factors

gowani wrote:break down 450 to prime factors...(5^2)(3^2)(2) then look at the powers and 1 to each and then sum it up

3 + 3 + 2 = 8

i'm saying C

The method starts off well, but there are a few mistakes at the end. If you want to work out how many positive divisors a number has:

-prime factorize
-look only at the powers
-add one to each power
-multiply what you get (don't add!)

So,
-450 = (2^1)*(3^2)*(5^2)
-the powers are 1, 2 and 2
-add one to each: 2, 3 and 3
-multiply: 2*3*3 = 18

450 has 18 different positive divisors, including 1 and itself.

Why does this work? Because any number that looks like (2^a)*(3^b)*(5^b) is a divisor of 450 = (2^1)*(3^2)*(5^2) as long as:

a = 0 or 1 (two choices)
b = 0, 1 or 2 (three choices)
c = 0, 1 or 2 (three choices)

and we multiply just as we would in any mathematical counting problem to work out the total number of choices for the exponents a, b and c.

Now, after all that, 18 is not the answer to the posted question. The question asks only for odd divisors. If a divisor is to be odd, it must not have a 2 in its prime factorization. Thus, the exponent on 2 must be 0:

a = 0 (one choice)
b = 0, 1 or 2 (three choices)
c = 0, 1 or 2 (three choices)

1*3*3 = 9 odd divisors.