2000 AIME II Problems/Problem 4
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Problem
What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors?
Solution
We use the fact that the number of divisors of a number 
is 
. If a number has 
factors, then it can have at most 
distinct primes in its factorization.
Dividing the greatest power of 
from 
, we have an odd integer with six positive divisors, which indicates that it either is (
) a prime raised to the 
th power, or two primes, one of which is squared. The smallest example of the former is 
, while the smallest example of the latter is 
.
Suppose we now divide all of the odd factors from ; then we require a power of with 
factors, namely 
. Thus, our answer is 
.
See also
| 2000 AIME II (Problems • Resources) | ||
| Preceded by Problem 3 | Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
