2021 AMC Fall 10A Problem 1

1. Do the subtraction inside the parentheses first: $2112 - 2021 = 91$. So the problem asks for $\dfrac{91^2}{169}$.
2. Before squaring $91$, look at the bottom. $169 = 13 \times 13$. That is a hint that $13$ matters.
3. Check whether $91$ is a multiple of $13$: $13 \times 7 = 91$. It is.
4. So $91^2 = (7 \times 13) \times (7 \times 13)$ and $169 = 13 \times 13$. The two $13$s on top cancel with the two $13$s on the bottom:
$$\frac{91^2}{169} = \frac{7 \times 13 \times 7 \times 13}{13 \times 13} = 7 \times 7 = 49.$$

The answer is $\boxed{\textbf{(C)}\ 49}$.

A square divided by a square is the square of the division: $\dfrac{91^2}{13^2} = \left(\dfrac{91}{13}\right)^2$. When the bottom of a fraction is a perfect square, divide first and square last, so the numbers stay small. It also helps to know the small multiples of $13$ by heart: $91 = 7 \times 13$ and $169 = 13 \times 13$ show up often.

First the subtraction: $2112 - 2021 = 91$.

Now notice that the denominator is a perfect square, $169 = 13^2$, and that $91 = 7 \cdot 13$. So instead of squaring $91$, rewrite the whole fraction as a square:
$$\frac{91^2}{13^2} = \left(\frac{91}{13}\right)^2 = 7^2 = 49.$$

The answer is $\boxed{\textbf{(C)}\ 49}$.