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>? | What is that largest [[positive integer]] <math>n</math> for which <math>n^3+100</math> is [[divisible]] by <math>n+10</math>? | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/zfChnbMGLVQ?t=1458 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== Solution 1 == | == Solution 1 == |
Latest revision as of 01:16, 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.