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Prove that if $n$ is a positive integer, then $7^n-1$ is a multiple of $6$.

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  • $\begingroup$ do you mean $$n^7-1$$ is divisible by $6$? $\endgroup$ – Dr. Sonnhard Graubner Mar 23 '17 at 7:19
  • $\begingroup$ If he does it's not true. $2^7-1=127$ is not divisible by $6$. Perhaps you mean $7^n-1$. But in that case it's trivial. $\endgroup$ – Mathematician 42 Mar 23 '17 at 7:24
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I think you meant this $$7^n-1$$ is divisible by $6$

This can be factored like this $$(7-1)(7^{n-1}+7^{n-2}+...+1)$$ which is obviously divisible by $6$.

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