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 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
Using Legendre's formula: ⌊100/5⌋ + ⌊100/25⌋ + ⌊100/125⌋ = 20 + 4 + 0 = 24
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.
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}.
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).
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.
Using Legendre's formula: ⌊27/3⌋ + ⌊27/9⌋ + ⌊27/27⌋ = 9 + 3 + 1 = 13.
Numbers are 7, 14, 21, 28. 7 = 7, 14 = 2 × 7, 21 = 3 × 7, 28 = 2² × 7. LCM = 2² × 3 × 7 = 4 × 3 × 7 × 2 = 1680.
Required number divides (247-7), (328-7), (409-7) = 240, 321, 402. HCF(240, 321, 402) = 120.