where is this question from?
1) q^4 is not a multiple of 128
to be a multiple of 64, q^4 needs to be a multiple of 2^6.
if q^4 is not a multiple of 128, then q^4 does not contain 7 2's in its factorization. Since we can only go down in sets of 4 of any factor (since we have q^4), if q^4 contains any 2's, it must contain either 2^4, 2^8, etc, since we know it contains less than 2^7, we can conclude it doesn't contain 2^6 either (since 2^4 is the next possible number of 2s)
1 is sufficient
2) q^2 has 27 factors, 7 of which are less than or equal to 10
to show this is insuff, we need to show 2 cases, one in which 2) is met and is a multiple of 64, and one in which 2) is met and is not a multiple of 64
q^2 has 27 factors, with 7 less than 10, means that we could have the 7 factors all be 2's, in which case we know it is a factor of 64 (regardless of the other 20 factors).
On the otherhand, if all 7 factors are 3's (and the other 20 factors are 11's), then we know it is not a multiple of 64.
2 is insuff
Therefore the answer is A
