2. Given are sets $A, B$, and $C$ such that:

- $A \cup B \cup C=\{1,2,3, \ldots, 100\}$;
- $A$ is the set of all natural numbers not greater than 100 that are divisible by 2;
- $B$ is the set of all natural numbers not greater than 100 that are divisible by 3;
- $B \cap C$ is the set of all natural numbers not greater than 100 whose sum of digits is equal to 9;
- $(A \cap C) \backslash B$ is the set of all two-digit numbers that are not divisible by 3 and whose unit digit is equal to 4.

Determine the number of elements in the set $C$.
2. We have $|A|=50$ and $|B|=33$, and since the numbers in the set $A \cap B$ are exactly the numbers divisible by 6, we have $|A \cap B|=16$. Furthermore, $B \cap C=\{9,18,27,36,45,54,63,72,81,90\}$, so we have $|B \cap C|=10$, and $(A \cap C) \backslash B=\{14,34,44,64,74,94\}$, so we have $|(A \cap C) \backslash B|=6$. Now we calculate

$$
\begin{aligned}
|(A \cup B) \backslash C| & =|A \cup B|-|B \cap C|-|(A \cap C) \backslash B| \\
& =|A|+|B|-|A \cap B|-|B \cap C|-|(A \cap C) \backslash B| \\
& =50+33-16-10-6=51
\end{aligned}
$$

from which we get $|C|=100-|(A \cup B) \backslash C|=49$.