<p>Let the sum be N and <i>x</i> = [<i>x</i>] + {<i>x</i>}. The number of values of N not greater than 1000 such that N = 20I + [2f] + [4f] + [6f] + [8f] ≤ 1000, where I ∈ Z₀⁺ and 0 ≤ f < 1, is:</p>
Step-by-Step Solution
Key Concept: Recognize that [2f] + [4f] + [6f] + [8f] depends only on the fractional part f, and this sum takes specific integer values based on which unit intervals f falls into. The constraint N = 20I + S(f) ≤ 1000 means we count valid pairs (I, f) where S(f) is the step function value.
<p><strong>Step 1:</strong> Analyze S(f) = [2f] + [4f] + [6f] + [8f] for f ∈ [0,1).</p><p>The function jumps at f = 0, 1/8, 1/6, 1/4, 1/3, 3/8, 1/2, 5/8, 2/3, 3/4, 5/6, 7/8.</p><p><strong>Step 2:</strong> Compute S(f) values in each subinterval:</p><ul><li>f ∈ [0, 1/8): S(f) = 0</li><li>f ∈ [1/8, 1/6): S(f) = 1</li><li>f ∈ [1/6, 1/4): S(f) = 2</li><li>f ∈ [1/4, 1/3): S(f) = 3</li><li>f ∈ [1/3, 3/8): S(f) = 4</li><li>f ∈ [3/8, 1/2): S(f) = 5</li><li>f ∈ [1/2, 5/8): S(f) = 6</li><li>f ∈ [5/8, 2/3): S(f) = 7</li><li>f ∈ [2/3, 3/4): S(f) = 8</li><li>f ∈ [3/4, 5/6): S(f) = 9</li><li>f ∈ [5/6, 7/8): S(f) = 9</li><li>f ∈ [7/8, 1): S(f) = 10</li></ul><p><strong>Step 3:</strong> S(f) takes values {0,1,2,3,4,5,6,7,8,9,10} with S(f)=9 achieved twice.</p><p><strong>Step 4:</strong> For N = 20I + S(f) ≤ 1000: when I = 50, S(f) ≤ 0 (only S(f)=0 works); when I ≤ 49, all 11 values of S(f) work.</p><p><strong>Step 5:</strong> Total valid N values = 50 × 11 + 1 = <strong>551</strong></p><p>∴ Answer: A,D</p>
Correct Answer: A,D