2.But is this rigorous enough? I think it is, because the probabilistic method shows existence.Alternatively, maybe we can use a combinatorial argument. For each silent student, decide whether to include them or not in a way that maximizes |X|.But I think the probabilistic method is sufficient here.Wait, but the problem says "the teacher can invite at least half the class." So, it's about existence, not construction. Therefore, the probabilistic method suffices.But I need to make sure that the conditions are met: for every talker in X, the number of their silent friends in X is odd.Wait, in the probabilistic method, we choose S' randomly, and then set T' as the set of talkers with an odd number of friends in S'. Then, X = S' ∪ T'. For each talker in T', the number of silent friends in X is odd, because it's the same as the number in S', which is odd. For talkers not in T', they are not in X, so it's okay.Therefore, X satisfies the condition that all talkers in X are silent. And by expectation, |X| is at least half the class.Therefore, such a subset X exists.Wait, but I need to make sure that |X| is at least half, not just that the expectation is half. Because expectation is half, but some subsets might have less, but there must exist one with at least half.Yes, because if all subsets had |X| < half, then the expectation would be less than half, which contradicts the expectation being exactly half. Therefore, there must exist a subset with |X| ≥ half.Therefore, the teacher can invite at least half the class in such a way that all talkers are silent.I think this works. So, the key idea is to use the probabilistic method to show that such a subset exists by considering the expected number of talkers with an odd number of silent friends in a random subset of silent students.

💡Okay, so I have this problem about a sequence of natural numbers (a_1, a_2, ldots, a_n, ldots). The condition is that for each (n), the quadratic equation (a_{n+2}x^2 + a_{n+1}x + a_n = 0) has a real root. I need to figure out if the sequence can have exactly 10 terms or if it can be infinite.First, I remember that for a quadratic equation (ax^2 + bx + c = 0) to have real roots, the discriminant must be non-negative. The discriminant is (b^2 - 4ac). So, for each (n), the discriminant of the equation (a_{n+2}x^2 + a_{n+1}x + a_n = 0) must satisfy:[(a_{n+1})^2 - 4a_n a_{n+2} geq 0]This inequality needs to hold for all (n). Let me tackle part (a) first: Can the sequence have exactly 10 terms?Hmm, so if the sequence has 10 terms, that means (a_1, a_2, ldots, a_{10}). Then, for each (n) from 1 to 8 (since (n+2) must be less than or equal to 10), the equation must have a real root. So, I need to ensure that for (n = 1) to (n = 8), the discriminant condition holds.I wonder if there's a specific sequence of natural numbers that satisfies this condition for all (n) up to 8. Maybe I can construct such a sequence. Let me think about a geometric sequence or something similar.Wait, if I consider a geometric sequence where each term is a multiple of the previous term, maybe I can satisfy the discriminant condition. Suppose (a_{n+1} = k a_n) for some constant (k). Then, the discriminant becomes:[(k a_n)^2 - 4a_n (k^2 a_n) = k^2 a_n^2 - 4k^2 a_n^2 = -3k^2 a_n^2]Oh, that's negative, which doesn't satisfy the discriminant condition. So, a simple geometric sequence won't work.Maybe I need a different approach. Perhaps a sequence where each term is related in a way that the discriminant is non-negative. Let me think about setting (a_{n+1}^2 = 4a_n a_{n+2}). This would make the discriminant zero, which is acceptable since the equation would have a real root (a repeated root).So, if I set (a_{n+1}^2 = 4a_n a_{n+2}), then each term is determined by the previous two terms. This is similar to a recurrence relation. Let me write it down:[a_{n+2} = frac{a_{n+1}^2}{4a_n}]Since we're dealing with natural numbers, (a_{n+1}^2) must be divisible by (4a_n). So, each term must be chosen such that this division results in a natural number.Let me try constructing such a sequence. Let's start with some initial terms. Suppose (a_1 = 1). Then, for (a_2), I need to choose a natural number such that when I compute (a_3), it's also a natural number.Let me choose (a_2 = 2^{10}). Then, (a_3 = frac{(2^{10})^2}{4 cdot 1} = frac{2^{20}}{4} = 2^{18}). Okay, that works.Next, (a_4 = frac{(2^{18})^2}{4 cdot 2^{10}} = frac{2^{36}}{4 cdot 2^{10}} = frac{2^{36}}{2^{12}} = 2^{24}).Continuing this pattern:- (a_5 = frac{(2^{24})^2}{4 cdot 2^{18}} = frac{2^{48}}{4 cdot 2^{18}} = frac{2^{48}}{2^{20}} = 2^{28})- Wait, but 2^{28} is not in the sequence I initially thought of. Hmm, maybe I made a mistake.Wait, let me check the calculations again.Wait, (a_3 = 2^{18}), (a_4 = 2^{24}), then (a_5 = frac{(2^{24})^2}{4 cdot 2^{18}} = frac{2^{48}}{4 cdot 2^{18}} = frac{2^{48}}{2^{20}} = 2^{28}). So, (a_5 = 2^{28}).But I initially thought of a sequence that goes up to (2^{26}). Maybe I need to adjust my initial terms.Wait, perhaps I should choose the initial terms such that the exponents form a specific pattern. Let me think about the exponents.If I let the exponents be 0, 10, 18, 24, 26, 26, 24, 18, 10, 0. Wait, that's 10 terms. Let me see:- (a_1 = 2^0 = 1)- (a_2 = 2^{10})- (a_3 = 2^{18})- (a_4 = 2^{24})- (a_5 = 2^{26})- (a_6 = 2^{26})- (a_7 = 2^{24})- (a_8 = 2^{18})- (a_9 = 2^{10})- (a_{10} = 2^0 = 1)Now, let's check the discriminant condition for each (n) from 1 to 8.For (n=1):[(a_2)^2 - 4a_1 a_3 = (2^{10})^2 - 4 cdot 1 cdot 2^{18} = 2^{20} - 4 cdot 2^{18} = 2^{20} - 2^{20} = 0]Good, discriminant is zero.For (n=2):[(a_3)^2 - 4a_2 a_4 = (2^{18})^2 - 4 cdot 2^{10} cdot 2^{24} = 2^{36} - 4 cdot 2^{34} = 2^{36} - 2^{36} = 0]Again, discriminant is zero.For (n=3):[(a_4)^2 - 4a_3 a_5 = (2^{24})^2 - 4 cdot 2^{18} cdot 2^{26} = 2^{48} - 4 cdot 2^{44} = 2^{48} - 2^{46} cdot 4 = 2^{48} - 2^{48} = 0]Still zero.For (n=4):[(a_5)^2 - 4a_4 a_6 = (2^{26})^2 - 4 cdot 2^{24} cdot 2^{26} = 2^{52} - 4 cdot 2^{50} = 2^{52} - 2^{52} = 0]Zero again.For (n=5):[(a_6)^2 - 4a_5 a_7 = (2^{26})^2 - 4 cdot 2^{26} cdot 2^{24} = 2^{52} - 4 cdot 2^{50} = 2^{52} - 2^{52} = 0]Zero.For (n=6):[(a_7)^2 - 4a_6 a_8 = (2^{24})^2 - 4 cdot 2^{26} cdot 2^{18} = 2^{48} - 4 cdot 2^{44} = 2^{48} - 2^{46} cdot 4 = 2^{48} - 2^{48} = 0]Zero.For (n=7):[(a_8)^2 - 4a_7 a_9 = (2^{18})^2 - 4 cdot 2^{24} cdot 2^{10} = 2^{36} - 4 cdot 2^{34} = 2^{36} - 2^{36} = 0]Zero.For (n=8):[(a_9)^2 - 4a_8 a_{10} = (2^{10})^2 - 4 cdot 2^{18} cdot 1 = 2^{20} - 4 cdot 2^{18} = 2^{20} - 2^{20} = 0]Zero.So, for all (n) from 1 to 8, the discriminant is zero, which means the quadratic equation has a real root (a repeated root). Therefore, this sequence of 10 terms satisfies the condition.Okay, so part (a) is possible.Now, part (b): Can the sequence be infinite?Hmm, if the sequence is infinite, we need to ensure that for every (n), the discriminant condition holds. Let's think about the recurrence relation again:[a_{n+2} = frac{a_{n+1}^2}{4a_n}]Since (a_{n+2}) must be a natural number, (a_{n+1}^2) must be divisible by (4a_n). This suggests that the sequence is determined by the previous two terms, and each term is a multiple of the previous term squared divided by four times the term before that.But if we try to extend the sequence beyond 10 terms, let's see what happens. After (a_{10} = 1), we would need to compute (a_{11}):[a_{11} = frac{a_{10}^2}{4a_9} = frac{1^2}{4 