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

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

Latest revision as of 17:39, 9 February 2023

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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