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Multiple a^n

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Manager
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20 Dec 2005, 09:14
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I dont get that:

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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20 Dec 2005, 10:45
sandalphon wrote:
I dont get that:

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

B.

1) a could be 2, 4, or 6.... INSUFF

2) Since 8! is a multiple of a^6, a must be even, and the only possible value for a is 2.... SUFF

Can someone please describe a faster, more logical way of proving that a must be 2? Thanks.
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20 Dec 2005, 11:04
yb wrote:
sandalphon wrote:
I dont get that:

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

B.

1) a could be 2, 4, or 6.... INSUFF

2) Since 8! is a multiple of a^6, a must be even, and the only possible value for a is 2.... SUFF

Can someone please describe a faster, more logical way of proving that a must be 2? Thanks.

How have you calculated 2, because at first glance it could be also 4 or 6, if one does not know that for instance 4^6 isn't a factor of 8!

Last edited by allabout on 20 Dec 2005, 11:58, edited 1 time in total.
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20 Dec 2005, 11:31
yb wrote:
sandalphon wrote:
I dont get that:

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

B.

1) a could be 2, 4, or 6.... INSUFF

2) Since 8! is a multiple of a^6, a must be even, and the only possible value for a is 2.... SUFF

Can someone please describe a faster, more logical way of proving that a must be 2? Thanks.

How have you calculated 2, because at first glance it could be also 4 or 6, if one does not know that for instance 4^6 isn't a multiple of 8!

Well I'm not sure if this way is correct, but this is how i went about it.

8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

= (2*2*2) * 7 * (3*2) * 5 * (2*2) 3 * 2 * 1
..... 2^7 is a factor of 8!, so 2^6 must be too

..... 4^6 = 2^12... you can see that the highest power of two in 8! is 2^7, so 2^12 or 4^6 is not a factor...

(same logic for all even numbers, a, for which 2^7 is not divisibly by a^6)

Hence, 2^6 is the only a^6 factor of 8!

(Can someone correct me if I am wrong)
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20 Dec 2005, 13:39
I dont get that:

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

hi i believe that the answer to this question has to be C
As A provides us with no concrete by which we can jump to the solution as it can be 2^6 or 4^3, both will give us the same solution
Selecting b dosen t not solve our question
the answer lies in C where in we can add up both the parts and arrive at asolution..
Do correct me if I am wrong
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20 Dec 2005, 20:19
I think it is D

2^6 = 64
4^3 = 64
8^2 = 64

Everything comes out to be 2^6.
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20 Dec 2005, 20:24
a and n -> positive
1*2*3*4*5*6*7*8 is a multiple of n

1*2*3*4*5*6*7*8 =
1 * 2 * 3 * 2^2 * 5 * 2 * 3 * 7 * 2^3 = 1 * 2^7 * 3^2 *5 * 7

(1) a^n = 64
Sufficient. If a is 2, so a^n = 2^6 which is a factor of 1 * 2^7 * 3^2 *5 * 7. But if a = 4, then a^n = 4^3 which is also a factor of 1 * 2^7 * 3^2 *5 * 7. So we do not know the value of a.

(2) n = 6 tells us a = 2, so a^n = 2^6 which is a factor of 1 * 2^7 * 3^2 *5 * 7. Sufficient.

Ans: B
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20 Dec 2005, 20:27
ywilfred wrote:

(1) a^n = 64
Sufficient. If a is 2, so a^n = 2^6 which is a factor of 1 * 2^7 * 3^2 *5 * 7. But if a = 4, then a^n = 4^3 which is also a factor of 1 * 2^7 * 3^2 *5 * 7. So we do not know the value of a.

Ans: B

I lost track that we were finding a. So according to A. a could be 2, 4 or 8. But according to B it is only 2.

Thanks makes sense. B should be it then.
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20 Dec 2005, 20:41
So my explanation was not that off
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20 Dec 2005, 22:26
yb wrote:
So my explanation was not that off

You rock yb
20 Dec 2005, 22:26
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