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# If the integers a and n are greater than 1 and the product

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If the integers a and n are greater than 1 and the product [#permalink]

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10 Jul 2007, 17:28
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If the integers a and n are greater than 1 and the product of the first 8 positive integers is a multiple of a^n, what is the value of a?

(1) a^n=64
(2) n=6
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10 Jul 2007, 17:48
1*2*3*4*5*6*7*8 = 1 * 2^7 * 3^2 * 5 * 7

a^n = x(1 * 2^7 * 3^2 * 5 * 7) where x is an integer

St1:
64 = x(1 * 2^7 * 3^2 * 5 * 7)
could be 2^6, 4^3..etc Insufficient.

St2:
a^6 = x(1 * 2^7 * 3^2 * 5 * 7)
Can only be a = 2. Sufficient.

Ans B
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10 Jul 2007, 17:51
ywilfred wrote:
1*2*3*4*5*6*7*8 = 1 * 2^7 * 3^2 * 5 * 7

a^n = x(1 * 2^7 * 3^2 * 5 * 7) where x is an integer

St1:
64 = x(1 * 2^7 * 3^2 * 5 * 7)
could be 2^6, 4^3..etc Insufficient.

St2:
a^6 = x(1 * 2^7 * 3^2 * 5 * 7)
Can only be a = 2. Sufficient.

Ans B

In #2, can't x=3*3*3*3 such that a=3? This is where I was confused...
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10 Jul 2007, 18:02
hmm... in fact i think it's not B. let me rework the problem.
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10 Jul 2007, 18:20
The OA is B. I just don't really see it.
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11 Jul 2007, 09:51
briks123 wrote:
The OA is B. I just don't really see it.

1: a^n could be 2^6, 4^3 and 8^2. nsf.
2. a^n = 2^6 only. so suff.

briks123 wrote:
If the integers a and n are greater than 1 and the product of the first 8 positive integers is a multiple of a^n, what is the value of a?

(1) a^n=64
(2) n=6
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11 Jul 2007, 11:20
briks123 wrote:
The OA is B. I just don't really see it.

S1 is clearly insufficient bceause we can have 2^6 or 4^3, where the value of a is different in each case.

S2 says that n=6, so we know that a^6 has to be a factor of 8!

it might be difficult to see why S2 is sufficient at first, but try out different numbers. 3^6 is not a factor of 8! because you need 6 3's in 8x7x6x5x...etc in order for it to be a factor. but 8! only has 2 6's in it...try it out for 4 and 5, and you'll quickly see that the only possibility for a is 2.
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11 Jul 2007, 12:33
yep, i see now. thank you!
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12 Jul 2007, 02:10
My approach:

8! = 2^7 * 3^2 * 5 * 7

Stmt1: a^n =64.
a^n can be 2^6, 4^3 , 8^2 etc..
So INSUFF

Stmt2: n=6
Only 2 has more than 6 multiples. So a =2
So SUFF

Hence 'B'
12 Jul 2007, 02:10
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# If the integers a and n are greater than 1 and the product

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