# Question

Can the "trick" taught in the videos be used to find the number of odd positive divisors, including 1? After prime factorization of 540, I'm left with 2^2 * 3^3 * 5^1. Now to get odd divisors, I understand we must remove 2 and consider 3^3 * 5^1. Can I use the 'trick' which says the number of positive divisors = (power+1)(power+1)? In this case it'll be (3 + 1)(1 + 1) which gives the correct answer 8. Is that just a coincidence or is the trick valid even for finding odd divisors after removing 2?

# Answer

It turns out that the "trick" can be used the way you mentioned to find *the number of odd positive divisors, including 1*.

So to find the **total number of factors**, we can add one to each to power in the prime factorization of an integer, then multiply all the (power + 1)s together. For 540, we would have (2 + 1)(3 + 1)(1 + 1) = 24 factors.

To find the **number of odd factors** (which includes 1), we can exclude any power of 2 and do the same. For 540, we have (3 + 1)(1 + 1) = 8 odd positive factors.

To find the **number of even factors**, we can multiply the number of odd factors by the power of 2 (not the power of 2 + 1!!!). For 540, we have (3 + 1)(1 + 1)(2) = 16 even factors.

Of course, also note that the total number of factors = the number of even factors + the number of odd factors.

Let's take 6300 as another example, for which the prime factorization is:

2^2 * 3^2 * 5^2 * 7

We have (2 + 1)(2 + 1)(2 + 1)(1 + 1) = 54 total factors

We have (2 + 1)(2 + 1)(1 + 1) = 18 odd factors

We have 18 * (2) = 36 even factors

# Link

http://gre.magoosh.com/questions/309

## Comments