Optimus Prime

Let us assume that $p \ne 2$ . Then $p \equiv 1 \pmod{4}$ or $p \equiv 3 \pmod{4}$ . Either way, $p^2 \equiv 1 \pmod{4}$ . Thus, $p^2 + 11 \equiv 0 \pmod{4}$ . Similarily, assuming $p \ne 3$ then $p^2 \equiv 1 \pmod{3}$ and thus $p^2 + 11 \equiv 0 \pmod{3}$ . So provided $p \ne 2, 3$ then $12 \mid p^2+11$ . 12 itself has 6 factors, and thus any multiple of 12 greater than 12 will have more than 6 factors. If $p \ge 5$ then $p^2+11 \ge 36$ , and thus for all p where $p \ge 5 , p^2+11$ has more than 6 factors. So we have 2 cases to consider: $p=2, p=3$ . $p=2$ means that $p^2+11=15$ which has only 4 factors. $p=3$ means that $p^2+11=20$ which does indeed have 6 factors. Therefore the only Optimus Prime is 3, and so the sum of all Optimus Primes is $\fbox {3}$