Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2024 AMC 10A Problems/Problem 24"
Line 11: Line 11:
 
Thus, <math>29 \mid \varphi(p) = p - 1</math>, so the only possible prime is <math>59</math>, and the sum of the digits is <math>\boxed{\textbf{(D)}~14}</math>.
 
Thus, <math>29 \mid \varphi(p) = p - 1</math>, so the only possible prime is <math>59</math>, and the sum of the digits is <math>\boxed{\textbf{(D)}~14}</math>.
  
−
Note that <math>\frac{4^{28} - 15}{59} = \cdot 1{,}221{,}315{,}153{,}185{,}219</math> and <math>\frac{3^{28} - 20}{59} = 387{,}742{,}244{,}999</math>.
+
Note that <math>\frac{4^{28} - 15}{59} = 1{,}221{,}315{,}153{,}185{,}219</math> and <math>\frac{3^{28} - 20}{59} = 387{,}742{,}244{,}999</math>.
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=A|num-b=23|num-a=25}}
 
{{AMC10 box|year=2024|ab=A|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:29, 20 March 2025

Problem

There exists a unique two-digit prime number $p$ such that both $4^{28} - 15$ and $3^{28} - 20$ are divisible by $p$. What is the sum of the digits of $p$?

$\textbf{(A)}~10\qquad\textbf{(B)}~11\qquad\textbf{(C)}~13\qquad\textbf{(D)}~14\qquad\textbf{(E)}~16$

Solution

Let $k$ be the residue such that $k \equiv \tfrac{4}{3} \pmod{p}$. Note that $4^{28} \equiv 15 \pmod{p}$ and $3^{28} \equiv 20 \pmod{p}$. Dividing yields

\[k^{28} = k^{-1} \pmod{p} \implies k^{29} \equiv 1 \pmod{p}.\]

Thus, $29 \mid \varphi(p) = p - 1$, so the only possible prime is $59$, and the sum of the digits is $\boxed{\textbf{(D)}~14}$.

Note that $\frac{4^{28} - 15}{59} = 1{,}221{,}315{,}153{,}185{,}219$ and $\frac{3^{28} - 20}{59} = 387{,}742{,}244{,}999$.

See also

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10A_Problems/Problem_24&oldid=245622"