Remainder Problem

[Hussain15]

The remainder when 1+3+3^2+3^3+..........+3^200 is divided by 13 is

A.12
B.7
C.0
D.5
E.3

A detailed explanation will be appreciated.

[punitkaur]

I hink the answer should be 0. (C). This is how I did it, although there may be a better way!

1+3+9+27+81+..3^200.

Take the terms in groups of 3 numbers
first group 1+3+9=13. divisible by 13.

Next group.

27+81+243 = 27(1+3+9), divisible by 13.

Since there are 201 numbers in the sequence, 201/3 = 67 groups.

Each of these groups is divisible by 13. So remainder is 0.

Whats the OA.

[palvarez]

(3^201 - 1)/2 is the sum

3 = 3
3^2 = 9
3^3 = 1

3^201 = 1 = 3^198

(3^198 . 3^3 - 1)/2 = (1*27 -1)/2 = 13 = 0

[punitkaur]

hi palvarez,

i dont understand ur last step. how r u multiplying the remainder of 3 ^198 by 27?

can u explain your solution in detail?

What is the property u r using there?

Thanks

[Abdulla]

(3^201 - 1)/2 is the sum

3 = 3
3^2 = 9
3^3 = 1

3^201 = 1 = 3^198

(3^198 . 3^3 - 1)/2 = (1*27 -1)/2 = 13 = 0

Hi palvarez, pls explain your steps.

[palvarez]

3^201 = 3^198 * 3^3 = 27

Thats what I did. Well, in this case, this step is unnecessary, since 3^201 = 1 (mod 13). 3^201 - 1 = 0 (mod 13). Since 2 and 13 have no factors in common, we can say that (3^201 - 1 )/2 = 0