Question

Joey、Chloe 和他们的女儿 Zoe 都同一天生日。Joey 比 Chloe 大 $1$ 岁，Zoe 今天正好 $1$ 岁。今天是 Chloe 的年龄会是 Zoe 年龄整数倍的 $9$ 个生日中的第一个。下一次 Joey 的年龄是 Zoe 年龄的整数倍时，Joey 年龄的两个数字之和是多少？

Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?

$7$

$8$

$9$

$10$

$11$

答案：E

知识点：因数个数整除性数字

难度评级：1990

解答：

设 Chloe 今天 $n$ 岁，Zoe 今天 $1$ 岁。 $t$ 年后，她们年龄之比为 $1+t$ ，所以它为整数当且仅当 $n-1$ 整除 $n-1$ 。 $9$ $\dfrac{n+t}{1+t}=1+\dfrac{n-1}{1+t},$

这样的生日数等于 $p^8$ 的正因子个数。共有九个这样的生日，所以 $p^2q^2$ 有 $9$ 个正因子；由题目所保证的两位数年龄可知，唯一可能是 $2^2\cdot3^2=36$ 。因此 Chloe 今天 $37$ 岁，Joey 今天 $38$ 岁。

现在 Joey 的年龄 $38+t$ 是 $1+t$ （即 Zoe 年龄）的倍数，当且仅当 $1+t$ 整除 $37$ 。所以下一次发生在 $t=36$ ，此时 Joey $74$ 岁，数字和为 $7+4=11$ 。正确答案是 E。

Let Chloe be $n$ today; Zoe is $1.$ In $t$ years their age ratio is $\dfrac{n+t}{1+t}=1+\dfrac{n-1}{1+t},$ which is an integer exactly when $1+t$ divides $n-1.$ Thus $n-1$ has exactly $9$ positive divisors.

A number with $9$ divisors has the form $p^8$ or $p^2q^2.$ The only two-digit possibility is $2^2\cdot3^2=36,$ so Chloe is $37$ and Joey is $38.$

Joey's age $38+t$ is a multiple of Zoe's age $1+t$ exactly when $1+t$ divides $37.$ The next time is $t=36,$ when Joey is $74.$ Its digit sum is $7+4=11.$ Thus, E is the correct answer.

2018 AMC 10B 第 19 题

答案：E

难度评级：1990

解答：

其他年份的第 19 题