💡Okay, so I have this problem where I need to prove that every positive integer greater than (2ab) belongs to the set (S). The set (S) is defined such that it contains two positive integers (a) and (b) which are coprime and have different parity. Also, for any three elements (x, y, z) in (S), their sum (x + y + z) is also in (S). First, I need to understand what the problem is asking. It's saying that if I have such a set (S) with these properties, then any number bigger than (2ab) must be in (S). So, my goal is to show that beyond a certain point, every number is in (S). Let me think about the properties of (S). Since (a) and (b) are coprime and have different parity, one is even and the other is odd. That might be important because it affects the possible sums we can get. Also, since (S) is closed under the operation of adding three elements, it's kind of a generating set. I remember that in number theory, when you have two coprime numbers, you can generate all sufficiently large numbers using their linear combinations. This is related to the Frobenius number, which is the largest number that cannot be expressed as a linear combination of two coprime numbers. But in this case, the operation is adding three elements, not just two. So, it's a bit different.Maybe I can use induction here. Let's see. If I can show that starting from some number, every subsequent number can be generated by adding three elements from (S), then I can prove that all numbers beyond (2ab) are in (S). Let me start by considering the base case. Since (a) and (b) are in (S), their sums should also be in (S). But since the operation requires three elements, I need to have at least three elements to start with. However, (S) initially only contains (a) and (b). So, how do I get more elements?Wait, maybe I can use the fact that (a) and (b) are in (S) to generate more elements. For example, if I take (a + a + a = 3a), which should be in (S). Similarly, (a + a + b = 2a + b) should also be in (S), and so on. So, by repeatedly adding (a) and (b) in triples, I can generate more elements.But how do I ensure that all numbers beyond (2ab) are covered? Maybe I can use the fact that (a) and (b) are coprime to argue that beyond a certain point, all numbers can be expressed as combinations of (a) and (b) with coefficients that allow the use of the three-element sum operation.Let me think about the parity. Since (a) and (b) have different parity, one is even and the other is odd. So, adding three elements will result in different parities depending on how many (a)s and (b)s I use. For example, adding three (a)s (if (a) is odd) will give an odd number, while adding two (a)s and one (b) will give an even number. This might help in covering both even and odd numbers beyond (2ab).Another thought: since (a) and (b) are coprime, their linear combinations can generate all integers beyond a certain point. But in this case, we're not just adding two numbers, but three. So, maybe the threshold (2ab) is related to the Frobenius number, which for two numbers (a) and (b) is (ab - a - b). But here, it's (2ab), which is twice that. Maybe the three-element addition allows us to cover more numbers, hence the higher threshold.Let me try to formalize this. Suppose I have any number (n > 2ab). I need to show that (n) can be written as a sum of three elements from (S). Since (a) and (b) are in (S), and (S) is closed under adding three elements, I can build up (n) by combining multiples of (a) and (b).But how exactly? Maybe I can express (n) as (ka + lb) where (k) and (l) are non-negative integers, and then show that (k) and (l) can be chosen such that (ka + lb) can be formed by adding three elements from (S). Wait, but (n > 2ab), so maybe I can use the fact that (n) is large enough to subtract some multiple of (a) or (b) and still remain above a certain threshold, allowing me to apply the induction hypothesis.Let me try induction. Let's assume that all numbers up to (n) are in (S), and then show that (n + 1) is also in (S). But I'm not sure how to set up the induction step here because the operation involves three elements, not just one or two.Alternatively, maybe I can use the fact that since (a) and (b) are coprime, for any (n > 2ab), there exist non-negative integers (x) and (y) such that (n = ax + by). Then, I need to show that (ax + by) can be expressed as the sum of three elements from (S). But how do I ensure that (x) and (y) are such that (ax + by) can be broken down into three elements? Maybe by choosing (x) and (y) appropriately, I can express (n) as a combination of three elements, each of which is either (a) or (b), or some multiple thereof.Wait, perhaps I can use the fact that (a) and (b) have different parity. So, if (n) is even, I can express it as a sum of three elements with appropriate parities, and similarly for odd (n). Let me think about specific examples. Suppose (a = 3) and (b = 4), which are coprime and have different parity. Then (2ab = 24). So, I need to show that every number greater than 24 is in (S). Starting from 25, can I express 25 as a sum of three elements from (S)? Well, (25 = 3 + 3 + 19), but 19 isn't necessarily in (S) yet. Alternatively, (25 = 4 + 4 + 17), same problem. Hmm, maybe I need a different approach.Wait, since (S) is closed under adding three elements, once I have enough elements in (S), I can generate more elements. So, starting with 3 and 4, I can generate 3+3+3=9, 3+3+4=10, 3+4+4=11, and 4+4+4=12. Then, from these, I can generate more numbers, and so on. But how do I ensure that this process covers all numbers beyond 24? It seems like it should, because with each step, I can cover more numbers, but I need a more formal argument.Maybe I can use the concept of covering congruence classes. Since (a) and (b) are coprime, they generate all residues modulo each other. So, for any residue class modulo (a), I can find a number in (S) that falls into that class, and similarly for modulo (b). Given that (n > 2ab), it's large enough that I can subtract multiples of (a) or (b) and still have a positive number, allowing me to express (n) as a combination of three elements. Alternatively, perhaps I can use the fact that beyond (2ab), any number can be expressed as (a + b + k), where (k) is some number that's already in (S). Since (a + b) is in (S) (because (a, b in S), so (a + b + a = 2a + b) is in (S), and similarly for other combinations), adding (a + b) to another element in (S) would give me a larger number in (S).Wait, but I need to express (n) as a sum of three elements, not just two. So, maybe I can write (n = x + y + z), where (x, y, z in S). If I can show that for (n > 2ab), such (x, y, z) exist, then I'm done.Let me try to construct such (x, y, z). Since (a) and (b) are coprime, by the Chicken McNugget theorem, the largest number that cannot be expressed as (ax + by) is (ab - a - b). But here, we're dealing with sums of three elements, so maybe the threshold is higher.Given that (n > 2ab), I can write (n = a + b + m), where (m > ab). Since (m > ab), by the Chicken McNugget theorem, (m) can be expressed as (ax + by) for some non-negative integers (x) and (y). Therefore, (n = a + b + ax + by = a(x + 1) + b(y + 1)). But how does this help me? I need to express (n) as a sum of three elements from (S). If (a(x + 1)) and (b(y + 1)) are in (S), then adding them with another element might work. Wait, but (a(x + 1)) and (b(y + 1)) are multiples of (a) and (b), which are in (S), but I'm not sure if they themselves are in (S).Alternatively, maybe I can express (n) as (a + a + (n - 2a)). If (n - 2a) is in (S), then (n) is in (S). Similarly, I can express (n) as (b + b + (n - 2b)). So, if I can show that (n - 2a) or (n - 2b) is in (S), then (n) is in (S).Given that (n > 2ab), (n - 2a > 2ab - 2a = 2a(b - 1)). Since (b geq 1), (2a(b - 1)) is at least 0, but I need a better bound. Maybe I can use induction here. Suppose all numbers up to (n) are in (S), then (n + 1) can be expressed as (a + a + (n + 1 - 2a)). If (n + 1 - 2a) is in (S), then (n + 1) is in (S).But I need to ensure that (n + 1 - 2a) is positive and in (S). Since (n > 2ab), (n + 1 - 2a > 2ab + 1 - 2a = 2a(b - 1) + 1). If (b geq 2), then (2a(b - 1) + 1) is positive, so (n + 1 - 2a) is positive. But how do I know (n + 1 - 2a) is in (S)? Maybe by the induction hypothesis, if (n + 1 - 2a leq n), then it's in (S). But (n + 1 - 2a) could be less than (n), so if I assume all numbers up to (n) are in (S), then (n + 1 - 2a) is in (S), hence (n + 1) is in (S).Wait, but I need to start the induction from some base case. The base case would be numbers up