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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]

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13 Mar 2004, 15:32
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61% (02:01) correct 39% (01:11) wrong based on 868 sessions

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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]

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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.

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

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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]

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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]

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16 Mar 2012, 14:54
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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]

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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]

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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]

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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.

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]

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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]

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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]

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24 Jun 2015, 09:28
1
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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

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

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27 Jun 2015, 01:35
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

9^45 = (6+3)^45 .. leaves us with 3^45..which will always give us a remainder of 3 when divided by 6 (27/6, 81/6, 9/6).. final answer B..

is this approach correct ?
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Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is [#permalink]

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19 Feb 2016, 12:54
There is a flaw you are stating that (6+3)^45 = 6^45 + 3^45 which is incorrect.

LaxAvenger 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

9^45 = (6+3)^45 .. leaves us with 3^45..which will always give us a remainder of 3 when divided by 6 (27/6, 81/6, 9/6).. final answer B..

is this approach correct ?

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

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13 Mar 2016, 23:54
Here The key o such kind of questions is to find any pattern
here sum of odd terms yields remainder of 3
and sum of even => remainder =0
since there are 9 terms involved => remainder = 3
Hence B
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Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is [#permalink]

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17 Mar 2016, 04:42
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The easiest way here is to find he pattern
here 9^1/6=> remainder =3
9^1+9^2/6=> reminder = 0
9^1+9^2+9^3/6=> remainder =3
hence the cyclicity is 2
so the number of terms are odd => remainder =3
hence B
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Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is [#permalink]

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31 Mar 2016, 09:20
An arithmetics question here, isn't 9^1 + 9^2 + 9^3 +...+ 9^9 the same as 9^11? like factor all the common nines, it will give you 9^2 (nine nines) then add them to the given 9^9 and get 9^11. Does this make sense?... We are given a sum, so Im not sure this logic works..

Thank you!
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Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is [#permalink]

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31 Mar 2016, 11:19
iliavko wrote:
An arithmetics question here, isn't 9^1 + 9^2 + 9^3 +...+ 9^9 the same as 9^11? like factor all the common nines, it will give you 9^2 (nine nines) then add them to the given 9^9 and get 9^11. Does this make sense?... We are given a sum, so Im not sure this logic works..

Thank you!

No, this does not make sense. Not sure how you are getting this... You CANNOT factor out 9^2 out of 9^1 + 9^2 + 9^3 +...+ 9^9, you can only factor out 9.
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Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is   [#permalink] 31 Mar 2016, 11:19
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