cdot 2^{10}} = frac{1}{4 cdot 1024} = frac{1}{4096}]But (a_{11}) is not a natural number. It's a fraction. So, this sequence cannot be extended beyond 10 terms while keeping all terms as natural numbers.Wait, maybe there's another way to construct an infinite sequence. Perhaps choosing different initial terms that allow the sequence to continue indefinitely.Let me think about choosing (a_1 = 1) and (a_2 = 2). Then:[a_3 = frac{a_2^2}{4a_1} = frac{4}{4} = 1][a_4 = frac{a_3^2}{4a_2} = frac{1}{8}]Again, (a_4) is not a natural number. Hmm.What if I choose (a_1 = 1) and (a_2 = 4):[a_3 = frac{16}{4} = 4][a_4 = frac{16}{16} = 1][a_5 = frac{1}{4}]Still, (a_5) is not natural.Wait, maybe choosing (a_1 = 2) and (a_2 = 4):[a_3 = frac{16}{8} = 2][a_4 = frac{4}{8} = 0.5]Not natural.Hmm, seems like it's difficult to get an infinite sequence because eventually, the terms become fractions. Maybe if I choose larger initial terms?Let me try (a_1 = 4) and (a_2 = 16):[a_3 = frac{256}{16} = 16][a_4 = frac{256}{64} = 4][a_5 = frac{16}{16} = 1][a_6 = frac{1}{4}]Again, (a_6) is not natural.It seems that no matter what initial terms I choose, eventually, the sequence will produce a non-natural number. This suggests that it's impossible to have an infinite sequence of natural numbers satisfying the given condition.Wait, but maybe there's a way to have a periodic sequence where the terms repeat in a way that keeps them natural. For example, a sequence that cycles through certain values.Suppose I have a sequence where (a_{n+2} = a_n). Then, the recurrence becomes:[a_{n+2} = frac{a_{n+1}^2}{4a_n} implies a_n = frac{a_{n+1}^2}{4a_n} implies 4a_n^2 = a_{n+1}^2 implies a_{n+1} = 2a_n]So, if (a_{n+1} = 2a_n), then (a_{n+2} = a_n). Let's see:- (a_1 = k)- (a_2 = 2k)- (a_3 = frac{(2k)^2}{4k} = frac{4k^2}{4k} = k)- (a_4 = frac{k^2}{4 cdot 2k} = frac{k^2}{8k} = frac{k}{8})But (a_4) must be natural, so (k) must be a multiple of 8. Let (k = 8):- (a_1 = 8)- (a_2 = 16)- (a_3 = 8)- (a_4 = frac{8^2}{4 cdot 16} = frac{64}{64} = 1)- (a_5 = frac{1^2}{4 cdot 8} = frac{1}{32})Not natural.Hmm, even with (k = 8), (a_5) is not natural. Maybe choosing (k) as a higher multiple?Let (k = 8 times 2 = 16):- (a_1 = 16)- (a_2 = 32)- (a_3 = 16)- (a_4 = frac{16^2}{4 cdot 32} = frac{256}{128} = 2)- (a_5 = frac{2^2}{4 cdot 16} = frac{4}{64} = frac{1}{16})Still not natural.It seems that no matter how I choose (k), eventually, the sequence will produce a non-natural term. Therefore, it's impossible to have an infinite sequence of natural numbers satisfying the given condition.So, for part (b), the answer is no.

💡Okay, so I have this problem where I need to find the smallest positive angle θ in degrees such that cos θ equals sin 60° + cos 42° - sin 12° - cos 6°. Hmm, that's a bit of a mouthful, but let's break it down step by step.First, I know that cos θ is equal to some combination of sine and cosine terms. Maybe I can simplify the right-hand side using some trigonometric identities. Let me recall some identities that might help here.I remember that sin x can be written as cos(90° - x). So, sin 60° is the same as cos(30°). That might be useful. Let me write that down:sin 60° = cos(30°)Similarly, sin 12° can be written as cos(78°), because sin 12° = cos(90° - 12°) = cos(78°). So, -sin 12° would be -cos(78°). Hmm, maybe I can express all these terms in terms of cosine functions to make it easier to combine them.So, rewriting the original equation:cos θ = cos 30° + cos 42° - cos 78° - cos 6°Now, I have all terms in cosine. Maybe I can combine these terms somehow. I remember that there are sum-to-product identities for cosines, which might help here. Let me recall those:cos A + cos B = 2 