<p>Since 3.14 < π < 3.142, what is the largest natural number n for which \(\left\lfloor \dfrac{\pi}{2} + \dfrac{n}{100} \right\rfloor = 1\)?</p>
Step-by-Step Solution
Key Concept: The floor function equals 1 when the argument is in [1, 2). Use the given bounds on π to find the maximum n where π/2 + n/100 < 2.
<p><strong>Step 1:</strong> Set up the floor function constraint.</p><p>We need ⌊π/2 + n/100⌋ = 1, which means:</p><p>1 ≤ π/2 + n/100 < 2</p><p><strong>Step 2:</strong> Find the lower bound (always satisfied).</p><p>Since π > 3.14, we have π/2 > 1.57, so π/2 + n/100 > 1.57 > 1 ✓</p><p><strong>Step 3:</strong> Find the upper bound (the critical constraint).</p><p>We need: π/2 + n/100 < 2</p><p>Therefore: n/100 < 2 - π/2</p><p>So: n < 100(2 - π/2) = 200 - 50π</p><p><strong>Step 4:</strong> Use the upper bound on π.</p><p>Since π < 3.142, we have 50π < 157.1</p><p>Therefore: n < 200 - 50π > 200 - 157.1 = 42.9</p><p><strong>Step 5:</strong> Use the lower bound to find exact maximum.</p><p>Since 3.14 < π < 3.142:</p><p>200 - 50(3.142) < 200 - 50π < 200 - 50(3.14)</p><p>42.9 < 200 - 50π < 43</p><p>Thus n < (a value between 42.9 and 43), so the largest natural number is n = 42.</p><p><strong>Verification:</strong> With n = 42: π/2 + 0.42 ∈ [1.57 + 0.42, 1.571 + 0.42] = [1.99, 1.991] ✓ (equals 1)</p><p>With n = 43: π/2 + 0.43 would approach 2.001 > 2 ✗ (would equal 2)</p><p>∴ Answer: B (n = 42)</p>
Correct Answer: B