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# For all n > 0, 7^6n - 6^6n is divisible by (1) 13

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For all n > 0, 7^6n - 6^6n is divisible by (1) 13  [#permalink]

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05 Jan 2005, 11:45
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For all n > 0, 7^6n - 6^6n is divisible by

(1) 13
(2) 127
(3) 559
(4) All of the above
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05 Jan 2005, 11:50
No idea how to solve it besides punching my calculator...

Senior Manager
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05 Jan 2005, 14:45
POE would give 127, but I'm still trying to solve it algebraically.
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05 Jan 2005, 16:18
I'm not sure but here is what I did. I considered 7^6n - 6^6n to be a differnce of squares. So 7^6n=(7^3n)^2 and the same for 6^6n. So I got (7^3n - 6^3n)* (7^3n + 6^3n). If you use n=1 you get 7^3=343 and 6^3=216. Their difference is 127 and their sum is 559 so I know that at least for n=1 those are factors. However I have no desire to try and test n=2, n=3, etc to see what happends. There has to be a better way that I'm not seeing.
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05 Jan 2005, 16:37
I think another way to do this would be split 6n as 7n - n ans use exponents and get 7^n in the denominator i am trying to fig out how that will help
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06 Jan 2005, 08:52
Hi,

The difference of squares approach was very helpful; I think it is sufficient to arrive at the answer.
For n=1, as was stated, product=127*559

Since 559/13=43,

The answer is All of the above.

I don't think we need to go any further.
difference of squares   [#permalink] 06 Jan 2005, 08:52
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