AMC12 2022 B

AMC12 2022 B · Q15

AMC12 2022 B · Q15. It mainly tests Remainders & modular arithmetic, Powers & residues.

One of the following numbers is not divisible by any prime number less than $10.$ Which is it?

以下数字中有一个不能被小于 $10$ 的任何质数整除。是哪一个？

(A) 2^{606}-1 2^{606}-1

(B) 2^{606}+1 2^{606}+1

(D) 2^{607}+1 2^{607}+1

(E) 2^{607}+3^{607} 2^{607}+3^{607}

Answer

Correct choice: (C)

正确答案：（C）

Solution

For $\textbf{(A)}$ modulo $3,$ \begin{align*} 2^{606} - 1 & \equiv (-1)^{606} - 1 \\ & \equiv 1 - 1 \\ & \equiv 0 . \end{align*} Thus, $2^{606} - 1$ is divisible by $3.$ For $\textbf{(B)}$ modulo $5,$ \begin{align*} 2^{606} + 1 & \equiv 2^{{\rm Rem} ( 606, \phi(5) )} + 1 \\ & \equiv 2^{{\rm Rem} ( 606, 4 )} + 1 \\ & \equiv 2^2 + 1 \\ & \equiv 0 . \end{align*} Thus, $2^{606} + 1$ is divisible by $5.$ For $\textbf{(D)}$ modulo $3,$ \begin{align*} 2^{607} + 1 & \equiv (-1)^{607} + 1 \\ & \equiv - 1 + 1 \\ & \equiv 0 . \end{align*} Thus, $2^{607} + 1$ is divisible by $3.$ For $\textbf{(E)}$ modulo $5,$ \begin{align*} 2^{607} + 3^{607} & \equiv 2^{607} + (-2)^{607} \\ & \equiv 2^{607} - 2^{607} \\ & \equiv 0 . \end{align*} Thus, $2^{607} + 3^{607}$ is divisible by $5.$ Therefore, the answer is $\boxed{\textbf{(C) }2^{607} - 1}.$

对于 $\textbf{(A)}$ 模 $3$， \begin{align*} 2^{606} - 1 & \equiv (-1)^{606} - 1 \\ & \equiv 1 - 1 \\ & \equiv 0 . \end{align*} 因此，$2^{606} - 1$ 能被 $3$ 整除。对于 $\textbf{(B)}$ 模 $5$， \begin{align*} 2^{606} + 1 & \equiv 2^{{\rm Rem} ( 606, \phi(5) )} + 1 \\ & \equiv 2^{{\rm Rem} ( 606, 4 )} + 1 \\ & \equiv 2^2 + 1 \\ & \equiv 0 . \end{align*} 因此，$2^{606} + 1$ 能被 $5$ 整除。对于 $\textbf{(D)}$ 模 $3$， \begin{align*} 2^{607} + 1 & \equiv (-1)^{607} + 1 \\ & \equiv - 1 + 1 \\ & \equiv 0 . \end{align*} 因此，$2^{607} + 1$ 能被 $3$ 整除。对于 $\textbf{(E)}$ 模 $5$， \begin{align*} 2^{607} + 3^{607} & \equiv 2^{607} + (-2)^{607} \\ & \equiv 2^{607} - 2^{607} \\ & \equiv 0 . \end{align*} 因此，$2^{607} + 3^{607}$ 能被 $5$ 整除。因此，答案是 $\boxed{\textbf{(C) }2^{607} - 1}$。

Topics