What is the remainder when (1^5 + 2^5.....+100^5) is divided by 4

Answer:if the base (i.e. in 2^5, 2 is the base.) is even, we can skip them since their remainder must be 0. 1^5 = 1 (mod 4) 3^5 = 3 (mod 4) 5^5 = 1 (mod 4) 7^5 = 3 (mod 4) 9^5 = 1 (mod 4) ... ... (there is a sequence we can know that 1 and 3 show up alternatively.) (1 x 5 x 2 + 1 x 5 x 3) + (3 x 5 x 3 + 3 x 5 x 2) = 100 = 0 (mod 4) ∴ the remainder is 0.