Let's consider 120 numbers $a_1,a_2,...,a_{120}\in \mathbb{N}$ such that $a_i\geq 1\ \forall i\in\{1,...,120\}$ and $\sum_{i=1}^{120}a_i\leq 180$. Prove that there exist indices $i, j$ such that $\sum_{k=i}^j a_k = 59$.
I think that I might be able to use the pigeonhole principle somehow, but I am not sure what would correspond to the pigeons here... The nests could possibly be 60 and the pigeons 59, solely by intuition, but I am not sure of anything else. Any help?
You can indeed use the pigeonhole principle. Suppose the statement doesn't hold. If there are two sums divisible by 59, then these sum to 118 and 177, whose difference is 59, a contradiction. Then, there are 119 initial sums partitioned to 59 remainders modulo 59. Thus, three leave the same remainder, let's call them $a<b<c$. If $c-a\leq 118$, then we are done. Thus, it must be that $c-a=177$ and $b-a=118$. However, that gives $c-b=59$. We get a contradiction, and the proof is complete.