Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $1000$.

share|cite|improve this question
I wish it were $1000$, too! – The Chaz 2.0 Mar 27 '12 at 2:18
up vote 12 down vote accepted

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 $890$

share|cite|improve this answer
Note that there is no need to compute the quotient of the polynomials - only the remainder. See my answer. – Bill Dubuque Mar 27 '12 at 3:25

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. $\rm\ mod\ n+10\!:\ n\equiv -10\ \Rightarrow\ f(n)\equiv f(-10)$

share|cite|improve this answer

$$\begin{array}{r r c} \rm m=n+10: & \rm (m-10)^3+100 & \rm \equiv0 \;\bmod{m} \\ & \rm (-10)^3+100 & \rm \equiv0\; \bmod m \\ \times (-1) & 900 & \rm \equiv 0 \;\bmod m \\ \\ \hline \end{array}$$

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.