Basic Mathematics & Logarithm
Greatest Integer Function
Grade 11
Question:
<p>If 'k' is the number of positive integers \(x\) which satisfy the condition \(\left[\dfrac{x}{99}\right] = \left[\dfrac{x}{101}\right]\), then k is. Here [x] is greatest integer function.</p>
<p>(a) a perfect square</p>
<p>(b) 3 an odd number</p>
<p>(c) less than 2500</p>
<p>(d) divisible by 3</p>
Step-by-Step Solution
Key Concept: The condition [x/99] = [x/101] holds when both fractions lie in the same integer interval [n, n+1). For a given integer n, find the range of x where both x/99 and x/101 are in [n, n+1), then count all valid positive integers across all possible n values.
<p><strong>Step 1:</strong> Let [x/99] = [x/101] = n for some non-negative integer n.</p><p>This means: n ≤ x/99 < n+1 AND n ≤ x/101 < n+1</p><p><strong>Step 2:</strong> From n ≤ x/99 < n+1: 99n ≤ x < 99(n+1)</p><p>From n ≤ x/101 < n+1: 101n ≤ x < 101(n+1)</p><p><strong>Step 3:</strong> Both conditions must hold simultaneously, so take the intersection:</p><p>max(99n, 101n) ≤ x < min(99(n+1), 101(n+1))</p><p>This gives: 101n ≤ x < 99(n+1) = 99n + 99</p><p><strong>Step 4:</strong> For valid x to exist: 101n < 99n + 99, which gives 2n < 99, so n ≤ 49</p><p><strong>Step 5:</strong> For each n from 0 to 49, the number of integers x is: (99n + 99) - 101n = 99 - 2n</p><p><strong>Step 6:</strong> Total count k = Σ(99 - 2n) for n = 0 to 49</p><p>= Σ99 - 2Σn = 99(50) - 2·(49·50/2) = 4950 - 2450 = 2500</p><p>∴ Answer: A (k = 2500)</p>
Correct Answer: A