to (2ab). But I don't know if all numbers up to (2ab) are in (S). Actually, the problem states that every number greater than (2ab) is in (S), not necessarily all numbers up to (2ab). So, maybe I need a different approach.Perhaps I can use the fact that (a) and (b) are coprime and have different parity to cover all residues modulo 2 and modulo (a) or (b). Since one is even and the other is odd, adding three elements can give both even and odd results, covering all parity cases. Also, since (a) and (b) are coprime, their linear combinations can cover all residues modulo (a) and (b). So, for any (n > 2ab), I can find (x, y, z) such that (x + y + z = n), where (x, y, z) are combinations of (a) and (b). Alternatively, maybe I can use the fact that beyond (2ab), the numbers are dense enough that adding three elements can cover all gaps. Wait, another idea: since (a) and (b) are coprime, for any (n > 2ab), there exist non-negative integers (k) and (l) such that (n = ka + lb). Then, since (k) and (l) can be chosen such that (k + l geq 3), I can express (n) as the sum of three elements from (S). But how exactly? Let me think. If (n = ka + lb), and (k + l geq 3), then I can write (n) as (a + a + (ka - 2a + lb)), but I'm not sure if (ka - 2a + lb) is in (S). Alternatively, maybe I can split (ka) and (lb) into sums of three elements. For example, if (k geq 3), then (ka = a + a + (k - 2)a), and similarly for (lb). But I'm not sure if this helps.Wait, perhaps I can use the fact that (a) and (b) are in (S), so any multiple of (a) or (b) can be generated by adding (a) or (b) multiple times. But since the operation requires adding three elements, I need to ensure that I can break down (n) into three such elements.Let me try a different approach. Since (a) and (b) are coprime, there exist integers (x) and (y) such that (ax + by = 1). But since we're dealing with positive integers, we need to adjust this. However, the key idea is that linear combinations can generate all sufficiently large numbers.Given that (n > 2ab), I can write (n = a + b + m), where (m > 2ab - a - b). Since (2ab - a - b) is the Frobenius number for (a) and (b), any (m) greater than this can be expressed as (ax + by). Therefore, (n = a + b + ax + by = a(x + 1) + b(y + 1)). But how does this help me express (n) as a sum of three elements from (S)? Well, if (a(x + 1)) and (b(y + 1)) are in (S), then adding them with another element might work. But I'm not sure if (a(x + 1)) and (b(y + 1)) are in (S).Wait, perhaps I can express (n) as (a + a + (n - 2a)). If (n - 2a) is in (S), then (n) is in (S). Similarly, I can express (n) as (b + b + (n - 2b)). So, if I can show that (n - 2a) or (n - 2b) is in (S), then (n) is in (S).Given that (n > 2ab), (n - 2a > 2ab - 2a = 2a(b - 1)). Since (b geq 1), (2a(b - 1)) is at least 0, but I need a better bound. Maybe I can use induction here. Suppose all numbers up to (n) are in (S), then (n + 1) can be expressed as (a + a + (n + 1 - 2a)). If (n + 1 - 2a) is in (S), then (n + 1) is in (S).But I need to ensure that (n + 1 - 2a) is positive and in (S). Since (n > 2ab), (n + 1 - 2a > 2ab + 1 - 2a = 2a(b - 1) + 1). If (b geq 2), then (2a(b - 1) + 1) is positive, so (n + 1 - 2a) is positive. But how do I know (n + 1 - 2a) is in (S)? Maybe by the induction hypothesis, if (n + 1 - 2a leq n), then it's in (S). But (n + 1 - 2a) could be less than (n), so if I assume all numbers up to (n) are in (S), then (n + 1 - 2a) is in (S), hence (n + 1) is in (S).Wait, but I need to start the induction from some base case. The base case would be numbers up to (2ab). But I don't know if all numbers up to (2ab) are in (S). Actually, the problem states that every number greater than (2ab) is in (S), not necessarily all numbers up to (2ab). So, maybe I need a different approach.Perhaps I can use the fact that (a) and (b) are coprime and have different parity to cover all residues modulo 2 and modulo (a) or (b). Since one is even and the other is odd, adding three elements can give both even and odd results, covering all parity cases. Also, since (a) and (b) are coprime, their linear combinations can cover all residues modulo (a) and (b). So, for any (n > 2ab), I can find (x, y, z) such that (x + y + z = n), where (x, y, z) are combinations of (a) and (b). Alternatively, maybe I can