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 (Simple)

Let $n+10=k$, then $n=k-10$. Then $n^3+100 = k^3-30k^2+300k-900$ Therefore, $900$ must be divisible by $k$, which is largest when $k=900$ and $n=\boxed{890}$

Solution 3

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 4

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

See also

1986 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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