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 $900 \mid 890^3+100$ .

1986 AIME (Problems • Answer Key • Resources)
Preceded by [[1986 AIME Problems/Problem {{{num-b}}}\|Problem {{{num-b}}}]]	Followed by Problem 4
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1986 AIME Problems/Problem 5

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