Difference between revisions of "2024 AIME I Problems/Problem 13"

(Solution)
Line 15: Line 15:
 
{{AIME box|year=2024|n=I|num-b=12|num-a=14}}
 
{{AIME box|year=2024|n=I|num-b=12|num-a=14}}
  
 +
[[Category:Intermediate Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:44, 2 February 2024

Problem

Let $p$ be the least prime number for which there exists a positive integer $n$ such that $n^{4}+1$ is divisible by $p^{2}$. Find the least positive integer $m$ such that $m^{4}+1$ is divisible by $p^{2}$.

Solution

$n^4+1\equiv 0\pmod{p^2}\implies n^8 \equiv 1\pmod{p^2}\implies p_{min}=17$

From there, we could get $n\equiv \pm 2, \pm 8\pmod{17}$

By doing binomial expansion bash, the four smallest $n$ in this case are $110, 134, 155, 179$, yielding $\boxed{110}$

~Bluesoul

See also

2024 AIME I (ProblemsAnswer KeyResources)
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. AMC logo.png