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

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<center><math>(a+1)(b+1)(c+1)=2^2\cdot 3\cdot 167.</math></center>   
 
<center><math>(a+1)(b+1)(c+1)=2^2\cdot 3\cdot 167.</math></center>   
  
We can think of this as [[partition]]ing the exponents to <math>a+1,</math>  <math>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.
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We can think of this as [[partition]]ing the exponents to <math>a+1,</math>  <math>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 three ways.  So we have <math>6\cdot 3\cdot 3 = 054</math> as our answer.
  
 
== See also ==
 
== See also ==

Revision as of 20:43, 27 July 2010

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 three ways. 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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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