Question

Joey、Chloe 和他们的女儿 Zoe 生日相同。Joey 比 Chloe 大 $1$ 岁，Zoe 今天正好 $1$ 岁。今天是 Chloe 的年龄将成为 Zoe 年龄整数倍的 $9$ 个生日中的第一个。下次 Joey 的年龄是 Zoe 年龄的倍数时，Joey 年龄的两位数字之和是多少？

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

难度评级：1870

解答：

设 Chloe 今天 $n$ 岁，那么她比 Zoe 大 $n-1$ 岁。 $y$ 年后，Chloe 的年龄 $n+y$ 是 Zoe 年龄 $1+y$ 的倍数，当且仅当 $1+y$ 整除 $n-1.$ 有 $9$ 个这样的生日，意味着 $n-1$ 恰有 $9$ 个因数。

恰有 $9$ 个因数的数形如 $p^2q^2$ ，其中 $p,q,$ 是不同质数，或形如 $p^8.$ 因为在所问的未来生日，Joey 的年龄是两位数，所以 $n-1\lt99;$ 唯一可能是 $2^2\cdot3^2=36.$ 因此 Chloe 是 $37$ 岁，Joey 是 $38.$ 岁。

Joey 的年龄 $38+y$ 是 $1+y$ 的倍数，当且仅当 $1+y$ 整除 $37.$ 下一次发生在 $y=36,$ 此时 Joey 是 $74,$ 岁，数位和为 $7+4=11.$

因此，正确答案是 E。

Let Chloe be $n$ today, so she is $n-1$ years older than Zoe. In $y$ years Chloe's age $n+y$ is a multiple of Zoe's age $1+y$ exactly when $1+y$ divides $n-1.$ Having $9$ such birthdays means $n-1$ has exactly $9$ divisors.

A number with exactly $9$ divisors has the form $p^2q^2$ for distinct primes $p,q,$ or $p^8.$ Because Joey's age at the requested future birthday has two digits, $n-1\lt99;$ the only possibility is $2^2\cdot3^2=36.$ So Chloe is $37$ and Joey is $38.$

Joey's age $38+y$ is a multiple of $1+y$ exactly when $1+y$ divides $37.$ The next time is $y=36,$ making Joey $74,$ with digit sum $7+4=11.$

Thus, the correct answer is E.

2018 AMC 12B 第 14 题

答案：E

难度评级：1870

解答：

其他年份的第 14 题