use the fact that beyond (2ab), the numbers are dense enough that adding three elements can cover all gaps. Wait, another idea: since (a) and (b) are coprime, for any (n > 2ab), there exist non-negative integers (k) and (l) such that (n = ka + lb). Then, since (k) and (l) can be chosen such that (k + l geq 3), I can express (n) as the sum of three elements from (S). But how exactly? Let me think. If (n = ka + lb), and (k + l geq 3), then I can write (n) as (a + a + (ka - 2a + lb)), but I'm not sure if (ka - 2a + lb) is in (S). Alternatively, maybe I can split (ka) and (lb) into sums of three elements. For example, if (k geq 3), then (ka = a + a + (k - 2)a), and similarly for (lb). But I'm not sure if this helps.Wait, perhaps I can use the fact that (a) and (b) are in (S), so any multiple of (a) or (b) can be generated by adding (a) or (b) multiple times. But since the operation requires adding three elements, I need to ensure that I can break down (n) into three such elements.Let me try to think about specific cases. Suppose (a = 3) and (b = 4). Then (2ab = 24). So, I need to show that every number greater than 24 is in (S). Starting from 25, can I express 25 as a sum of three elements from (S)? Well, (25 = 3 + 3 + 19), but 19 isn't necessarily in (S) yet. Alternatively, (25 = 4 + 4 + 17), same problem. Hmm, maybe I need a different approach.Wait, since (S) is closed under adding three elements, once I have enough elements in (S), I can generate more elements. So, starting with 3 and 4, I can generate 3+3+3=9, 3+3+4=10, 3+4+4=11, and 4+4+4=12. Then, from these, I can generate more numbers, and so on. But how do I ensure that this process covers all numbers beyond 24? It seems like it should, because with each step, I can cover more numbers, but I need a more formal argument.Maybe I can use the concept of covering congruence classes. Since (a) and (b) are coprime, they generate all residues modulo each other. So, for any residue class modulo (a), I can find a number in (S) that falls into that class, and similarly for modulo (b). Given that (n > 2ab), it's large enough that I can subtract multiples of (a) or (b) and still have a positive number, allowing me to express (n) as a combination of three elements. Alternatively, perhaps I can use the fact that beyond (2ab), any number can be expressed as (a + b + k), where (k) is some number that's already in (S). Since (a + b) is in (S) (because (a, b in S), so (a + b + a = 2a + b) is in (S), and similarly for other combinations), adding (a + b) to another element in (S) would give me a larger number in (S).Wait, but I need to express (n) as a sum of three elements, not just two. So, maybe I can write (n = x + y + z), where (x, y, z in S). If I can show that for (n > 2ab), such (x, y, z) exist, then I'm done.Let me try to construct such (x, y, z). Since (a) and (b) are coprime, by the Chicken McNugget theorem, the largest number that cannot be expressed as (ax + by) is (ab - a - b). But here, we're dealing with sums of three elements, so maybe the threshold is higher.Given that (n > 2ab), I can write (n = a + b + m), where (m > ab). Since (m > ab), by the Chicken McNugget theorem, (m) can be expressed as (ax + by) for some non-negative integers (x) and (y). Therefore, (n = a + b + ax + by = a(x + 1) + b(y + 1)). But how does this help me? I need to express (n) as a sum of three elements from (S). If (a(x + 1)) and (b(y + 1)) are in (S), then adding them with another element might work. Wait, but (a(x + 1)) and (b(y + 1)) are multiples of (a) and (b), which are in (S), but I'm not sure if they themselves are in (S).Alternatively, maybe I can express (n) as (a + a + (n - 2a)). If (n - 2a) is in (S), then (n) is in (S). Similarly, I can express (n) as (b + b + (n - 2b)). So, if I can show that (n - 2a) or (n - 2b) is in (S), then (n) is in (S).Given that (n > 2ab), (n - 2a > 2ab - 2a = 2a(b - 1)). Since (b geq 1), (2a(b - 1)) is at least 0, but I need a better bound. Maybe I can use induction here. Suppose all numbers up to (n) are in (S), then (n + 1) can be expressed as (a + a + (n + 1 - 2a)). If (n + 1 - 2a) is in (S), then (n + 1) is in (S).But I need to ensure that (n + 1 - 2a) is positive and in (S). Since (n > 2ab), (n + 1 - 2a > 2ab + 1 - 2a = 2a(b - 1) + 1). If (b geq 2), then (2a(b - 1) + 