Difference between revisions of "2004 AIME II Problems/Problem 8"
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== Problem == | == Problem == | ||
| − | + | What is the 2004th positive odd integer? | |
== Solution == | == Solution == | ||
Revision as of 17:40, 18 July 2018
Problem
What is the 2004th positive odd integer?
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. 
