Modulo of Division of Two Numbers

Yes, but it's different:

(a - b) mod p = ((a mod p - b mod p) + p) mod p

(a / b) mod p = ((a mod p) * (b^(-1) mod p)) mod p

Where b^(-1) mod p is the modular inverse of b mod p. For p = prime, b^(-1) mod p = b^(p - 2) mod p.

Edit:

(N*(N^2+5)/6)%P

You don't need any modular inverses from this. Just simplify the fraction: N or N^2+5 will be divisible by 2 and 3. So divide them and then you have (a*b) mod P.

Vlad's answer is correct:

(a - b) mod p = ((a mod p - b mod p) + p) mod p
(a / b) mod p = ((a mod p) * (b^(-1) mod p)) mod p

These and some other operations are outlined here in the Equivalencies section.

Just want to let you know that this will work not only for prime number p. The first one will work for any p. The second one will work for any p, where b^(-1) or modular inverse is defined.

The modular inverse can be calculated with extended Euclidian algorithm