Six consecutive prime numbers have sum $p$. Given that $p$ is also prime, determine all possible values of $p$.
We consider two cases.

- If 2 is one of the six prime numbers, then the only possible sum is $p=2+3+5+7+11+13=41$, which is indeed a prime.
- If 2 is not one of the six prime numbers, then all six primes are odd, so their sum must be even. Moreover, $p>2$. Therefore $p$ is not prime.

So, the only possible value of $p$ is $p=41$.

Remark. In any problem which asks to "determine all", "find all', et cetera, the solution always requires two claims: first, one must show that all claimed values are achievable (here, we needed to show $p=41$ actually occurs), and secondly, to prove that the claimed values are the only ones possible (here, we needed to show no other $p$ could occur).