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Difference between revisions of "2004 AIME II Problems/Problem 8"

 
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== Solution ==
 
== Solution ==
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The [[prime factorization]] of 2004 is <math>2^2\cdot 3\cdot 167</math>.  Thus the prime factorization of <math>2004^{2004}</math> is <math>2^{4008}\cdot 3^{2004}\cdot 167^{2004}</math>.
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We can [[divisor function | count the number of divisors]] of a number by multiplying together one more than each of their [[exponent]]s.  For example, the number of divisors of <math>2004=2^2\cdot 3^1\cdot 167^1</math> is <math>(2+1)(1+1)(1+1)=4</math>.  A positive integer divisor of <math>2004^{2004}</math> will be of the form <math>2^a\cdot 3^b\cdot 167^c</math>.  Thus we need to find how many <math>(a,b,c)</math> satisfy <math>(a+1)(b+1)(c+1)=2^2\cdot 3\cdot 167</math>.  We can think of this as [[partition]]ing the exponents to <math>a+1, b+1,</math> and  <math>c+1</math>.  So let's partition the 2's first.  There are two 2's so this is equivalent to partitioning two items in three containers.  We can do this in <math>{4 \choose 2} = 6</math> ways.  We can partition the 3 in three ways and likewise we can partition the 167 in one way.  So we have <math>6\cdot 3\cdot 3 = 054</math> as our answer.
  
 
== See also ==
 
== See also ==
 
* [[2004 AIME II Problems]]
 
* [[2004 AIME II Problems]]
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[[Category:Intermediate Number Theory Problems]]
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[[Category:Intermediate Combinatorics Problems]]

Revision as of 10:28, 30 July 2006

Problem

How many positive integer divisors of $2004^{2004}$ are divisible by exactly 2004 positive integers?

Solution

The prime factorization of 2004 is $2^2\cdot 3\cdot 167$. Thus the prime factorization of $2004^{2004}$ is $2^{4008}\cdot 3^{2004}\cdot 167^{2004}$.

We can count the number of divisors of a number by multiplying together one more than each of their exponents. For example, the number of divisors of $2004=2^2\cdot 3^1\cdot 167^1$ is $(2+1)(1+1)(1+1)=4$. A positive integer divisor of $2004^{2004}$ will be of the form $2^a\cdot 3^b\cdot 167^c$. Thus we need to find how many $(a,b,c)$ satisfy $(a+1)(b+1)(c+1)=2^2\cdot 3\cdot 167$. We can think of this as partitioning the exponents to $a+1, b+1,$ and $c+1$. So let's partition the 2's first. There are two 2's so this is equivalent to partitioning two items in three containers. We can do this in ${4 \choose 2} = 6$ ways. We can partition the 3 in three ways and likewise we can partition the 167 in one way. So we have $6\cdot 3\cdot 3 = 054$ as our answer.

See also

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