davidlee05 wrote:How many factors does 36^2 have?

For any real number N, if the prime factorization of N is (p^a)*(q^b)*(r^c)*..., then the number of factors of N is given by (a + 1)(b + 1)(c + 1)...

Here, 36^2 = 2^4 * 3^4

So, number of factors of 36^2 = (4 + 1)(4 + 1) = 25

The correct answer is D.

davidlee05 wrote:Can you please expand on your method?

Please check my previous post.
I've edited the reply with the explanation.

To explain more, let us assume for some real number N the prime factorization of N is (p^a)*(q^b)*(r^c)*...

Now number of factors of N = (Number of ways to select the prime factor p)*(Number of ways to select the prime factor q)*(Number of ways to select the prime factor r)*... = (a + 1)(b + 1)(c + 1)...

Now, for each prime factor we may or may not select that factor. When none of the prime factors are selected we get the factor 1. Hence, for the prime factor p, number of ways to select p to include it in a factor of N is exponent of p in the prime factorization of N, i.e. a and number of ways to not select p is 1. That's why we are adding 1 to each of the exponents of the prime factors of N.

Hope that helps.