1. Interpret the sets

We have:

• $a \in \{2,4,6,\dots,100\}$ = all even numbers from $2$ to $100$.
• $b \in \{1,3,5,\dots,99\}$ = all odd numbers from $1$ to $99$.

Thus,

• $a=2x$ where $x=1,2,\dots,50$
• $b=2y-1$ where $y=1,2,\dots,50$

We want $a+b \equiv 2 \pmod{23}.$

2. Rewrite the condition modulo 23

Substitute: $a+b=2x+(2y-1)=2(x+y)-1.$

Condition: $2(x+y)-1 \equiv 2 \pmod{23}$ $2(x+y) \equiv 3 \pmod{23}.$

Now inverse of $2$ modulo $23$ is $12$, because $2\cdot 12=24\equiv 1 \pmod{23}.$

So, $x+y \equiv 3\cdot 12=36 \equiv 13 \pmod{23}.$

Hence we need the number of pairs $(x,y)$ with $1\le x,y\le 50, \quad x+y\equiv 13 \pmod{23}.$

3. Count residues of $x$ and $y$ modulo 23

The numbers $1$ to $50$ distributed modulo $23$ are:

• residues $1,2,3,4$ occur $3$ times each,
• residues $5,6,\dots,22,0$ occur $2$ times each.

Reason: since $50=23\cdot 2+4,$ each residue appears at least $2$ times, and the first $4$ residues ($1,2,3,4$) appear one extra time.

Let $f(r)$ be the count of numbers in $\{1,2,\dots,50\}$ congruent to $r \pmod{23}$. Then $f(1)=f(2)=f(3)=f(4)=3,$ and for all other residues, $f(r)=2.$

4. Required residue pairing

We need $x+y \equiv 13 \pmod{23}.$

For each residue $r$ of $x$, residue of $y$ must be $13-r \pmod{23}$. So total number of pairs is $\sum_{r=0}^{22} f(r)f(13-r).$

Now compute systematically.

Special residues with frequency $3$ are $1,2,3,4$. Their matching residues are:

• $r=1 \Rightarrow 13-r=12$ with frequency $2$
• $r=2 \Rightarrow 11$ with frequency $2$
• $r=3 \Rightarrow 10$ with frequency $2$
• $r=4 \Rightarrow 9$ with frequency $2$

Contribution from these 4 residues: $4\times (3\cdot 2)=24.$

Now check if any target residue among $13-r$ is also in $\{1,2,3,4\}$. That happens when:

• $13-r=4 \Rightarrow r=9$
• $13-r=3 \Rightarrow r=10$
• $13-r=2 \Rightarrow r=11$
• $13-r=1 \Rightarrow r=12$

So residues $r=9,10,11,12$ each contribute $2\cdot 3=6,$ for total $4\times 6=24.$

All remaining $23-8=15$ residues pair with residues of frequency $2$, so each contributes $2\cdot 2=4.$ Thus contribution is $15\times 4=60.$

Therefore total number of pairs is $24+24+60=108.$

5. Match with options

The required number of ways is $\boxed{108}.$

So the correct option is:

• C: 108

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C (108)

They match.