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

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What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is [#permalink] New post 13 Mar 2004, 15:32
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A
B
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Difficulty:

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Question Stats:

65% (01:59) correct 35% (01:04) wrong based on 193 sessions
What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is divided by 6?

A. 0
B. 3
C. 2
D. 5
E. None of the above
[Reveal] Spoiler: OA

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Pls include reasoning along with all answer posts.
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Last edited by Bunuel on 20 Feb 2012, 22:43, edited 2 times in total.
Edited the question and added the OA
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Re: What is the remainder when 9^1 + 9^2 + 9^3 +....+ 9^9 is [#permalink] New post 20 Feb 2012, 22:40
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sunniboy007 wrote:
What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is divided by 6?

A. 0
B. 3
C. 2
D. 5
E. None of the above


30 sec approach:
Given: 9^1+(9^2+9^3+9^4+9^5+9^6+9^7+9^8+9^9). Notice that in the brackets we have the sum of 8 odd multiples of 3, hence the sum in the brackets will be even multiple of 3 (the sum of 8 odd numbers is even). So, the sum in the brackets is multiple of 6 (remainder is zero). So we are just left with the first term 9, which yields remainder of 3 upon division by 6.

Answer: B.
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Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is [#permalink] New post 21 Feb 2012, 12:50
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Don't really know if my approach is correct but this is how I approached it.

When divided by 6, 9^1 leaves a remainder of 3
When divided by 6, 9^2 leaves a remainder of 3
When divided by 6, 9^3 leaves a remainder of 3

You can check further if you want to, but at this point I had decided that all the terms individually leave a remainder of 3, so all the remainder added up would be 9*3=27 , and 27 divided by 6 leaves a remainder of 3 . Hence the answer should be B.

If I am correct, remainders can be added and then divided by the original number to come up with the remainder. For example, lets take two numbers, 11 and 13 and divide them by 4. 11 and 13 add up to 24 and 24 divided by 4 leaves a remainder of 0. 11 divided by 4 leaves a remainder of 3, 13 divided by 4 leaves a remainder of 1. Now when you add the remainders, 3+1=4, which leaves a remainder of 0 when divided by 4 or is divisible by 4.
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Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is [#permalink] New post 16 Mar 2012, 14:36
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Hi.. i did it like -
the last digits are -
9^1 = 9
9^2 = 8
9^3 = 7
9^4 = 6
till
9^9 = 1

=> adding no. from 1 to 9 = 45
45/6 = 3
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Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is [#permalink] New post 16 Mar 2012, 14:54
monalimishra wrote:
Hi.. i did it like -
the last digits are -
9^1 = 9
9^2 = 8
9^3 = 7
9^4 = 6
till
9^9 = 1

=> adding no. from 1 to 9 = 45
45/6 = 3


There is a flaw in your aproach... 9^2 = 81
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Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is [#permalink] New post 16 Mar 2012, 19:20
yup... realized my mistake sometime after posting... :-(
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Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is [#permalink] New post 26 Jun 2013, 01:24
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Re: What is the remainder when 9^1 + 9^2 + 9^3 +....+ 9^9 is [#permalink] New post 12 Jul 2013, 02:45
Bunuel wrote:
sunniboy007 wrote:
What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is divided by 6?

A. 0
B. 3
C. 2
D. 5
E. None of the above


30 sec approach:
Given: 9^1+(9^2+9^3+9^4+9^5+9^6+9^7+9^8+9^9). Notice that in the brackets we have the sum of 8 odd multiples of 3, hence the sum in the brackets will be even multiple of 3 (the sum of 8 odd numbers is even). So, the sum in the brackets is multiple of 6 (remainder is zero). So we are just left with the first term 9, which yields remainder of 3 upon division by 6.

Answer: B.


Hi Bunnel,

I did it as below:
Sum = 9/8*(9^9-1)
Rem (s/6) = ?

9^9 - has units digit 9,
9-1 = 8/8 = 1
9^2 = 81-1 = 80/8 = 10
9^3 = 729-1 = 728/8 = 91

Sum = 9*Integer
Rem (s, 6) = 3

Here I could do this because the integral multiple of 9 in the sum is not a multiple of 6.

Can I use this method for other cases?
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Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is [#permalink] New post 06 Aug 2013, 09:10
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I searched for the patrons in the digit of nine, which resulted in 1,9,1,9,1,9..... after that I summed them up which was 49. 49 divided by 6 left a remainder of 3.
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Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is [#permalink] New post 06 Aug 2013, 10:10
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6 is an even multiple of 3. When any even multiple of 3 is divided by 6, it will leave a remainder of 0. Or in other words it is perfectly divisible by 6.

On the contrary, when any odd multiple of 3 is divided by 6, it will leave a remainder of 3. For e.g when 9 an odd multiple of 3 is divided by 6, you will get a remainder of 3.

9 is an odd multiple of 3. And all powers of 9 are odd multiples of 3.
Therefore, when each of the 9 powers of 9 listed above are divided by 6, each of them will leave a remainder of 3.

The total value of the remainder = 3 + 3 + .... + 3 (9 remainders) = 27.
27 is divisible by 6. Hence, it will leave remainder as 3.

Hence, the final remainder when the expression 9^1 + 9^2 + 9^3 + .... + 9^9 is divided by 6 will be equal to '3'.
and one more point to add if the expression is 9^1+9^2+...........+9^10 is divided by 6 then the remainter will be '0'

We can generalize it further:-
if (9^1+9^2+.......9^n) if n is odd then the remainder will always be 3 and if n is even then the remainder will always be '0'.

I hope people will like this explaination and if it helps you further please give Kudos to me.
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Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is [#permalink] New post 14 Aug 2014, 00:49
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Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is   [#permalink] 14 Aug 2014, 00:49
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