1968 AMC 12 第 15 题

Question

设

P

是任意三个连续正奇数的乘积。所有这样的

P

都能被整除的最大整数为：

Let

P

be the product of any three consecutive positive odd integers. The largest integer dividing all such

P

is:

15

6

5

3

1

考察三个连续奇数模

3

的余数

Among three consecutive odd integers, inspect residues modulo

3

用两个例子排除所有更大的公因数

Use two examples to rule out every larger common divisor

三个连续奇数模

3

的余数依次连续，所以其中一个能被

3

整除。因此每个这样的乘积都能被

3

整除。两个乘积

1\cdot3\cdot5=15

与

7\cdot9\cdot11=693

的最大公因数为

3

，所以不存在总能整除该乘积的更大整数。

因此，正确答案是 D。

Three consecutive odd integers occupy three consecutive residues modulo

3,

so one is divisible by

3.

Thus every product is divisible by

3.

The products

1\cdot3\cdot5=15

and

7\cdot9\cdot11=693

have greatest common divisor

3,

so no larger integer always divides the product.

Therefore, the correct answer is D.

其他年份的第 15 题