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 == | ||
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* [[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 17:20, 29 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 double-check manually and we find that indeed 
.
See also
| 1986 AIME (Problems • Answer Key • Resources) | ||
| Preceded by [[1986 AIME Problems/Problem {{{num-b}}}|Problem {{{num-b}}}]] |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
