# 3. The sum of two natural addends is 2016. If the last digit of one of them is erased - the second one is obtained. Find these numbers.
Let's denote the second number as A. Then the first number is $10 \mathrm{~A}+\mathrm{B}$, where $\mathrm{B}$ is its last digit. According to the condition, we get ( $10 \mathrm{~A}+\mathrm{B}$ ) $+\mathrm{A}=11 \mathrm{~A}+\mathrm{B}=2016$. Since the single-digit number B is less than 11, the last equality means that B is the remainder of the division of the number 2016 by 11. Dividing 2016 by 11 with a remainder, we get $2016=183 \cdot 11+3$, hence $B=3, A=183$ and $10 A+B==1833$.

Answer: the only possibility: $1833+183$.