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What is the largest positive $n$ for which $n^3+100$ is divisible by $n+10$

I tried to factorize $n^3+100$, but $100$ is not a perfect cube. I wish it were $1000$.

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    $\begingroup$ I wish it were $1000$, too! $\endgroup$ Mar 27 '12 at 2:18
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By division we find that $n^3 + 100 = (n + 10)(n^2 − 10n + 100)−900$.

Therefore, if $n +10$ divides $n^3 +100$, then it must also divide $900$. Since we are looking for largest $n$, $n$ is maximized whenever $n + 10$ is, and since the largest divisor of $900$ is $900$, we must have $n + 10 = 900 \Rightarrow n = 890$

The largest $n$ is therefore $890$

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    $\begingroup$ Note that there is no need to compute the quotient of the polynomials - only the remainder. See my answer. $\endgroup$ Mar 27 '12 at 3:25
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Hint $\rm\quad\ \ n\!+\!10\ \,|\,\ f(n) \:\iff\: n\!+\!10\ \,|\,\ f(-10),\, $ for any $\rm\:f(x)\in \mathbb Z[x],\, $ by the Factor Theorem

i.e. $\rm\ mod\,\ n\!+\!10\!:\,\ \color{#c00}{n\equiv -10}\ \Rightarrow\ f(\color{#c00}n)\equiv f(\color{#c00}{-10})\ $ by the Polynomial Congruence Rule.

Thus the largest such $\rm\,n\,$ occcurs when $\,\rm n\!+\!10\,$ is the largest divisor of $\rm\,f(-10)$.

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  • $\begingroup$ nice solution.. I didn't know about polynomial congruence rule.. learnt something new :)! $\endgroup$ Oct 28 '20 at 14:09
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$$\begin{array}{r r c} \rm m=n+10: & \rm (m-10)^3+100 & \rm \equiv0 \;\bmod{m} \\ & \rm (-10)^3+100 & \rm \equiv0\; \bmod m \\ \times (-1) & 900 & \rm \equiv 0 \;\bmod m \\ \\ \hline \end{array}$$

$$\rm \max_m \{m:m|900\,\}=900 \implies n=890. $$

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  • $\begingroup$ answer is 890 but.?? $\endgroup$
    – Tony
    Aug 24 '20 at 19:07

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