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sqrt(111...1 (200 digits) - 222...2 (100 digits)) equals which of the following?

1. sqrt(1313...13) (100-digit number with repeating 13)
2. sqrt(33...3) (100 digits)
3. sqrt(2323...23) (100-digit number with repeating 23)
4. sqrt(333...3) (100 digits)

Correct answer: sqrt(333...3) (100 digits)

Solution

Let Rₙ denote the repunit with n ones = (10ⁿ-1)/9. R₂₀₀ = (10²⁰⁰-1)/9 = (10¹⁰⁰-1)(10¹⁰⁰+1)/9 = 9*R₁₀₀*(10¹⁰⁰+1)/9... Let me compute carefully: R₂₀₀ = (10²⁰⁰-1)/9. 2*R₁₀₀ = 2*(10¹⁰⁰-1)/9. R₂₀₀ - 2*R₁₀₀ = [(10²⁰⁰-1) - 2*(10¹⁰⁰-1)]/9 = [10²⁰⁰ - 1 - 2*10¹⁰⁰ + 2]/9 = [10²⁰⁰ - 2*10¹⁰⁰ + 1]/9 = [(10¹⁰⁰-1)²]/9 = [(10¹⁰⁰-1)/3]² = [3*R₁₀₀]². Wait: (10¹⁰⁰-1)/3 = 3*R₁₀₀ only if 3|R₁₀₀. (10¹⁰⁰-1)/9 = R₁₀₀, so (10¹⁰⁰-1) = 9*R₁₀₀. (10¹⁰⁰-1)/3 = 3*R₁₀₀. So R₂₀₀ - 2*R₁₀₀ = (3*R₁₀₀)² = 9*(R₁₀₀)². Therefore sqrt(R₂₀₀ - 2*R₁₀₀) = 3*R₁₀₀ = 333...3 (100 digits).

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