2018 AMC 10B Problem 19

The age gap between a parent and a child never changes. Work with the gap.

1. Set up. Let the gap between Chloe and Zoe be $d$. When Zoe is $n$ years old, Chloe is $n + d$.
2. When is Chloe's age a multiple of Zoe's? Try a gap of $d = 6$. When Zoe is $2$, Chloe is $8$, a multiple of $2$. When Zoe is $4$, Chloe is $10$, not a multiple of $4$. In general, $n + d$ is a multiple of $n$ exactly when $d$ is a multiple of $n$, because $n$ always divides the $n$ part. So it happens exactly on the birthdays when Zoe's age $n$ is a divisor of $d$.
3. Nine birthdays means nine divisors. Zoe's ages that work are the divisors of $d$, starting with $1$ (today). So $d$ has exactly $9$ divisors.
4. Which gap has 9 divisors? Divisors come in pairs that multiply to $d$, like $2 \times 18$ for $36$. An odd number of divisors means one divisor is paired with itself, so $d$ is a perfect square. A realistic gap between a mother and her baby is between about $15$ and $60$. The squares there:
- $16$: divisors $1, 2, 4, 8, 16$, only $5$.
- $25$: divisors $1, 5, 25$, only $3$.
- $36$: divisors $1, 2, 3, 4, 6, 9, 12, 18, 36$, exactly $9$.
- $49$: divisors $1, 7, 49$, only $3$.

So $d = 36$. Chloe is $37$ and Joey is $38$ today.
5. Now Joey. His gap with Zoe is $38 - 1 = 37$. By the same reasoning, his age is a multiple of Zoe's when Zoe's age divides $37$. Since $37$ is prime, that happens only when Zoe is $1$ (today) and when Zoe is $37$.
6. Next time. When Zoe is $37$, Joey is $37 + 37 = 74$. The digit sum is $7 + 4 = 11$.

The answer is $\boxed{\textbf{(E)}\ 11}$.

"The older age is a multiple of the younger age" turns into "the younger age divides the gap," and the gap is fixed. That changes a question about many birthdays into a question about counting divisors of one number. Only perfect squares have an odd number of divisors, which narrows the search quickly.

Use the divisor-count rule. If $d = p^a q^b$ with different primes $p$ and $q$, it has $(a+1)(b+1)$ divisors. For $9 = 3 \times 3$ or $9 = 9$, the number is $p^2 q^2$ or $p^8$. The smallest options are $2^2 \times 3^2 = 36$, then $2^2 \times 5^2 = 100$, $2^2 \times 7^2 = 196$, $3^2 \times 5^2 = 225$ and $2^8 = 256$. Only $36$ gives a sensible parent's age.