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Question Stats:
45% (02:57) correct
54% (01:58) wrong based on 0 sessions
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]
 05 Nov 2009, 13:21
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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]
05 Nov 2009, 13:54
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]
05 Nov 2009, 18:29
Good work AKProdigy87
The answer is indeed 0
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Re: Remainder Problem [#permalink]
06 Nov 2009, 00:40
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.
.
.
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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]
09 Nov 2009, 03:31
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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]
11 Nov 2009, 21:18
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]
11 Nov 2009, 21:50
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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 |
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Re: Remainder Problem [#permalink]
12 Nov 2009, 05:59
the answer is 0 ....
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Re: Remainder Problem [#permalink]
12 Nov 2009, 06:42
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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Re: Remainder Problem [#permalink]
12 Nov 2009, 20:10
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. 
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Re: Remainder Problem [#permalink]
13 May 2011, 19:17
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]
13 May 2011, 19:17
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