Kane Diep

We see the last digit of the 5^23 is 5 always. The last digit of 7^17 followed the sequence 7, 9, 3, 1, 7, ... so every multiple of 4 exponents, its last digit will be 1, so 7^17 = (7^16)*7 have 7 as it last digit. The sum 5 + 7 for the last digit is 2 th...