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

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

解答：

其他年份的第 19 题