I love prime numbers! 3

It is easy enough to spot that $p=5$ works upon substituting the respective values of the first few odd numbers. This is because

$5, 7, 11, 13, 17, 19$

are all prime numbers.

However, how can we make sure that there isn't any other possibilities?

* *here's what I did **

Well, let's consider the number $p$ itself. It is evident that $p$ must end with an odd number, or else some (if not all) the values $p, p+2, p+6, p+8, p+12, p+14$ will not be prime.

So $p$ must either end on $1, 3, 5, 7$ or $9$ .

Let's assume $p$ ends with $1$ . Then, the values of $p+2, p+6, p+8, p+12, p+14$ will end with

$3, 7, 9, 3, 5$

If $p$ ends with $3$ , then the values of $p+2, p+6, p+8, p+12, p+14$ will end with

$5, 9, 1, 5, 7$

Out of all the cases, only when $p$ ends with $5$ , the values of $p+2, p+6, p+8, p+12, p+14$ doesn't end with a $5$ . If a number ends with the digit $5$ , it automatically becomes a multiple of $5$ . Since none of the numbers $p+2, p+6, p+8, p+12, p+14$ which ends with $5$ are the number $5$ itself (they will be too big!) , we can conclude that $p$ must end with the digit $5$ .

However, if $p$ is not a one-digit number, then $p$ itself will be a multiple of $5$ too, therefore the only possibility is when $p = 5$ .

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Other prime number problems by me:

I love prime numbers! 1

I love prime numbers! 2

It is easy to see that $p =5$ works. To show uniqueness, let's consider the remainders when the given set of numbers are divided by $5$ . The set of remainders are $\{ p \mod 5, p+2 \mod 5, p+6 \mod 5, p+8 \mod 5, p+12 \mod 12, p+14 \mod 14\}$ which is the same as the following set $\{ p \mod 5, p+1 \mod 5, p +2 \mod 5, p+3 \mod 5, p+4 \mod 5 \} = \{0,1,2,3,4\}.$ Hence, there must be at least one number which must be divisible by $5$ . Since the given numbers are given to be all primes, either $p$ must be equal to $5$ or $p+2$ must be equal to $5$ (the rest of the numbers are trivially larger than $5$ ). But $p+2=5$ implies $p+6=9$ which is not a prime. Thus the only possibility is that $p=5$ .

What a solution.....😊😊