Difference between revisions of "2024 AIME I Problems/Problem 13"
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~Bluesoul | ~Bluesoul | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | If | ||
+ | |||
+ | For integer | ||
+ | \begin{equation*} | ||
+ | p\mid\gcd\left(n^{p-1}-1,n^8-1\right)=n^{\gcd(p-1,8)}-1. | ||
+ | \end{equation*} | ||
+ | Here, | ||
+ | So we have to find the smallest positive integer | ||
+ | \begin{array}{|c|cccccccccccccccc|} | ||
+ | \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)^2+1\bmod{17}&2&0&14&2&14&5&5&0&0&5&5&14&2&14&0&2\\hline | ||
+ | \end{array} | ||
+ | So | ||
+ | \begin{align*} | ||
+ | 0&\equiv(17k+2)^4+1\equiv\mathrm C_4^1(17k)(2)^3+2^4+1=17(1+32k)\pmod{17^2}\[3pt] | ||
+ | \implies0&\equiv1+32k\equiv1-2k\pmod{17}. | ||
+ | \end{align*} | ||
+ | So the smallest possible | ||
+ | |||
+ | If | ||
+ | \begin{align*} | ||
+ | 0&\equiv(17k-2)^4+1\equiv\mathrm C_4^1(17k)(-2)^3+2^4+1=17(1-32k)\pmod{17^2}\[3pt] | ||
+ | \implies0&\equiv1-32k\equiv1+2k\pmod{17}. | ||
+ | \end{align*} | ||
+ | So the smallest possible | ||
+ | |||
+ | If | ||
+ | \begin{align*} | ||
+ | 0&\equiv(17k+8)^4+1\equiv\mathrm C_4^1(17k)(8)^3+8^4+1=17(241+2048k)\pmod{17^2}\[3pt] | ||
+ | \implies0&\equiv241+2048k\equiv3+8k\pmod{17}. | ||
+ | \end{align*} | ||
+ | So the smallest possible | ||
+ | |||
+ | If | ||
+ | \begin{align*} | ||
+ | 0&\equiv(17k-8)^4+1\equiv\mathrm C_4^1(17k)(-8)^3+8^4+1=17(241-2048k)\pmod{17^2}\[3pt] | ||
+ | \implies0&\equiv241+2048k\equiv3+9k\pmod{17}. | ||
+ | \end{align*} | ||
+ | So the smallest possible | ||
+ | |||
+ | In conclusion, the smallest possible | ||
+ | |||
+ | <font size=2>Solution by Quantum-Phantom</font> | ||
==Video Solution 1 by OmegaLearn.org== | ==Video Solution 1 by OmegaLearn.org== |
Revision as of 19:58, 3 February 2024
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
From there, we could get
By doing binomial expansion bash, the four smallest in this case are , yielding
~Bluesoul
Solution 2
If
For integer
If
If
If
In conclusion, the smallest possible
Solution by Quantum-Phantom
Video Solution 1 by OmegaLearn.org
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.