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 | + | 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 == | == Solution 2 == | ||
Revision as of 19:40, 17 February 2019
Contents
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
In a similar manner, we can apply synthetic division. We are looking for 
. Again, 
must be a factor of 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
