The number of ways of selecting two numbers $$a$$ and $$b$$, $$a \in \{2, 4, 6, \ldots, 100\}$$ and $$b \in \{1, 3, 5, \ldots, 99\}$$ such that $$2$$ is the remainder when $$a + b$$ is divided by $$23$$ is

We want the number of ways to select $$a \in \{2, 4, 6, \ldots, 100\}$$ and $$b \in \{1, 3, 5, \ldots, 99\}$$ such that $$a + b$$ leaves remainder 2 when divided by 23.

Since $$a$$ is even and $$b$$ is odd, the sum $$a + b$$ is always odd. We require $$a + b \equiv 2 \pmod{23}$$. The smallest possible sum is $$2 + 1 = 3$$ and the largest is $$100 + 99 = 199$$. The odd numbers in this range that are congruent to 2 modulo 23 are $$25, 71, 117, 163$$.

For $$a + b = 25$$, we have $$a \in \{2, 4, \ldots, 24\}$$ and $$b = 25 - a$$. Each even $$a$$ from 2 to 24 produces an odd $$b$$ between 23 and 1, giving 12 pairs.

For $$a + b = 71$$, we have $$a \in \{2, 4, \ldots, 70\}$$ and $$b = 71 - a$$. There are 35 even values of $$a$$ from 2 to 70, so this yields 35 pairs.

For $$a + b = 117$$, the condition $$b = 117 - a \le 99$$ implies $$a \ge 18$$, and $$b \ge 1$$ implies $$a \le 116$$. Since $$a$$ must be even and at most 100, it runs through $$\{18, 20, \ldots, 100\}$$, which gives 42 pairs.

For $$a + b = 163$$, the inequality $$b = 163 - a \le 99$$ gives $$a \ge 64$$, and $$b \ge 1$$ gives $$a \le 162$$. Among the even values up to 100, this means $$a \in \{64, 66, \ldots, 100\}$$, yielding 19 pairs.

Adding these counts gives $$12 + 35 + 42 + 19 = 108$$.

The answer is Option C: 108.