Let numbers be 3k and 5k where HCF(3k, 5k) = k = 4. Numbers are 12 and 20. Sum = 32.

LCM(7, 11, 13) = 1001. Next multiple is 1001. 1001 - 1000 = 1. Actually 1000 ÷ 1001 remainder = 1000. Need 1001 - 1000 = 1. Recheck: 1000 mod 1001 = 1000, so add 1 gives 1001. Add 12 gives 1012 = 1001 + 11.

For divisibility by 11: alternate sum of digits. 12321: (1+3+1) - (2+2) = 5 - 4 = 1. Recheck: (1+3+1) - (2+2) = 5-4=1. Actually 1-2+3-2+1 = 1. Try: 1-2+3-2+1 = 1. Check: 12321/11 = 1120.09... Correct: (2+2) - (1+3+1) = 4-5 = -1, still divisible.

By Wilson's theorem, (p-1)! ≡ -1 (mod p) for prime p. So 16! ≡ -1 (mod 17). 16! = 13! × 14 × 15 × 16. -1 ≡ 13! × 14 × 15 × 16 (mod 17).

Let number be x. x + 1/x = 2.5. Multiply by x: x² - 2.5x + 1 = 0. x = (2.5 ± √(6.25-4))/2 = (2.5 ± 1.5)/2. x = 2 or 0.5.

6^x ÷ 216 = 6. 216 = 6³. So 6^x ÷ 6³ = 6. Therefore 6^(x-3) = 6¹. Thus x - 3 = 1, so x = 4.

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