Number of divisors = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24.

20 = 2² × 5. Divisors are: 1, 2, 4, 5, 10, 20. Sum = 1 + 2 + 4 + 5 + 10 + 20 = 42.

For n ≡ 3 (mod 7): possible numbers are 3, 10, 17, 24, 31, 38, 45, 52, 59... For n ≡ 5 (mod 11): possible numbers are 5, 16, 27, 38, 49, 60... Common number is 58. Check: 58 = 7(8) + 2... Let me recheck: 58/7 = 8 rem 2, not 3. Try 38: 38/7 = 5 rem 3 ✓, 38/11 = 3 rem 5 ✓. Answer is A=38.

We need LCM(6, 8, 9). 6 = 2 × 3, 8 = 2³, 9 = 3². LCM = 2³ × 3² = 8 × 9 = 72. Smallest multiple of 72 greater than 100: 72 × 2 = 144.

Divisors of 20: 1, 2, 4, 5, 10, 20. Proper divisors (excluding 20): 1, 2, 4, 5, 10. Sum = 1 + 2 + 4 + 5 + 10 = 22, not 48.

We find the pattern of powers of 2 mod 7: 2^1≡2, 2^2≡4, 2^3≡1 (mod 7). The cycle repeats every 3 terms. Since 100 = 33×3 + 1, we have 2^100 ≡ 2^1 ≡ 2 (mod 7).

Using Euclidean algorithm: 144 = 96×1 + 48, 96 = 48×2 + 0. Therefore HCF = 48. Alternatively, 144 = 2^4×3^2 and 96 = 2^5×3. HCF = 2^4×3 = 48.

We need LCM(6, 8). 6 = 2×3, 8 = 2³. LCM = 2³×3 = 24. Therefore, 24 is the smallest number divisible by both 6 and 8.

Unit digits of powers of 7 follow pattern: 7^1→7, 7^2→9, 7^3→3, 7^4→1, 7^5→7. Cycle = 4. Since 2019 = 504×4 + 3, unit digit of 7^2019 = unit digit of 7^3 = 3. Wait, let me recalculate: 7^3 = 343 (unit 3), but the cycle shows 7,9,3,1. For 2019 mod 4 = 3, so 7^3 has unit digit 3. Actually checking: option answer is B(7), but calculation shows 3. There may be a typo in options.

Using the property: HCF × LCM = Product of two numbers. 6 × 60 = 12 × x. 360 = 12x. x = 30.