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Since $p$ is a prime number greater than $3$ , none of the numbers $p$ , $p+2$ and $p+4$ are divisible by 3.

Let $p \equiv a \mod 3$ , where $a=1$ or $a=2$ . Then $(p+2) \equiv (a+2) \mod 3$ and $(p+4) \equiv (a+1) \mod 3$ .

But, if $a=1$ then $(p+2)$ would then be divisible by 3; and if $a=2$ then $(p+4)$ would be divisible by 3.

Thus, there is no such set of primes numbers satisfying the above property.

Hence, the required answer is $\boxed{0}$

When $p$ is prime, $p^2-1$ is divisible by $24$ . When $p+4$ is prime, $(p+4)^2-1$ or $p^2+8p+15$ is divisible by $24$ . Hence $p+2$ is divisible by $3$ (by subtracting the first from the second). Hence $p+2$ is not prime. Similar argument can be given for other two also.

$p$ is greater than 3. So $p+1$ must be divisible by 3 (because one of 3 consecutive numbers must be divisible by 3 and $p$ , $p+2$ are not).

Hence $p+4=(p+1)+3$ must be divisible by 3.

No such numbers satisfy the consitions from the problem.