I arrived at B.
Quick little tip: both statements together essentially ask the test-taker to solve a^6 = 64, which also completely ignores the information given in the question. This makes it highly unlikely that the answer is C. It's just too easy. Why would they even ask it? And, obviously, it's not E.
Anyway.
(1) gives a^n = 64. Since the problem statement only gives information relating to a^n, and not any of its component parts, we are unable to ascertain specific values of a and n from this statement, as 64 = 2^6 = 4^3 = 8^2. Insufficient.
(2) is where the real question is. We have to find all the prime factors of a^n inside of 8!. Can we find 2^6 in there? In that factorial, we have 2 * 4 * 6 * 8, which is a total of seven 2's, or 2^7. So, this number is necessarily a multiple of 2^6. Are there any other possibilities? What about 3^6? We need six 3's, but I only see 3 * 6, which is only two 3's. How about 4^6? That's twelve 2's, and we know we don't have that many. It only gets worse from here. 5^6? Get real. Since a = 1 is not an option, we know that a must equal 2. Sufficient.
B.
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Feep wrote:I arrived at B.
Quick little tip: both statements together essentially ask the test-taker to solve a^6 = 64, which also completely ignores the information given in the question. This makes it highly unlikely that the answer is C. It's just too easy. Why would they even ask it? And, obviously, it's not E.
Anyway.
(1) gives a^n = 64. Since the problem statement only gives information relating to a^n, and not any of its component parts, we are unable to ascertain specific values of a and n from this statement, as 64 = 2^6 = 4^3 = 8^2. Insufficient.
(2) is where the real question is. We have to find all the prime factors of a^n inside of 8!. Can we find 2^6 in there? In that factorial, we have 2 * 4 * 6 * 8, which is a total of seven 2's, or 2^7. So, this number is necessarily a multiple of 2^6. Are there any other possibilities? What about 3^6? We need six 3's, but I only see 3 * 6, which is only two 3's. How about 4^6? That's twelve 2's, and we know we don't have that many. It only gets worse from here. 5^6? Get real. Since a = 1 is not an option, we know that a must equal 2. Sufficient.
B.
Feep,
What about if you consider 0 or negative as part of the set of the first integers? the constraint for being greater than 1 is only for a and n, not for any members of the first integers. is it?
Thanks.
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Feep
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I completely agree. The question is worded vaguely in that respect and probably would never appear as such on a real GMAT exam, but I could figure out the meaning from context. Since the question was one of very high quality otherwise, I decided to just let it slide. = Dthetenor wrote: Feep,
What about if you consider 0 or negative as part of the set of the first integers? the constraint for being greater than 1 is only for a and n, not for any members of the first integers. is it?
Thanks.
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Agreed - it's a GMATPrep question, but the wording of the original is slightly different than what was posted above:Feep wrote: I completely agree. The question is worded vaguely in that respect and probably would never appear as such on a real GMAT exam, but I could figure out the meaning from context. Since the question was one of very high quality otherwise, I decided to just let it slide. = D
If the integers a and n are greater than 1 and the product of the first eight positive integers is a multiple of a^n, what is a?
(1) a^n = 64
(2) n = 6
With the insertion of the word 'positive', the meaning of the question is unambiguous.
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