89) How many factors

ern5231 wrote:Asked the 2^n*3^3 factor and 2^51 are the same, ask n

Is there a quick way of soving this?

Very unclear wordings, I am just guessing and answering whatever I could think as true in the stem

I interpret it as if the total number of factors of the number 2^n*3^3 is same as that of another number 2^51, then find n.

Well, the total number of factors of the number 2^n*3^3 is given by (n + 1)*(3 + 1) and that of another number 2^51 is equal to (51 + 1) = 52

hence, 4 (n + 1) = 52 or [spoiler]n = 12[/spoiler].

sanju09 wrote:

ern5231 wrote:Asked the 2^n*3^3 factor and 2^51 are the same, ask n

Is there a quick way of soving this?

Very unclear wordings, I am just guessing and answering whatever I could think as true in the stem

I interpret it as if the total number of factors of the number 2^n*3^3 is same as that of another number 2^51, then find n.

Well, the total number of factors of the number 2^n*3^3 is given by (n + 1)*(3 + 1) and that of another number 2^51 is equal to (51 + 1) = 52

hence, 4 (n + 1) = 52 or [spoiler]n = 12[/spoiler].

how did you get this ---> Well, the total number of factors of the number 2^n*3^3 is given by (n + 1)*(3 + 1)

M811 wrote:

sanju09 wrote:

ern5231 wrote:Asked the 2^n*3^3 factor and 2^51 are the same, ask n

Is there a quick way of soving this?

Very unclear wordings, I am just guessing and answering whatever I could think as true in the stem

I interpret it as if the total number of factors of the number 2^n*3^3 is same as that of another number 2^51, then find n.

Well, the total number of factors of the number 2^n*3^3 is given by (n + 1)*(3 + 1) and that of another number 2^51 is equal to (51 + 1) = 52

hence, 4 (n + 1) = 52 or [spoiler]n = 12[/spoiler].

how did you get this ---> Well, the total number of factors of the number 2^n*3^3 is given by (n + 1)*(3 + 1)

This is a rule that all GMATians must know, to find the total number of factors of any positive integer K, split K into its prime factors, say a, b, c, ... and rewrite K = a^p*b^q*c^r..., where p, q, r, ... are the positive integers representing the appearances of their corresponding primes in the spelled out prime product of K. Then add 1 to each exponent on primes and get the product of the proceeds as the total number of factors for K.