1. Five consecutive natural numbers were written on the board, and then one number was erased. It turned out that the sum of the remaining four numbers is 2015. Find the smallest of these four numbers.
Answer: 502

Solution. If the smallest number is K, then the sum of all numbers is not less than $\mathrm{K}+(\mathrm{K}+1)+(\mathrm{K}+2)+ (\mathrm{K}+3)=4 \mathrm{~K}+6$ and not more than $\mathrm{K}+(\mathrm{K}+2)+(\mathrm{K}+3)+(\mathrm{K}+4)=4 \mathrm{~K}+9$. Therefore, $4 \mathrm{~K}+6 \leq 2015 \leq 4 \mathrm{~K}+9$, from which $\mathrm{K}=502$.