2. Positive Integer P has exactly 2 positive prime factors, 5 and 11. If P has a total of 8 positive factors, including 1 and P, what is the value of P?

(1) 5^3 is a factor of P
(2) 11^2 is not a factor of P


We can break down the prime factorization of P as follows:

P = 5^a * 11^b (where a and b are positive integers)

There's a handy little trick whereby you can find the number of factors of any number using its prime factorization. Simply bump each exponent up by 1 and multiply the results:

(a+1)*(b+1) = 8

There are only two ways in which this could work:

a=1, b=3

OR

a=3, b=1

Therefore, P is either 5^3 * 11 or 5 * 11^3.

Statement (1) says 5^3 is a factor of P. Therefore, P = 5^3 * 11. SUFFICIENT

Statement (2) says 11^2 is not a factor of P. That means P cannot be 5 * 11^3, and thus it must be 5^3 * 11. SUFFICIENT

If you're interested in learning more about the prime-factorization number-of-factors technique, check out this blog post I wrote:

https://blog.knewton.com/2010/05/06/gmat ... d-factors/

Hope this helps!

raz1024 wrote:2. Positive Integer P has exactly 2 positive prime factors, 5 and 11. If P has a total of 8 positive factors, including 1 and P, what is the value of P?

(1) 5^3 is a factor of P
(2) 11^2 is not a factor of P


We can break down the prime factorization of P as follows:

P = 5^a * 11^b (where a and b are positive integers)

There's a handy little trick whereby you can find the number of factors of any number using its prime factorization. Simply bump each exponent up by 1 and multiply the results:

(a+1)*(b+1) = 8

There are only two ways in which this could work:

a=1, b=3

OR

a=3, b=1

Therefore, P is either 5^3 * 11 or 5 * 11^3.

Statement (1) says 5^3 is a factor of P. Therefore, P = 5^3 * 11. SUFFICIENT

Statement (2) says 11^2 is not a factor of P. That means P cannot be 5 * 11^3, and thus it must be 5^3 * 11. SUFFICIENT

If you're interested in learning more about the prime-factorization number-of-factors technique, check out this blog post I wrote:

https://blog.knewton.com/2010/05/06/gmat ... d-factors/

Hope this helps!

Thanks a ton!

This trick is indeed handy and it helps. I will look into other problems like this so that I can apply this kind of approach.