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

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CEO
Joined: 15 Aug 2003
Posts: 3454

Kudos [?]: 919 [0], given: 781

What is the remainder when 9^1 + 9^2 + 9^3 + .... + 9^8 is [#permalink]

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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.

Kudos [?]: 919 [0], given: 781

VP
Joined: 21 Sep 2003
Posts: 1057

Kudos [?]: 82 [0], given: 0

Location: USA

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07 Oct 2003, 19:07
From my lengthy calculation, I get 4 as the remainder. But it's not in any of the answers

Kudos [?]: 82 [0], given: 0

Intern
Joined: 12 Sep 2003
Posts: 31

Kudos [?]: [0], given: 0

Location: Peru
Re: PS : Remainder # 2 [#permalink]

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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.

Kudos [?]: [0], given: 0

SVP
Joined: 03 Feb 2003
Posts: 1603

Kudos [?]: 303 [0], given: 0

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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.

Kudos [?]: 303 [0], given: 0

07 Oct 2003, 23:23
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