Find the largest positive integer $n$ such that $n^3 + 100$ is divisible by $n+10$

Asked Sep 05 '16 at 06:19

Active Sep 05 '16 at 06:19

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I'm still kinda new to number theory so could someone tell me if my solution is right?

Reducing $\mod n +10$ $$n \equiv -10 \pmod{n+10} \implies n^3 + 100 \equiv -900 \pmod{n+10}$$ So we want $(n+10) \mid 900$ and the largest number would be $n = 890$

asked Sep 05 '16 at 06:19

Airdish

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5

$\frac{n^3+100}{n+10}=n^2-10n+100-\frac{900}{n+10}$. So yes, $890$ is the largest integer value of $n$ such that the entire expression is also integer. – barak manos Sep 05 '16 at 06:22
@barakmanos Thank you! – Airdish Sep 05 '16 at 06:25
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See http://math.stackexchange.com/questions/124927/what-is-the-largest-positive-n-for-which-n3100-is-divisible-by-n10 – lab bhattacharjee Sep 05 '16 at 07:52

Find the largest positive integer $n$ such that $n^3 + 100$ is divisible by $n+10$

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