A Simple Modular congruence

Since $256 = 2^{8}$ we have that $256^{379} = (2^{8})^{379} = 2^{3032}$ . Now $2^{5} = 32 \equiv 1 \pmod {31}$ , so $2^{3032} = (2^{3030})*(2^{2}) \equiv 1 * 4 \pmod {31} \equiv 4 \pmod{31}$ . Thus the desired remainder is $\boxed{4}$ .

256^(379)=2^(8 x 379)=2^(3032) To find 2^(3032)MOD31 we can use the Euler Totient Function, which for 31 is 30. A totiative of a natural number 'b' is basically any natural number less than b which is coprime to 'b'. Thus the Totient Function of a number is the cardinality of the set of all totiatives of that number or simply the number of totiatives. It can be generated as Phi(n) = n(1-(1/p)) (1-(1/q)) (........ where p,q... are the prime factors of n. This has been proved by induction on Wolfram MathWorld's page on the same and also on doc.ic.ac.uk . Thus we can observe that for any prime p, Phi(p) will be (p-1). Now, the property is that while taking the modulo of a powered base with respect to a number, the remainder will be the same as when the modulo is taken of the same index raised to the remainder we get when the original index is divided by the Euler Totient of the number with whose respect the original modulo has to be taken. Confusing? Let's say we need to find (N^b)MODp. Then define Phi(p) = q. Now , (N^b)MODp ~(couldn't get the triple equivalent symbol) {N^(bMODq)MODp}. Thus it helps reduce the ordinality of the index, thus simplifying the original modulo. Here, let N=2 b=3032 and p=31 , then q=30. So, (2^3032)MOD31= {2^(3032MOD30)MOD31} = (2^2)MOD31 = 4. Taking more examples will help in understanding it thoroughly. Hope it helped.😊