February 23, 2019 February 23, 2019 07:00 AM PST 09:00 AM PST Learn reading strategies that can help even nonvoracious reader to master GMAT RC. Saturday, February 23rd at 7 AM PT February 24, 2019 February 24, 2019 07:00 AM PST 09:00 AM PST Get personalized insights on how to achieve your Target Quant Score.
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 19 Nov 2007
Posts: 179

The remainder when 1+3+3^2+3^3+..........+3^200 is divided 1
[#permalink]
Show Tags
05 Nov 2009, 10:52
Question Stats:
60% (02:21) correct 40% (02:40) wrong based on 171 sessions
HideShow timer Statistics
The remainder when 1+3+3^2+3^3+..........+3^200 is divided 13. A. 12 B. 7 C. 0 D. 5 E. 3
Official Answer and Stats are available only to registered users. Register/ Login.




Manager
Joined: 11 Sep 2009
Posts: 130

Re: Remainder Problem
[#permalink]
Show Tags
05 Nov 2009, 12:21
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.




Senior Manager
Affiliations: PMP
Joined: 13 Oct 2009
Posts: 253

Re: Remainder Problem
[#permalink]
Show Tags
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
_________________
Thanks, Sri  keep uppp...ing the tempo...
Press +1 Kudos, if you think my post gave u a tiny tip



Manager
Joined: 19 Nov 2007
Posts: 179

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



Intern
Affiliations: CA  India
Joined: 27 Oct 2009
Posts: 42
Location: India
Schools: ISB  Hyderabad, NSU  Singapore

Re: Remainder Problem
[#permalink]
Show Tags
05 Nov 2009, 23:40
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. . . . . 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: 32

Re: Remainder Problem
[#permalink]
Show Tags
09 Nov 2009, 02:31
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: 95

Re: Remainder Problem
[#permalink]
Show Tags
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



Intern
Joined: 28 Apr 2009
Posts: 32

Re: Remainder Problem
[#permalink]
Show Tags
11 Nov 2009, 20:50
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



Intern
Joined: 19 Feb 2009
Posts: 46
Schools: INSEAD,Nanyang Business school, CBS,

Re: Remainder Problem
[#permalink]
Show Tags
12 Nov 2009, 04:59
the answer is 0 ....
_________________
Working without expecting fruit helps in mastering the art of doing faultfree action !



Manager
Joined: 29 Oct 2009
Posts: 195

Re: Remainder Problem
[#permalink]
Show Tags
12 Nov 2009, 05:42
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 \(3^{x1}\) 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.
_________________
Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilationoftipsandtrickstodealwithremainders86714.html#p651942
Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/wordproblemsmadeeasy87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/workwordproblemsmadeeasy87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distancespeedtimewordproblemsmadeeasy87481.html



Intern
Joined: 11 Jul 2009
Posts: 40

Re: Remainder Problem
[#permalink]
Show Tags
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.



Retired Moderator
Joined: 16 Nov 2010
Posts: 1406
Location: United States (IN)
Concentration: Strategy, Technology

Re: Remainder Problem
[#permalink]
Show Tags
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
_________________
Formula of Life > Achievement/Potential = k * Happiness (where k is a constant)
GMAT Club Premium Membership  big benefits and savings



Director
Joined: 03 Aug 2012
Posts: 710
Concentration: General Management, General Management
GMAT 1: 630 Q47 V29 GMAT 2: 680 Q50 V32
GPA: 3.7
WE: Information Technology (Investment Banking)

Re: Remainder Problem
[#permalink]
Show Tags
09 Aug 2013, 04:57
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



Senior Manager
Joined: 10 Jul 2013
Posts: 313

Re: Remainder Problem
[#permalink]
Show Tags
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
_________________
Asif vai.....



Director
Joined: 13 Mar 2017
Posts: 703
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)

Re: The remainder when 1+3+3^2+3^3+..........+3^200 is divided 1
[#permalink]
Show Tags
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
_________________
CAT 2017 (98.95) & 2018 (98.91) : 99th percentiler UPSC Aspirants : Get my app UPSC Important News Reader from Play store.
MBA Social Network : WebMaggu
Appreciate by Clicking +1 Kudos ( Lets be more generous friends.) What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish".



Director
Joined: 13 Mar 2017
Posts: 703
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)

Re: The remainder when 1+3+3^2+3^3+..........+3^200 is divided 1
[#permalink]
Show Tags
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
_________________
CAT 2017 (98.95) & 2018 (98.91) : 99th percentiler UPSC Aspirants : Get my app UPSC Important News Reader from Play store.
MBA Social Network : WebMaggu
Appreciate by Clicking +1 Kudos ( Lets be more generous friends.) What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish".



NonHuman User
Joined: 09 Sep 2013
Posts: 9896

Re: The remainder when 1+3+3^2+3^3+..........+3^200 is divided 1
[#permalink]
Show Tags
11 Feb 2019, 19:35
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: The remainder when 1+3+3^2+3^3+..........+3^200 is divided 1
[#permalink]
11 Feb 2019, 19:35






