2006 AMC 12B 第 19 题

Question

Jones 先生有八个年龄各不相同的孩子。在一次家庭旅行中，他最大的孩子

9

岁，看到一块车牌上的

4

位数，其中两个数字各出现两次。她喊道：“看，爸爸！那个数能被我们每个孩子的年龄整除！” Jones 先生回答：“没错，而且最后两位数字刚好是我的年龄。”下列哪一个不是 Jones 先生某个孩子的年龄？

Mr. Jones has eight children of different ages. On a family trip his oldest child, who is

9,

spots a license plate with a

4

-digit number in which each of two digits appears two times. “Look, daddy!” she exclaims. “That number is evenly divisible by the age of each of us kids!” “That’s right,” replies Mr. Jones, “and the last two digits just happen to be my age.” Which of the following is not the age of one of Mr. Jones’s children?

4

5

6

7

8

答案：B

知识点：整除性数字分类讨论

难度评级：1920

小提示：

最大的孩子

9

岁，所以该数能被

9

整除，迫使两个数字之和为

9

。

The oldest child is

9,

so the number is divisible by

9,

forcing the two digits to sum to

9

大提示：

一个以两位年龄结尾的数不能以

0

结尾，因此思考它必须避开哪个较小的年龄。

A number ending in a two-digit age cannot end in

0,

so think about which small ages it must avoid

解答：

这个数的形式为

aabb

，

abab

，或

baab

。能被

9

整除说明

2(a + b)

是

9

的倍数，所以

a + b = 9

。

八个不同年龄是从

1

到

9

的九个整数中的八个，所以年龄

4

和

8

中至少出现一个。因此这个数能被

4

整除。可能性变为

1188, 2772, 3636

，

5544, 6336, 7272, 9900

。

因为末两位是 Jones 先生的年龄，

9900

不可能，其余数也都不是

5

的倍数。所以孩子们的年龄不能包含

5

。事实上，

5544

能被

1, 2, 3, 4, 6, 7, 8, 9

整除。

所以正确答案是 B。

The number has the form

aabb,

abab,

or

baab.

Divisibility by

9

means

2(a + b)

is a multiple of

9,

so

a + b = 9.

The eight distinct ages are eight of the nine integers from

1

through

9,

so at least one of ages

4

and

8

occurs. Therefore the number is divisible by

4.

The possibilities become

1188, 2772, 3636,

5544, 6336, 7272, 9900.

Since the last two digits are Mr. Jones’s age,

9900

is impossible, and none of the others is a multiple of

5.

So the children’s ages cannot include

5.

Indeed

5544

is divisible by

1, 2, 3, 4, 6, 7, 8, 9.

Thus, the correct answer is B.

2006 AMC 12B 第 19 题

答案：B

难度评级：1920

小提示：

大提示：

解答：

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