Let $$p$$ be the smallest prime number such that the numbers $$(p + 6)$$, $$(p + 8)$$, $$(p + 12)$$ and $$(p + 14)$$ are also prime. Then the remainder when $$p^2$$ is divided by 4 is

Trying the smallest primes, $$p = 2$$ fails because $$2 + 6 = 8$$ is not prime, and $$p = 3$$ fails because $$3 + 6 = 9$$ is not prime. For $$p = 5$$ the four numbers are 11, 13, 17 and 19, all of which are prime, so $$p = 5$$. Then $$p^2 = 25$$ and $$25 = 4 \times 6 + 1$$, so the remainder is 1.