cos[(A+B)/2] cos[(A-B)/2]cos A - cos B = -2 sin[(A+B)/2] sin[(A-B)/2]So, perhaps I can pair these terms and apply these identities. Let me see how to pair them. I have four terms: cos 30°, cos 42°, -cos 78°, and -cos 6°. Maybe I can pair cos 30° with -cos 6°, and cos 42° with -cos 78°.Let's try that.First pair: cos 30° - cos 6°Using the identity cos A - cos B = -2 sin[(A+B)/2] sin[(A-B)/2]So, cos 30° - cos 6° = -2 sin[(30° + 6°)/2] sin[(30° - 6°)/2] = -2 sin(18°) sin(12°)Second pair: cos 42° - cos 78°Again, using the same identity: cos 42° - cos 78° = -2 sin[(42° + 78°)/2] sin[(42° - 78°)/2] = -2 sin(60°) sin(-18°)Wait, sin(-x) is -sin x, so this becomes -2 sin(60°)(-sin 18°) = 2 sin(60°) sin(18°)So, putting it all together, the original expression becomes:cos θ = [ -2 sin(18°) sin(12°) ] + [ 2 sin(60°) sin(18°) ]Let me factor out the common terms. I see that both terms have 2 sin(18°), so:cos θ = 2 sin(18°) [ -sin(12°) + sin(60°) ]Now, let's compute the expression inside the brackets: -sin(12°) + sin(60°)I know that sin(60°) is √3/2, which is approximately 0.8660, and sin(12°) is approximately 0.2079. So, -sin(12°) + sin(60°) ≈ -0.2079 + 0.8660 ≈ 0.6581.But maybe I can find an exact expression instead of using approximate values. Let's see.I recall that sin A - sin B can be expressed as 2 cos[(A+B)/2] sin[(A-B)/2]. Wait, but here I have -sin(12°) + sin(60°), which is the same as sin(60°) - sin(12°). So, applying the identity:sin(60°) - sin(12°) = 2 cos[(60° + 12°)/2] sin[(60° - 12°)/2] = 2 cos(36°) sin(24°)So, substituting back into the expression for cos θ:cos θ = 2 sin(18°) [ 2 cos(36°) sin(24°) ] = 4 sin(18°) cos(36°) sin(24°)Hmm, that seems a bit complicated, but maybe I can simplify it further.I remember that sin(24°) can be expressed as 2 sin(12°) cos(12°), using the double-angle identity sin(2x) = 2 sin x cos x. So, sin(24°) = 2 sin(12°) cos(12°). Let me substitute that in:cos θ = 4 sin(18°) cos(36°) [ 2 sin(12°) cos(12°) ] = 8 sin(18°) cos(36°) sin(12°) cos(12°)Hmm, this is getting more complex. Maybe there's another approach. Let me think.Alternatively, perhaps instead of pairing cos 30° with -cos 6°, and cos 42° with -cos 78°, I could pair cos 30° with cos 42°, and -cos 78° with -cos 6°. Let's try that.First pair: cos 30° + cos 42°Using the sum identity: cos A + cos B = 2 cos[(A+B)/2] cos[(A-B)/2]So, cos 30° + cos 42° = 2 cos[(30° + 42°)/2] cos[(30° - 42°)/2] = 2 cos(36°) cos(-6°)Since cosine is even, cos(-6°) = cos(6°), so this becomes 2 cos(36°) cos(6°)Second pair: -cos 78° - cos 6°This can be written as -(cos 78° + cos 6°)Again, using the sum identity: cos A + cos B = 2 cos[(A+B)/2] cos[(A-B)/2]So, cos 78° + cos 6° = 2 cos[(78° + 6°)/2] cos[(78° - 6°)/2] = 2 cos(42°) cos(36°)Therefore, -(cos 78° + cos 6°) = -2 cos(42°) cos(36°)Putting it all together, the original expression becomes:cos θ = [2 cos(36°) cos(6°)] + [ -2 cos(42°) cos(36°) ] = 2 cos(36°) [ cos(6°) - cos(42°) ]Now, let's compute cos(6°) - cos(42°). Using the identity cos A - cos B = -2 sin[(A+B)/2] sin[(A-B)/2]So, cos(6°) - cos(42°) = -2 sin[(6° + 42°)/2] sin[(6° - 42°)/2] = -2 sin(24°) sin(-18°)Since sin(-x) = -sin x, this becomes -2 sin(24°)(-sin 18°) = 2 sin(24°) sin(18°)Substituting back into the expression for cos θ:cos θ = 2 cos(36°) [ 2 sin(24°) sin(18°) ] = 4 cos(36°) sin(24°) sin(18°)Hmm, this still seems a bit complicated. Maybe I can use another identity to simplify sin(24°) sin(18°). I recall that sin A sin B can be expressed as [cos(A - B) - cos(A + B)] / 2.So, sin(24°) sin(18°) = [cos(6°) - cos(42°)] / 2Wait, but we already have cos(6°) - cos(42°) in our previous steps, which we simplified to 2 sin(24°) sin(18°). So, this might not help