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.
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.
Put them into five subsets according to their remainder mod 5. You have [0],[1],[2],[3],[4].
If all five subsets contain a number, take one from each subset.
If only four subsets contain a number, then one contains at least three. Suppose [a] contains three or more, [b] contains none. Take three of [a], none of [b],none of [2a-b], and one each of the other two. (Imagine we start with one from each group, but replace [b] and [2a-b] by two extra [a], the sum (mod 5) remains the same.)
If two or one subsets contain a number, take five from one subset.
Suppose three subsets [a],[b],[c] contain a number, but not [d] or [e]. Draw a regular pentagon with [0],[1],[2],[3],[4] in order at the corners. By symmetry, you can relabel a,b,c so that [d+e]=[a+b]=[2c]. If possible, pick a+a+b+b+c, or a+b+c+c+c. If that is impossible, then there is only one in [a], and also fewer than three in [c]. So there is at least six in [b], and you pick five from [b].
Just an idea, haven't yet finalized it.
$c_0$ - count numbers with remainder 0 mod 5
$c_1$ - count numbers with remainder 1 mod 5
$c_2$ - count numbers with remainder 2 mod 5
$c_3$ - count numbers with remainder 3 mod 5
$c_4$ - count numbers with remainder 4 mod 5
It's clear that $c_k < 5$ for all $k$ (if it's not true, we'll pick the 5 numbers with same residue and problem is solved).
Also it's clear that at least one of $c_k$ is zero (if it's not we'll pick 1 number of each residue and problem is solved).
Using these two observations, look at all possible cases about which $c_k$ exactly is zero. See if you can pick 5 numbers from the others which get the job done.
... let me think further if we can further cut more of the logical branches ...
This can be solved using pigeonhole principle. Here is an idea :
You can group the pairs of numbers ending in different digits like $(1,4),(2,3),(3,2),(4,1)$ Now whenever you pick any 5 numbers at random, it is gurranteed that any one complete pair will be definitley picked (pigeonhole principle) $\implies$ the number will end in 5 $\implies$ divisible by 5.
Note it's just an idea, I will try it in more mathematical language.
