# 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

7 comments

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!

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...

Can someone let me know why this method work?

Thanks

Why are you thinking as Chile, first you talk about prime factors then you take into account 3^0...etc...1 is not a prime factor so for 540=2x2x3x3x3x5. These are prime factors... You can not take as.. 540=2x2x3x3x3x5x1.. Leave 1 here... I think you got my point.. Thanks

Hi ,

2=2^1 ,4=2^2 ,8=2^3 etc

No. of factors =(1+1)=2

The rule for counting number of odd factors for N=2^n does not hold true

Can u say me 18 odd and even factors and 600 also

With solution

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