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

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== Problem ==
 
== Problem ==
 
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What is that largest positive integer <math>n</math> for which <math>n^3+100</math> is divisible by <math>n+10</math>?
 
== Solution ==
 
== Solution ==
 
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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 check manually and we find that indeed <math>900 \mid 890^3+100</math>.
 
== See also ==
 
== See also ==
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* [[1986 AIME Problems/Problem 4 | Previous Problem]]
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* [[1986 AIME Problems/Problem 6 | Next Problem]]
 
* [[1986 AIME Problems]]
 
* [[1986 AIME Problems]]

Revision as of 09:48, 27 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 check manually and we find that indeed $900 \mid 890^3+100$.

See also

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