Primetime!

$2$ is the only even prime number, the other primes are odd. But if $p=2$ then $p+2;p+6;p+8;p+12; p+14$ are obviously not prime. So the last digit of any prime number which is greater than $2$ must be $1,3, 5, 7$ or $9$ . If the last digit is $1$ , $p+14$ is divisible by $5$ . If the last digit of $p$ is $3$ , $p+2$ and $p+12$ are divisible by $5$ . If the last digit of $p$ is 5, $p$ is prime only if $p=5$ . $5, 7, 11, 13, 17, 19$ are all prime. $p=5$ is a solution. If the last digit of $p$ is $7$ , $p+8$ is divisible by $5$ and if the last digit is $9$ , $p+6$ is divisible by 5. The only solution is $\boxed{p=5}$

There is only one such number, namely p = 5.

We can find that, the required property does not hold for p < 5. For p = 5, we obtain primes 5, 7, 11, 13, 17, and 19.

We know that any number when divided by 5 yields 5 remainders :0,1,2,3 and 4. So there are 5 types of numbers greater than 5.

• If p > 5 and p = 5k with some positive integer k, then p is composite. If p = 5k+l, then p+14 is divisible by 5, hence ,composite.

• If p = 5k+2, then p+8 is divisible by 5, hence composite.

• If p = 5k+3, then 5/p+12 and p+12 is composite.

• Finally, if p = 5k+4, then 5/p+6, and p+6 is composite.

Friends , I did this question by hit and trial but can anybody suggest a better method for this question .

since p is prime, so from 1-10 it can be 3,5,7.(if p=2 then p+2 will not be prime).For any number above 10 if unit digit is:

1 $\rightarrow$ p+14 is divisible by 5

2 $\rightarrow$ p is divisible by 2

3 $\rightarrow$ p+2 is divisible by 5

4 $\rightarrow$ p is divisible by 2

5 $\rightarrow$ p is divisible by 5

6 $\rightarrow$ p is divisible by 2

7 $\rightarrow$ p+8 is divisible by 5

8 $\rightarrow$ p is divisible by 2

9 $\rightarrow$ p+6 is divisible by 5

0 $\rightarrow$ p is divisible by 2.

Therefore, only 5 is number which satisfy the conditions.

Answer: $\rightarrow$ $\boxed{5}$

Consider the set modulo 5. Note that no prime can be congruent to 0 mod 5 except for 5 itself.

If $p\equiv 1\pmod{5}, \implies p+14\equiv0\pmod{5}\text{ and }p+14>5$

If $p\equiv 2\pmod{5}, \implies p+8\equiv0\pmod{5}\text{ and }p+8>5$

If $p\equiv 3\pmod{5}, \implies p+12\equiv0\pmod{5}\text{ and }p+12>5$

If $p\equiv 4\pmod{5}, \implies p+6\equiv0\pmod{5}\text{ and }p+6>5$

Therefore, the only option left is $p\equiv0\pmod{5}$ , so $p=5$ .

only 5 will fit in the question.