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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.

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    Well this expression is just $2^{n+1}-1$, then use Fermat Theorem to find a number such that $2^{n+1}-1$ is divisible by 3 and 37 – 2017-01-28
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    @arberavdullahu : $3\times 37\neq 101$ but $101$ is prime so $n=99$ does the job – 2017-01-28

2 Answers 2

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Because $2^{n+1}-1$ divided by $101$ for $n+2=101$ by Fermat.

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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 $101$.