Difficult Math Problem #96 - Arithmetic

[800guy]

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 these


OA coming when people reply with explanations...from difficult math doc

[Neo2000]

Many ways to solve this problem

Consider the first 2 terms
9^1 + 9^2 = 9+81 = 90 which when divided by 6 = 0
The third term i.e. 9^3 = 729 when divided by 6 = remainder3

Hence the last term 9^9 when divided by 6 = remainder 3
The first 8terms can be split into 4groups of 2 each (9 to the odd power+9 to the even power) which when divided by 6 = 0

Hence Option 2

The Remainder is 3

[800guy]

oa from the diff math doc:

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.

Another way: We get the value as 9*odd/6 = 3*odd/2 Since 3*odd = odd; odd/2 = XXXX.5
so something divided by 6, gives XXXX.5, hence remainder is 6*0.5 = 3

[LalaB]

the unit digit of 9^1 + 9^2 + 9^3 + ...... + 9^9 is 9
so, 9/6 .remainder is 3