2001 AMC 10 第 12 题

Question

假设

n

是三个连续整数的乘积，并且

n

能被

7

整除。下列哪一个不一定是

n

的因数？

Suppose that

n

is the product of three consecutive integers and that

n

is divisible by

7.

Which of the following is not necessarily a divisor of

n?

6

14

21

28

42

答案：D

知识点：整除性质因数分解反例

难度评级：1370

小提示：

三个连续整数中，一定有一个偶数，也一定有一个

3

的倍数。

Among three consecutive integers, one is even and one is a multiple of

3

大提示：

寻找一个需要两个因数

2

的选项；三个连续整数的乘积不一定含有两个这样的因数。

Look for a choice requiring two factors of

2;

three consecutive integers need not provide both

解答：

三个连续整数中，一定有一个是

3

的倍数，且乘积

n

也有一个偶数因子，因此能被

6

整除。再加上题设的因数

7

，它一定能被

6

、

14

、

21

和

42

整除。

但

28=2^2\cdot7

需要两个因数

2

，这不一定保证；例如

5\cdot6\cdot7=210

能被

7

整除，却不能被

28

整除。

所以正确答案是 D。

Among three consecutive integers, at least one is even and one is a multiple of

3,

so

n

is divisible by

6.

With the given factor of

7,

it is divisible by

6,

14,

21,

and

42.

But

28=2^2\cdot7

requires two factors of

2,

which is not guaranteed:

5\cdot6\cdot7=210

is divisible by

7

but not by

28.

Thus, the correct answer is D.

2001 AMC 10 第 12 题

答案：D

难度评级：1370

小提示：

大提示：

解答：

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