If n is a multiple 18 and m is a multiple of 30, is mn a multiple of 1

n is a multiple 18 and m is a multiple of 30, is mn a multiple of 120?

1) n is a multiple of 6

If n= 6 and m=30, then mn= 180 and 120 is not a factor of 180. Ans is NO.
If n= 12 and m=30, then mn= 360 and 120 is a factor of 360. Ans is YES.
We aren't getting a unique answer.

Statement I alone is insufficient.



2) m is a multiple of 12

n=18, m=60 (LCM of 12 & 30), nm= 1080 and 1080/120= 9.

Answer is YES.

Statement II alone is sufficient.

Option B is the answer

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If n is a multiple of 18 (provided in question)
and
n is a multiple of 6 (as per statement 1)
and
m is a multiple of 30 (as per statement 2), then how are we taking n = 6 as a number for checking if the condition is satisfied?
You took n to be 6 when you tried to analyze statement 1. But n cannot be 6 as n has to be a multiple of 18 (18, 36, 54 ....) as mentioned in the question

In my opinion, it should be B because if n is a multiple of 18 (2 x 3 x 3) and m is a multiple of 30 (2 x 3 x 5), then mn must be a multiple of 90, which is not a multiple of 12

avigutman gmatophobia

I used prime boxing to solve this question - could you validate my approach please?

According to the question stem:

n = multiple of 18 = 2 x 3^2 x some integer

m = multiple of 30 = 2 x 3 x 5 x some integer

From this info, we know that mn = 2^2 x 3^3 x 5 x some integer. But because 120 = 2^3 x 3 x 5, we need to find out if the additional multiple of 2 hiding within "some integer" of m or n.

S1. n = 2 x 3 x some integer
Redundant. We already know that n contains 2 x 3^2 at the very least.
INSUFFICIENT.

S2. m = 2^2 x 3 x some integer
This tells us that m does indeed contain the additional multiple of 2 that we need to make mn a multiple of 120.
SUFFICIENT.

Yes, well done achloes!