Question

The remainder when 59^28 is divided by 7

Answer 1

as 56 is completely divisible by 7

so                 = 3^28 mod 7

                    =(3^2)^14 mod 7

                    =9^14 mod 7

                    =(7+2)^14 mod 7

similarly

                    = 2^14 mod 7    

                   = 4^7 mod 7

                    =4*(4^2)^3 mod 7