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Step-by-Step Solution
Step 1: Understand the Problem
We have two sets:
• Even numbers, denoted by $a$, where $a \in \{2,4,6,\ldots,100\}$.
• Odd numbers, denoted by $b$, where $b \in \{1,3,5,\ldots,99\}$.
We want the total number of pairs $(a,b)$ such that when $a + b$ is divided by 23, the remainder is 2. Mathematically,
$$
a + b \equiv 2 \; (\bmod \; 23).
$$
Step 2: Express the Condition Algebraically
Write the given condition as
$$
a + b = 23\lambda + 2,
$$
where $\lambda$ is an integer. Because $a$ is even and $b$ is odd, $a + b$ must be odd. Notice that $23$ is odd, so for $23\lambda + 2$ to be odd, $\lambda$ itself must be odd.
Step 3: Determine Possible Values of $\lambda$
The smallest possible sum is $2 + 1 = 3$, and the largest possible sum is $100 + 99 = 199$. So we need
$$
3 \;\le\; 23\lambda + 2 \;\le\; 199.
$$
Simplifying:
$$
1 \;\le\; 23\lambda \;\le\; 197 \;\;\Longrightarrow\;\; \frac{1}{23} \;\le\; \lambda \;\le\; \frac{197}{23}.
$$
Since $\lambda$ must be a positive odd integer within that range,
$$
\lambda \in \{1, 3, 5, 7\}.
$$
($\lambda=9$ would give $23 \cdot 9 + 2 = 209 > 199,$ thus not valid.)
Step 4: Count the Valid Pairs for Each $\lambda$
For each valid $\lambda$, count the number of ways to pick $a$ (even) and $b$ (odd) such that $a + b = 23\lambda + 2$:
$\lambda = 1$: $a + b = 23\cdot 1 + 2 = 25.$ Number of valid pairs = 12.
$\lambda = 3$: $a + b = 23\cdot 3 + 2 = 71.$ Number of valid pairs = 35.
$\lambda = 5$: $a + b = 23\cdot 5 + 2 = 117.$ Number of valid pairs = 42.
$\lambda = 7$: $a + b = 23\cdot 7 + 2 = 163.$ Number of valid pairs = 19.
Step 5: Sum the Counts
Adding up all valid pairs gives:
$$
12 + 35 + 42 + 19 = 108.
$$
Final Answer
The number of ways to select the numbers $a$ and $b$ so that $a + b$ leaves a remainder of 2 when divided by 23 is
108.
