Difference between revisions of "1986 AIME Problems/Problem 5"
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== Solution 1 == | == Solution 1 == | ||
− | 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)</math> <math>= \gcd(100n+100,n+10)</math> <math>= \gcd(-900,n+10)</math>, so <math>n+10</math> must divide <math>900</math>. The greatest [[integer]] <math>n</math> for which <math>n+10</math> divides <math>900</math> is <math>\boxed{890}</math>; 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)</math> <math>= \gcd(100n+100,n+10)</math> <math>= \gcd(-900,n+10)</math>, so <math>n+10</math> must divide <math>900</math>. The greatest [[integer]] <math>n</math> for which <math>n+10</math> divides <math>900</math> is <math>\boxed{890}</math>; we can double-check manually and we find that indeed <math>900\mid 890^3+100</math>. |
== Solution 2 (Simple) == | == Solution 2 (Simple) == |
Revision as of 23:43, 24 October 2020
Problem
What is that largest positive integer for which is divisible by ?
Solution 1
If , . Using the Euclidean algorithm, we have , so must divide . The greatest integer for which divides is ; we can double-check manually and we find that indeed .
Solution 2 (Simple)
Let , then . Then Therefore, must be divisible by , which is largest when and
Solution 3
In a similar manner, we can apply synthetic division. We are looking for . Again, must be a factor of .
Solution 4
The key to this problem is to realize that for all . Since we are asked to find the maximum possible such that , we have: . This is because of the property that states that if and , then . Since, the largest factor of 900 is itself we have:
~qwertysri987
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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