<p>Find the minimum natural number n, such that the equation \(\left[\dfrac{10^n}{x}\right] = 1989\) has integer solution x.</p>
Step-by-Step Solution
Key Concept: For [10^n/x] = 1989 to have an integer solution, we need 1989 ≤ 10^n/x < 1990, which means 10^n/1990 < x ≤ 10^n/1989. An integer x exists in this interval if and only if the interval length exceeds 1, or equivalently, when 10^n/1989 - 10^n/1990 ≥ 1.
<p><strong>Step 1:</strong> From [10^n/x] = 1989, we have 1989 ≤ 10^n/x < 1990</p><p><strong>Step 2:</strong> Rearranging: 10^n/1990 < x ≤ 10^n/1989</p><p><strong>Step 3:</strong> For an integer x to exist in interval (10^n/1990, 10^n/1989], the interval must contain at least one integer. The interval length is:</p><p>10^n/1989 - 10^n/1990 = 10^n(1/1989 - 1/1990) = 10^n · 1/(1989·1990)</p><p><strong>Step 4:</strong> For guaranteed integer existence, we need: 10^n/(1989·1990) ≥ 1</p><p>Therefore: 10^n ≥ 1989·1990 = 3,948,110</p><p><strong>Step 5:</strong> Since 10^6 = 1,000,000 < 3,948,110 and 10^7 = 10,000,000 > 3,948,110</p><p>The minimum value is n = 7</p><p><strong>Verification:</strong> When n = 7, x can be chosen in (10^7/1990, 10^7/1989] = (5025.126..., 5030.015...], which contains integers 5026, 5027, 5028, 5029, 5030</p><p>∴ Answer: A (n = 7)</p>
Correct Answer: A