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What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is

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What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is [#permalink] New post 18 Mar 2005, 12:57
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What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6

Last edited by banerjeea_98 on 18 Mar 2005, 13:36, edited 1 time in total.
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 [#permalink] New post 18 Mar 2005, 13:29
i get 2

It took me a minute to get to 950 as the sum but after that it is easy.

Add the digits of 950 so 9+5+0=14 divide by = 2 with remainder 2.

Remainders when the sum of the digits is divided by 3 and 6 are the same as the actual number. so 452: 4+5+2= 11 divide 11 by 3 and get 3 with 2 remainder.

divide 452 get 150 with 2 remainder :smoke
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 [#permalink] New post 18 Mar 2005, 13:36
You don't need to add the numbers individually.

if you list them out individually:

91 92 93 94 95 96 97 98 99

you can see the average of these numbers is 95. since there are 10 numbers total, 95*10 = 950.

950 / 6 = 158, remainder 2.
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 [#permalink] New post 18 Mar 2005, 13:37
guys try it again, I have updated the ques, there was an error before.
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Re: PS [#permalink] New post 18 Mar 2005, 14:25
banerjeea_98 wrote:
What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6


Let A = 9^9 + 9^8 + ... + 9^1

9A = 9^10 + (9^9 + 9^8 + ... + 9^2 + 9^1) - 9^1
9A = 9^10 + A - 9
8A = 9^10 - 9
A = (9^10 - 9)/8
= 9( 9^9 -1)/8

Since 9^9 = (9^3)^3

A = 9 ((9^3)^3 - 1)/8
= 9 (729^3 -1)/8

Since a^3 - b^3 = (a-b)^3 + 3ab(a-b)

A = 9/8[(729-1)^3 + 3*729(729-1)]
= 9/8[728^3 + 3*729*728]
= 9[91*728^2 + 3*729*91]

Dividing this by 6, we have this remainder
9[ 1*2*2 + 3*3*1 ] = 9*13 = 117 which leaves 3 as the remainder.

Thus I'd go with 3 as the answer.

Hope that helps.
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Re: PS [#permalink] New post 18 Mar 2005, 21:41
banerjeea_98 wrote:
What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6


=9^1 + 9^2 + 9^3 + ...... + 9^9
sum of 1st term and 2nd term is even and when divided by 6 leaves 0 as reminder. similarly, sum of 3rd and 4th also result in even number which is divided by 6 and leaves 0 as reminders. so on for 5th and 6th, and 7th and 8th term. the only remaining term is 9th. the reminder is 3 no mater the value of 9^n.
Re: PS   [#permalink] 18 Mar 2005, 21:41
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