How many pos. odd divisors does 540 have?

jab wrote:Hey guys

Just had a chat with a friend of mine about a GMAT question in his exam. We both weren't quite sure about the way to tackle the following question:
How many pos. odd divisors does 540 have?

First I would factorize it to 3^3 * 2^2 * 5^1 so that the number of all possible divisors is 4*3*2 = 24. Now the "interesting" bit: Since we know that odd*odd = odd and odd*even = even there must not be any divisor of 540 divisible by 2. Does this leave us with 4*3 = 12 possible positive odd divisors for 540?

Thanks a lot, beatthegmat has been awesome so far!

Hi jab, indeed BTG is the great source for GMAT.

you started correctly by factoring 540

540=2^2 + 3^3 + 5^1

the number of all factors (divisors) for 540, 2+1=3, 3+1=4, 1+1=2 => 3*4*2=24 factors (divisors)

note that there 2^2 primes, 2 is even => a number is even if it is divided by 2 => number of the evens is 2 times more

e.g. for 1 odd there are 2 evens

24 factors=16 even+8 odd

you can test this with many numbers

say 12

/2 => 6
/2 => 3
/3 => 1

2^2 + 3^1 => (2+1)*(1+1)=6 factors => 2 evens for 1 odd

4 evens {2,4,6,12} and 2 odds {3,1}

..............

number 270 has 16 factors <=> 2^1 + 3^3 + 5^1 => 2*4*2=16

there is one even present => 1 even for 1 odd => 8 evens and 8 odds

..........

in your original question the answer is 540 contains 16 even and 8 odd positive factors (divisors)