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

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For 2, 2, 501, there are three choices to give a power of 500 to and the rest get a power of 1.
 
For 2, 2, 501, there are three choices to give a power of 500 to and the rest get a power of 1.
  
−
Therefor we have <math>5 * 6</math> normal multiples and <math>3 *2</math> "half" multiples. Sum to get <math>\boxed{54}</math> as our answer.
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Therefore we have <math>5 * 6</math> normal multiples and <math>3 *2</math> "half" multiples. Sum to get <math>\boxed{54}</math> as our answer.
  
 
== See also ==
 
== See also ==

Revision as of 18:56, 27 June 2020

Problem

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

Solution 1

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 $6\cdot 3\cdot 3 = \boxed{54}$ as our answer.


Solution 2 (bash)

Clearly we need to find a group of numbers that multiply to 2004. We can list them all out since we know that 2004 is only $167 * 2^2 * 3$.

167, 2, 2, 3

4, 3, 167

12, 167

4, 501

2, 1004

2, 3, 334

2, 2, 501*

6, 2, 167

3, 668

6, 334

2004*

To begin, the first multiple doesn't work because there are only 3 prime divisors of 2004. We can apply all multiples because the prime factorization of $2004^2004$ is 2^4008 * 3^2004 * 167^2004. Every multiple has six ways of distributing numbers to become powers of 167, 3, and 2, except for the ones with a star. For a single power of 2004, we have three choices (2, 3, and 167) to give a power of 2003 to. For 2, 2, 501, there are three choices to give a power of 500 to and the rest get a power of 1.

Therefore we have $5 * 6$ normal multiples and $3 *2$ "half" multiples. Sum to get $\boxed{54}$ as our answer.

See also

2004 AIME II (Problems • Answer Key • Resources)
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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