last digit
What is the last digit in this calculation involving powers?
What is the last digit of $7^{2016}+8^{2016}$?
This problem is adapted from the World Mathematics Championships
Answer: 7
7$^1$ = 7 8$^1$ = 8
7$^2$ = 49 8$^2$ = 64
7$^3$ = ...3 8$^3$ = ...2 (last digit comes from multiplying last digits)
7$^4$ = ...1 8$^4$ = ...6
7$^5$ = ...7 8$^5$ = ...8 (back to where we began)
7$^6$ = ...9 8$^6$ = ...4
7$^7$ = ...3 8$^7$ = ...2
7$^8$ = ...1 8$^8$ = ...6
The last digits will keep looping in 4s
7$^{2000}$ = ...1 8$^{2000}$ = ...6
7$^{2016}$ = ...1 8$^{2016}$ = ...6
$\therefore$ 7$^{2016}$ + 8$^{2016}$ = ...7