Difference between revisions of "2024 AIME I Problems/Problem 13"
Wescarroll (talk | contribs) m (Added link to Fermat's Little Theorem) |
Mathophobia (talk | contribs) (→Solution 3 (Easy, given specialized knowledge)) |
||
Line 59: | Line 59: | ||
~Aaryabhatta1 | ~Aaryabhatta1 | ||
+ | |||
+ | ==Solution 4 (Uses only elementary modulus)== | ||
+ | |||
+ | The eqution is <math>n^4 \equiv -1 \pmod{p^2}</math> and we check p = 2 and it fails. We know <math>p^2</math> is odd meaning n is even and <math>n^4</math> is divisible by 16, substituting in to the equation above and taking the equation mod 16, we see that <math>p^2 \equiv 1 \pmod{16}</math>. We have that p = 1,7,9,15 (mod 16) giving p = 7,17,23,... Testing p = 7, <math>n^4 \equiv 48 \pmod{49}</math> | ||
+ | <math>n^2 \equiv 6 \pmod{7}</math> which is impossible. Next we try 17 and precede like solution 1 to get | ||
+ | ~mathophobia | ||
==Video Solution 1 by OmegaLearn.org== | ==Video Solution 1 by OmegaLearn.org== |
Revision as of 12:45, 28 February 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 (Uses only elementary modulus)
The eqution is and we check p = 2 and it fails. We know is odd meaning n is even and is divisible by 16, substituting in to the equation above and taking the equation mod 16, we see that . We have that p = 1,7,9,15 (mod 16) giving p = 7,17,23,... Testing p = 7,
which is impossible. Next we try 17 and precede like solution 1 to get
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.