Difference between revisions of "2004 AIME II Problems/Problem 8"
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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>. | 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 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 [[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>. |
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| + | 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> | ||
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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 one way. So we have <math>6\cdot 3\cdot 3 = 054</math> as our answer. | ||
== See also == | == See also == | ||
Revision as of 10:32, 30 July 2006
Problem
How many positive integer divisors of 
are divisible by exactly 2004 positive integers?
Solution
The prime factorization of 2004 is 
. Thus the prime factorization of is 
.
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 
is 
.
A positive integer divisor of will be of the form 
. Thus we need to find how many 
satisfy
We can think of this as partitioning the exponents to 
and 
. 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 
ways. We can partition the 3 in three ways and likewise we can partition the 167 in one way. So we have 
as our answer.
