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

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If <math>n+10 \mid n^3+100</math>, <math>\gcd(n^3+100,n+10)=n+10</math>. Using the [[Euclidean algorithm]], we have <math>\gcd(n^3+100,n+10)= \gcd(-10n^2+100,n+10)= \gcd(100n+100,n+10)= \gcd(-900,n+10)</math>, so <math>n+10</math> must divide 900. The greatest [[integer]] <math>n</math> for which <math>n+10</math> divides 900 is 890; we can double-check manually and we find that indeed <math>900 \mid 890^3+100</math>.
 
If <math>n+10 \mid n^3+100</math>, <math>\gcd(n^3+100,n+10)=n+10</math>. Using the [[Euclidean algorithm]], we have <math>\gcd(n^3+100,n+10)= \gcd(-10n^2+100,n+10)= \gcd(100n+100,n+10)= \gcd(-900,n+10)</math>, so <math>n+10</math> must divide 900. The greatest [[integer]] <math>n</math> for which <math>n+10</math> divides 900 is 890; we can double-check manually and we find that indeed <math>900 \mid 890^3+100</math>.
 
== See also ==
 
== See also ==
* [[1986 AIME Problems/Problem 4 | Previous Problem]]
 
* [[1986 AIME Problems/Problem 6 | Next Problem]]
 
 
* [[1986 AIME Problems]]
 
* [[1986 AIME Problems]]
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{{AIME box|year=1986|num-a=4|num-b|6}}
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]

Revision as of 18:20, 29 January 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 $900 \mid 890^3+100$.

See also

1986 AIME (ProblemsAnswer KeyResources)
Preceded by
[[1986 AIME Problems/Problem {{{num-b}}}|Problem {{{num-b}}}]]
Followed by
Problem 4
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