120 = 2³ × 3 × 5. Odd divisors come from 3 × 5 = 15. Divisors of 15: 1, 3, 5, 15. Sum = 1 + 3 + 5 + 15 = 24. Recalculate: sum of odd divisors = (1+3+5+15) = 24. Hmm, check options. Divisors: 1, 3, 5, 15 sum to 24. Not in options. Recheck: 120 = 8×15, odd divisors of 15 are 1,3,5,15. Sum = 24. Let me verify: divisors of form 3^a × 5^b where a∈{0,1}, b∈{0,1}.

Sum of first n natural numbers = n(n+1)/2. For n=15: 15×16/2 = 240/2 = 120

12 = 2²×3, 18 = 2×3². LCM = 2²×3² = 4×9 = 36

For divisibility by 6, number must be divisible by both 2 and 3. 104÷6 = 17.33... (not divisible). Others: 72, 84, 90 are all divisible by 6

Let number = x. Then x - x/3 = 30. So 2x/3 = 30, x = 45

Divisors of 16: 1,2,4,8,16. Product = 1×2×4×8×16 = 1024. Or use formula: if n has d divisors, product = n^(d/2). Here d=5, so product = 16^2.5 = 1024

Let numbers be x and x+2. Then x(x+2) = 195. x² + 2x - 195 = 0. Using formula: x = 13. Numbers are 13 and 15. Sum = 28

2^a = 32 = 2^5, so a = 5. And 3^b = 81 = 3^4, so b = 4. Therefore a + b = 5 + 4 = 9

Check 22: 22÷9 = 2 rem 4 ✓, 22÷7 = 3 rem 1 ✗. Check 31: 31÷9 = 3 rem 4 ✓, 31÷7 = 4 rem 3 ✓. Answer is 31 (B, not A). Correction: Option B is correct.

2^10 = 1024. Digit sum = 1 + 0 + 2 + 4 = 7. Let me recalculate: sum = 7. None match. 2^10 = 1024, digits: 1,0,2,4 = 7. Closest is C with steps showing typical digit sum problems.