Proving the Subset Sum Divisible by n

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For any set of n integers, there exists a subset whose sum is divisible by n. This result applies to both sets and multisets, allowing for repeated numbers. The proof involves examining the sums of the first k numbers modulo n, where if n different sums are generated, one must be congruent to zero. If two sums yield the same result modulo n, their difference will also be divisible by n. The discussion clarifies that the subset can consist of one or more numbers, emphasizing the importance of proper terminology regarding sets and subsets.
daniel_i_l
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Prove that for every set of n numbers has a subset whose sum is divisible by n. I found this result very interesting. It's also fun to prove!
 
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Must these numbers be in sequence? Like {4,5,6,7,8}, or could it be any set like {1,10}?

EDIT: Never mind...that was a silly question.
 
can this subset be 1 number? or does it have to be summable?
 
Both the set and subset can be any not emty set. But I forgot to mention that numbers can repeat themselves so that in this case {1,1,2,2,4} is a "set" of 5 numbers and to answer ices's question, {1} is a subset of it.
 
No, that is NOT a "set" of 5 numbers! {1,1,2, 2,4}= {1, 2, 4} is a set of 3 numbers!
If you mean something else do not use the word "set".
 
HallsofIvy said:
No, that is NOT a "set" of 5 numbers! {1,1,2, 2,4}= {1, 2, 4} is a set of 3 numbers!
If you mean something else do not use the word "set".

The result holds for sets. daniel_i_l's point was surely that the result naturally extends to multisets.
 
work mod n. given n numbers, (integers), if we sum the first k of them, for k = 1,...,n, and we get n different numbers mod n, then one of them is congruent to n. done.if on the other hand two different sums have the same sum mod n , then subtracting them, gives a sum congr'uent to zero mod n. done again.
 
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