<p>If 'K' is the total number of integers n between 1 and 10000 (both inclusive) such that n is divisible by \([\sqrt{n}]\), then Here [] is G.I.F</p>
<p>(a) K is more than 198</p>
<p>(b) K is an even no.</p>
<p>(c) K is less than 300</p>
<p>(d) K is more than 200</p>
Step-by-Step Solution
Key Concept: For each value of k = [√n], find all integers n in the range [k², (k+1)²) where k divides n. Count all such n up to 10000 using the constraint that if [√n] = k, then k² ≤ n < (k+1)².
<p><strong>Step 1:</strong> If [√n] = k, then k² ≤ n < (k+1)². For n to be divisible by [√n], we need k | n.</p><p><strong>Step 2:</strong> For each k, count multiples of k in the range [k², min((k+1)² - 1, 10000)]. The multiples are k·⌈k²/k⌉, k·⌈k²/k⌉ + k, ..., k·⌊min((k+1)² - 1, 10000)/k⌋.</p><p><strong>Step 3:</strong> Since k² ≤ n, the smallest multiple is k·k = k². For k from 1 to 99, the range is [k², (k+1)² - 1], giving (k+1)² - k² = 2k + 1 multiples per interval.</p><p><strong>Step 4:</strong> Sum for k = 1 to 99: Σ(2k + 1) = 2·(99·100/2) + 99 = 9900 + 99 = 9999.</p><p><strong>Step 5:</strong> For k = 100: [√n] = 100 when 10000 ≤ n ≤ 10000 (only n = 10000). Since 100 | 10000, add 1.</p><p><strong>Step 6:</strong> K = 9999 + 1 = 10000.</p><p>∴ Answer: A</p>
Correct Answer: A