"Let S be a set of n pair-wise distinct positive integers. Prove that there exists a subset T in S, such that the sum of elements in T is divisible by n." For example, suppose we have 5 numbers: 3,6,11,8,7 Then we have 7+3=10 which is divisible by 5

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"Let S be a set of n pair-wise distinct positive integers. Prove that there exists a subset T in S, such that the sum of elements in T is divisible by n." For example, suppose we have 5 numbers: 3,6,11,8,7 Then we have 7+3=10 which is divisible by 5

Mathematics
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Actually this question is of @Traxter
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If it is wrong PLZ tell me as I will work on it again..... Since @Traxter is not online for a long time I want u guys to check it.
@sauravshakya what is pair wise distinct positive numbers, heard that for the first time
I guess it means no integer is repeated...... all elements of the set are unique.
I'm not quite sure of this, but I'll show you my solution: Let's try and create a set S with \(n\) elements such that subset T as described doesn't exist. For the first element, we can choose \(a_{1}\) as long as \(a_{1}\not=0modn \). So there is one choice mod n we are not allowed to take. Set so far: \(S=(a_{1})\) For the second element, we can choose \(a_{2}\) as long as \(a_{2}\not= 0modn\) and \(a_{2}\not= (-a_{1})modn\), since otherwise \((a_{1}+a_{2})modn=0modn\). So there are 2 choices mod n we are not allowed to take. Set so far: \(S=(a_{1}, a_{2})\) For the third element, we can choose \(a_{3}\) as long as: \(a_{3}\not= 0modn\) \(a_{3}\not= (-a_{1})modn\) \(a_{3}\not= (-a_{2})modn\) \(a_{3}\not= -(a_{1}+a_{2})modn\) So there are 4 choices mod n we aren't allowed to take. As we continue, we are not allowed to take elements which are equal to the additive inverse of subsets of the set we have so far. So when picking the k'th element, where \( 0\le k \le n\), we will have: \[n-2^k\] choices of elements to choose from which would not lead to the formation of a subset T as described in the question. We can keep choosing elements as long as: \[n-2^k>0\] But will run out of choices as soon as: \[n-2^k\le0\] So we will run out of choices on the k'th turn when k satisfies: \(n \le 2^k \) This will be satisfied for some \( 0\le k \le n\), thus it is impossible to create such a set S.
Thats better.

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