1986 AIME Problems/Problem 5

Problem

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

Solution 1

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 $\boxed{890}$ ; we can double-check manually and we find that indeed $900 \mid 890^3+100$ .

Solution 2

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 $900 \Longrightarrow n = \boxed{890}$ .

1986 AIME (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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1986 AIME Problems/Problem 5

Contents

Problem

Solution 1

Solution 2

See also