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Re: Remainder Problem [#permalink]
05 Nov 2009, 12:21
4
This post received KUDOS
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
Re: Remainder Problem [#permalink]
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
Re: Remainder Problem [#permalink]
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!!
Re: Remainder Problem [#permalink]
12 Nov 2009, 05:42
1
This post received KUDOS
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.
Re: Remainder Problem [#permalink]
09 Aug 2013, 04:57
1
This post received KUDOS
1
This post was BOOKMARKED
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 _________________
Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________
Re: The remainder when 1+3+3^2+3^3+..........+3^200 is divided 1 [#permalink]
05 Oct 2014, 10:21
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