Any other approach for this the solution says
1+9+9^2+.....9^8 is odd which is hard to determine so any other approach
What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6?
(1) 0
(2) 3
(3) 4
(4) 2
(5) None of the Remainder of 9*odd /6 is 3
remainder of 9*even/6 is 0
9^1 + 9^2 + 9^3 + ...... + 9^9=9*(1+9+9^2+.....9^8)
1+9+9^2+.....9^8 is odd.
Thus we obtain 3 as a remainder when we divide 9*(1+9+9^2+.....9^8) by 6.
1+9+9^2+.....9^8 is odd which is hard to determine so any other approach
What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6?
(1) 0
(2) 3
(3) 4
(4) 2
(5) None of the Remainder of 9*odd /6 is 3
remainder of 9*even/6 is 0
9^1 + 9^2 + 9^3 + ...... + 9^9=9*(1+9+9^2+.....9^8)
1+9+9^2+.....9^8 is odd.
Thus we obtain 3 as a remainder when we divide 9*(1+9+9^2+.....9^8) by 6.
