2019 AMC 10A Problem 3

Everyone gets one year older each year, so one letter for Bonita's age is enough.

1. Name this year's ages. Say Bonita is $b$ years old now. Ana's age is the square of that, so Ana is $b^2$.
2. Go back one year. Last year Bonita was $b - 1$ and Ana was $b^2 - 1$.
3. Use "5 times as old".
$$b^2 - 1 = 5(b - 1).$$
4. Spot a shared factor. The left side is a difference of squares: $b^2 - 1 = (b - 1)(b + 1)$. So
$$(b - 1)(b + 1) = 5(b - 1).$$
5. Divide by $b - 1$. This is allowed because $b$ cannot be $1$: then Bonita would have been $0$ last year and both girls would be $1$ now, the same age, but they are born in different years. Dividing leaves
$$b + 1 = 5, \qquad b = 4.$$
6. Find the ages. Bonita is $4$ and Ana is $4^2 = 16$.
7. Check last year. Bonita was $3$ and Ana was $15 = 5 \cdot 3$. It works.

The question asks for the gap between them: $n = 16 - 4 = 12$.

The answer is $\boxed{\textbf{(D)}\ 12}$.

Age problems need just one unknown, because the gap between two people never changes. The "square" condition looks like it makes the equation hard, but both sides share the factor $b - 1$, and cancelling it leaves a one-step equation. Whenever you see $b^2 - 1$ next to $b - 1$, try the difference of squares.

Try the answer choices. If the gap is $n$, then Ana's age $b + n$ must equal $b^2$.

• $n = 12$: $b + 12 = b^2$ is true for $b = 4$ (since $4 + 12 = 16$). Last year the ages were $3$ and $15$, and $15 = 5 \cdot 3$. This works.
• The other choices fail. For example, $n = 3$ gives no whole-number $b$, and $n = 15$ gives no whole-number $b$ either (try $b = 4$: $19 \ne 16$; $b = 5$: $20 \ne 25$).