# Problem 4. (3 points)

Given three numbers: a five-digit number, a four-digit number, and a three-digit number. Each of them consists of identical digits (each from its own set). Can it be such that their sum is a five-digit number consisting of five different digits?
Answer: Yes

## Solution:

Actually, one example is enough, but let's explain how it is obtained.

If there are no carries during addition, the last three digits of the sum will be the same. Therefore, we need to ensure that there are carries from the units place to the tens place and from the tens place to the hundreds place, and that these carries are different.

This can be achieved if the sum of our three digits equals 19. Then, in the tens place, with the carry of 1, it becomes 20 instead of 19, and a 2 is carried over to the hundreds place. This way, the sum of the numbers will end in 109, and a 2 will be carried over to the ten-thousands place. It remains to ensure that the sum of the digits of the five-digit and four-digit numbers does not end in 7, 8, or 9 (so that when added to 2, it does not result in 9, 10, or 11), and that the sum is a five-digit number with the first digit different from the others.

Example: $55555 + 6666 + 888 = 63109$