Question

下列数中，有一个不能被任何小于 $10$ 的质数整除。是哪一个？

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

$2^{606}-1$

$2^{606}+1$

$2^{607}-1$

$2^{607}+1$

$2^{607}+3^{607}$

答案：C

知识点：因式分解模运算整除性

难度评级：1820

解答：

使用事实： $a^n-b^n$ 能被 $a-b$ 整除。

选项 A 是 $2^{606}-1=4^{303}-1$ ，能被 $4-1=3$ 整除。

选项 B 是 $2^{606}+1=4^{303}-(-1)^{303}$ ，能被 $4-(-1)=5$ 整除。

选项 D 是 $2^{607}+1$ 。因为 $2^{606}-1$ 能被 $3$ 整除，乘以 $2$ 得到 $2^{607}-2$ 能被 $3$ 整除，从而 $2^{607}+1$ 也能被 $3$ 整除。

选项 E 是 $3^{607}+2^{607}=3^{607}-(-2)^{607}$ ，能被 $3-(-2)=5$ 整除。

对于选项 C， $2^{607}-1$ 是奇数。并且 $2^{607}\equiv2\pmod3$ 、 $2^{607}\equiv3\pmod5$ 、 $2^{607}\equiv2\pmod7$ ，所以 $2^{607}-1$ 不能被 $3,5$ 或 $7$ 整除。

所以答案是 C。

Use the fact that $a^n-b^n$ is divisible by $a-b.$

Choice A is $2^{606}-1=4^{303}-1,$ which is divisible by $4-1=3.$

Choice B is $2^{606}+1=4^{303}-(-1)^{303},$ which is divisible by $4-(-1)=5.$

Choice D is $2^{607}+1.$ Since $2^{606}-1$ is divisible by $3,$ multiplying by $2$ gives $2^{607}-2$ divisible by $3,$ so $2^{607}+1$ is divisible by $3.$

Choice E is $3^{607}+2^{607}=3^{607}-(-2)^{607},$ which is divisible by $3-(-2)=5.$

For choice C, $2^{607}-1$ is odd. Also $2^{607}\equiv2\pmod3,$ $2^{607}\equiv3\pmod5,$ and $2^{607}\equiv2\pmod7,$ so $2^{607}-1$ is not divisible by $3,5,$ or $7.$

Thus, our answer is C .

2022 AMC 10B 第 17 题

答案：C

难度评级：1820

解答：

其他年份的第 17 题