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$A$ is a set of $n$ arbitrary natural numbers.

We know that $|A|=n$, so $\forall a_i\in A; a_i=nq+j(i)\ \ \ \ (0\le j(i)\lt n)$.

If there exists an $a_k$ such that $j(k)=0$, then the subset $\{a_k\}$ is the answer.

If not, then $\forall a_i\in A; a_i=nq+j(i)\ \ \ \ (0\lt j(i)\lt n)$.

Then I can say there exist at least two members of $A$ such that $n \mod a_a=n\mod a_b$.

But I can't go further. Help needed.

Guys sorry. It's "divisible by $n$". :-shame

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    Is it that there is a subset whose sum is $n$, or a subset whose sum is *divisible* by $n$? – 2017-01-10
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    Do we know anything else about $A$? For instance, if $A = \{n+1, n+2, \ldots, 2n\}$ this does not hold. – 2017-01-10
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    Assuming you mean "divisible by $n$", hint: let $S_i=\{a_1,\cdots,a_i\}$. If one of the $S_i$ works, you are done. If none of them do, show that two of them must sum to the same thing $\pmod n$ and use that. – 2017-01-10
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    You can't mean "equal to $n$". Just take the set $\{4,5\}$. $n=2$ but no subset sums to $2$. – 2017-01-10
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    I meant "divisible" @Alqatrkapa sorry – 2017-01-10
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    I'd suggest pigeonhole principle – 2017-01-10
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    @AHB no problem :) – 2017-01-10
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    @JorgeFernándezHidalgo I don't think so. At the first sight, at least, they aren't much similar. – 2017-01-10
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    you're right, that one is harder, let me find an exact duplicate – 2017-01-10

2 Answers 2

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Hint 1

If the $n$ numbers $a_{1}, a_{1} + a_{2}, \dots, a_{1} + \cdots + a_{n}$ are distinct modulo $n$, then one of them is congruent to $0 \pmod{n}$.

Hint 2

If two of them are congruent modulo $n$, consider the second one (longer sum) minus the first one (shorter sum).

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    I need the sum. not the subtraction. – 2017-01-10
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    It's a set subtraction ... – 2017-01-10
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    What happens if you do $a_{1} + \dots + a_{k} - (a_{1} + \dots + a_{h})$, for $k > h$? Please learn to read the answers in full, it's a form of respect for those who invest their time in answering. – 2017-01-10
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    No. It's not the matter of reading. If I re-read it many times for a day. I can't understand why. It's not obvious for me. I am incapable of it. I don't know about congruency. – 2017-01-10
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Hint 1:

Consider $a_1,a_1+a_2,...$

Hint 2:

Use the Pigeonhole Principle.