1986 AIME Problems/Problem 5: Difference between revisions
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== Problem == | == Problem == | ||
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 == | ||
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 == | ||
* [[1986 AIME Problems/Problem 4 | Previous Problem]] | |||
* [[1986 AIME Problems/Problem 6 | Next Problem]] | |||
* [[1986 AIME Problems]] | * [[1986 AIME Problems]] | ||
Revision as of 08:48, 27 January 2007
Problem
What is that largest positive integer 
for which 
is divisible by 
?
Solution
If 
, 
. Using the Euclidean Algorithm, we have 
, so must divide 900. The greatest integer for which divides 900 is 890; we can check manually and we find that indeed 
.
