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

Solution 3

The key to this problem is to realize that $n+10 \mid n^3 +1000$ for all $n$ . Since we are asked to find the maximum possible $n$ such that $n+10 \mid n^3 +100$ , we have: $n+10 \mid ((n^3 +1000) - (n^3 +100) \longrightarrow n+10 \mid 900$ . This is because of the property that states that if $a \mid b$ and $a \mid c$ , then $a \mid b \pm c$ . Since, the largest factor of 900 is itself we have: $n+10=900 \Longrightarrow \boxed{n = 890}$

~qwertysri987

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

Solution 3

See also