2000 AIME II Problems/Problem 4

Problem

What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors?

Solution

If a number has twice as many even divisors as odd divisors, then the number has two factors of 2 in it.(a.k.a. 4 is a factor of the number, but not 8.)

So we have $2^2*whatever$ .

Since $whatever$ has 6 factors(as stated in the problem), $whatever$ is either the product of a prime and a perfect cube, or a perfect 5th power. We see which gives out the smallest:

$3*5^3=375$

$3^3*5=135$

$3^5=243$

Therefore, the smallest value of $whatever$ is 135.

$135*4=\boxed{540}$

2000 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2000 AIME II Problems/Problem 4

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