1. The sum of two natural numbers is 2015. If the last digit of one of them is erased, the result is the second number. Find all such numbers.
Solution: Let b be the smaller number. Then the larger number a $=10 \mathrm{~b}+\mathrm{c}$, where c is the crossed-out digit, $0 \leq \mathrm{c} \leq 9$. According to the condition, $2015=\mathrm{a}+\mathrm{b}=11 \mathrm{~b}+\mathrm{c}$, so b is the quotient, and c is the remainder when 2015 is divided by 11, i.e., $b=183, \mathrm{c}=2$. The required numbers are 1832 and 183.

Criteria. Correct answer --- 1 point.

It is stated that the numbers are structured as b and $10 b+c$ --- 1 point.

This fact can be stated in another way. For example, that one of the numbers is $\bar{x}$, and the other is $\bar{x}$. Correct solution --- 7 points.