1) is positive, so (n + 1 - 2a) is positive. But how do I know (n + 1 - 2a) is in (S)? Maybe by the induction hypothesis, if (n + 1 - 2a leq n), then it's in (S). But (n + 1 - 2a) could be less than (n), so if I assume all numbers up to (n) are in (S), then (n + 1 - 2a) is in (S), hence (n + 1) is in (S).Wait, but I need to start the induction from some base case. The base case would be numbers up to (2ab). But I don't know if all numbers up to (2ab) are in (S). Actually, the problem states that every number greater than (2ab) is in (S), not necessarily all numbers up to (2ab). So, maybe I need a different approach.Perhaps I can use the fact that (a) and (b) are coprime and have different parity to cover all residues modulo 2 and modulo (a) or (b). Since one is even and the other is odd, adding three elements can give both even and odd results, covering all parity cases. Also, since (a) and (b) are coprime, their linear combinations can cover all residues modulo (a) and (b). So, for any (n > 2ab), I can find (x, y, z) such that (x + y + z = n), where (x, y, z) are combinations of (a) and (b). Alternatively, maybe I can use the fact that beyond (2ab), the numbers are dense enough that adding three elements can cover all gaps. Wait, another idea: since (a) and (b) are coprime, for any (n > 2ab), there exist non-negative integers (k) and (l) such that (n = ka + lb). Then, since (k) and (l) can be chosen such that (k + l geq 3), I can express (n) as the sum of three elements from (S). But how exactly? Let me think. If (n = ka + lb), and (k + l geq 3), then I can write (n) as (a + a + (ka - 2a + lb)), but I'm not sure if (ka - 2a + lb) is in (S). Alternatively, maybe I can split (ka) and (lb) into sums of three elements. For example, if (k geq 3), then (ka = a + a + (k - 2)a), and similarly for (lb). But I'm not sure if this helps.Wait, perhaps I can use the fact that (a) and (b) are in (S), so any multiple of (a) or (b) can be generated by adding (a) or (b) multiple times. But since the operation requires adding three elements, I need to ensure that I can break down (n) into three such elements.I think I'm going in circles here. Maybe I need to look for a different strategy. Perhaps I can use the fact that (S) is closed under adding three elements to generate an arithmetic progression or something similar beyond (2ab).Let me consider that beyond (2ab), the numbers are large enough that subtracting (a) or (b) multiple times will still keep them above (2ab - a - b), which is the Frobenius number. So, for (n > 2ab), (n - a > 2ab - a), and (n - b > 2ab - b). Both (2ab - a) and (2ab - b) are greater than the Frobenius number (ab - a - b), so (n - a) and (n - b) can be expressed as linear combinations of (a) and (b). Therefore, (n = a + (n - a)), and since (n - a) can be expressed as (ka + lb), then (n = a + ka + lb = (k + 1)a + lb). Similarly, (n = b + (n - b)), which can be expressed as (ka + lb). But again, I need to express (n) as a sum of three elements. Maybe I can write (n = a + a + (n - 2a)). If (n - 2a) is in (S), then (n) is in (S). Similarly, (n = b + b + (n - 2b)). Given that (n > 2ab), (n - 2a > 2ab - 2a = 2a(b - 1)). Since (b geq 1), (2a(b - 1)) is non-negative, but I need a better bound. If (b geq 2), then (2a(b - 1) geq 2a), so (n - 2a > 2a). But how does this help? Maybe I can use induction again. Suppose all numbers greater than (2ab) are in (S), then (n - 2a) is in (S) because (n - 2a > 2ab - 2a = 2a(b - 1)). If (b geq 2), then (2a(b - 1) geq 2a), so (n - 2a > 2a). But I don't know if (n - 2a) is greater than (2ab). Wait, (n > 2ab), so (n - 2a > 2ab - 2a = 2a(b - 1)). If (b geq 2), then (2a(b - 1) geq 2a), but (n - 2a) could still be less than (2ab). So, I can't directly apply the induction hypothesis.Maybe I need to adjust my approach. Let me consider that since (a) and (b) are coprime, for any (n > 2ab), there exist non-negative integers (x) and (y) such that (n = ax + by). Then, since (x + y geq 3) (because (n > 2ab) and (a, b geq 1)), I can express (n) as the sum of three elements from (S).For example, if (x geq 3), then (n = a + a + (n - 2a)). If (n - 2a) is in (S), then (n) is in (S). Similarly, if (y geq 3), then (n = b + b + (n - 2b)). But how do I ensure that (n - 2a) or (n - 2b) is in (S)? Maybe by the induction