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What is the remainder when 9^1 + 9^2 + 9^3 + .... + 9^8 is

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What is the remainder when 9^1 + 9^2 + 9^3 + .... + 9^8 is [#permalink] New post 07 Oct 2003, 17:03
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What is the remainder when 9^1 + 9^2 + 9^3 + .... + 9^8 is divided by 6?

1) 3
2) 2
3) 0
4) 5

Last edited by Praetorian on 07 Oct 2003, 19:40, edited 1 time in total.
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 [#permalink] New post 07 Oct 2003, 19:07
From my lengthy calculation, I get 4 as the remainder. But it's not in any of the answers :roll:
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Re: PS : Remainder # 2 [#permalink] New post 07 Oct 2003, 22:39
Even multiples of 9 divided by 6 leave 0 as their remainder
Odd multiples of 9 divided by 6 leave 3 as their remainder

There are eight terms in the addition, each one odd, but as the number of terms is even, the result is even, so 0 should be the remainder.
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 [#permalink] New post 07 Oct 2003, 23:23
9^N=6z+R (0<=R<=5) both parts should be divisible by 3, so R=3

9^1 mod 6 leaves 3
9^2 mod 6 leaves 3
9^3 mod 6 leaves 3
...... and so on

the sum of remainders=8*3=24, divisible by 6 evenly, i.e. the final remainder is 0.
  [#permalink] 07 Oct 2003, 23:23
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