Explanation

Modulo 10: 1^a ≡ 1; 2^b cycles through {2, 4, 8, 6}; 3^c through {3, 9, 7, 1}; 4^d through {4, 6}. The sum is 1 + even + odd + even, always even, and each of 0, 2, 4, 6, 8 occurs (for instance 1+2+3+4 = 10 gives 0; 1+8+3+6 = 18 gives 8). That is 5 possible remainders, option (c).