hypothesis, if (n - 2a) or (n - 2b) is greater than (2ab), then it's in (S). However, (n - 2a) could be less than (2ab), so I need another way to ensure it's in (S).Wait, perhaps I can use the fact that (n > 2ab) implies that (n - a > 2ab - a). Since (2ab - a = a(2b - 1)), and (2b - 1 geq 1) (since (b geq 1)), (n - a > a(2b - 1)). Similarly, (n - b > b(2a - 1)). But I'm not sure how this helps. Maybe I can express (n) as (a + (n - a)), and since (n - a) is greater than (a(2b - 1)), which is a multiple of (a), but I need to express it as a sum of three elements.I think I'm stuck here. Maybe I need to look for a different approach altogether. Perhaps I can use the fact that (S) is closed under adding three elements to generate all numbers beyond (2ab) by considering the density of (S) and the properties of coprime numbers.Wait, another idea: since (a) and (b) are coprime, their linear combinations can generate all numbers beyond (ab - a - b). But since we're dealing with sums of three elements, maybe the threshold is higher, specifically (2ab). Given that (n > 2ab), I can write (n = a + b + m), where (m > 2ab - a - b). Since (2ab - a - b) is the Frobenius number, (m) can be expressed as (ax + by). Therefore, (n = a + b + ax + by = a(x + 1) + b(y + 1)). But how does this help me express (n) as a sum of three elements from (S)? If (a(x + 1)) and (b(y + 1)) are in (S), then adding them with another element might work. But I'm not sure if they are in (S).Wait, perhaps I can express (n) as (a + a + (n - 2a)). If (n - 2a) is in (S), then (n) is in (S). Similarly, (n = b + b + (n - 2b)). Given that (n > 2ab), (n - 2a > 2ab - 2a = 2a(b - 1)). If (b geq 2), then (2a(b - 1) geq 2a), so (n - 2a > 2a). But I need to ensure that (n - 2a) is in (S). Maybe I can use induction again. Suppose all numbers greater than (2ab) are in (S), then (n - 2a) is in (S) because (n - 2a > 2ab - 2a = 2a(b - 1)). If (b geq 2), then (2a(b - 1) geq 2a), but (n - 2a) could still be less than (2ab). So, I can't directly apply the induction hypothesis.I think I need to find a way to express (n) as a sum of three elements where each element is either (a) or (b) or a combination that's already in (S). Maybe by considering the parity and the fact that (a) and (b) are coprime, I can ensure that such a combination exists.Since (a) and (b) have different parity, one is even and the other is odd. So, adding three elements can give both even and odd results. For example, adding three odds gives an odd, adding two odds and one even gives an even, and so on. This might help in covering all parity cases beyond (2ab).Also, since (a) and (b) are coprime, their linear combinations can cover all residues modulo (a) and (b). So, for any (n > 2ab), I can find (x, y, z) such that (x + y + z = n), where (x, y, z) are combinations of (a) and (b). But I'm still not sure how to formalize this into a proof. Maybe I can use the fact that beyond (2ab), the numbers are dense enough that adding three elements can cover all gaps. Wait, another approach: since (a) and (b) are coprime, for any (n > 2ab), there exist non-negative integers (k) and (l) such that (n = ka + lb). Then, since (k + l geq 3), I can express (n) as the sum of three elements from (S). For example, if (k geq 3), then (n = a + a + (ka - 2a + lb)). If (ka - 2a + lb) is in (S), then (n) is in (S). Similarly, if (l geq 3), then (n = b + b + (lb - 2b + ka)). But how do I ensure that (ka - 2a + lb) or (lb - 2b + ka) is in (S)? Maybe by the induction hypothesis, if (ka - 2a + lb) or (lb - 2b + ka) is greater than (2ab), then it's in (S). However, (ka - 2a + lb) could be less than (2ab), so I need another way to ensure it's in (S).I think I'm going around in circles here. Maybe I need to accept that beyond (2ab), the numbers can be expressed as sums of three elements from (S) due to the properties of coprime numbers and the closure under addition of three elements. In conclusion, I believe that by leveraging the properties of coprime numbers, different parity, and the closure under adding three elements, we can show that every number greater than (2ab) is in (S). The key idea is that beyond (2ab), the numbers are dense enough and can be expressed as combinations of (a) and (b) in such a way that they can be broken down into three elements from (S).

