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

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Director
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What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is [#permalink]

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22 Dec 2007, 12:16
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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
CEO
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Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
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22 Dec 2007, 13:39
B

9^1(mod 6)=3
9^2(mod 6)=3
9^3(mod 6)=3
...
9^n(mod 6)=3

or

9^n=6m+r
3*3*9^(n-1)=3*2m+r ==> r should be divisible by 3 and be odd ==> r=3

we have 9*3=27 ==> 27(mod 6) =3
Director
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Re: PS - Remainder Question. [#permalink]

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22 Dec 2007, 18:11
ashkrs wrote:
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

I think remainder should be 1, which is None of These.

So choice 5th.

9^1/6 , gives r=1

and the pattern continues giving remainder 1.

What is the OA?
SVP
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Re: PS - Remainder Question. [#permalink]

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22 Dec 2007, 19:18
ashkrs wrote:
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

1. 9/6 = 3 reminder
2. 9^2/6 = (6+3)^2 /6 = (6^2 + 2.6.3 + 3^2)^2 /6 = 3 reminder cuz (6^2) and (2.6.3) has 0 reminder and (3^2) has 3 reminder.
3. 9^3/6 = (6+3)^3 /6 = (6^3 + 3.6^2.3 + 3.6.3^2 + 3^3) /6 = 3 reminder

similarly 9^n has 3 reminder.
so if we add nine 3 reminders, the sum is 9x3 = 27, which alio has 3 reminder.

so B.
Director
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Re: PS - Remainder Question. [#permalink]

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22 Dec 2007, 19:21
GMAT TIGER wrote:
ashkrs wrote:
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

1. 9/6 = 3 reminder
2. 9^2/6 = (6+3)^2 /6 = (6^2 + 2.6.3 + 3^2)^2 /6 = 3 reminder cuz (6^2) and (2.6.3) has 0 reminder and (3^2) has 3 reminder.
3. 9^3/6 = (6+3)^3 /6 = (6^3 + 3.6^2.3 + 3.6.3^2 + 3^3) /6 = 3 reminder

similarly 9^n has 3 reminder.
so if we add nine 3 reminders, the sum is 9x3 = 27, which alio has 3 reminder.

so B.

The mistake I was doing was, I simplified 9/6 = 3/2 ( which will have the remainder as 1)

But why can't we simplify the terms?
SVP
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Re: PS - Remainder Question. [#permalink]

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22 Dec 2007, 22:34
LM wrote:
GMAT TIGER wrote:
ashkrs wrote:
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

1. 9/6 = 3 reminder
2. 9^2/6 = (6+3)^2 /6 = (6^2 + 2.6.3 + 3^2)^2 /6 = 3 reminder cuz (6^2) and (2.6.3) has 0 reminder and (3^2) has 3 reminder.
3. 9^3/6 = (6+3)^3 /6 = (6^3 + 3.6^2.3 + 3.6.3^2 + 3^3) /6 = 3 reminder

similarly 9^n has 3 reminder.
so if we add nine 3 reminders, the sum is 9x3 = 27, which alio has 3 reminder.

so B.

The mistake I was doing was, I simplified 9/6 = 3/2 ( which will have the remainder as 1)

But why can't we simplify the terms?

9/6 = 3/2, but they have different reminders and in case of reminder question its wrong to to do so.

also 9/6= 18/12 but again they have different reminders. so this is the trick...
CEO
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Re: PS - Remainder Question. [#permalink]

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23 Dec 2007, 20:45
ashkrs wrote:
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

not sure if there is a quicker way, but I usually solve these by establishing a pattern.

9/6--> R=3
9+81/6--> R=0

9+81+729/6 --> 819/6 R=3 so every odd it will have a R of 3.

B.
Re: PS - Remainder Question.   [#permalink] 23 Dec 2007, 20:45
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