Help please

[navdeepbajwa]

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.

[uttam.albela]

Whenever you divide a series of numbers a particular number a pattern is established.

For example:

5^1 / 8 remainder = 5

5^2 / 8 remainder = 1

5^ / 8 remainder = 5

this 5 , 1, 5, 1 ..... repeats

9^1 / 6 remainder = 3

9^2 / 6 remainder = 3

so the pattern is 3,3,3,3,3,3, whatever may be the power of 9.

so 3 is remainder each time. So if there are 9 terms, then sum of remainder = 9*3=27
Which gives remainder of 3 when divided by 6.

[navdeepbajwa]

Thanks

[viju9162]

My understanding is as follows:

9^1 + 9^2 + 9^3 + ...... + 9^9 ..the pattern is of 9,1,9,1,9,1,...

Hence there are 5 9's and 4 1's. Therefore, the unit digit will be 9. Now divide the number by 6, it any probability the reminder is not 2,0 or 4. reminder can be 3.