AMC10 2006 B

AMC10 2006 B · Q25

AMC10 2006 B · Q25. It mainly tests Casework, Divisibility & factors.

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?

Jones先生有八个不同年龄的孩子。在家庭旅行中，他最大的孩子9岁，看到一个四位数车牌号，其中两个数字各出现两次。“看，爸爸！”她叫道。“这个数字能被我们每个孩子的年龄整除！”“没错，”Jones先生回答，“而且最后两位数字正好是我的年龄。”以下哪项不是Jones先生孩子之一的年龄？

(A) 4 4

(B) 5 5

(D) 7 7

(E) 8 8

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) The 4-digit number on the license plate has the form $aabb$ or $abab$ or $baab$, where $a$ and $b$ are distinct integers from 0 to 9. Because Mr. Jones has a child of age 9, the number on the license plate is divisible by 9. Hence the sum of the digits, $2(a+b)$, is also divisible by 9. Because of the restriction on the digits $a$ and $b$, this implies that $a+b=9$. Moreover, since Mr. Jones must have either a 4-year-old or an 8-year-old, the license plate number is divisible by 4. These conditions narrow the possibilities for the number to 1188, 2772, 3636, 5544, 6336, 7272, and 9900. The last two digits of 9900 could not yield Mr. Jones’s age, and none of the others is divisible by 5, so he does not have a 5-year-old. Note that 5544 is divisible by each of the other eight non-zero digits.

(B) 车牌上的四位数形如 $aabb$ 或 $abab$ 或 $baab$，其中 $a$ 与 $b$ 是 0 到 9 之间互不相同的整数。由于琼斯先生有一个 9 岁的孩子，车牌号码能被 9 整除。因此其数字和 $2(a+b)$ 也能被 9 整除。结合对数字 $a$、$b$ 的限制，可推出 $a+b=9$。另外，由于琼斯先生一定还有一个 4 岁或 8 岁的孩子，车牌号码还必须能被 4 整除。由此将可能的号码缩小为 1188、2772、3636、5544、6336、7272 和 9900。9900 的后两位不能给出琼斯先生孩子的年龄，而其余数都不能被 5 整除，所以他没有 5 岁的孩子。注意 5544 能被其余八个非零数字中的每一个整除。

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