Question 147490
I am thinking this:
We are being asked to find the largest value of n such that {{{12^10/2^n}}} is an integer

{{{12^10/2^n}}} =
{{{(4*3)^10/2^n}}}
{{{(4^10*3^10)/2^n}}}
{{{((2^2)^10*3^10)/2^n}}}
{{{(2^20*3^10)/2^n}}}

2 will never factor into 3, since both are prime. So the largest number for n is the largest one that still results in {{{(2^20)/2^n}}} being an integer. What is the smallest number that still is a positive integer? well, that's 1. So what value of n results in {{{(2^20)/2^n}}} is 1? n = 20