Numbers divisible by both 6 and 9 must be divisible by LCM(6,9) = 18. Between 1-100: 18, 36, 54, 72, 90. Count = 5.
Digit sum = 9 + 9 + 9 + 9 = 36. Note: A number and its digit sum have the same remainder when divided by 9.
Let numbers be a and b. (a+b)² - 4ab = (a-b)². So (a-b)² = 400 - 384 = 16, thus |a-b| = 4.
Using Legendre's formula: ⌊27/3⌋ + ⌊27/9⌋ + ⌊27/27⌋ = 9 + 3 + 1 = 13.
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.