# 2020 AIME I Problems/Problem 12

Note: Please do not post problems here until after the AIME.

## Problem

Let $n$ be the least positive integer for which $149^n-2^n$ is divisible by $3^3\cdot5^5\cdot7^7.$ Find the number of positive integer divisors of $n.$

## Solution

Lifting the Exponent shows that $$v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1$$ so thus, $3^2$ divides $n$. It also shows that $$v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2$$ so thus, $7^5$ divides $n$.

Now, multiplying $n$ by $4$, we see $v_5(149^{4n}-2^{4n}) = v_5(149^{4n}-16^{n})$ $and since$145^{4} \equiv 1 \pmod{25} $and$16^1 \equiv 16 \pmod{25} $then$v_5(149^{4n}-2^{4n})=1 $meaning that we have that by LTE,$4 \cdot 5^4 $divides$n$. Since$ (Error compiling LaTeX. ! Missing $inserted.)3^2 $,$7^5 $and$4\cdot 5^4 $all divide$n $, the smallest value of$n $working is their LCM, also$3^2 \cdot 7^5 \cdot 4 \cdot 5^4 = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5 $. Thus the number of divisors is$(2+1)(2+1)(4+1)(5+1) = \boxed{270}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 