2. The natural numbers from 1 to 100 were written in a row without spaces. Then, plus signs were placed between some of the digits. (For example, $1234567+891011 \ldots 15+1617$... 99100.) Can the resulting sum be divisible by 111?
Solution. The remainder of a number divided by three is equal to the remainder of the sum of its digits divided by three. Therefore, the sum of the remainders of all numbers divided by three, separated by plus signs, is equal to the remainder of the sum of all digits of numbers from 1 to 100 divided by 3. In turn, this sum of digits gives the same remainder when divided by 3 as the sum $1+\cdots+100=5050$. Since 5050 is not divisible by 3, the sum in the problem's condition is also not divisible by 3. If a number is not divisible by 3, then it is not divisible by 111 either.

Answer: no