If the integers a and n are greater than 1 and the product of the first 8 integers is a multiple of a^n, what is the value of a?
1) a^n=64
2) n=6
The OA is b, but I'm not certain how that is sufficient. Any insight would be really appreciated! Thanks.
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gabriel
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the question seems incomplete to me ... what is the darkened part supposed to mean .. which integers are we talking about .. plz take a look at the question again and repost ..mindruna wrote:If the integers a and n are greater than 1 and the product of the first 8 integers is a multiple of a^n, what is the value of a?
1) a^n=64
2) n=6
The OA is b, but I'm not certain how that is sufficient. Any insight would be really appreciated! Thanks.
product of the first 8 integers is a multiple of a^nmindruna wrote:If the integers a and n are greater than 1 and the product of the first 8 integers is a multiple of a^n, what is the value of a?
1) a^n=64
2) n=6
The OA is b, but I'm not certain how that is sufficient. Any insight would be really appreciated! Thanks.
so, ( 1*2*3*...*8 )/(a^n) = k => where a, n, and k are all integers.
(1)
4^3 = 64 => a=4, n=3
8^2 = 64 => a=8, n=2
2^6 = 64 => a=2, n=6
INSUFF
(2)
n=6 => only a=2 can satisfy this.
SUFF
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gabriel
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Anonymous wrote:blah45 wrote:product of the first 8 integers is a multiple of a^nmindruna wrote:If the integers a and n are greater than 1 and the product of the first 8 integers is a multiple of a^n, what is the value of a?
1) a^n=64
2) n=6
The OA is b, but I'm not certain how that is sufficient. Any insight would be really appreciated! Thanks.
so, ( 1*2*3*...*8 )/(a^n) = k => where a, n, and k are all integers.
(1)
4^3 = 64 => a=4, n=3
8^2 = 64 => a=8, n=2
2^6 = 64 => a=2, n=6
INSUFF
(2)
n=6 => only a=2 can satisfy this.
SUFF
Well, the above means the answer should be [D], not . We are looking to justify . Good thinking though. Thanks
actually that means the answer is C .. anyway, that does not matter bcoz u cant reach the conclusion in blah45's solution without knowing which "first 8 integers" is supposed to be meant over here ...
Anonymous wrote:Anonymous wrote: Well, the above means the answer should be [D], not . We are looking to justify . Good thinking though. Thanks
How is it (D)? I mean it's clear from statement (1) 'a' could have three possible values. ONLY statement (2) indicates that a=2 and n=6 work.
I am sorry, my bad. The answer would be , in fact. The original answer is , which is correct. Here is how:
product of first 8 integers:
8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
taking (2) into consideration, n = 6... then prime factorize the left hand side:
(2^3) x 7 x (2 x 3) x 5 x (2 x 2) x 3 x 2 x 1 = (a^6) k
(2^7) x (3^2) x 5 x 7 = (a^6) k
The only value of 'a' that satisfies the above is: a = 2
Hence (2) alone is SUFFICIENT... (1) alone is NOT SUFFICIENT...
