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# Difference between revisions of "1995 AIME Problems/Problem 6"

## Problem

Let $n=2^{31}3^{19}.$ How many positive integer divisors of $n^2$ are less than $n_{}$ but do not divide $n_{}$?

## Solution

We know that $n^2 = 2^{62}3^{38}$ must have $(62+1)\times (38+1)$ factors by its prime factorization. If we group all of these factors (excluding $n$) into pairs that multiply to $n^2$, then one factor per pair is less than $n$, and so there are $\frac{63\times 39-1}{2} = 1228$ factors of $n^2$ that are less than $n$. There are $32\times20-1 = 639$ factors of $n$, which clearly are less than $n$, but are still factors of $n^2$. Therefore, there are $1228-639=\boxed{589}$ factors of $n$ that do not divide $n^2$.