JEE Advanced Maths: Number Theory / Sequences questions with solutions

Q1. Consider the following numbers: a = 111...1 (20 digits), b = 444...4 (20 digits), c = 777...7 (10 digits), d = 999...9 (10 digits). Evaluate (441*a - 81*c²) / (14*d).

1. 3
2. 5
3. 6
4. 7

Answer: 7

Using repunit notation: a = R₂₀ = (10²⁰-1)/9, c = 7*R₁₀, d = 9*R₁₀. Numerator: 441*a - 81*c² = 441*(10²⁰-1)/9 - 81*49*(R₁₀)² = 49*(10²⁰-1) - 3969*(R₁₀)². Since 10²⁰-1 = 9*(R₂₀) and R₂₀ = R₁₀*(10¹⁰+1), numerator = 441*R₁₀*(10¹⁰+1) - 81*49*(R₁₀)² = R₁₀*[441*(10¹⁰+1) - 3969*R₁₀]. With R₁₀ = (10¹⁰-1)/9: 3969*R₁₀ = 441*(10¹⁰-1), so numerator = R₁₀*[441*10¹⁰+441-441*10¹⁰+441] = R₁₀*882. Denominator: 14*d = 14*9*R₁₀ = 126*R₁₀. Result = 882/126 = 7. Wait — re-evaluating: 441*a - 81*c² where a has 20 digits and c has 10 digits. a=(10²⁰-1)/9, c=7*(10¹⁰-1)/9. 441*a=441*(10²⁰-1)/9=49*(10²⁰-1). 81*c²=81*49*(10¹⁰-1)²/81=49*(10¹⁰-1)². Numerator=49*[(10²⁰-1)-(10¹⁰-1)²]=49*[10²⁰-1-10²⁰+2*10¹⁰-1]=49*(2*10¹⁰-2)=98*(10¹⁰-1). Denominator=14*9*(10¹⁰-1)/9=14*(10¹⁰-1). Result=98*(10¹⁰-1)/(14*(10¹⁰-1))=7.