Help please?

$3(6x+1) \equiv 4 \pmod{p}\\ 18x + 3 \equiv 4 \pmod{p}\\ 18x \equiv 1 \pmod{p}\\ 3^3 \cdot 2 x \equiv 1 \mod{p}\\ \text{It should be clear any integer }x \text{ will cause }\\ 18x \equiv 0 \pmod{2} \text{ and }18x \equiv 0 \pmod{3} \\ \text{there will be an integer solution for all other primes}\\ 2+3=5$