Difference between revisions of "1986 AIME Problems/Problem 5"
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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 == | + | == Video Solution == |
+ | https://youtu.be/zfChnbMGLVQ?t=1458 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | == 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>. | ||
+ | |||
+ | == Solution 2 (Simple) == | ||
+ | Let <math>n+10=k</math>, then <math>n=k-10</math>. Then <math>n^3+100 = k^3-30k^2+300k-900</math> Therefore, <math>900</math> must be divisible by <math>k</math>, which is largest when <math>k=900</math> and <math>n=\boxed{890}</math> | ||
+ | |||
+ | == Solution 3 == | ||
+ | In a similar manner, we can apply synthetic division. We are looking for <math>\frac{n^3 + 100}{n + 10} = n^2 - 10n + 100 - \frac{900}{n + 10}</math>. Again, <math>n + 10</math> must be a factor of <math>900 \Longrightarrow n = \boxed{890}</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | The key to this problem is to realize that <math>n+10 \mid n^3 +1000</math> for all <math>n</math>. Since we are asked to find the maximum possible <math>n</math> such that <math>n+10 \mid n^3 +100</math>, we have: <math>n+10 \mid ((n^3 +1000) - (n^3 +100) \longrightarrow n+10 \mid 900</math>. This is because of the property that states that if <math>a \mid b</math> and <math>a \mid c</math>, then <math>a \mid b \pm c</math>. Since, the largest factor of 900 is itself we have: <math>n+10=900 \Longrightarrow \boxed{n = 890}</math> | ||
+ | |||
+ | ~qwertysri987 | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=1986|num-b=4|num-a=6}} | |
+ | |||
+ | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 02:16, 17 January 2021
Contents
Problem
What is that largest positive integer for which is divisible by ?
Video Solution
https://youtu.be/zfChnbMGLVQ?t=1458
~ pi_is_3.14
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.