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What is the remainder when you divide 2^200 by 7?

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What is the remainder when you divide 2^200 by 7?  [#permalink]

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New post Updated on: 17 Oct 2012, 04:22
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What is the remainder when you divide 2^200 by 7?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

my approach :

2^x has a cyclicity of 4
Therefore, Rem(200/4) = 0

Rem(2^0/7) =1

Am i missing something here?

OA is D

Originally posted by g3kr on 16 Oct 2012, 20:42.
Last edited by Bunuel on 17 Oct 2012, 04:22, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Remainder problem.  [#permalink]

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New post 16 Oct 2012, 21:46
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g3kr wrote:
I am confused . Please help!

What is the remainder when you divide 2^200 by 7?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

my approach :

2^x has a cyclicity of 4
Therefore, Rem(200/4) = 0

Rem(2^0/7) =1

Am i missing something here?

OA is D


I think you are getting confused between cyclicity of last digit and cyclicity of remainders.

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64

If you see, the last digits are 2, 4, 8, 6 i.e. cyclicity of 4.

On the other hand,

2^1/7 Rem = 2
2^2/7 Rem = 4
2^3/7 Rem = 1
2^4/7 Rem = 2
2^5 / 7 Rem = 4
2^6/7 Rem = 1

Here the cyclicity is 3.
\(2^{198}\) will give a remainder of 1. \(2^{200}\) gives a remainder of 4.

Or, you can easily use binomial theorem here.
\(\frac{2^{200}}{7} = 2*2*\frac{2^{198}}{7} = 4*\frac{8^{66}}{7} = 4*\frac{(7 + 1)^{66}}{7}\)

Remainder must be 4. (Check out this link: http://www.veritasprep.com/blog/2011/05 ... ek-in-you/)
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Re: Remainder problem.  [#permalink]

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New post 16 Oct 2012, 21:18
4
1
\(2^{200} = (2^{5})^{40}\)

32 =28 +4

\(= (M7+4)^{40}\) ; M7 is a multiple of 7

= M7+4

so the remainder is 4.

Hope it helps.
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Re: Remainder problem.  [#permalink]

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New post 16 Oct 2012, 21:11
1
2
Actually the cyclicity is 3:

(2^0)/7 = 0 R 1
(2^1)/7 = 0 R 2
(2^2)/7 = 0 R 4
(2^3)/7 = 1 R 1
(2^4)/7 = 2 R 2
(2^5)/7 = 4 R 4
...

The pattern is such that the remainder is 4 for every third one and we know there are 201 numbers between 0 and 200. Since 201 is a multiple of 3, D is in fact correct.
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Re: Remainder problem.  [#permalink]

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New post 16 Oct 2012, 21:19
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Re: Remainder problem.  [#permalink]

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New post 17 Oct 2012, 01:17
1
mindmind wrote:
\(2^{200} = (2^{5})^{40}\)

32 =28 +4

\(= (M7+4)^{40}\) ; M7 is a multiple of 7

= M7+4

so the remainder is 4.

Hope it helps.


\((M7+4)^{40}=M7+4^{40}\) then you have to find the remainder of \(4^{40}\) when divided by 7.

Instead of taking \(32 = 28 +4\), look for a power of 2 which gives a remainder of 1 when divided by 7.
The smallest one is \(8 = 7 + 1\).

Therefore, \(2^{200}=(2^3)^{66}\cdot{2^2}=8^{66}\cdot{4}=(7+1)^{66}\cdot{4}=(M7+1)\cdot4=M7+4\).
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Re: Remainder problem.  [#permalink]

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New post 17 Oct 2012, 01:29
EvaJager wrote:
mindmind wrote:
\(2^{200} = (2^{5})^{40}\)

32 =28 +4

\(= (M7+4)^{40}\) ; M7 is a multiple of 7

= M7+4

so the remainder is 4.

Hope it helps.


\((M7+4)^{40}=M7+4^{40}\) then you have to find the remainder of \(4^{40}\) when divided by 7.

Instead of taking \(32 = 28 +4\), look for a power of 2 which gives a remainder of 1 when divided by 7.
The smallest one is \(8 = 7 + 1\).

Therefore, \(2^{200}=(2^3)^{66}\cdot{2^2}=8^{66}\cdot{4}=(7+1)^{66}\cdot{4}=(M7+1)\cdot4=M7+4\).


Please help me understand better :

The remainder of 4^40 would be same as that of 4 .
Is my approach wrong? Can you provide similar examples, where it is not.
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Re: Remainder problem.  [#permalink]

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New post Updated on: 17 Oct 2012, 03:48
1
mindmind wrote:
EvaJager wrote:
mindmind wrote:
\(2^{200} = (2^{5})^{40}\)

32 =28 +4

\(= (M7+4)^{40}\) ; M7 is a multiple of 7

= M7+4

so the remainder is 4.

Hope it helps.


\((M7+4)^{40}=M7+4^{40}\) then you have to find the remainder of \(4^{40}\) when divided by 7.

Instead of taking \(32 = 28 +4\), look for a power of 2 which gives a remainder of 1 when divided by 7.
The smallest one is \(8 = 7 + 1\).

Therefore, \(2^{200}=(2^3)^{66}\cdot{2^2}=8^{66}\cdot{4}=(7+1)^{66}\cdot{4}=(M7+1)\cdot4=M7+4\).


