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

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== Problem ==
 
== Problem ==
How many [[positive integer]] [[divisor]]s of <math> 2004^{2004} </math> are [[divisibility | divisible]] by exactly 2004 positive integers?
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How many positive integer divisors of <math>2004^{2004}</math> are divisible by exactly 2004 positive integers?
  
== Solution ==
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== Solution 1 ==
 
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>.
 
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>.
  
We can [[divisor function | count the number of divisors]] of a number by multiplying together one more than each of the [[exponent]]s of the prime factors in its prime factorization.  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)=12</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 the [[exponentiation | exponents]] of the prime factors in its prime factorization.  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)=12</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  
 
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  
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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 = \boxed{54}</math> as our answer.
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==Solution 2 (bash)==
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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 <math>167 * 2^2 * 3</math>.
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167, 2, 2, 3
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4, 3, 167
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12, 167
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4, 501
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2, 1004
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2, 3, 334
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2, 2, 501*
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6, 2, 167
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3, 668
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6, 334
 +
 
 +
2004*
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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 <math>2004^{2004}</math> is <math>2^{4008} * 3^{2004} * 167^{2004}.</math> Every multiple has six ways of distributing numbers to become powers of 167, 3, and 2, except for the ones with a star.
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For a single power of 2004, we have three choices (2, 3, and 167) to give a power of 2003 to.
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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.
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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 ==
* [[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]]
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{{MAA Notice}}

Latest revision as of 17:20, 1 August 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 (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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