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.

Using the division algorithm: Number = (Divisor × Quotient) + Remainder. Number = (15 × 23) + 8 = 345 + 8 = 353.

Using the property: GCD(a,b) × LCM(a,b) = a × b. Therefore: 12 × 144 = 36 × b. So 1728 = 36b, which gives b = 48.

Let number be 10a + b. Reversed number is 10b + a. Given: (10b + a) - (10a + b) = 27, so 9b - 9a = 27, thus b - a = 3. Also a + b = 9. Solving: b = 6, a = 3. Original number = 36.

Let three consecutive odd numbers be x, x+2, x+4. Their sum: x + (x+2) + (x+4) = 51. So 3x + 6 = 51, thus 3x = 45, and x = 15.