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