If a, b, c, d are natural numbers, then how many possible remainders are there when 1^a + 2^b + 3^c + 4^d is divided by 10 ?

1^a always = 1. 2^b mod 10 cycles through {2, 4, 8, 6}. 3^c mod 10 cycles through {3, 9, 7, 1}. 4^d mod 10 cycles through {4, 6}. The number of possible sums mod 10 depends on combinations. After enumeration of all possible last-digit sums (1 + {2,4,8,6} + {3,9,7,1} + {4,6}) mod 10, there are 6 distinct remainders.