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，看到一个 $4$ 位数车牌，其中两个数字各出现两次。她喊道：“看，爸爸！这个数能被我们每个孩子的年龄整除！”Jones 先生回答：“没错，而且最后两位数正好是我的年龄。”下列哪一项不是 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：整除性分类讨论

难度评级：2120

解答：

因为有一个孩子是 $9,$ 岁，车牌号能被 $9,$ 整除，所以它的各位数字之和 $2(a+b)$ 是 $9,$ 的倍数，这迫使 $a+b=9.$

八个不同年龄是从 $1$ 到 $9$ 的九个整数中的八个，所以 $4$ 和 $8$ 中至少有一个是孩子的年龄。因此车牌号能被 $4.$ 整除。不计两个数字互换，排列形式为 $aabb,$ $abab,$ 或 $baab.$ 把这些形式与 $a+b=9$ 及能被 $4$ 整除结合，剩下 $\begin{gathered} 1188,\ 2772,\ 3636,\ 5544,\\ 6336,\ 7272,\ 9900. \end{gathered}$

最后一个候选会使 Jones 先生的年龄为 $00,$ 所以不可能。其余六个候选都不能被 $5,$ 整除，因此 $5$ 不可能是孩子的年龄。条件可以达到： $5544$ 能被 $\{1,2,3,4,6,7,8,9\}$ 中的每个年龄整除，而末两位给出 Jones 先生的年龄 $44.$

所以正确答案是 B。

Since a child is $9,$ the number is divisible by $9,$ so its digit sum $2(a+b)$ is a multiple of $9,$ which forces $a+b=9.$

The eight distinct ages are eight of the nine integers from $1$ through $9,$ so at least one of $4$ and $8$ is an age. Hence the plate number is divisible by $4.$ Up to interchanging the two digits, its pattern is $aabb,$ $abab,$ or $baab.$ Combining these patterns with $a+b=9$ and divisibility by $4$ leaves $\begin{gathered} 1188,\ 2772,\ 3636,\ 5544,\\ 6336,\ 7272,\ 9900. \end{gathered}$

The last candidate would make Mr. Jones's age $00,$ so it is impossible. None of the other six candidates is divisible by $5,$ so $5$ cannot be one of the children's ages. The conditions are attainable: $5544$ is divisible by each age in $\{1,2,3,4,6,7,8,9\}$ and its last two digits give Mr. Jones's age as $44.$

Thus, the correct answer is B.

2006 AMC 10B 第 25 题

答案：B

难度评级：2120

解答：

其他年份的第 25 题