3. The sum of two natural numbers is 2015. If the last digit of one of them is removed, the other number is obtained. Determine all such pairs of numbers.
Solution. Let $X$ and $Y$ be the given numbers and let $Y<X$. Then $X=10 Y+k$, where $k$ is the missing digit. From the condition of the problem, we have: $X+Y=2015$ and $X+Y=11 Y+k$ and hence $11 Y+k=2015$. Therefore, $k$ is the remainder when 2015 is divided by 11, i.e., $k=2$. Then $11 Y+2=2015$, so $Y=183$ and $X=1832$.