1971 AHSME Problems/Problem 14
Problem
The number 
is exactly divisible by two numbers between 
and 
. These numbers are
Solution
Factor.
![\[2^{48}-1 = (2^{24}+1)(2^{12}+1)(2^{6}+1)(2^{3}+1)(2^{3}-1)\]](http://latex.artofproblemsolving.com/3/6/c/36cd52625be9634ce8b3d02db45465a8e160e42b.png)
We only care about two terms: 
and 
. These simplify to 
and 
The answer is 
-edited by coolmath34
