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# 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

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### Comments

• yang xu

Hello David, back to the odd divisor of 540 question.

you said 4 options on number 3 and 2 options on number 5

but it looks to me that the exponents of 3 and 5 cannot be both 0, right? if they are 0 the same time, 3^0 * 5^0 = 1, then 1 * 2^2, you will get an even divisor.

so it should be 7 odd divisors:  3, 3^2, 3^3, 3*5, 3^2*5, 3^3*5, 5

correct me if I am wrong. thanks!

Edited by yang xu
• Erika K Fennelly

Can someone please answer the last comment? I'm also wondering if you're double or multiple-counting 1 as a divisor if you're counting 3^0 and 5^0...