I need help on how to show that for each positive integer $n$, $2009^n − 209^n − 839^n + 92^n$ is divisible by $117$.
I have tried divisible rule but couldn't come up with anything meaningful.
Any help?
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I will recommend a proof by induction... – Anton Vrdoljak May 15 '20 at 19:59
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2Hint: what are $2009\bmod 117,209\bmod 117,$ and $839\bmod 117$? – TonyK May 15 '20 at 20:03
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3Hint: $2009 \equiv 839 \mod 117$ and $209 \equiv 92 \mod 117$. – Robert Israel May 15 '20 at 20:03
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It might be easier to prove divisibility by 9 and 13 (because $117=9\cdot 13$). – richrow May 15 '20 at 20:08
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@Robert, pearls before swine... – TonyK May 15 '20 at 20:31
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See also https://math.stackexchange.com/a/3623463/589 – lhf May 15 '20 at 20:43
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By using $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1}),$$ we obtain: $$2009^n-209^n-839^n+92^n=$$ $$=(2009-839)(2009^{n-1}+...+839^{n-1})-(209-92)(209^{n-1}+...+92^{n-1}).$$ Can you end it now?
Michael Rozenberg
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Since $117=9\cdot 13$
So, it's sufficient to show that $9,13$ are divisors
$2009\equiv 2\mod 9$
$209\equiv 2\mod 9$
$839\equiv 2\mod 9$
$92\equiv 2\mod 9$
Hence,
$2009^n − 209^n − 839^n + 92^n\equiv 0\mod 9$
$2009\equiv 7\mod 13$
$209\equiv 1\mod 13$
$839\equiv 7\mod 13$
$92\equiv 1\mod 13$
Thus,
$2009^n − 209^n − 839^n + 92^n\equiv 0\mod 13$
Finally,
$2009^n − 209^n − 839^n + 92^n\equiv 0\mod 117$
ole
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