2024 AMC 10A Problems/Problem 24

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 $\frac{3^{28} - 20}{59} = 387{,}742{,}244{,}999$ .

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2024 AMC 10A Problems/Problem 24

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