2004 AMC 10B Problem 17

1. A two-digit age like $31$ means $3$ tens and $1$ one, so $31 = 10 \times 3 + 1$. Call Jack's digits $a$ (tens) and $b$ (ones). Then Jack is $10a + b$ and Bill, with the digits reversed, is $10b + a$. Jack will be the older one, so $a$ is bigger than $b$.
2. The "twice as old" rule is about five years from now, so add $5$ to both ages first:
- Jack then: $10a + b + 5$.
- Bill then: $10b + a + 5$.
3. Jack will be twice Bill:
$$10a + b + 5 = 2(10b + a + 5) = 20b + 2a + 10.$$
4. Gather the $a$'s on the left and everything else on the right: $10a - 2a = 20b - b + 10 - 5$, so
$$8a = 19b + 5.$$
5. Now $a$ and $b$ are single digits, so try small values of $b$:

Bigger $b$ only makes $a$ bigger, so $a = 3$, $b = 1$ is the only answer.
6. Jack is $31$ and Bill is $13$. Check: in five years they are $36$ and $18$, and $36 = 2 \times 18$.
7. The difference in their ages is $31 - 13 = 18$.

The answer is $\boxed{\textbf{(B)}\ 18}$.

"Digits reversed" is a signal to write the numbers as $10a + b$ and $10b + a$. That gives one equation with two unknowns, but since each unknown must be a digit from $0$ to $9$, a short table finishes the job. In age problems, always move everyone to the right year before using the "twice as old" rule.

Use the fact that an age gap never changes. In five years Jack is twice Bill, so the gap then (and now) equals Bill's age in five years. Try each choice as the gap:

• Gap $18$: Bill will be $18$, so he is $13$ now and Jack is $13 + 18 = 31$. Those digits are reversed. It works.
• Gap $9$: Bill is $4$ now, a one-digit age. Gap $27$: Bill is $22$ and Jack $49$, not reversed. Gap $36$: Bill $31$, Jack $67$, not reversed. Gap $45$: Bill $40$, Jack $85$, not reversed.