Question

Find all three consecutive positive integers such that the first is divisible by 3, the second is divisible by 5 and the third is divisible by 53.

Answer 1

Let the required integers be n, n + 1, n + 2 given that n is divisible by 3 implies n is a multiple of 3 therefore n = 3a also given that n + 1 divisible by s and n + 2 divisible by 53 so we assume that n = 3a n = -1 + 56 n = -2 + 53 c n = 0 mod 3 n + 1 = 0 mod 5 n + 2 = 0 mod 53 \r\n n = 0 mod 3 n = -1 mod 5 n = -2 mod 53 n = 0 mod 3 n = 4 mod 5 n = 51 mod 53 Thus we get a system of 3 equation whose modulus are relatively prime to each other. so we can apply Chinese remainder theorem. m = 3 times 5 times 53 = 795 M_1 = 5 times 53 = 265 m_1 = 3 M_2 = 3 times 53 = 159 m_2 = 5 M_3 = 3 times 5 = 15 m_3 = 53