Three Primes

Note modulo $3$ , $p+10\equiv p+1, p+14\equiv p+2$ . Therefore, one of $p,p+10,p+14$ is divisible by $3$ . Since each number is prime, we conclude one of $p, p+10,p+14$ is equal to $3$ . Since $p$ is positive, we must have $p=3$ .

We easily verify that $p=3$ does in fact yield $3$ prime numbers (namely $3,13,17$ ). Therefore, there is one such value of $p$ , so the answer is $\boxed{1}$ .

As p+10 and p+14 are prime numbers,they cannot be divisible by 3,so the sum of their digits cannot be a multiple of 3,so p cannot has his sum equal to 2 otherwise the sum of digits of p+10 would be a multiple of 3,and its sum of digits cannot be neither 2,5,8,11... P cannot has the sum of digits equal to 1 otherwise p +14 would be a multiple of 3.p cannot has the sum equal to 1,4,7,10... Therefore the sum of digits of p can only be 3,6,9....but if the sum of digits of p is equal to such a multiple of 3 p wouldn't be prime,except if p is 3 Therefore p can only be 3 Sorry by my english errors,i am br,hu3hu3 gibe money and points