Mock AIME 1 2005-2006/Problem 7

Problem

Let $f(n)$ denote the number of divisors of a positive integer $n$. Evaluate $f(f(2006^{6002}))$.

Solution

$2006$ = $2*17*59$, so $f(2006^{6002})$ has $6003^3$ positive divisors. $6003^3$ = $(3^6)(23^3)(29^3)$ so $6003^3$ has $(6+1)(3+1)(3+1)$, or $\boxed {112}$ divisors.

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