## 8. Lokot

Marko can't remember the three-digit code consisting of digits from 0 to 9. He remembers that two of the digits are the same, and the remaining digit is different from them. He also remembers that either all three digits are even or all three are odd. Zero is an even digit.

Result:
Solution.

Considering which two digits are the same, we distinguish 3 cases:

$\overline{a a b}, \overline{a b a}$ and $\overline{b a a}$ where $a, b \in\{0,1, \ldots, 9\}, a \neq b$.

Let's consider the first case, $\overline{a a b}$. If all digits are even, the digit $a$ can be $0,2,4,6$ or 8, which means there are 5 possibilities. For each of these possibilities, the digit $b$ can be chosen in 4 ways since it can be any except the one chosen for digit $a$. This results in a total of $5 \cdot 4=20$ possibilities for the first case.

For the second and third cases, $\overline{a b a}$ and $\overline{b a a}$, there will also be 20 possibilities each in the case of even digits. Therefore, for even digits, there are a total of $3 \cdot 20=60$ possibilities.

For odd digits, in a similar manner, 60 possibilities are obtained, which means there are a total of 120 possibilities for the locker code.