Maths Olympiad Prep

Track / Stage 4 / 107 of 340 #367 of 1964

Problem 367

AMC 12 late, AIME early
Number theory Difficulty 4.7 Find the answer

Example 9 Calculate
(2+1)(22+1)(24+1)(22n+1). (2+1)\left(2^{2}+1\right)\left(2^{4}+1\right) \cdots \cdot\left(2^{2^{n}}+1\right) .

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

=1×(2+1)(22+1)(24+1)(22n+1)=(21)(2+1)(22+1)(24+1)(22n+1)=(221)(22+1)(24+1)(22n+1)==(22n1)(22n+1)=22n+11. \begin{array}{l} =1 \times(2+1)\left(2^{2}+1\right)\left(2^{4}+1\right) \cdots \cdots\left(2^{2^{n}}+1\right) \\ =(2-1)(2+1)\left(2^{2}+1\right)\left(2^{4}+1\right) \cdots \cdot\left(2^{2^{n}}+1\right) \\ =\left(2^{2}-1\right)\left(2^{2}+1\right)\left(2^{4}+1\right) \cdots \cdot\left(2^{2^{n}}+1\right) \\ =\cdots \cdots \\ =\left(2^{2^{n}}-1\right)\left(2^{2^{n}}+1\right) \\ =2^{2^{n+1}}-1 . \end{array}

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