Using Legendre's formula: ⌊27/3⌋ + ⌊27/9⌋ + ⌊27/27⌋ = 9 + 3 + 1 = 13.
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.
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).
To find the sum of all odd divisors of 120, first extract the odd part by removing all factors of 2, then find the sum of divisors of that odd part.
Step 1: Prime factorization of 120
Step 2: Identify the odd part
The odd divisors of 120 come only from the odd prime factors (ignoring all powers of 2):
Step 3: Find all divisors of 15
The divisors of 15 are: \(1, 3, 5, 15\)
These are exactly the odd divisors of 120 (since multiplying any of these by powers of 2 gives even divisors).
Step 4: Calculate the sum
Alternatively, using the divisor sum formula for \(3^1 \times 5^1\):
Answer: The sum of all odd divisors of 120 is \(24\) (Option C)
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.
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
We need LCM(6, 8, 9). 6 = 2 × 3, 8 = 2³, 9 = 3². LCM = 2³ × 3² = 8 × 9 = 72. Smallest multiple of 72 greater than 100: 72 × 2 = 144.
Divisors of 20: 1, 2, 4, 5, 10, 20. Proper divisors (excluding 20): 1, 2, 4, 5, 10. Sum = 1 + 2 + 4 + 5 + 10 = 22, not 48.
