DS: If q is an integer, is q^4 a multiple of 64?

[dataiwandude]

I found this question on http://gmat-maths.blogspot.com. I had gotten the right answer at the time of practice but realized that my explanation might have holes in it. I am looking for help to further explain only choice (1).

Here is the question:

If q is an integer, is q^4 a multiple of 64?

(1) q^4 is not a multiple of 128.
(2) q^2 has 27 factors, 7 of which are less than or equal to 10

The answer is: (A), choice (1) alone is sufficient.

Here is how I initially solved it for choice (1):

1. If q^4 is not a multiple of 128

2. Then q^4 certainly will not contain the following numbers in its prime factorization series: 2^6, 2, and some N (N being the multiplier of 128)

3. From the question, for q^4 to be a multiple of 64 q^4 would need to contain 2^6 in its prime factorization series.

4. Since I just proved that q^4 does not contain 2^6, then I can deduce that q^4 will never be a multiple of 64.

Here are the holes I see in my explanation:

What if q^4 is missing the 2 from the prime factorization series of 2^6, 2, and N rather than 2^6? That q is raised to the fourth power migh have something to do with the restriction that missing factors would never come from 2^6. But, I have not been able to establish that proof in my head.

Any thoughts from anyone?

Thanks!!!

[StaceyKoprince]

Hi - when I click on that link I get "404 not found"

Also, it isn't sufficient to cite a location as the source of the question - we actually have to specify the author of the question, not just where else on the web it might be posted. If the author of that blog wrote the question, then cite that person as the author of the question.

Please check into it and let us know the author - then we'd be happy to help you!