Difference between revisions of "2024 AIME I Problems/Problem 13"
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+ | ==Solution 4== | ||
+ | These kind of problems, are by nature, elementary. We get:<cmath> m^4\equiv(-1) mod p^2</cmath>, thus , m is even , and <cmath>p^2=16k+1</cmath> | ||
+ | or <cmath>p=16k+1</cmath>, since p is prime. Therefore the smallest possible such p is 17. Again, | ||
+ | <cmath>m^4\equiv(16) mod 17</cmath>. This is where it gets a bit tricky. | ||
+ | <cmath>(m^2-1)*(m^2+1)\equiv (0) mod 17</cmath> or <cmath>m^2\equiv 13 mod 17</cmath> Giving rise to:<cmath>m\equiv(8) mod 17</cmath> | ||
+ | Now m lies in the series 42, 76, 110, 144..... . It is easy to see that the smallest value of m is 110 as neither 42 nor 76 satisfy all criteria. | ||
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+ | ~Grammaticus | ||
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==Video Solution== | ==Video Solution== |
Revision as of 02:29, 6 June 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
Solution 4
These kind of problems, are by nature, elementary. We get:, thus , m is even , and or , since p is prime. Therefore the smallest possible such p is 17. Again, . This is where it gets a bit tricky. or Giving rise to: Now m lies in the series 42, 76, 110, 144..... . It is easy to see that the smallest value of m is 110 as neither 42 nor 76 satisfy all criteria.
~Grammaticus
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.