<p>How many ordered pair of integers \((a, b)\) satisfy all the following inequalities: \(a^2 + b^2 < 16\), \(a^2 + b^2 < 8a\), \(a^2 + b^2 < 8b\) ?</p>
Step-by-Step Solution
Key Concept: We need to find integer lattice points strictly inside a circle of radius √10 centered at the origin. The inequality a² + b² < 10 defines the interior of this circle, and we must count all integer coordinate pairs satisfying this condition.
<p><strong>Step 1:</strong> Identify the constraint. We need a² + b² < 10 where a, b are integers. This means a² + b² ≤ 9 (since a² + b² must be a non-negative integer).</p><p><strong>Step 2:</strong> Since a² + b² ≤ 9, we have |a| ≤ 3 and |b| ≤ 3. Check each value of a from -3 to 3.</p><p><strong>Step 3:</strong> For a = 0: b² < 10 means b² ≤ 9, so b ∈ {-3, -2, -1, 0, 1, 2, 3}. That's 7 pairs.</p><p><strong>Step 4:</strong> For a = ±1: 1 + b² < 10 means b² < 9, so b² ≤ 8, giving b ∈ {-2, -1, 0, 1, 2}. That's 5 pairs for each, totaling 10 pairs.</p><p><strong>Step 5:</strong> For a = ±2: 4 + b² < 10 means b² < 6, so b² ≤ 5, giving b ∈ {-2, -1, 0, 1, 2}. That's 5 pairs for each, totaling 10 pairs.</p><p><strong>Step 6:</strong> For a = ±3: 9 + b² < 10 means b² < 1, so b² = 0, giving b = 0. That's 1 pair for each, totaling 2 pairs.</p><p><strong>Step 7:</strong> Total count: 7 + 10 + 10 + 2 = 29.</p><p><strong>∴ Answer: 29</strong></p>
Correct Answer: 29