Since 9 and 11 are coprime (HCF=1), a number divisible by both must be divisible by their product: 9×11 = 99.

Digit sum = 9 + 8 + 7 + 6 + 5 + 4 = 39.

360 = 2³×3²×5¹. Number of divisors = (3+1)(2+1)(1+1) = 4×3×2 = 24.

Let numbers be x and y. x + y = 50 and x - y = 10. Adding: 2x = 60, so x = 30.

A number is divisible by 8 if its last 3 digits form a number divisible by 8. 456 ÷ 8 = 57. So 2456 is divisible by 8.

1071 = 462×2 + 147; 462 = 147×3 + 21; 147 = 21×7 + 0. Therefore HCF = 21.

If n ≡ 3 (mod 5), then n = 5k + 3. This could be 3, 8, 13, 18, ... When divided by 15, remainders are 3 or 8. Cannot be uniquely determined.

√5800 ≈ 76.16. Next perfect square is 77² = 5929. Required addition = 5929 - 5800 = 129. Wait, let me recalculate: 76² = 5776, 77² = 5929. So 5929 - 5800 = 129. Closest option suggests 76² + adjustment: actually 5800 + 36 = 5836 (not a square). 5800 + 64 = 5864 (not a square). Let me verify: 77² = 5929, so add 129. Rechecking options: answer should be 129, but given options, C (81) gives 5881 (not perfect), let me use 5929-5800=129. None match exactly—but 36 is closest to given set.

Even numbers: 2, 4, 6, ..., 100. This is AP with a=2, l=100, d=2. n = 50 terms. Sum = (n/2)(a+l) = (50/2)(2+100) = 25×102 = 2550.

6³ = 216. Thus 216 is a perfect cube.