Task 4. Masha wrote a three-digit number on the board, and Vera wrote the same number next to it, but she swapped the last two digits. After that, Polina added the obtained numbers and got a four-digit sum, the first three digits of which are 195. What is the last digit of this sum? (The answer needs to be justified.)

Answer: 4.
Solution. Let Masha write the number $100 x+10 y+z$. Then Vera wrote the number $100 x+10 z+y$, and the sum of these numbers is $200 x+11 y+11 z$. For $x \leqslant 8$ this expression does not exceed 1798, and therefore cannot start with 195. Thus, $x=9$. Then $11(y+z)$ is a three-digit number starting with 15. Among three-digit numbers starting with 15, only 154 is divisible by 11, so the last digit of the sum is 4.

Note. Masha could have written the numbers 959, 968, 977, 986, or 995.

## Criteria

1 p. The correct answer is provided.

2 p. It is determined that Masha's number starts with the digit 9, but there is no further progress.

2 p. An example of a three-digit number that satisfies the condition of the problem is provided.

3 p. It is determined that Masha's number starts with the digit 9, and an example of a three-digit number that satisfies the condition of the problem is provided.

4 p. The correct answer and justification are provided.