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

3

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

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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

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+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

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