Using Legendre's formula: ⌊100/5⌋ + ⌊100/25⌋ + ⌊100/125⌋ = 20 + 4 + 0 = 24

Let numbers be a and b. a + b = 15 and a² + b² = 117. We know (a+b)² = a² + 2ab + b². So 225 = 117 + 2ab, thus 2ab = 108, ab = 54

By Fermat's Little Theorem, since 13 is prime and gcd(7,13)=1, we have 7^12 ≡ 1 (mod 13). Since 100 = 12×8 + 4, we calculate 7^4 = 2401 = 13×184 + 9, so 7^100 ≡ 7^4 ≡ 9 (mod 13). Answer should be C, but using theorem: 7^12 ≡ 1, so 7^100 = 7^96 × 7^4 ≡ 1 × 9 = 9

Let the number be x. According to the problem: 5x - 3 = 47. Therefore, 5x = 50, so x = 10.

12 = 2² × 3, 18 = 2 × 3², 24 = 2³ × 3. LCM = 2³ × 3² = 8 × 9 = 72.

We need to find x where x² = 169. Taking square root: x = √169 = 13.

We need a number of the form 8k + 5. Check: 29 = 8(3) + 5 = 24 + 5. ✓

Let the integers be x, x+1, x+2. Then x + (x+1) + (x+2) = 45. So 3x + 3 = 45, thus 3x = 42, x = 14.

144 = 2⁴ × 3². Odd divisors come only from 3² = (2+1) = 3 odd divisors: 1, 3, 9.

We know that GCD × LCM = Product of the two numbers. So 4 × LCM = 120. Therefore, LCM = 120/4 = 30.