Jana had to calculate the product of two six-digit numbers for her homework. When copying from the board, she omitted one digit from one of the numbers, and instead of a six-digit number, she wrote only 85522. When she got home, she realized her mistake. However, she remembered that the number she had copied incorrectly was divisible by three. She decided to try to determine what the original number could have been.

Determine how many such six-digit numbers exist.

(M. Dillingerová)
An integer is divisible by three if and only if the sum of its digits is divisible by three. The sum of the digits of the given number is

$$
8+5+5+2+2=22
$$

which is not a number divisible by three. Therefore, the missing digit was not 0, and the digit sum of the original number must be greater. The next larger numbers divisible by three are 24, 27, 30, 33, etc. If the digit sum were 24, 27, or 30, the missing digit would be 2, 5, or 8, respectively. (A digit sum of 33 or more cannot be achieved by adding a single digit.) Now, we need to determine on which positions these digits can be added to form different numbers. The simplest way to do this is to list all possibilities:

- $285522,825522,852522,855222$,
- $585522,855522,855252,855225$,
- $885522,858522,855822,855282,855228$.

In total, there are 13 possibilities.

Scoring. 2 points for identifying the missing digit and justification; 3 points for listing all possible numbers; 1 point for the number of possibilities.