how can I prove a set with n numbers contains subset that the sum of whose elements is a multiple of n
I'm not sure if I can use EGZ theory here
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If we allow the emptyset, then the statement is trivial since everything divides zero. We instead prove the stronger statement for a nonempty subset.
Let the set be $\{a_1,a_2,\dots,a_n\}$
Consider the set of numbers $\{b_1,b_2,\dots,b_n\}$ where $b_i := a_1+a_2+\dots+a_i$
Applying the pigeon-hole principle, let our set $\{b_1,\dots,b_n\}$ be our set of pigeons and the equivalence classes modulo $n$ be our holes.
One of two things will occur: either there exist $b_i,b_j$ with $i<j$ such that $b_i\equiv b_j\pmod{n}$, or every equivalence class has exactly one entry $b_i$ in it, in particular the equivalence class of $0$.
In the first case, notice that $b_j-b_i = (a_1+a_2+\dots+a_j) - (a_1+a_2+\dots+a_i) = a_{i+1}+a_{i+2}+\dots+a_j$
That $b_j\equiv b_i\pmod{n}$ implies $b_j-b_i$ is a multiple of $n$, and therefore the subset $\{a_{i+1},a_{i+2},\dots,a_j\}$ has the sum a multiple of $n$.
In the second case, you have some $i$ for which $b_i\equiv 0\pmod{n}$ which implies $a_1+a_2+\dots+a_i$ is a multiple of $n$.
JMoravitz
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