Question

If \( 4x^2 + y^2<52 \), where \( x, y \in \mathbb{Z} \), then the number of ordered pairs \( (x,y) \) is:

• A. \(67\)
• B. \(87\)
• C. \(77\)
• D. \(38\)

Hint:

For inequalities involving integers:
First restrict one variable using bounds
Then count valid integers for the other variable Symmetry about axes simplifies counting.

Answer 1

The Correct Option is C

Concept: We are asked to count the number of \emph{integer lattice points} lying strictly inside the region: \[ 4x^2 + y^2<52 \] This represents an ellipse centered at the origin. We count all integer values of \(x\) and corresponding integer values of \(y\) satisfying the inequality.
Step 1: Find possible integer values of \(x\). \[ 4x^2<52 \Rightarrow x^2<13 \] \[ x = -3, -2, -1, 0, 1, 2, 3 \]
Step 2: Count integer values of \(y\) for each \(x\).
\(x = 0:\quad y^2<52 \Rightarrow |y| \le 7 \Rightarrow 15 \text{ values}\)
\(x = \pm1:\quad y^2<48 \Rightarrow |y| \le 6 \Rightarrow 13 \times 2 = 26\)
\(x = \pm2:\quad y^2<36 \Rightarrow |y| \le 5 \Rightarrow 11 \times 2 = 22\)
\(x = \pm3:\quad y^2<16 \Rightarrow |y| \le 3 \Rightarrow 7 \times 2 = 14\)
Step 3: Add all possible ordered pairs. \[ 15 + 26 + 22 + 14 = 77 \] \[ \boxed{\text{Number of ordered pairs } = 77} \]