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CAT 2025Slot 1QAQues & Sol

AlgebraHard

Question

The number of distinct pairs of integers (x, y) satisfying the inequalities $x>y\geq3 $ and $x+y<14$ is

Solution

1. Concept Used

  • Topic: Integer Inequalities & Systematic Case Enumeration
  • Formula: $$ x > y \geq 3 \quad \text{and} \quad x + y < 14 $$

2. Calculation

We are given two constraints: $x > y \geq 3$ and $x + y < 14$, where both $x$ and $y$ are integers.

Since $y \geq 3$ and $x > y$, we have $x \geq y + 1$. Also, since $x + y < 14$, we get $x < 14 - y$. So for a given $y$, the valid range of $x$ is: $y + 1 \leq x \leq 13 - y$

For this range to have at least one valid integer, we need $y + 1 \leq 13 - y$, which gives $2y \leq 12$, i.e., $y \leq 6$.

Combined with $y \geq 3$, we get $y \in {3, 4, 5, 6}$.

Now we count the number of valid $x$ values for each $y$:

Case 1: $y = 3$ — $x$ ranges from $4$ to $10$, giving $10 - 4 + 1 = 7$ values.

Case 2: $y = 4$ — $x$ ranges from $5$ to $9$, giving $9 - 5 + 1 = 5$ values.

Case 3: $y = 5$ — $x$ ranges from $6$ to $8$, giving $8 - 6 + 1 = 3$ values.

Case 4: $y = 6$ — $x$ ranges from $7$ to $7$, giving $7 - 7 + 1 = 1$ value.

Notice the pattern forms an odd number sequence: $7, 5, 3, 1$.

Total number of valid pairs: $7 + 5 + 3 + 1 = 16$

This is the sum of the first 4 odd numbers: $\sum_{k=1}^{4}(2k-1) = 4^2 = 16$

3. Solution

Answer = 16 ✅

The total number of distinct integer pairs $(x, y)$ satisfying both constraints is 16.