directly.Alternatively, maybe I can compute the numerical values to approximate cos θ and then find θ.Let me try that approach. Let's compute each term numerically:First, compute sin 60°, cos 42°, sin 12°, and cos 6°:sin 60° ≈ √3/2 ≈ 0.8660cos 42° ≈ 0.7431sin 12° ≈ 0.2079cos 6° ≈ 0.9945Now, plug these into the original equation:cos θ = sin 60° + cos 42° - sin 12° - cos 6° ≈ 0.8660 + 0.7431 - 0.2079 - 0.9945Let's compute step by step:0.8660 + 0.7431 = 1.60911.6091 - 0.2079 = 1.40121.4012 - 0.9945 = 0.4067So, cos θ ≈ 0.4067Now, to find θ, we take the arccosine of 0.4067.Using a calculator, arccos(0.4067) ≈ 66°Wait, that's interesting. So, θ ≈ 66°. But let me verify this because sometimes when dealing with multiple angles, there might be another angle in the range of 0° to 360° that also satisfies the equation, but since we're looking for the smallest positive angle, 66° seems likely.But let me cross-check my calculations because sometimes approximations can be misleading.Let me compute the exact value step by step more accurately.Compute sin 60°: exactly √3/2 ≈ 0.8660254038Compute cos 42°: let's use a calculator for more precision. cos(42°) ≈ 0.7431448255Compute sin 12°: sin(12°) ≈ 0.2079116918Compute cos 6°: cos(6°) ≈ 0.9945218954Now, plug these into the equation:cos θ = 0.8660254038 + 0.7431448255 - 0.2079116918 - 0.9945218954Compute step by step:0.8660254038 + 0.7431448255 = 1.60917022931.6091702293 - 0.2079116918 = 1.40125853751.4012585375 - 0.9945218954 = 0.4067366421So, cos θ ≈ 0.4067366421Now, compute arccos(0.4067366421). Let's use a calculator for this.Using a calculator, arccos(0.4067366421) ≈ 66°Yes, that's consistent. So, θ ≈ 66°. But let me check if 66° is indeed the correct angle.Wait, let me compute cos(66°) to see if it's approximately 0.4067.cos(66°) ≈ 0.4067366431Wow, that's almost exactly the value we got. So, cos θ ≈ 0.4067366421, which is very close to cos(66°). Therefore, θ ≈ 66°.But wait, let me think again. Is there a possibility that θ could be 360° - 66° = 294°, which is also a solution for cos θ = 0.4067? But since we're looking for the smallest positive angle, 66° is indeed the smallest.Alternatively, could there be another angle between 0° and 66° that also satisfies cos θ ≈ 0.4067? Let me check cos(60°) = 0.5, which is larger than 0.4067, and cos(70°) ≈ 0.3420, which is smaller. So, 66° is between 60° and 70°, and since cosine decreases as the angle increases from 0° to 90°, 66° is the correct angle.Therefore, the smallest positive angle θ is 66°.

💡First, I need to determine the ratio of Ashok's investment to Pyarelal's investment. According to the problem, Ashok's capital is ( frac{1}{9} ) of Pyarelal's capital. This means the ratio of their investments is ( 1:9 ).Next, I calculate the total parts of the ratio by adding Ashok's and Pyarelal's parts together:[1 + 9 = 10 text{ parts}]The total loss incurred is Rs 670. To find Pyarelal's share of the loss, I determine what fraction of the total loss corresponds to Pyarelal's 9 parts out of the total 10 parts:[text{Pyarelal's share of the loss} = left( frac{9}{10} right) times 670]Performing the multiplication:[frac{9}{10} times 670 = 603]Therefore, Pyarelal's share of the loss is Rs 603.

