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The remainder when 1+3+3^2+3^3+..........+3^200 is divided 1 [#permalink]
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05 Nov 2009, 10:52
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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
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Re: Remainder Problem [#permalink]
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05 Nov 2009, 12:21
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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.



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Re: Remainder Problem [#permalink]
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05 Nov 2009, 12:54
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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Re: Remainder Problem [#permalink]
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05 Nov 2009, 17:29
Good work AKProdigy87
The answer is indeed 0



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Re: Remainder Problem [#permalink]
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05 Nov 2009, 23:40
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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!!



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Re: Remainder Problem [#permalink]
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09 Nov 2009, 02:31
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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..



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Re: Remainder Problem [#permalink]
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11 Nov 2009, 20:18
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



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Re: Remainder Problem [#permalink]
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11 Nov 2009, 20:50
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It is always easy if u remember that the coefficient only term can be written as (coefficient) 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



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Re: Remainder Problem [#permalink]
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12 Nov 2009, 04:59
the answer is 0 ....
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Re: Remainder Problem [#permalink]
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12 Nov 2009, 05:42
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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 \(3^{x1}\) 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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Re: Remainder Problem [#permalink]
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12 Nov 2009, 19:10
mbaquestionmark wrote: It is always easy if u remember that the coefficient only term can be written as (coefficient) 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.



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Re: Remainder Problem [#permalink]
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13 May 2011, 18:17
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 Answer  C
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Re: Remainder Problem [#permalink]
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09 Aug 2013, 04:57
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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^n1)/(r1) where a=First term and r =common ration Hence, 1*(3^201 1 )/(31) Rem of (3^2011)/2 divided by 13 3^201 1 /26 WKT, 3^3 = 27 = 26+1 {(26+1)^67  1}/26 {11}/26 =>0
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Re: Remainder Problem [#permalink]
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10 Aug 2013, 13:39
jade3 wrote: 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^(2011) = (3^201  1)/(31) = (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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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]
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21 Aug 2017, 01:09
jade3 wrote: 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 Answer C
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Re: The remainder when 1+3+3^2+3^3+..........+3^200 is divided 1 [#permalink]
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16 Sep 2017, 01:44
jade3 wrote: 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 = (3^2011)/(31) = (3^2011)/2 Now (1+3+3^2+3^3+..........+3^200)/13 = (3^2011)/26 = [(3^3)^67 1]/26 =[(1+26)^67 1]/26 ~ (1^67 1)/26 ~ (11)/26 ~ 0/26 Remainder = 0 Answer C
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Re: The remainder when 1+3+3^2+3^3+..........+3^200 is divided 1
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