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

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For all n>0,7^6n-6^6n is divisible by....?? a)13  [#permalink] New post 19 Dec 2003, 14:14
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A
B
C
D
E

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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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 [#permalink] New post 19 Dec 2003, 16:11
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
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Re: PS [#permalink] New post 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
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 [#permalink] New post 20 Dec 2003, 10:10
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.
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 [#permalink] New post 20 Dec 2003, 15:12
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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 [#permalink] New post 20 Dec 2003, 16:35
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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D is right.. [#permalink] New post 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.
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 [#permalink] New post 20 Dec 2003, 21:40
mantha is exactly right-- why not just choose 1/6 for n, plug, and be done?
  [#permalink] 20 Dec 2003, 21:40
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