sum of one hundred numbers

Solution 1:

This result is a special case of the famous EGZ or Erdős Ginzburg Ziv theorem, which states that any set of $2n-1$ integers, must have some subset of size $n$ whose sum is a multiple of $n$. Hence we can even throw out one of the 200 numbers; out of any 199 integers, some 100 must sum to a multiple of 100.

You can find some lovely proofs of EGZ on mathoverflow.

Solution 2:

The EGZ theorem can be proved through the Cauchy-Davenport theorem:

If $A$ and $B$ are subsets of $\mathbb{Z}_{/100\mathbb{Z}}$, $$ |A+B| \geq \min(100,|A|+|B|-1) $$

or the Kneser's theorem:

If $A,B$ are subsets of $\mathbb{Z}_{/100\mathbb{Z}}$ and $C$ is the restricted sumset $\{a+b:(a,b)\in A\times B, a\neq b\}$, then: $$ |C| \geq \min(100,|A|+|B|-3). $$

Both theorems can be proved through the Dirichlet box principle with quite sophisticated proofs, or through the combinatorial nullstellensatz as shown by Noga Alon in the eighties.

In any case, among $199$ integers there are for sure $100$ integers having sum $\equiv 0\pmod{100}$.