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# For all n>0,7^6n-6^6n is divisible by....?? a)13

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Senior Manager
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For all n>0,7^6n-6^6n is divisible by....?? a)13  [#permalink]  19 Dec 2003, 14:14
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For all n>0,7^6n-6^6n is divisible by....??

a)13
b)127
c)559
d)none
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shubhangi

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Joined: 15 Dec 2003
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I believe A is the answer

Let's take 6n=2 or in other words, n=1/3
The difference between 7^2 and 6^2 = 49-36 = 13
Therefore, as we increase n, the difference between the 2 terms will increase by a base of 13 exponent any number. Thus, the answer is that the solution is divisible by 13
CEO
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Re: PS [#permalink]  19 Dec 2003, 18:14
shubhangi wrote:
For all n>0,7^6n-6^6n is divisible by....??

a)13
b)127
c)559
d)none

factor 13 = 13*1

numerator does not have a 13 at all...out it goes

factor 127=> i think its prime. again..no luck

factor 559 => again prime

i think its just D
Senior Manager
Joined: 12 Oct 2003
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Why would you want to factor 13? Or am I just missing the point here?

I also did not follow ..
Quote:
If n=1, 7^6n-6^6n=1^6=1 so d)

Also, in the question stem, should n be associated with 6 or with 7?

Paul wrote:
Quote:
Let's take 6n=2 or in other words, n=1/3
The difference between 7^2 and 6^2 = 49-36 = 13
Therefore, as we increase n, the difference between the 2 terms will increase by a base of 13 exponent any number

What if n = 1/6??? the difference comes up to 1 which is not divisible by 13.
Senior Manager
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Location: dallas , tx
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Well the correct answer is D .. but i didnt understand the official approach. Praet's way was quite logical and easy .
Thanks
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shubhangi

GMAT Instructor
Joined: 07 Jul 2003
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Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
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BG wrote:
If n=1, 7^6n-6^6n=1^6=1 so d)

I'm sorry, but x^z - y^z is NOT equal to (x-y)^z.
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Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Manager
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D is right.. [#permalink]  20 Dec 2003, 17:22
Let n = 1/6 then 7-6 =1. It will be an interesting problem if n was an integer > 0.
Director
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mantha is exactly right-- why not just choose 1/6 for n, plug, and be done?
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