TITLE: Let $p$ be an odd prime number. How many $p$-element subsets of $\{1,2,3,4, \ldots, 2p\}$ have sums divisible by $p$?
QUESTION [8 upvotes]: Since there are $2p$ elements in the set, there are exactly $p$ distinct remainders (including $0$) when divided by $p$.
We can write it as :
$1$ and $p+1 \equiv 1 \pmod{p}$
$2$ and $p+2 \equiv 2 \pmod{p}$
$3$ and $p+3 \equiv 3 \pmod{p}$
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$p-1$ and $2p-1 \equiv p - 1 \pmod{p}$
$p$ and $2p \equiv 0 \pmod{p}$
I noticed that we can make $p$-element subsets considering each from same remainder set
for eg: Subsets can be = ($1,2,3,...p$), ($p+1,p+2,3,5,..2p$).
Therefore, number of ways choosing such $p$-element subsets is $2^p$. And I believe there are few more. Hope someone could help?

REPLY [7 votes]: I remember an elegant complex number proof from an Olympiad book (apparently this solution was given by a Bulgarian contestant who won an award for creativity).
Let $\omega$ be a $p$-th root of unity. Consider the sum:
$$F(\omega) = \sum_{1 \leq i_1 < i_2 ... < i_p \leq 2p} \omega^{i_1}\omega^{i_2}...\omega^{i_p}.$$
We observe that $\omega, \omega^{2}, ... , \omega^{2p}$ are the roots of $(x^p - 1)^2 = x^{2p} - 2x^p + 1$. By the relation between elementary symmetric functions of roots and coefficients, we get $F(\omega) = 2$.
But we can write $\omega^{i_1}\omega^{i_2}...\omega^{i_p} =  \omega^{i_1+i_2+...+i_p}$ and read the powers of $\omega$ modulo $p$. So another way to write the sum is $$F(\omega) = \sum_{0 \leq k \leq p-1}a_k \omega ^k.$$
Therefore equating the two different ways of writing the sum, we get:
$$(a_o - 2) + a_1 \omega + a_2 \omega^2 ... + a_{p-1} \omega^{p-1} = 0.$$
But this means $$a_0 - 2 = a_1 = a_2 = ... = a_{p-1}.$$ Clearly $\sum_k a_k$ counts the total number of terms in the summation $F(\omega)$ which is $\binom{2p}{p}$.
Therefore $$\binom{2p}{p} - 2 = (a_0 - 2) + a_1 + a_2 + ...+ a_{p-1} = p(a_0 - 2).$$
Thus we get $$a_0 = \frac{\binom{2p}{p} - 2}{p} + 2.$$
Note that this solves the problem since $a_0$ counts the number of terms in the sum with the property $i_1 + i_2 + ... + i_p = 0$ modulo $p$.