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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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Updated on: 20 Feb 2012, 23:43
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Question Stats:

61% (01:31) correct 39% (01:47) wrong based on 1227 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

Originally posted by sunniboy007 on 13 Mar 2004, 16:32.
Last edited by Bunuel on 20 Feb 2012, 23:43, 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, 23:40
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16
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, 13:50
21
8
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, 15:36
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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, 15:54
1
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, 20:20
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, 02:24
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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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, 03:45
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, 10:10
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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, 11:10
6
1
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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27 Jun 2015, 02:35
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, 13:54
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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14 Mar 2016, 00:54
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, 05:42
1
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, 10:20
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, 12:19
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]

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15 Jul 2017, 20:35
Here was what I did, might be a little faster.

What is remainder of

9^1+9^2....9^8+9^9/6.

Since 9 is 3 more than 6, every multiple of 9 divided 6 will have a remainder of 3.

Therefore 3x9 (for the 9 numbers)= 27/6= 4 with r3.

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

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16 Jul 2017, 16:47
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

I used cyclicity, but I'm not sure it's correct, though it gave me the correct answer.

$$9^1$$ = 9
$$9^2$$ = 81
$$9^3$$ = ..9
$$9^4$$= ...1

Odd powers of nine have a units digit of 9, even powers of nine have a units digit of 1.

There are 5 odd powers of nine (1,3,5,7,9). Units digit is 9, 9*5 = 45, so odd powers' sum will have units digit of 5.

There are four even powers of nine(2,4,6,8). Each ends in 1. 4*1 = 4. Even powers' sum will have a units digit of 4.

Add the two groups: 5 + 4 = 9.

9/6 = remainder 3.

You could also pair four odd powers that end in 9 with four even powers that end in 1: 9,1,9,1,9,1,9,1. Each pair sums to 9 + 1 = 10, with a last digit of 0.

Add one more odd power of 9, which will end in 9 (we have five odd powers, we've only used four). 0 + 9 = 9. 9/6 has R3.

I took 33 seconds to answer. I'm a little nervous.

Bunuel - are these methods 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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30 Jan 2019, 05:22
Solution:

Given: $$9^1+ 9^2+9^3+ …………………………….+9^9$$

Approach: Let’s take $$N = 9^1+ 9^2+9^3+ …………………………….+9^9$$

We have to find the remainder when N is divided by 6.

$$N = 9^1+ (9^2+9^3+ 9^4+ 9^5+ 9^6+ 9^7+ 9^8+9^9)$$
If we look at the terms in the brackets we have the sum of 8 odd numbers which are multiples of 3. We have to note here that sum of 8 ODD numbers will result in EVEN. [ODD + ODD = EVEN].
So, the sum in brackets is multiple of 6 and obviously, the remainder will be ZERO.
So, we are left with the first term 9, the remainder will be “3” when divided by 6.

The correct answer option is “B”.

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Re: What is the remainder when 9^1 + 9^2 + 9^3 +...+ 9^9 is   [#permalink] 30 Jan 2019, 05:22
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