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Manager  Joined: 19 Nov 2007
Posts: 137
The remainder when 1+3+3^2+3^3+..........+3^200 is divided 1  [#permalink]

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4
22 00:00

Difficulty:   75% (hard)

Question Stats: 61% (02:25) correct 39% (02:41) wrong based on 185 sessions

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The remainder when 1+3+3^2+3^3+..........+3^200 is divided 13.

A. 12
B. 7
C. 0
D. 5
E. 3
Manager  Joined: 11 Sep 2009
Posts: 117
Re: Remainder Problem  [#permalink]

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7
4
I'm not sure if there is a simpler way to approach this problem, but this is how I solved it. It relies on taking the remainder of each term individually, and from that, determining the remainder of the sum of the terms.

Remainder of 1/13 = 1
Remainder of 3/13 = 3
Remainder of (3^2)/13 = (Remainder of 3)*(Remainder of 3) / 13 = Remainder of 3*3/13 = 9
Remainder of (3^3)/13 = (Remainder of 3^2)*(Remainder of 3) / 13 = Remainder of 9*3/13 = 1
Remainder of (3^4)/13 = (Remainder of 3^3)*(Remainder of 3) / 13 = Remainder of 1*3/13 = 3

And so on....

Essentially, the question can be reduced to:

What is the remainder of: 1 + 3 + 9 + 1 + 3 + 9 + 1... (with 200 + 1 terms) when divided by 13

= (1+3+9) * (201/3) mod 13 (i.e. remainder)
= 13 * (some integer) mod 13
= 0

Therefore, the answer should be 0.
General Discussion
Manager  Affiliations: PMP
Joined: 13 Oct 2009
Posts: 210
Re: Remainder Problem  [#permalink]

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1
AKProdigy87 wrote:
I'm not sure if there is a simpler way to approach this problem, but this is how I solved it. It relies on taking the remainder of each term individually, and from that, determining the remainder of the sum of the terms.

Remainder of 1/13 = 1
Remainder of 3/13 = 3
Remainder of (3^2)/13 = (Remainder of 3)*(Remainder of 3) / 13 = Remainder of 3*3/13 = 9
Remainder of (3^3)/13 = (Remainder of 3^2)*(Remainder of 3) / 13 = Remainder of 9*3/13 = 1
Remainder of (3^4)/13 = (Remainder of 3^3)*(Remainder of 3) / 13 = Remainder of 1*3/13 = 3

And so on....

Essentially, the question can be reduced to:

What is the remainder of: 1 + 3 + 9 + 1 + 3 + 9 + 1... (with 200 + 1 terms) when divided by 13

= (1+3+9) * (201/3) mod 13 (i.e. remainder)
= 13 * (some integer) mod 13
= 0

Therefore, the answer should be 0.

Good way to do it +1 Kudos
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Manager  Joined: 19 Nov 2007
Posts: 137
Re: Remainder Problem  [#permalink]

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Good work AKProdigy87

The answer is indeed 0
Intern  Affiliations: CA - India
Joined: 27 Oct 2009
Posts: 39
Location: India
Schools: ISB - Hyderabad, NSU - Singapore
Re: Remainder Problem  [#permalink]

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2
i tried the problem with similar method:

3^0/13= remainder 1.
3^0/13+3^1/13= remainder 4.
3^0/13+3^1/13+3^2/13= remainder 0.
3^3/13= remainder 1.
3^3/13+3^4/13= remainder 4.
3^3/13+3^4/13+3^5/13= remainder 0.
.
.
.
.
Hence, sum of last 3 digits in the given equation must also give remainder 0. Some times, pattern of similar answers saves time, i guess.. works for me!!
Intern  Joined: 28 Apr 2009
Posts: 27
Re: Remainder Problem  [#permalink]

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1
worked it out like this..

1+3+3^2 = 14 => divisible by 13

3^3+3^4+3^5 = 3^2(1+3+3^2) = 9x13 => divisible by 13..

starting from 1 the sum of every three terms is a multiple of 13..

so upto the power of 200, there are 201 terms.. which is a multiple of 3.. so the sum must me a multiple of 13..

Remainder would be therefore 0..
Manager  Joined: 23 Jun 2009
Posts: 73
Re: Remainder Problem  [#permalink]

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mbaquestionmark wrote:
worked it out like this..

1+3+3^2 = 14 => divisible by 13

3^3+3^4+3^5 = 3^2(1+3+3^2) = 9x13 => divisible by 13..

starting from 1 the sum of every three terms is a multiple of 13..

so upto the power of 200, there are 201 terms.. which is a multiple of 3.. so the sum must me a multiple of 13..

Remainder would be therefore 0..

I worked it out the same way.. but was a bit confused between 200 and 201 terms
Intern  Joined: 28 Apr 2009
Posts: 27
Re: Remainder Problem  [#permalink]

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4
It is always easy if u remember that the co-efficient only term can be written as (co-efficient) x (the variable raised to 0).

so it is 3^0 to 3^200.. i.e. 0 to 200 so 201 terms.

cheers.

Casinoking wrote:
mbaquestionmark wrote:
worked it out like this..

