1995 AIME Problems/Problem 6

Revision as of 17:58, 8 February 2007 by Anirudh (talk | contribs) (Solution)

Problem

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

Solution

We know that $n^2$ must have $63\times 39$ factors by its prime factorization. There are $\frac{63\times 39-1}{2} = 1228$ factors of $n^2$ that are less than $n$, because if they form pairs $a$, then there is one factor per pair that is less than $n$. There are $32\times20-1 = 639$ factors of $n$ that are less than $n$ itself. These are also factors of $n^2$. Therefore, there are $1228-639=\fbox{539}$ factors of $n$ that does not divide $n$.

See also