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

(Solution)
(Problem)
Line 2: Line 2:
 
== Problem ==
 
== Problem ==
  
Let <math>f(n)</math> denote the number of divisors of a positive integer n. Evaluate f(f(<math>2006^{6002}</math>)).
+
Let <math>f(n)</math> denote the number of divisors of a positive integer <math>n</math>. Evaluate <math>f(f(2006^{6002}</math>))$.
 
 
  
 
== Solution ==
 
== Solution ==
  
 
<math>2006</math> = <math>2*17*59</math>, so <math>f(2006^{6002})</math> has <math>6003^3</math> positive divisors. <math>6003</math> = <math>(3^2)(667)</math> so <math>6003^3</math> has <math>(6+1)(2+1)</math>, or <math>\boxed {021}</math> divisors.
 
<math>2006</math> = <math>2*17*59</math>, so <math>f(2006^{6002})</math> has <math>6003^3</math> positive divisors. <math>6003</math> = <math>(3^2)(667)</math> so <math>6003^3</math> has <math>(6+1)(2+1)</math>, or <math>\boxed {021}</math> divisors.

Revision as of 23:39, 17 April 2009

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.