Difference between revisions of "2024 AIME I Problems/Problem 13"
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\vphantom{\tfrac11}x\bmod{17}&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16\\hline | \vphantom{\tfrac11}x\bmod{17}&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16\\hline | ||
− | \vphantom{\dfrac11}\left(x^4\right) | + | \vphantom{\dfrac11}\left(x^4\right)+1\bmod{17}&2&0&14&2&14&5&5&0&0&5&5&14&2&14&0&2\\hline |
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Revision as of 18:30, 23 May 2024
Contents
[hide]Problem
Let be the least prime number for which there exists a positive integer such that is divisible by . Find the least positive integer such that is divisible by .
Solution 1
If
For integer
If
If
If
In conclusion, the smallest possible
Solution by Quantum-Phantom
Solution 2
We work in the ring
Solution 3 (Easy, given specialized knowledge)
Note that means The smallest prime that does this is and for example. Now let be a primitive root of The satisfying are of the form, So if we find one such , then all are Consider the from before. Note by LTE. Hence the possible are, Some modular arithmetic yields that is the least value.
~Aaryabhatta1
Video Solution
https://www.youtube.com/watch?v=_ambewDODiA
~MathProblemSolvingSkills.com
Video Solution 1 by OmegaLearn.org
Video Solution 2
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2024 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.