2021 AMC 10A Problem 3

1. A number divisible by $10$ ends in $0$. Try an example: erasing the $0$ from $470$ leaves $47$, which is $470 \div 10$. So erasing the last digit here is the same as dividing by $10$.
2. Call the smaller number $n$. The larger number is then $10n$.
3. Their sum is $10n + n = 11n$, so
$$11n = 17{,}402, \qquad n = 17{,}402 \div 11 = 1582.$$
4. The larger number is $15{,}820$. Check: $15{,}820 + 1582 = 17{,}402$.
5. The question asks for the difference: $15{,}820 - 1582 = 14{,}238$.

The answer is $\boxed{\textbf{(D)}\ 14{,}238}$.

Erasing a final zero sounds like a digit trick, but it is ordinary division by $10$. Once both numbers are written using the one letter $n$, the sum is $11n$ and the difference is $9n$. So the difference is always $\frac{9}{11}$ of the sum, and $9 \times 1582 = 14{,}238$ gives the same answer.

A number divisible by $10$ ends in $0$. Erasing that final $0$ is the same as dividing by $10$. So if the smaller number is $n$, the larger one is $10n$.

Their sum is $10n + n = 11n = 17{,}402$, giving $n = 1582$.

The difference is $10n - n = 9n = 9 \cdot 1582 = 14{,}238$.

The answer is $\boxed{\textbf{(D)}\ 14{,}238}$.