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 three ways. So we have <math>6\cdot 3\cdot 3 = \boxed{ | + | 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. |
== See also == | == See also == | ||
Revision as of 10:19, 10 July 2019
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 the exponents of the prime factors in its prime factorization. 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 three ways. So we have 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
