To find 𝑥 we solve the system of congruences. Starting with x=3k+2, we substitute into the second congruence 3k+2 ≡ 3(mod5), giving 3𝑘 ≡ 1(mod5).
Solving for 𝑘, we find 𝑘 ≡ 2(mod5), so 𝑘 = 5m+2. Substituting back, 𝑥 = 3(5m+2)+2=15m+8, and the smallest positive 𝑥 is 8.
The prime factorization of 360 is 22 ⋅ 32 ⋅ 5 The number of divisors is (3+1)(2+1)(1+1)=4×3×2=24.
To find the GCD, we can use the Euclidean algorithm. We divide 252 by 198, then 198 by the remainder, and repeat until the remainder is 0. The GCD is the last non-zero remainder.
We need to solve the system of congruences. Let 𝑛 = 4k+1. Substituting into the second congruence, 4k+1≡2 (mod5), we get 4𝑘 ≡ 1 (mod5). Since 4 and 5 are coprime, we find 𝑘 (mod5), so 𝑘 = 5m+4.
Substituting into the third congruence, 𝑛 = 4(5m+4)+1=20m+17, we solve 20m+17 ≡ 3(mod6), giving 20m ≡ −14 ≡ 4(mod6). Thus 2m ≡ 2(mod3), so 𝑚 ≡ 1 (mod3). The smallest positive 𝑚 is 1, so 𝑘 = 5(3p+1)+4. Substituting back, we get 𝑛 = 20×3×p+17, and the smallest positive 𝑛 is 58.
Relatively prime numbers have a GCD of 1, meaning they share no common factors other than 1.
