If $n\in \mathbb{Z}$ and follow the property
$$n+10| n^3+100$$
then what is the largest value of $n$
Please help me to solve this!!!
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Hint: $(n+10)(n^2 - 10n + 100)$ is pretty close to $n^3 + 100$. How can $n+10$ divide both of these? – Ravi Fernando Dec 18 '16 at 05:26
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Hint: Alternatively worded to ravi's hint, $n^3+100=n^3+10n^2-10n^2-100n+100n+1000-1000+100$ – JMoravitz Dec 18 '16 at 05:27
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problems like these almost always fall to the euclidean algorithm on both polynomials – Asinomás Dec 18 '16 at 05:30
3 Answers
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Let $m=n+10$ then $n=m-10$ $$n^3+100=(m-10)^3+100$$ $$=m^3-30m^2+300m-900$$ condition $$(n+10)|(n^3+100)$$ now takes the form $$m|(m^3-30m^2-300m-900)$$ Since $m$ is divisor of each of the quantities $m^3,30m^2,300m$
It follows that $$m|900$$ The largest $m$ for which this is true is $m=900$, so it follows that the largest $n$ with the given property is $$n=890$$
John
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I like how this is systematic, without needing to appeal to memory to remember some expansions – GuPe Dec 18 '16 at 05:39
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Note that $$ n+10|n^3+10n^2\implies n+10|(n^3+10n^2)-(n^3+100)=10n^2-100. $$ Similarly, \begin{aligned} &\quad n+10|(10n^2+100n)-(10n^2-100)=100n+100\\ &\implies n+10|(100n+1000)-(100n+100)=900. \end{aligned} So the largest possible $n$ is $900-10=\boxed{890}$. Verify that this works: $$ \frac{890^3+100}{890+10}=783299\in\mathbb{Z}. $$
yurnero
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Notice that: $n+10| n^3+100$ is same as $$n+10| n^3+1000-900$$ $$\implies n+10| (n^3+1000)-900$$
$$\implies n+10| (n+10)(n^2-10n+100)-900\implies n+10|900$$.
The greatest integer dividing $900$ is $900$ itself. So, for maximum $n$, $$n+10=900\implies n=900-10=890$$
Vidyanshu Mishra
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