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1986 AIME Problems/Problem 5

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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 substitution. 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}.

See also

1986 AIME (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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