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Problem 799

AIME late

Number theory Difficulty 5.5 Prove it

1. Prove that the number $A=2^{1}+2^{2}+\ldots+2^{1985}$ is divisible by 31.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Solution. We have:

$\begin{aligned} A & =\left(2+2^{2}+2^{3}+2^{4}+2^{5}\right)+\left(2^{6}+2^{7}+2^{8}+2^{9}+2^{10}\right)+\ldots+\left(2^{1981}+2^{1982}+2^{1983}+2^{1984}+2^{1985}\right) \\ & =2\left(1+2+2^{2}+2^{3}+2^{4}\right)+2^{6}\left(1+2+2^{2}+2^{3}+2^{4}\right)+\ldots+2^{1981}\left(1+2+2^{2}+2^{3}+2^{4}\right) \\ & =31\left(2+2^{6}+2^{11}+\ldots+2^{1981}\right) \end{aligned}$

from which it follows that the number $A$ is divisible by 31.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.