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I came across this step in an inductive proof, but my algebra skills seem a bit rusty..

$7(7^{k+2}) + 64(8^{2k+1}) = 7(7^{k+2}+8^{2k+1})+57(8^{2k+1})$

How did they do this?

Note: The point was to show that the expressions are divisible by 57.

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2 Answers

up vote 4 down vote accepted

$64=7+57$, so $64\cdot 8^{2k+1}=(7+57)\cdot 8^{2k+1}=7\cdot 8^{2k+1}+57 \cdot 8^{2k+1}$.

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Wow. Embarrassed that I didn't see that... Thank you. –  James Sep 20 '12 at 21:16
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Hint $\ $ With $\rm\: A = 7^{\,k+2}\:$ and $\rm\: B = 8^{\,2k+1}\:$ it is more easy to comprehend

$$\rm 7\, A + 64\ B\ =\ 7\,(A + B) + 57\ B $$

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