# 1986 AIME Problems/Problem 5

## Problem

What is that largest positive integer $n$ for which $n^3+100$ is divisible by $n+10$?

## Solution

If $n+10 \mid n^3+100$, $\gcd(n^3+100,n+10)=n+10$. Using the Euclidean algorithm, we have $\gcd(n^3+100,n+10)= \gcd(-10n^2+100,n+10)$ $= \gcd(100n+100,n+10)$ $= \gcd(-900,n+10)$, so $n+10$ must divide 900. The greatest integer $n$ for which $n+10$ divides 900 is 890; we can double-check manually and we find that indeed $900 \mid 890^3+100$.

In a similar manner, we can apply synthetic division. We are looking for $\frac{n^3 + 100}{n + 10} = n^2 - 10n - 100 - \frac{900}{n + 10}$. Again, $n + 10$ must be a factor of $900 \Longrightarrow n = \boxed{890}$.

## Solution 2

After applying long division, we see that $\frac{n^3+100}{n+10} = n^2 - 10n + 100 - \frac{900}{n+10}$. Thus, $n+10$ must be a factor of $900$, and if we want the largest value of $n$, we have that $n+10 = 900 \Longrightarrow n = \boxed{890}$.

-ilovepi3.14

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