2018 AMC 10B Problem 19

Let Chloe be $c$ years old today, so the age gap between Chloe and Zoe is $d = c - 1$ and stays constant. When Zoe is $n$, Chloe is $n + d$, and
$$n \mid n + d \iff n \mid d.$$
So the birthdays on which Chloe's age is a multiple of Zoe's are exactly the values $n$ that divide $d$. There are $9$ of them (today, $n = 1$, being the first), so $d$ has exactly $9$ positive divisors.

A number with $9$ divisors has the form $p^8$ or $p^2 q^2$. The candidates are $2^8 = 256$, $2^2 3^2 = 36$, $2^2 5^2 = 100$, and larger; only $d = 36$ gives a plausible parent's age, $c = 37$. Then Joey is $38$ today.

For Joey, the gap with Zoe is $38 - 1 = 37$, and Joey's age is a multiple of Zoe's when $n \mid 37$. Since $37$ is prime, this happens for $n = 1$ (today) and next for $n = 37$, when Joey is $37 + 37 = 74$. The digit sum is $7 + 4 = 11$.

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

"Parent's age is a multiple of child's age" is a divisibility statement about the constant age difference, since $n \mid n + d$ exactly when $n \mid d$. That turns "9 such birthdays" into "9 divisors," a standard prime-factorization count, and the realism of ages picks out $36$. The same trick applies to Joey with difference $37$.

Once Chloe is $37$, one can list: Joey $38$ is a multiple of Zoe's $1$ today; next, Joey $= 37 + n$ must be divisible by $n$, so $n \mid 37$, and $n = 37$ is the only option, giving $74$. Also, one may sanity check that $36$ indeed has divisors $1, 2, 3, 4, 6, 9, 12, 18, 36$, nine in all.

• Looking for a number with 9 divisors among Chloe's possible ages instead of the age difference, or picking 256 or 100 (also 9 divisors) as the difference without noticing they make Chloe unrealistically old.
• Answering with Joey's age $74$ or Chloe's next multiple instead of the digit sum $11$.