Is there at least one prime number in 20 consecutive natural numbers right Don't forget to say the reason
This is wrong
Let 21! = 1 * 2 * 3 *... * 20 * 21, then 20 consecutive natural numbers 21! + 2,21! + 3,... 21! + 21 are not prime numbers: for example, 21! + 3 = 3 * (1 * 2 * 4 * 5 *... * 21) + 3 = 3 * (1 * 2 * 4 * 5 *... * 21 + 1). The same method can prove that for any n, all consecutive n natural numbers are not prime numbers
What is the difference between even numbers that are not prime numbers in natural numbers 1 to 20
In natural numbers 1 to 20, even numbers that are not prime numbers are 4,6,8,10,12,14,16,18,20
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 in contradiction with 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 is in contradiction with the proposition The number can only be 1