Mock AIME 1 2005-2006/Problem 7: Difference between revisions

Revision as of 05:50, 28 August 2013

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 $(1+1)(2+1)$ , or $\boxed {006}$ divisors.

Revision as of 05:48, 28 August 2013 view source PlatinumFalcon (talk \| contribs) 4 edits →Solution ← Older edit		Revision as of 05:50, 28 August 2013 view source PlatinumFalcon (talk \| contribs) 4 edits →Solution Newer edit →
Line 1:		Line 1:

	== Problem ==		== Problem ==

Page

Toolbox

Search

Mock AIME 1 2005-2006/Problem 7: Difference between revisions

Revision as of 05:50, 28 August 2013

Problem

Solution