Brent@GMATPrepNow wrote:How many odd positive divisors does 9000 have?

A) 6
B) 8
C) 9
D) 12
E) 15

Source: GMATPrepNow.com
Difficulty level: 650 - 700

Answer: D

Brent original!

This question exploits the following property: if the prime factorization of some value is 2^x * 3^y * 5^z, we can calculate the number of factors this value contains by finding (x+1)(y+1)(z+1).

To see why this is true, consider a simple number, such as 100. 100 has 9 factors: 1, 2,4, 5, 10, 20, 25, 50, 100.
The prime factorization of 100 is 2^2 * 5^2. Notice that we can produce any of the aforementioned factors by raising our bases, 2 and 5, to 0, 1, or 2.

1 = 2^0 * 5^0
2 = 2^1 * 5^0
4 = 2^2 * 5^0
5 = 2^0 * 5^1
10 = 2^1 * 5^1
20 = 2^2 * 5
25 = 2^0 * 5^2
50 = 2^1 = 5^2
100 = 2^2 * 5^2

Because each base can be raised to 0, 1, or 2, we have three options for each. That's why we add 1 to each exponent when doing the calculation. If we know the rule, we can simply see that if 100 = 2^2 * 5^2, then 100 must have (2+1)(2+1) = 3*3 = 9 factors.

Now to the question. 9000 = 2^3 * 3^2 * 5^3. If we only want odd numbers, we only care about the number of ways we can combine the odd bases: 3^2 *5^3. (2+1)(3+1) = 3*4 = 12. So the answer is D (The logic: 3 can be raised to 0,1, or 2, giving us 3 options. 5 can be raised to 0, 1, 2, or 3, giving us 4 options.)