TITLE: Given any nine numbers, prove there exists a subset of five numbers such that its sum is divisible by $5$.
QUESTION [7 upvotes]: Given any nine numbers, prove there exists a subset of five numbers such that its sum is divisible by $5$.
I tried to take the numbers in the format $5k+1$, $5l+2$, and so on.  However, I am stuck in choosing ANY five from them...also,the numbers included in the subset may or may not belong to same format.

REPLY [3 votes]: We can prove the following more general result, which is a classical problem. The proof below is a classical one.
Lemma If $p$ is prime, then among any $2p-1$ numbers you can find $p$ whose sum is divisible by $p$.
Proof: Let $a_1,.., a_{2p-1}$ be the numbers. 
Consider all $n=\binom{2p-1}{p}$ subsets of $p$ numbers and denote by $S_1,...,S_n$ the sums of each subset.
Let 
$$S:= S_1^{p-1}+...+S_n^{p-1}$$
Let us observe first that $p|S$. 
Note that 
$$S_j^{p-1} = (a_{j_1}+...+a_{j_p})^{p-1}$$
 Any term in this sum has the form $a_{k_1}^{b_1}a_{k_2}^{b_2}...a_{k_l}^{b_l}$ with the coefficient being $\binom{p-1}{b_1,..,b_k}$.
Therefore $S$ is the sum of terms of this form. 
Now, let us check the coefficient of such a term. Whenever when this appears in some $S_j^{p-1}$ the coefficient is exactly $\binom{p-1}{b_1,..,b_l}$.
So we need to figure how many times does this appear in $S_j^{p-1}$. In order for this to happen, $a_1,a_2,..., a_l$ need to be $l$ of the $p-1$ terms, the other $p-l$ can be anything. Therefore, this appears in exactly $\binom{2p-1-l}{p-l}$ sums.  But since $p$ is prime, 
$$p| \frac{(2p-1-l)!}{(p-l)!(p-1)!}=\binom{2p-1-l}{p-l}$$
Now, if we assume by contradiction that none of $S_j$ is divisible by $p$, by Fermat Little Theorem we have
$$S= S_1^{p-1}+...+S_n^{p-1} \equiv 1+1+...+1 \equiv n \pmod{p}$$
But $n \not\equiv 0 \pmod{p}$ contradiction.