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. $6003$ = $(3^2)(667)$ so $6003^3$ has $(6+1)(2+1)$ , or $\boxed {021}$ divisors.

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Mock AIME 1 2005-2006/Problem 7

Problem

Solution