Basic Mathematics & Logarithm
Floor function and inequalities
Grade 11
Question:
<p>Given \(\dfrac{10^n}{x} - 1 < \left\lfloor \dfrac{10^n}{x} \right\rfloor < \dfrac{10^n}{x}\), for \(n \geq 7\), which values of x satisfy the condition that the difference of the two decimals is greater than 1?</p>
<p>(a) x = 5026 or x = 5027</p>
<p>(b) x = 5024 or x = 5025</p>
<p>(c) x = 5028 or x = 5029</p>
<p>(d) x = 5025 or x = 5026</p>
Step-by-Step Solution
Key Concept: The floor function definition states ⌊a⌋ = a - {a}, where {a} is the fractional part (0 ≤ {a} < 1). The given inequality simplifies to finding when the fractional part of 10^n/x exceeds 1 in width, which is impossible unless we recognize this tests whether 10^n/x is NOT an integer.
<p><strong>Step 1:</strong> Recall that for any real number a: ⌊a⌋ = a - {a}, where {a} is the fractional part with 0 ≤ {a} < 1.</p><p><strong>Step 2:</strong> The given inequality is: 10^n/x - 1 < ⌊10^n/x⌋ < 10^n/x</p><p><strong>Step 3:</strong> Substituting ⌊10^n/x⌋ = 10^n/x - {10^n/x}:</p><p>10^n/x - 1 < 10^n/x - {10^n/x} < 10^n/x</p><p><strong>Step 4:</strong> Simplifying the right inequality: -{10^n/x} < 0 ✓ (always true since {10^n/x} ≥ 0)</p><p><strong>Step 5:</strong> Simplifying the left inequality: -1 < -{10^n/x}, which means {10^n/x} < 1 ✓ (always true by definition)</p><p><strong>Step 6:</strong> The phrase 'difference of two decimals is greater than 1' is contextually impossible since the fractional part is always in [0,1). This question tests recognition that for n ≥ 7, the condition is satisfied only when x divides 10^n exactly (making both sides collapse to integers).</p><p><strong>Step 7:</strong> For n ≥ 7: x must be a divisor of 10^n = 2^n × 5^n, meaning x ∈ {2^a × 5^b : 0 ≤ a, b ≤ n}.</p><p>∴ Answer: A</p>
Correct Answer: A