Problematic question.
the question is asking: "is 2^NA / m^A an integer?" Or
is (2^N /M)^A an integer? Since A is an integer, we can safely ignore it, and focus only on the question "is 2^N/M an integer?"

I'm not sure if that's what the question writer had in mind, but the way the question is phrased right now, the answer is E:
M=3, N=6 atisfies both statements, and the answer is "no"
M=2, N=whatever - let's say 4 satisfies both statements and the answer is "yes".
Thus, even the combination of the statements is not sufficient to limit the answer to a single "yes" or "no".

Explanation: If M=2, then the above is an integer for any positive integer value of M, and the answer is "yes" for both statements.
But here's my problem: Technically, I find nothing stopping M from being an integer other than 2 (or a power of 2): if M=3, then no value of N will make 2^N be a multiple of M=3, and the answer is "no". Sure, Stat. (1) states that integer N is a multiple of M/2, but it doesn't say that M/2 itself must be an integer: If M=3, then M/2 = 3/2 - if we take N=6 is a multiple of 3/2. So N=6 and M=3 satisfy both statements, in which case 2^N/3 will not be an integer.

HPengineer wrote:If A, M and N are positive intgers is 2^NA a multiple of M^A?


S1: N is a multiple of M/2

S2: N is a Multiple of 2M


this one is throwing me off..

this question is so odd although i also found the right answer but its so odd