2004 AMC 10B Problem 17

Let Jack be $10a + b$ years old and Bill $10b + a$, where $a$ and $b$ are digits. Since Jack ends up older, $a > b$.

In five years Jack is $10a + b + 5$ and Bill is $10b + a + 5$, and the first is twice the second:
$$10a + b + 5 = 2(10b + a + 5) = 20b + 2a + 10.$$
Collecting terms: $8a = 19b + 5$.

Now test small digits $b$. For $b = 1$: $8a = 24$, so $a = 3$. For $b = 2$: $8a = 43$, not divisible. For $b = 3$: $8a = 62$, not divisible. For $b \ge 4$: $8a \ge 81$, so $a > 9$, impossible.

So Jack is $31$ and Bill is $13$. Check: in five years, $36 = 2 \cdot 18$. The difference in their ages is $31 - 13 = 18$.

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

"Digits reversed" is an invitation to write both numbers as $10a + b$ and $10b + a$; the resulting equation is linear in two digit-unknowns, and the digit bounds $0 \le a, b \le 9$ make a short search finish it. Age problems almost always hinge on applying the ratio at the correct moment in time, so add the years before setting up the ratio.

Reversing digits gives an age difference of $9(a - b)$, a multiple of $9$, and the difference never changes. In five years Jack is twice Bill, so the difference then equals Bill's future age. The choices are the only candidates: a difference of $18$ makes Bill $13$ now and Jack $31$, which works; $27$ would make Bill $22$ (same as Jack), $36$ would make Bill $31$ (older than Jack's $13$), and $45$ or $9$ give Bill $40$ or $4$, whose reversals fail.

• Applying 'twice as old' to the current ages rather than the ages in five years, or dropping the +5 from one side.
• Reporting the difference in five years (also $18$, luckily) by accident, or answering an age ($31$ or $13$) instead of the difference.