2022 AMC 10B 第 17 题

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

小提示：

对剩余选项检验能否被

2

、

3

、

5

和

7

整除

For the remaining choice, test divisibility by

2,

3,

5,

and

7

大提示：

找出小质因数，从而排除四个选项

Eliminate four choices by finding a small prime divisor

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

小提示：

大提示：

解答：

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