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All even numbers greater than 2 are composite because they are all divisible by 2. Therefore, from 3 onwards, all prime numbers are odd.

Consider three consecutive odd numbers. They must be of the form 2n+1, 2n+3 and 2n+5 where n is an integer.

Since n is an integer, n will leave a remainder of 0, 1 or 2 when it is divided by 3.

Suppose n leaves a remainder of 0 when divided by 3. Therefore n = 3k for some integer k. Then 2n+3 = 2*(3k) + 3 = 6k + 3 = 3*(2k+1). That is, middle of the three consecutive odd numbers is divisible by 3 and so it is not a prime.

Now, suppose n leaves a remainder of 1 when divided by 3. Therefore n = 3k+1 for some integer k. Then 2n+1 = 2*(3k+1) + 1 = 6k+2+1 = 6k+3 = 3*(2k+1). That is, first of the three consecutive odd numbers is divisible by 3 and so it is not a prime.

Finally, suppose n leaves a remainder of 2 when divided by 3. Therefore n = 3k+2 for some integer k. Then 2n+5 = 2*(3k+2) + 5 = 6k+4+5 = 6k+9 = 3*(2k+3). That is, last of the three consecutive odd numbers is divisible by 3 and so it is not a prime.

Thus for any three consecutive odd numbers greater than 3, one of them is divisible by 3 and therefore the three cannot all be prime.

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