If n is a natural number, N + 3 and N + 7 are prime numbers, find the remainder of N divided by 3

If n is a natural number, N + 3 and N + 7 are prime numbers, find the remainder of N divided by 3

(1) if the remainder is 0, that is, n = 3K (k is a non negative integer, the same below), then n + 3 = 3K + 3 = 3 (K + 1), so 3|n + 3 and 3 ≠ n + 3, so n + 3 is not a prime number, which contradicts the proposition. (2) if the remainder is 2, and N = 3K + 2, then n + 7 = 3K + 2 + 7 = 3 (K + 3), so 3|n + 7 and N + 7 are not prime numbers, which contradicts the proposition It can only be 1

If n is a natural number, N + 3 and N + 7 are prime numbers, find the remainder of N divided by 3

If the remainder is 0, that is, n = 3K (k is a non negative integer, the same below), then n + 3 = 3K + 3 = 3 (K + 1), so 3|n + 3, and 3 ≠ n + 3, so n + 3 is not a prime number, which is contradictory to the problem

If natural numbers n + 3 and N + 7 are prime numbers, find the remainder of N divided by 6

Let's divide n into six categories, n = 6K, n = 6K + 1 When n = 6K, N + 3 = 6K + 3 (2k + 1) and N + 3 are prime contradictions; when n = 6K + 1, N + 3 = 6K + 4 = 2 (3K + 2) and N + 3 are prime contradictions; when n = 6K + 2, N + 7 = 6K + 9 = 3 (2k + 3) and N + 7 are prime contradictions; when n = 6K + 3, N + 3 = 6K + 6 = 6 (K + 1) and N + 3 are prime contradictions; when n = 6K + 5, N + 7 = 6K + 12 = 6 (K + 2) and N + 7 are prime contradictions. So only n = 6K + 4, that is n division The remainder of 6 is 4