1+3+3^2 = 14 => divisible by 13

3^3+3^4+3^5 = 3^2(1+3+3^2) = 9x13 => divisible by 13..

starting from 1 the sum of every three terms is a multiple of 13..

so upto the power of 200, there are 201 terms.. which is a multiple of 3.. so the sum must me a multiple of 13..

Remainder would be therefore 0..

I worked it out the same way.. but was a bit confused between 200 and 201 terms
Intern  Joined: 19 Feb 2009
Posts: 43
Re: Remainder Problem  [#permalink]

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the answer is 0 ....
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Manager  Joined: 29 Oct 2009
Posts: 178
GMAT 1: 750 Q50 V42 Re: Remainder Problem  [#permalink]

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1
Hey guys,

This is how I worked it out:

If $$3^x$$ is a number such that x is evenly divisible by 3 (ie. it leaves remainder of 0), then the sum of numbers from $$3^0$$ to $$3^{x-1}$$ will always be evenly divisible by 13.

Now, we know that 201 is divisible by 3. Therefore, 200 = 201 - 1 (which satisfies our condition)

Hence sum of the numbers from $$3^0$$ to $$3^{200}$$ will be divisible by 13.

Thus answer is 0.
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Intern  Joined: 11 Jul 2009
Posts: 32
Re: Remainder Problem  [#permalink]

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mbaquestionmark wrote:
It is always easy if u remember that the co-efficient only term can be written as (co-efficient) x (the variable raised to 0).

so it is 3^0 to 3^200.. i.e. 0 to 200 so 201 terms.

cheers.

Casinoking wrote:
mbaquestionmark wrote:
worked it out like this..

1+3+3^2 = 14 => divisible by 13

3^3+3^4+3^5 = 3^2(1+3+3^2) = 9x13 => divisible by 13..

starting from 1 the sum of every three terms is a multiple of 13..

so upto the power of 200, there are 201 terms.. which is a multiple of 3.. so the sum must me a multiple of 13..

Remainder would be therefore 0..

I worked it out the same way.. but was a bit confused between 200 and 201 terms

Useful tip. I too used to get confused. Kudos. Retired Moderator B
Joined: 16 Nov 2010
Posts: 1259
Location: United States (IN)
Concentration: Strategy, Technology
Re: Remainder Problem  [#permalink]

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There are 201 numbers in series

(1/13 + 3/13 + 9/13) + (27/13 + 81/13 + 243/13)

(1+ 3 + 9) + (1 + 3 + 9) - Pattern of remainers

= 13 + 13 + ..

On dividing by 13 again

0 + 0 + 0

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Re: Remainder Problem  [#permalink]

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2
1
1+3+3^2+3^3+..................+3^200

Is a Geometric progression having common ratio as '3' and number of terms as '201'.

Since Sum to n terms in GP = a(r^n-1)/(r-1)

where a=First term and r =common ration

Hence,

1*(3^201 -1 )/(3-1)

Rem of (3^201-1)/2 divided by 13

3^201 -1 /26

WKT, 3^3 = 27 = 26+1

{(26+1)^67 - 1}/26

{1-1}/26

=>0
Senior Manager  Joined: 10 Jul 2013
Posts: 289
Re: Remainder Problem  [#permalink]

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The remainder when 1+3+3^2+3^3+..........+3^200 is divided 13.

A.12
B.7
C.0
D.5
E.3

1+3+3^2+3^3+..........+3^(201-1) = (3^201 - 1)/(3-1) = (3^201 - 1) / 2
Now by 13 = (3^201 - 1) / 26 = (3^3)^67 - 1 / 26 = (26+1)^67 - 1 / 26 = (26)^67/26....... (1^67/26 - 1/26)
= An integer quotient and from the last part 0 . so remainder = 0
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Re: The remainder when 1+3+3^2+3^3+..........+3^200 is divided 1  [#permalink]

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The remainder when 1+3+3^2+3^3+..........+3^200 is divided 13.

A. 12
B. 7
C. 0
D. 5
E. 3

1+3 +3^2 = 13
so let make the sets of 3 terms..all these sets will be divisible by 13

There are total 201 terms.. 201/3 is fully divisible. So we can consume all the terms in set of 3.

(1+3+3^2) + 3^3(1+3+3^2 ) + ......... + 3^198(1+3+3^2)
Now these sets are divisible by 13 .
Remainder = 0

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Re: The remainder when 1+3+3^2+3^3+..........+3^200 is divided 1  [#permalink]

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The remainder when 1+3+3^2+3^3+..........+3^200 is divided 13.

A. 12
B. 7
C. 0
D. 5
E. 3

First time I solved it by another method .. This time I am providing a different method to approach the same.

1+3+3^2+3^3+..........+3^200
= (3^201-1)/(3-1) = (3^201-1)/2

Now (1+3+3^2+3^3+..........+3^200)/13 = (3^201-1)/26 = [(3^3)^67 -1]/26 =[(1+26)^67 -1]/26 ~ (1^67 -1)/26 ~ (1-1)/26 ~ 0/26

Remainder = 0

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Re: The remainder when 1+3+3^2+3^3+..........+3^200 is divided 1  [#permalink]

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_________________ Re: The remainder when 1+3+3^2+3^3+..........+3^200 is divided 1   [#permalink] 11 Feb 2019, 20:35
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