<p>The total number of positive integers x for which \(\left\lfloor \dfrac{x}{99} \right\rfloor = \left\lfloor \dfrac{x}{101} \right\rfloor\) is given by:</p>
Step-by-Step Solution
Key Concept: For the floor values to be equal, x must lie in a range where both fractions have the same integer part. This occurs when x < 99 (both quotients < 1) or when x falls in specific intervals where the floor functions haven't yet jumped to different values.
<p><strong>Step 1:</strong> Let ⌊x/99⌋ = ⌊x/101⌋ = k for some non-negative integer k.</p><p><strong>Step 2:</strong> This means: 99k ≤ x < 99(k+1) AND 101k ≤ x < 101(k+1)</p><p><strong>Step 3:</strong> The intersection is: max(99k, 101k) ≤ x < min(99(k+1), 101(k+1))</p><p>This simplifies to: 101k ≤ x < 99(k+1) = 99k + 99</p><p><strong>Step 4:</strong> For a non-empty interval: 101k < 99k + 99, which gives 2k < 99, so k < 49.5, meaning k ≤ 49.</p><p><strong>Step 5:</strong> For k = 0: 0 ≤ x < 99 gives 99 positive integers (x = 1,2,...,99)</p><p><strong>Step 6:</strong> For k = 1,2,...,49: each gives (99k + 99) - 101k = 99 - 2k positive integers.</p><p><strong>Step 7:</strong> Total = 99 + Σ(99 - 2k) for k=1 to 49 = 99 + 49(99) - 2(1+2+...+49) = 99 + 4851 - 2(1225) = 99 + 4851 - 2450 = 2500</p><p>∴ Answer: <strong>2500</strong></p>
Correct Answer: A,B