Question

1的20次方➕2的20次方➕3的20次方➕………➕9的20次方除以3的余数是?

Answer 1

根据余数定理：
1^20 mod 3 = 1 mod 3 =1
2^20 mod 3 = (2 mod 3)^20 = (-1)^20 = 1
3^20 mod 3 = 0
4^20 mod 3 = (2^20 mod 3)^2 = 1^2 = 1
5^20 mod 3 = (5 mod 3)^20 mod 3 = 2^20 mod 3 = 1
6^20 mod 3 = (2^20 mod 3)*(3^20 mod 3) = 1*0 = 0
7^20 mod 3 = 4^20 mod 3 =1
8^20 mod 3 = (2^20 mod 3)^3 = 1^3 = 1
9^20 mod 3 = (3^20 mod 3)^2 = 0^2 = 0
所以，(1^20 + 2^20 + 3^20 + ....... + 9^20) mod 3
= (1+1+0+1+1+0+1+1+0) mod 3
= 6 mod 3
= 0