1951 AMC 12 第 15 题

Question

对于

n

的一切整数值，表达式

n^3-n

恒能被下列哪个最大的数整除？

The largest number by which the expression

n^3-n

is divisible for all possible integral values of

n

is:

2

3

4

5

6

将

n^3-n

因式分解

Factor

n^3-n

三个因数

n-1,n,n+1

是连续整数

The factors

n-1,n,n+1

are three consecutive integers

有

n^3-n=n(n-1)(n+1)\text{，}

这是三个连续整数的乘积。其中一个能被

3

整除，且至少一个是偶数，所以该乘积恒能被

6

整除。取

n=2

时，表达式的值恰为

6

，所以不存在更大的整数能恒整除该表达式。

因此，正确答案是 E。

We have

n^3-n=n(n-1)(n+1),

a product of three consecutive integers. One is divisible by

3

and at least one is even, so the product is always divisible by

6.

Taking

n=2

gives exactly

6,

so no larger integer always divides it.

Thus, the correct answer is E.

其他年份的第 15 题