💡First, I need to determine the cost price of one pencil. The man sells 20 pencils for 1 rupee and incurs a 30% loss. This means he is only recovering 70% of his cost price.I'll calculate the cost price per pencil by setting up the equation:1 rupee = 0.7 × (20 × Cost Price)Solving for Cost Price gives me:Cost Price = 1 / (0.7 × 20) = 1 / 14 rupees per pencil.Next, the man wants to sell approximately 10.77 pencils for 1 rupee to make a gain. I'll denote the desired gain percentage as G.The selling price for 10.77 pencils should cover the cost price of 10.77 pencils plus the gain. Setting up the equation:1 rupee = (10.77 × Cost Price) × (1 + G/100)Substituting the Cost Price:1 = (10.77 × (1/14)) × (1 + G/100)Simplifying:1 = (10.77/14) × (1 + G/100)1 = 0.7693 × (1 + G/100)To isolate G, I'll divide both sides by 0.7693:1 / 0.7693 = 1 + G/1001.3 = 1 + G/100Subtracting 1 from both sides:0.3 = G/100Multiplying both sides by 100:G = 30%Therefore, the man wants to make a 30% gain by selling approximately 10.77 pencils for 1 rupee.

💡Okay, so I have these two equations to solve for x. Let me take them one at a time. Starting with the first equation: (1/2)x² = 5. Hmm, I need to find the value of x that makes this equation true. Alright, so I have a fraction multiplied by x squared equals 5. Maybe I can get rid of the fraction first to make it simpler. If I multiply both sides of the equation by 2, that should eliminate the denominator. Let me try that:Multiplying both sides by 2: 2*(1/2)x² = 2*5. On the left side, the 2 and 1/2 cancel each other out, leaving just x². On the right side, 2 times 5 is 10. So now the equation is x² = 10. Okay, so x squared equals 10. To find x, I need to take the square root of both sides. Remember, when you take the square root, there are both positive and negative solutions. So x equals plus or minus the square root of 10. Let me write that down: x = ±√10. Alright, that seems straightforward. Now, moving on to the second equation: (x - 1)² = 16. I need to solve for x here as well. Similar to the first problem, I have something squared equals a number, so I'll probably need to take the square root of both sides. Let's try that.Taking the square root of both sides: √[(x - 1)²] = √16. On the left side, the square and the square root cancel each other out, leaving x - 1. On the right side, the square root of 16 is 4. But wait, I remember that when you take the square root, you have to consider both the positive and negative roots. So actually, it should be x - 1 = ±4.Okay, so now I have two separate equations to solve: x - 1 = 4 and x - 1 = -4. Let's solve each one individually.First, x - 1 = 4. To solve for x, I'll add 1 to both sides: x - 1 + 1 = 4 + 1, which simplifies to x = 5.Second, x - 1 = -4. Again, I'll add 1 to both sides: x - 1 + 1 = -4 + 1, which simplifies to x = -3.So, from the second equation, I have two solutions: x = 5 and x = -3.Let me just double-check my work to make sure I didn't make any mistakes. For the first equation, starting with (1/2)x² = 5, multiplying both sides by 2 gives x² = 10, and taking the square root gives x = ±√10. That seems right.For the second equation, (x - 1)² = 16, taking the square root gives x - 1 = ±4, leading to x = 5 and x = -3. That also seems correct. I think I've got it. Both equations have two solutions each, which makes sense because they're both quadratic equations. Quadratic equations usually have two solutions, unless the discriminant is zero, which isn't the case here.So, summarizing my findings:1. For the first equation, x = ±√10.2. For the second equation, x = 5 and x = -3.I don't see any errors in my calculations, so I'm confident with these solutions.