💡Okay, so I have this problem where I need to find the remainder when the sum ( T = sum_{k=0}^{80} binom{2024}{k} ) is divided by 2027. They mentioned that 2027 is a prime number, which might be useful. Hmm, let me think about how to approach this.First, I remember that when dealing with binomial coefficients modulo a prime, Lucas's Theorem can sometimes be helpful. But Lucas's Theorem is more about expressing the binomial coefficients in terms of their base-p digits. I'm not sure if that's directly applicable here because the upper index is 2024, which is less than 2027, so maybe Lucas's Theorem isn't necessary.Wait, another thought: since 2027 is prime, we can use properties related to binomial coefficients modulo primes. I recall that for a prime ( p ), the sum ( sum_{k=0}^{p-1} binom{n}{k} ) modulo ( p ) can sometimes be simplified, especially if ( n ) is related to ( p ) in some way.In this case, ( n = 2024 ) and ( p = 2027 ). Let me compute ( 2024 ) modulo ( 2027 ). That would be ( 2024 equiv -3 mod 2027 ). So, ( 2024 = -3 ) in modulo 2027. Maybe I can express the binomial coefficients ( binom{2024}{k} ) in terms of ( (-3) ) somehow.Let me write out the general form of a binomial coefficient:( binom{2024}{k} = frac{2024 times 2023 times dots times (2024 - k + 1)}{k!} )Since ( 2024 equiv -3 mod 2027 ), each term in the numerator can be expressed as ( -3, -4, -5, dots ). Specifically, the numerator becomes:( (-3) times (-4) times (-5) times dots times (-3 - k + 1) )Which simplifies to:( (-1)^k times (3 times 4 times 5 times dots times (k + 2)) )Wait, because each term is negative, so multiplying ( k ) terms gives ( (-1)^k ), and the product of the numbers is ( (k + 2)! / 2! ) because it's from 3 to ( k + 2 ). So, that would be ( frac{(k + 2)!}{2!} ).Therefore, the numerator is ( (-1)^k times frac{(k + 2)!}{2} ). The denominator is ( k! ), so putting it together:( binom{2024}{k} equiv (-1)^k times frac{(k + 2)!}{2 times k!} mod 2027 )Simplifying ( frac{(k + 2)!}{k!} ) gives ( (k + 1)(k + 2) ). So, we have:( binom{2024}{k} equiv (-1)^k times frac{(k + 1)(k + 2)}{2} mod 2027 )Which is the same as:( (-1)^k times binom{k + 2}{2} mod 2027 )So, the sum ( T ) becomes:( T equiv sum_{k=0}^{80} (-1)^k times binom{k + 2}{2} mod 2027 )Now, let's write out the first few terms to see the pattern:- For ( k = 0 ): ( (-1)^0 times binom{2}{2} = 1 times 1 = 1 )- For ( k = 1 ): ( (-1)^1 times binom{3}{2} = -1 times 3 = -3 )- For ( k = 2 ): ( (-1)^2 times binom{4}{2} = 1 times 6 = 6 )- For ( k = 3 ): ( (-1)^3 times binom{5}{2} = -1 times 10 = -10 )- For ( k = 4 ): ( (-1)^4 times binom{6}{2} = 1 times 15 = 15 )- And so on...So, the sequence is an alternating series of triangular numbers: 1 - 3 + 6 - 10 + 15 - 21 + ... up to ( k = 80 ).I need to find a way to sum this alternating series. Maybe there's a formula for the sum of such a series. Let me think about how to approach this.I recall that the sum of an alternating series can sometimes be expressed in terms of partial sums or by grouping terms. Let me try to see if there's a pattern or a closed-form expression for this sum.Let me denote the sum as:( S = sum_{k=0}^{n} (-1)^k times binom{k + 2}{2} )Where ( n = 80 ) in our case.Let me compute ( S ) for small values of ( n ) to see if I can spot a pattern.- For ( n = 0 ): ( S = 1 )- For ( n = 1 ): ( 1 - 3 = -2 )- For ( n = 2 ): ( 1 - 3 + 6 = 4 )- For ( n = 3 ): ( 1 - 3 + 6 - 10 = -6 )- For ( n = 4 ): ( 1 - 3 + 6 - 10 + 15 = 9 )- For ( n = 5 ): ( 1 - 3 + 6 - 10 + 15 - 21 = -12 )Hmm, the results are: 1, -2, 4, -6, 9, -12,...Looking at the absolute values: 1, 2, 4, 6, 9, 12,...These numbers seem familiar. They are the triangular numbers again, but arranged in a specific way.Wait, let's see:- For ( n = 0 ): 1 = (1)^2- For ( n = 1 ): |-2| = 2 = (2)^2 / 2- For ( n = 2 ): 4 = (2)^2- For ( n = 3 ): |-6| = 6 = (3)^2 + 3- For ( n = 4 ): 9 = (3)^2- For ( n = 5 ): |-12| = 12 = (4)^2 + 4Hmm, not sure if that's the right direction. Maybe another approach.Alternatively, let's consider the general term:( (-1)^k times binom{k + 2}{2} = (-1)^k times frac{(k + 2)(k + 1)}{2} )Let me write this as:( frac{(-1)^k (k^2 + 3k + 2)}{2} )So, the sum ( S ) becomes:( S = frac{1}{2} sum_{k=0}^{n} (-1)^k (k^2 + 3k + 2) )Which can be split into three separate sums:( S = frac{1}{2} left( sum_{k=0}^{n} (-1)^k k^2 + 3 sum_{k=0}^{n} (-1)^k k + 2 sum_{k=0}^{n} (-1)^k right) )So, I need to compute these three sums:1. ( A = sum_{k=0}^{n} (-1)^k k^2 )2. ( B = sum_{k=0}^{n} (-1)^k k )3. ( C = sum_{k=0}^{n} (-1)^k )I remember that there are known formulas for these alternating sums.Starting with ( C ):( C = sum_{k=0}^{n} (-1)^k )This is a finite geometric series with ratio ( -1 ). The sum is:( C = frac{1 - (-1)^{n + 1}}{1 - (-1)} = frac{1 - (-1)^{n + 1}}{2} )So, if ( n ) is even, ( (-1)^{n + 1} = -1 ), so ( C = frac{1 - (-1)}{2} = 1 ). If ( n ) is odd, ( C = frac{1 - 1}{2} = 0 ).Next, ( B = sum_{k=0}^{n} (-1)^k k )I recall that the sum ( sum_{k=0}^{n} (-1)^k k ) can be expressed as:If ( n ) is even, ( B = frac{n + 1}{2} )If ( n ) is odd, ( B = -frac{n + 1}{2} )Wait, let me verify this with small ( n ):- For ( n = 0 ): ( B = 0 ). Formula: ( (0 + 1)/2 = 0.5 ). Hmm, doesn't match.- For ( n = 1 ): ( 0 - 1 = -1 ). Formula: ( -(1 + 1)/2 = -1 ). Correct.- For ( n = 2 ): ( 0 - 1 + 2 = 1 ). Formula: ( (2 + 1)/2 = 1.5 ). Hmm, not matching.Wait, maybe I misremembered. Let me look for another approach.Alternatively, consider the generating function for ( B ):( G(x) = sum_{k=0}^{infty} (-1)^k k x^k )But since we have a finite sum, maybe it's better to use recursion or another method.Alternatively, notice that:( sum_{k=0}^{n} (-1)^k k = -frac{1}{4} times frac{(-1)^n (2n + 1) + 1}{1} )Wait, I think I need to derive it.Let me denote ( S = sum_{k=0}^{n} (-1)^k k )Consider ( S = 0 - 1 + 2 - 3 + 4 - 5 + dots + (-1)^n n )If ( n ) is even, say ( n = 2m ):( S = (0 - 1) + (2 - 3) + dots + (2m - 2 - (2m - 1)) + 2m )Each pair ( (2k - (2k + 1)) = -1 ), and there are ( m ) such pairs, plus the last term ( 2m ).So, ( S = -m + 2m = m )Since ( n = 2m ), ( m = n/2 ), so ( S = n/2 )Similarly, if ( n ) is odd, say ( n = 2m + 1 ):( S = (0 - 1) + (2 - 3) + dots + (2m - (2m + 1)) )Each pair is ( -1 ), and there are ( m + 1 ) pairs.So, ( S = -(m + 1) )Since ( n = 2m + 1 ), ( m = (n - 1)/2 ), so ( S = -((n - 1)/2 + 1) = -(n + 1)/2 )Therefore, summarizing:- If ( n ) is even, ( B = n/2 )- If ( n ) is odd, ( B = -(n + 1)/2 )Okay, that seems correct.Now, moving on to ( A = sum_{k=0}^{n} (-1)^k k^2 )This one is trickier. Let me see if I can find a pattern or a formula.Again, let me compute for small ( n ):- ( n = 0 ): ( 0^2 = 0 )- ( n = 1 ): ( 0 - 1 = -1 )- ( n = 2 ): ( 0 - 1 + 4 = 3 )- ( n = 3 ): ( 0 - 1 + 4 - 9 = -6 )- ( n = 4 ): ( 0 - 1 + 4 - 9 + 16 = 10 )- ( n = 5 ): ( 0 - 1 + 4 - 9 + 16 - 25 = -15 )Looking at these results: 0, -1, 3, -6, 10, -15,...These are the triangular numbers with alternating signs. Specifically, the absolute values are 0, 1, 3, 6, 10, 15,... which are the triangular numbers ( binom{k + 1}{2} ) for ( k = 0, 1, 2, 3, 4, 5 ).So, it seems that ( A = (-1)^n times binom{n + 1}{2} ) when ( n ) is odd, and ( A = binom{n}{2} ) when ( n ) is even? Wait, let's check:Wait, for ( n = 0 ): 0. Formula: If even, ( binom{0}{2} = 0 ). Correct.For ( n = 1 ): -1. Formula: If odd, ( (-1)^1 times binom{2}{2} = -1 ). Correct.For ( n = 2 ): 3. Formula: If even, ( binom{2}{2} = 1 ). Wait, 3 ≠ 1. Hmm, not matching.Wait, maybe another approach.Alternatively, perhaps ( A = (-1)^n times frac{n(n + 1)}{2} ) when ( n ) is odd, and ( A = frac{n(n - 1)}{2} ) when ( n ) is even.Let me test:For ( n = 0 ): Even, ( 0 times (-1)/2 = 0 ). Correct.For ( n = 1 ): Odd, ( (-1)^1 times 1 times 2 / 2 = -1 ). Correct.For ( n = 2 ): Even, ( 2 times 1 / 2 = 1 ). But the actual sum is 3. Not matching.Hmm, not quite.Wait, maybe the formula is different. Let me look for a general formula.I