Numbers divisible by 3: floor(100/3) = 33. Numbers divisible by both 3 and 5 (i.e., by 15): floor(100/15) = 6. Numbers divisible by 3 but not by 5 = 33 - 6 = 27. Wait, that's option B. Let me verify: 27 is correct.

3^1 ≡ 3, 3^2 ≡ 9, 3^3 ≡ 27 ≡ 5, 3^4 ≡ 15 ≡ 4, 3^5 ≡ 12 ≡ 1 (mod 11). Wait: 3×4 = 12 ≡ 1. So 3^5 ≡ 1 (mod 11). Answer should be A.

HCF takes minimum power of each prime: HCF = 2^min(4,3) × 3^min(3,2) × 5^min(1,2) = 2^3 × 3^2 × 5.

First multiple of 7 ≥ 50: 56 (7×8). Last multiple of 7 ≤ 150: 147 (7×21). Count = 21 - 8 + 1 = 14. Hmm, should be 14 not 15. Let me verify: 56, 63, 70, 77, 84, 91, 98, 105, 112, 119, 126, 133, 140, 147. That's 14 numbers.

Using Legendre's formula: floor(100/5) + floor(100/25) + floor(100/125) = 20 + 4 + 0 = 24.

By Fermat's Little Theorem, since 13 is prime and gcd(5,13)=1, we have 5^12 ≡ 1 (mod 13). 100 = 12×8 + 4. So 5^100 ≡ 5^4 (mod 13). 5^4 = 625 = 48×13 + 1 ≡ 1 (mod 13).

Let numbers be 3k and 5k. Since gcd(3,5)=1, LCM = 3k×5k/1 = 15k. Given LCM = 150, so 15k = 150, k = 10. Numbers are 30 and 50.

2^50 = 1,125,899,906,842,624. Sum of digits = 1+1+2+5+8+9+9+9+0+6+8+4+2+6+2+4 = 76. (Note: This requires calculation; the answer provided may vary based on computation.)

Using divisibility rule for 11: alternating sum of digits must be divisible by 11. For 121: (1-2+1) = 0, which is divisible by 11. Verification: 121 ÷ 11 = 11.

Smallest 4-digit number is 1000. For divisibility by 18, number must be divisible by both 2 and 9. 1000 ÷ 18 = 55.55... Next: 1008 ÷ 18 = 56. So 1008 is the answer.