Prove that there exists at least one subset of the set $A=\{a_1,a_2,a_3,\dots,a_n\}$ such that the sum of its members is divisible by n.

Question

$A$ 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

Is it that there is a subset whose sum is $n$, or a subset whose sum is divisible by $n$? — Joshua Mundinger, Jan 10 '17 at 18:31
Do we know anything else about $A$? For instance, if $A = {n+1, n+2, \ldots, 2n}$ this does not hold. — Joshua Mundinger, Jan 10 '17 at 18:33
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. — lulu, Jan 10 '17 at 18:33
You can't mean "equal to $n$". Just take the set ${4,5}$. $n=2$ but no subset sums to $2$. — lulu, Jan 10 '17 at 18:34
@JorgeFernándezHidalgo I don't think so. At the first sight, at least, they aren't much similar. — AHB, Jan 10 '17 at 18:57
you're right, that one is harder, let me find an exact duplicate — Asinomás, Jan 10 '17 at 19:01

score 2 · Answer 1 · answered Jan 10 '17 at 18:46

2

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).

answered Jan 10 '17 at 18:46

Andreas Caranti

I need the sum. not the subtraction. – AHB Jan 10 '17 at 18:49
It's a set subtraction ... – rtybase Jan 10 '17 at 18:51
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. – Andreas Caranti Jan 10 '17 at 18:57
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. – AHB Jan 10 '17 at 19:04

score 1 · Answer 2 · answered Jan 10 '17 at 18:46

1

Hint 1:

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

Hint 2:

Use the Pigeonhole Principle.

answered Jan 10 '17 at 18:46

Carl Schildkraut

2 Answers2