found a resource that says:( sum_{k=0}^{n} (-1)^k k^2 = (-1)^n frac{n(n + 1)}{2} ) when ( n ) is odd, and ( sum_{k=0}^{n} (-1)^k k^2 = frac{n(n + 1)}{2} ) when ( n ) is even.Wait, let's test this:For ( n = 2 ): Even, ( 2 times 3 / 2 = 3 ). Which matches the sum 3. Correct.For ( n = 3 ): Odd, ( (-1)^3 times 3 times 4 / 2 = -6 ). Correct.For ( n = 4 ): Even, ( 4 times 5 / 2 = 10 ). Correct.For ( n = 5 ): Odd, ( (-1)^5 times 5 times 6 / 2 = -15 ). Correct.Okay, so the formula is:- If ( n ) is even, ( A = frac{n(n + 1)}{2} )- If ( n ) is odd, ( A = -frac{n(n + 1)}{2} )So, in summary:( A = (-1)^n times frac{n(n + 1)}{2} )Wait, no. Because when ( n ) is even, it's positive, and when ( n ) is odd, it's negative. So, actually, ( A = (-1)^n times frac{n(n + 1)}{2} ). But wait, for ( n = 2 ), it's positive, which is ( (-1)^2 times 3 = 3 ). Correct. For ( n = 3 ), it's ( (-1)^3 times 6 = -6 ). Correct.So, yes, ( A = (-1)^n times frac{n(n + 1)}{2} )Wait, but hold on, for ( n = 0 ), it's 0, and the formula gives ( (-1)^0 times 0 times 1 / 2 = 0 ). Correct.Okay, so we have:( A = (-1)^n times frac{n(n + 1)}{2} )Wait, but earlier, when I computed ( A ) for ( n = 2 ), it was 3, which is ( (-1)^2 times 2 times 3 / 2 = 3 ). Correct.So, that seems to be the formula.Therefore, putting it all together:( S = frac{1}{2} (A + 3B + 2C) )Now, let's substitute ( A ), ( B ), and ( C ) based on whether ( n ) is even or odd.Given that in our problem, ( n = 80 ), which is even.So, let's compute each term:1. ( A = (-1)^{80} times frac{80 times 81}{2} = 1 times frac{6480}{2} = 3240 )2. ( B = frac{80}{2} = 40 )3. ( C = 1 ) (since ( n ) is even)Therefore:( S = frac{1}{2} (3240 + 3 times 40 + 2 times 1) )Compute step by step:- ( 3 times 40 = 120 )- ( 2 times 1 = 2 )- So, inside the parentheses: 3240 + 120 + 2 = 3362- Then, ( S = frac{1}{2} times 3362 = 1681 )Therefore, ( T equiv 1681 mod 2027 )Wait, but let me double-check my steps to make sure I didn't make a mistake.First, I expressed ( binom{2024}{k} ) in terms of ( (-1)^k times binom{k + 2}{2} ). That seems correct because ( 2024 equiv -3 mod 2027 ), and expanding the binomial coefficients gives that expression.Then, I recognized the sum as an alternating series of triangular numbers and tried to find a formula for the sum. I broke it down into three separate sums ( A ), ( B ), and ( C ), which I computed based on whether ( n ) is even or odd.Since ( n = 80 ) is even, I applied the formulas for even ( n ):- ( A = frac{n(n + 1)}{2} = 3240 )- ( B = frac{n}{2} = 40 )- ( C = 1 )Plugging these into the expression for ( S ):( S = frac{1}{2} (3240 + 3 times 40 + 2 times 1) = frac{1}{2} (3240 + 120 + 2) = frac{1}{2} times 3362 = 1681 )So, the remainder when ( T ) is divided by 2027 is 1681.I think this makes sense. Let me just verify with a smaller case to ensure my formula is correct.Suppose ( n = 2 ):- ( T = binom{2024}{0} + binom{2024}{1} + binom{2024}{2} )- Which is ( 1 + 2024 + binom{2024}{2} )- Modulo 2027, ( 2024 equiv -3 ), so ( binom{2024}{1} equiv -3 )- ( binom{2024}{2} equiv frac{(-3)(-4)}{2} = 6 )- So, ( T equiv 1 - 3 + 6 = 4 mod 2027 )- According to my formula, ( S = frac{1}{2} (A + 3B + 2C) )- For ( n = 2 ), ( A = 3 ), ( B = 1 ), ( C = 1 )- So, ( S = frac{1}{2} (3 + 3 times 1 + 2 times 1) = frac{1}{2} (3 + 3 + 2) = frac{8}{2} = 4 ). Correct.Another test with ( n = 4 ):- ( T = 1 - 3 + 6 - 10 + 15 = 9 mod 2027 )- Using the formula, ( A = frac{4 times 5}{2} = 10 ), ( B = 2 ), ( C = 1 )- ( S = frac{1}{2} (10 + 3 times 2 + 2 times 1) = frac{1}{2} (10 + 6 + 2) = frac{18}{2} = 9 ). Correct.So, the formula seems to hold for these test cases. Therefore, I feel confident that the result is correct.**Final Answer**boxed{1681}