MGMAT Question

x/144=integer <> x=144*integer
cubic_rt(x)=integer
by performing prime factorization of 144 we solve this q.
144(2)72
72(2)36
36(2)18
18(2)=9
9(3)=3
3(3)=1 -> 2^4*3^2, So cubic_root(integer*144)=integer, we are missing one 3 and one 2 is extra. There must be two 2s and one 3. This makes 12. So we know that 144*12 is (4*3)^3 and 12 is divisible by 4 and 12.

Strongt wrote:X is divisible by 144. If the cubed root of X is an integer, then which of the following is the cubed root of x definitely divisible by? Choose all that apply.

a) 4
b) 8
c) 9
d) 12


Answer: 4 and 12

is there a rule for this type of questions?

for many arithmetic type of questions the key is prime factorization just start from rooting this concept into the q. and drill in

X=144*K
Prime factorization of X will give
X=12*12*K (we can do it quickly since we know that 144 is the square of 12)
So for the cube root of X to be integer, K must be 12.Consequently, the cube root of X will be 12.
Among the answer choices, 4 & 12 are divisible by 12

Hope it helps.