The number of ways of selecting two numbers a and b, a ∈ {2, 4, 6, ..., 100} and b ∈ {1, 3, 5, ...,

Q: Wait, why do we reject the even values of $k$?

Great observation! We know $a$ is even and $b$ is odd, so $a + b$ must be odd. In our equation $a + b = 23k + 2$, the right side must be odd. Since $2$ is even, $23k$ must be odd. Because $23$ is odd, $k$ must be odd. If $k$ were even, $23k$ would be even, and $23k + 2$ would be even, which contradicts our parity requirement.

Q: Why do the bounds for $a$ change in the later cases?

That is a classic trap! In the first two cases, $a$ is limited only by its own set ($a \le 100$). However, in the later cases, the sum $a + b$ is large. Since $b$ cannot exceed $99$, $a$ must be large enough to make up the difference ($a = (a+b) - b \ge (a+b) - 99$). Always check the constraints on both variables!

Q: Is there a faster way than case-by-case analysis?

While you could potentially use generating functions or more advanced combinatorics, for a JEE Advanced problem with only four cases, the case-by-case method is the most reliable and least error-prone. It forces you to respect the boundary conditions, which is where most students lose marks.

Question

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

Options:

(A) 186

(B) 54

(C) 108

(D) 268

Sigma Percentile · Accepted Answer

### Defining the Playground

We are selecting $a$ from the set $A = \{2, 4, 6, \dots, 100\}$ and $b$ from the set $B = \{1, 3, 5, \dots, 99\}$. Both sets contain exactly $50$ elements.

We seek the number of pairs $(a, b)$ such that $a + b \equiv 2 \pmod{23}$. This condition is equivalent to the equation:
$$a + b = 23k + 2$$
where $k$ is an integer.

### The Parity Insight

We observe that $a$ is always even and $b$ is always odd. Consequently, their sum $a + b$ must be an odd integer.

In our equation $a + b = 23k + 2$, the left side is odd. Since $2$ is even, the term $23k$ must be odd for the right side to be odd. Because $23$ is odd, $k$ must be an odd integer.

### Finding the Boundaries

The range of the sum $S = a + b$ is determined by the minimum and maximum values of the sets. The minimum sum is $2 + 1 = 3$, and the maximum sum is $100 + 99 = 199$.

We solve the inequality:
$$3 \le 23k + 2 \le 199$$

Subtracting $2$ from all sides yields $1 \le 23k \le 197$. Dividing by $23$, we find:
$$0.04 \le k \le 8.56$$

Since $k$ must be an odd integer, the possible values for $k$ are $1, 3, 5,$ and $7$.

### The Case-by-Case Analysis

We now evaluate the number of valid pairs for each permitted value of $k$:

**Case 1: $k = 1 \implies a + b = 25$**
Since $b \ge 1$, we have $a = 25 - b \le 24$. Given $a$ is even and $a \ge 2$, $a \in \{2, 4, \dots, 24\}$. There are $\frac{24}{2} = 12$ values.

**Case 2: $k = 3 \implies a + b = 71$**
Since $b \ge 1$, we have $a = 71 - b \le 70$. Given $a$ is even and $a \ge 2$, $a \in \{2, 4, \dots, 70\}$. There are $\frac{70}{2} = 35$ values.

**Case 3: $k = 5 \implies a + b = 117$**
Since $b \le 99$, we have $a = 117 - b \ge 117 - 99 = 18$. Given $a \le 100$, $a$ ranges from $18$ to $100$. The number of even values is:
$$\frac{100 - 18}{2} + 1 = 41 + 1 = 42$$

**Case 4: $k = 7 \implies a + b = 163$**
Since $b \le 99$, we have $a = 163 - b \ge 163 - 99 = 64$. Given $a \le 100$, $a$ ranges from $64$ to $100$. The number of even values is:
$$\frac{100 - 64}{2} + 1 = 18 + 1 = 19$$

### The Final Summation

To find the total number of valid pairs, we sum the results from each case:
$$12 + 35 + 42 + 19 = 108$$

The total number of valid pairs $(a, b)$ is **108**.

Animated Solution for Mathematics - Permutations and Combinations: The number of ways of selecting two numbers a and b, $a \in {2, 4, 6, \dots, 100}$ and $b \in {1, 3, 5, \dots, 99}$ such that 2 is the remainder when $a + b$ is divided by 23 is

Select Answer:

Visualized Solution

Define the Input Sets

Analyze the Remainder Condition

Apply Parity Logic

Determine Range of $a + b$

Possible Values of $k$

Case 1: $a + b = 25$

Case 2: $a + b = 71$

Case 3: $a + b = 117$

Case 4: $a + b = 163$

Total Number of Ways

The Sigma Insight: Fundamental Principle of Counting

Defining the Playground

The Parity Insight

Finding the Boundaries

The Case-by-Case Analysis

The Final Summation

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