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

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== See also ==
 
== See also ==
* [[2004 AIME II Problems/Problem 7 | Previous problem]]
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{{AIME box|year=2004|n=II|num-b=7|num-a=9}}
* [[2004 AIME II Problems/Problem 9 | Next problem]]
 
* [[2004 AIME II Problems]]
 
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]

Revision as of 13:27, 19 April 2008

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 the exponents of the prime factors in its prime factorization. For example, the number of divisors of $2004=2^2\cdot 3^1\cdot 167^1$ is $(2+1)(1+1)(1+1)=12$.

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

2004 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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All AIME Problems and Solutions
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