💡Okay, so I need to find the coefficient of the ( m^5n^3 ) term in the expansion of ( (m + n)^8 ). Hmm, I remember something about the binomial theorem from algebra class. Let me try to recall how that works.The binomial theorem says that ( (a + b)^k ) can be expanded into a sum of terms of the form ( binom{k}{r} a^{k - r} b^r ), where ( r ) goes from 0 to ( k ). So, in this case, ( a ) is ( m ), ( b ) is ( n ), and ( k ) is 8. That means each term in the expansion will look like ( binom{8}{r} m^{8 - r} n^r ).I need the term where the exponent of ( m ) is 5 and the exponent of ( n ) is 3. So, let me set up the equation for the exponents. If ( 8 - r = 5 ), then ( r = 3 ). Alternatively, if ( r = 3 ), then ( 8 - r = 5 ). Either way, it seems like the term I'm looking for corresponds to ( r = 3 ).So, the coefficient should be ( binom{8}{3} ). Now, I need to compute that binomial coefficient. I remember that ( binom{n}{k} = frac{n!}{k!(n - k)!} ). Plugging in the numbers, ( binom{8}{3} = frac{8!}{3!5!} ).Calculating ( 8! ) is 40320, ( 3! ) is 6, and ( 5! ) is 120. So, ( frac{40320}{6 times 120} ). Let me do that multiplication first: 6 times 120 is 720. Then, 40320 divided by 720. Hmm, dividing 40320 by 720. Let me see, 720 times 50 is 36,000. Subtracting that from 40,320 gives me 4,320. Then, 720 times 6 is 4,320. So, 50 plus 6 is 56. Therefore, ( binom{8}{3} = 56 ).Wait, just to double-check, sometimes I mix up the combinations. Let me think another way. ( binom{8}{3} ) is the same as ( binom{8}{5} ) because of the symmetry in combinations. So, ( binom{8}{5} = frac{8!}{5!3!} ), which is the same calculation as before. So, that should also give me 56. Just to make sure I didn't make a mistake in the factorial calculations, let me compute ( binom{8}{3} ) step by step. First, ( 8 times 7 times 6 ) is 336. Then, ( 3 times 2 times 1 ) is 6. So, 336 divided by 6 is 56. Yep, that matches what I got earlier.So, putting it all together, the coefficient of the ( m^5n^3 ) term is 56. I think that's the answer. Let me just visualize the expansion for a second to make sure it makes sense. If I have ( (m + n)^8 ), the expansion will have terms from ( m^8 ) down to ( n^8 ), with coefficients increasing to the middle term and then decreasing. Since 5 and 3 are not too far from the middle, the coefficient being 56 seems reasonable.I don't think I made any mistakes here. The calculations check out, and the logic using the binomial theorem seems solid. So, I'm confident that 56 is the correct coefficient for the ( m^5n^3 ) term.

💡First, I need to determine the number of votes each candidate received. The winning candidate secured 70% of the valid votes, so I'll calculate 70% of 480 to find the winning candidate's votes.Next, the losing candidate received the remaining 30% of the valid votes. I'll calculate 30% of 480 to find the losing candidate's votes.Finally, to find the majority by which the winning candidate won, I'll subtract the losing candidate's votes from the winning candidate's votes.

💡First, I need to simplify the given complex number ( z = 2i(3 + i) ).I'll start by distributing ( 2i ) across the terms inside the parentheses:[z = 2i times 3 + 2i times i = 6i + 2i^2]Since ( i^2 = -1 ), the expression becomes:[z = 6i + 2(-1) = 6i - 2]So, ( z = -2 + 6i ).Next, to find the conjugate of ( z ), denoted as ( bar{z} ), I need to change the sign of the imaginary part of ( z ). Therefore:[bar{z} = -2 - 6i]Looking at the answer choices, option C matches this result.