How to prove $\exists n \in \Bbb{N} : (1 + 2 + 2^2 + 2^3 + \dots + 2^n) \mod 101 \equiv 0$? I know that this can be done using the Pigeonhole principle, but I can't understand how to do it.
I would be glad if there are any other solutions. 
 A: Because $2^{n+1}-1$ divided by $101$ for $n+2=101$ by Fermat.
A: If you specifically want to use the pigeon-hole principal, a standard argument is as follows.
We first note that since $2$ and $101$ are relatively prime, that if $101 \mid 2^k \cdot m$ for some $k$ and $m$, then $101 \mid m$.
Now consider the numbers $a_n$ given by
$$ a_n = 1 + 2 + 2^2 + 2^3 + \dots + 2^n. $$
If we consider the numbers $a_1, a_2, a_3, \dots, a_{102}$, then we have $102$ natural numbers, each of which has one of $101$ possible remainders modulo $101$. Thus by the pigeon-hole principal, two of the numbers $a_i$ and $a_j$ have the same remainder when divided by $101$.
We see that there are distinct natural numbers $i$ and $j$ (where we may assume that $j > i$ such that
$$ a_j - a_i \equiv 0 \pmod{101}. $$
We thus have that
$$ 2^{i+1} + 2^{i+2} + \dots + 2^{j} $$
is divisible by $101$, and hence
$$ 2^{i+1} \left( 1 + 2 + 2^2 + \dots + 2^{j-i-1} \right) $$
is divisible by $101$. By the observation at the start of the solution, this implies that
$$ 1 + 2 + 2^2 + \dots + 2^{j-i-1} $$
is also divisible by $101$.
