If n is a natural number, then what is the sum of all distinct remainders of 4ⁿ + 6ⁿ + 9ⁿ + 11ⁿ when divided by 10 for various values of n ?

Last digits: 4ⁿ cycles {4,6}; 6ⁿ always ends in 6; 9ⁿ cycles {9,1}; 11ⁿ always ends in 1. For n odd: 4+6+9+1 = 20, remainder 0. For n even: 6+6+1+1 = 14, remainder 4. Wait, need to also consider 4²=16 ends in 6, 9²=81 ends in 1. For n=1: 4+6+9+1=20, ends in 0. For n=2: 6+6+1+1=14, ends in 4. For n=3: same as n=1, ends in 0. So distinct remainders are 0 and 4. Hmm sum = 4, not 6. Re-checking: actually 6ⁿ ends in 6 always, 11ⁿ ends in 1 always. For n=1: 4+6+9+1 = 20 → 0. For n=2: 16+36+81+121 → 6+6+1+1=14 → 4. So distinct remainders {0, 4}, sum = 4. The answer is (b) 4.