AMC10 2018 B

AMC10 2018 B · Q19

AMC10 2018 B · Q19. It mainly tests Fractions, Counting divisors.

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?

乔伊、克洛伊和他们的女儿佐伊都有相同的生日。乔伊比克洛伊大 1 岁，佐伊今天正好 1 岁。今天是克洛伊年龄是佐伊年龄整数倍的 9 个生日中的第一个。下次乔伊年龄是佐伊年龄整数倍时，乔伊年龄的两数字之和是多少？

(A) 7 7

(B) 8 8

(D) 10 10

(E) 11 11

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Let Chloe be $n$ years old today, so she is $n-1$ years older than Zoe. For integers $y \ge 0$, Chloe’s age will be a multiple of Zoe’s age $y$ years from now if and only if $$ \frac{n+y}{1+y}=1+\frac{n-1}{1+y} $$ is an integer, that is, $1+y$ is a divisor of $n-1$. Thus $n-1$ has exactly 9 positive integer divisors, so the prime factorization of $n-1$ has one of the two forms $p^2q^2$ or $p^8$. There are no two-digit integers of the form $p^8$, and the only one of the form $p^2q^2$ is $2^2 \cdot 3^2 = 36$. Therefore Chloe is 37 years old today, and Joey is 38. His age will be a multiple of Zoe’s age in $y$ years if and only if $1+y$ is a divisor of $38-1=37$. The nonnegative integer solutions for $y$ are 0 and 36, so the only other time Joey’s age will be a multiple of Zoe’s age will be when he is $38+36=74$ years old. The requested sum is $7+4=11$.

答案（E）：设 Chloe 今天 $n$ 岁，则她比 Zoe 大 $n-1$ 岁。对整数 $y \ge 0$，当且仅当满足下式时，$y$ 年后 Chloe 的年龄会是 Zoe 年龄的整数倍： $$ \frac{n+y}{1+y}=1+\frac{n-1}{1+y} $$ 该式为整数，也就是 $1+y$ 是 $n-1$ 的一个因数。因此 $n-1$ 恰有 9 个正因数，所以 $n-1$ 的素因数分解形式只能是 $p^2q^2$ 或 $p^8$。不存在形如 $p^8$ 的两位整数，而形如 $p^2q^2$ 的唯一两位数是 $2^2\cdot 3^2=36$。因此 Chloe 今天 37 岁，Joey 38 岁。$y$ 年后他的年龄是 Zoe 年龄的整数倍，当且仅当 $1+y$ 是 $38-1=37$ 的因数。满足条件的非负整数 $y$ 为 0 和 36，所以 Joey 年龄再次成为 Zoe 年龄整数倍的唯一时刻是他 $38+36=74$ 岁时。所求的和为 $7+4=11$。

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