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Difference between revisions of "1986 AIME Problems/Problem 5"

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== See also ==
 
== See also ==
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* [[1986 AIME Problems]]
 
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{{AIME box|year=1986|num-b=4|num-a=6}}
 
{{AIME box|year=1986|num-b=4|num-a=6}}
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* [[AIME Problems and Solutions]]
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* [[American Invitational Mathematics Examination]]
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* [[Mathematics competition resources]]
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]

Revision as of 13:40, 6 May 2007

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 $900 \Longrightarrow n = 890$.

See also

1986 AIME (Problems • Answer Key • Resources)
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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