Actually, the meaning of life is 42, so now that we have that out of the way, let's solve the prime number and remainder problem. I will offer two solutions for this tasty little puzzler. I'll call them the Format Method and the Consecutive Integer Method.
Format Method. First of all we need an agreement that all integers are of the following formats: 6n, 6N + 1, 6N + 2, 6N + 3, 6N + 4, 6N + 5, where N is an integer. If we go to 6N + 6 this can be written 6(N+1) which is just another version of 6N. I claim 6N + 1 and 6N + 5 are the only possible primes (having exactly 2 divisors, 1 and itself). 6N is obviously divisible by 6. 6N + 2 is the sum of 2 even numbers thus even and divisible by 2, as is 6N + 4. 6N + 3 is the sum of two numbers divisible by 3 thus divisible by three itself. Therefore we only need consider 6N + 1 and 6N + 5.
Now onto our problem:
36N^2 + 12N + 1 or 36N^2 + 60N + 25
36N^2 + 12N + 18 or 36N^2 + 60N + 42
With a minimum of algebra we get:
12(3N^2 + N + 1) + 6 or 12(3N^2 + 5N + 3) + 6
Dividing by 12 we can see that both these numbers leave a remainder of 6.
Consecutive Integer Method. Let P be our prime number bigger than 3
Square it: P^2
Add 17: P^2 + 17
Rearranging algebraically we get: (P^2 - 1) + 18
Now factor. (P - 1)(P + 1) + 18
Here's where consecutive integers comes in.
P - 1, P, P + 1 are consecutive integers (for example if P is 11 the we get 10, 11, 12)
Therefore either P + 1 or P - 1 is divisible by 3
And because P itself is odd (all prime numbers bigger than 3 are odd) then both P - 1 and P + 1 are even and (P - 1)(P + 1) is divisible by 4. Thus (P - 1)(P + 1) is divisible by 12.
Sooooooo (P - 1)(P +1) + 18 when divided by 12 leaves a remainder of 6 (we need only consider what happen when we divide 18 by 12).