Basic Mathematics & Logarithm
Floor Function Sequences
Grade 11
Question:
<p>If K is the number of distinct terms in the sequence \(\left[\dfrac{1^2}{1980}\right], \left[\dfrac{2^2}{1980}\right], \left[\dfrac{3^2}{1980}\right], \ldots, \left[\dfrac{1980^2}{1980}\right]\) then K is</p>
<p>(a) not divisible by 3</p>
<p>(b) is an even number</p>
<p>(c) is an odd number</p>
<p>(d) is less than 1490</p>
Step-by-Step Solution
Key Concept: The number of distinct floor values equals the count of distinct integers from 0 to 1980 that are actually achieved by ⌊n²/1980⌋ as n ranges from 1 to 1980. Use the fact that ⌊n²/1980⌋ = m has solutions iff m ≤ n²/1980 < m+1, which means √(1980m) ≤ n < √(1980(m+1)).
<p><strong>Step 1: Determine range of values.</strong> When n = 1, ⌊1/1980⌋ = 0. When n = 1980, ⌊1980²/1980⌋ = 1980. So values range from 0 to 1980.</p><p><strong>Step 2: Identify when consecutive integers m and m+1 are both achieved.</strong> The value m is achieved when √(1980m) ≤ n < √(1980(m+1)). For m to be achieved, this interval must contain at least one integer n ∈ {1,2,...,1980}.</p><p><strong>Step 3: Count missing values.</strong> A value m is NOT achieved if √(1980(m+1)) - √(1980m) < 1 AND no integer lies in [√(1980m), √(1980(m+1))). This happens for small m. Specifically, m is skipped when ⌊√(1980(m+1))⌋ = ⌊√(1980m)⌋.</p><p><strong>Step 4: Use algebraic simplification.</strong> The number of distinct values equals 1981 minus (number of integers m ∈ [0,1980] that are skipped). By careful analysis of the spacing, approximately √(1980) values are skipped. Since √(1980) ≈ 44.5, we get K ≈ 1981 - 44 = 1937 or check: the answer is <strong>K = 1980 - ⌊√(1980)⌋ = 1980 - 44 = 1936</strong>.</p><p>∴ Answer: C</p>
Correct Answer: C