Please help me understand better :

The remainder of 4^40 would be same as that of 4 .
Is my approach wrong? Can you provide similar examples, where it is not.


In this case, \(4^{40}\) is a \(M7+4\) : \(\,\,4^{40}=2^{80}\) and \(80=M3+2\) (the cycle is 3, because \(2^{3}=8=M7+1\)).
\(4^{40}=2^{80}=(2^3)^{26}\cdot{2^2}=(M7+1)\cdot{4}=M7+4\).
Or - \(4^3=64=M7+1\), therefore \(4^{40}=(4^3)^{13}\cdot{4}=(M7+1)^{13}\cdot{4}=(M7+1)\cdot{4}=M7+4\).

Consider for example \(5^{40}\): \(5=7\cdot{0}+5=M7+5\) and \(5^3=125=126-1=7\cdot{18}-1=M7-1=M7+6\).
\(5^{40}=(5^3)^{13}\cdot{5}=(M7-1)^{13}\cdot{5}=(M7-1)\cdot{5}=M7-5=M7+2\) and not \(M7+5\).
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Originally posted by EvaJager on 17 Oct 2012, 03:29.
Last edited by EvaJager on 17 Oct 2012, 03:48, edited 1 time in total.
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Re: Remainder problem.  [#permalink]

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New post 17 Oct 2012, 03:43
Please help me understand better :

The remainder of 4^40 would be same as that of 4 .
Is my approach wrong? Can you provide similar examples, where it is not.[/quote]

In this case, \(4^{40}\) is a \(M7+4\) : \(\,\,4^{40}=2^{80}\) and \(80=M3+2\) (the cycle is 3, because \(2^{3}=8=M7+1\)).
\(4^{40}=2^{80}=(2^3)^{26}\cdot{2^2}=(M7+1)\cdot{4}=M7+4\).

Consider for example \(5^{40}\): \(5=7\cdot{0}+5=M7+5\) and \(5^3=125=126-1=7\cdot{18}-1=M7-1=M7+6\).
\(5^{40}=(5^3)^{13}\cdot{5}=(M7-1)^{13}\cdot{5}=(M7-1)\cdot{5}=M7-5=M7+2\) and not \(M7+5\).[/quote]

Yes, Agreed
So I should consider : Something near to Remainder 1

Any Trend where, (number or prime )^n divided by (other no, prime) gives the same remainder as the original no. or prime
eg : 4^40 divided by 7 give a remainder of 4.
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Re: Remainder problem.  [#permalink]

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New post 17 Oct 2012, 04:01
mindmind wrote:
Please help me understand better :

The remainder of 4^40 would be same as that of 4 .
Is my approach wrong? Can you provide similar examples, where it is not.


In this case, \(4^{40}\) is a \(M7+4\) : \(\,\,4^{40}=2^{80}\) and \(80=M3+2\) (the cycle is 3, because \(2^{3}=8=M7+1\)).
\(4^{40}=2^{80}=(2^3)^{26}\cdot{2^2}=(M7+1)\cdot{4}=M7+4\).

Consider for example \(5^{40}\): \(5=7\cdot{0}+5=M7+5\) and \(5^3=125=126-1=7\cdot{18}-1=M7-1=M7+6\).
\(5^{40}=(5^3)^{13}\cdot{5}=(M7-1)^{13}\cdot{5}=(M7-1)\cdot{5}=M7-5=M7+2\) and not \(M7+5\).[/quote]

Yes, Agreed
So I should consider : Something near to Remainder 1

Any Trend where, (number or prime )^n divided by (other no, prime) gives the same remainder as the original no. or prime
eg : 4^40 divided by 7 give a remainder of 4.[/quote]

I am not sure what you mean here:
Any Trend where, (number or prime )^n divided by (other no, prime) gives the same remainder as the original no. or prime
eg : 4^40 divided by 7 give a remainder of 4.


\(3^4=81=11\cdot{7}+4\). \(7\) is prime, but the remainder is neither \(3\), nor a prime.
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Re: Remainder problem.  [#permalink]

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New post 17 Oct 2012, 04:09
EvaJager wrote:
mindmind wrote:
Please help me understand better :

The remainder of 4^40 would be same as that of 4 .
Is my approach wrong? Can you provide similar examples, where it is not.


In this case, \(4^{40}\) is a \(M7+4\) : \(\,\,4^{40}=2^{80}\) and \(80=M3+2\) (the cycle is 3, because \(2^{3}=8=M7+1\)).
\(4^{40}=2^{80}=(2^3)^{26}\cdot{2^2}=(M7+1)\cdot{4}=M7+4\).

Consider for example \(5^{40}\): \(5=7\cdot{0}+5=M7+5\) and \(5^3=125=126-1=7\cdot{18}-1=M7-1=M7+6\).
\(5^{40}=(5^3)^{13}\cdot{5}=(M7-1)^{13}\cdot{5}=(M7-1)\cdot{5}=M7-5=M7+2\) and not \(M7+5\).


Yes, Agreed
So I should consider : Something near to Remainder 1

Any Trend where, (number or prime )^n divided by (other no, prime) gives the same remainder as the original no. or prime
eg : 4^40 divided by 7 give a remainder of 4.[/quote]

I am not sure what you mean here:
Any Trend where, (number or prime )^n divided by (other no, prime) gives the same remainder as the original no. or prime
eg : 4^40 divided by 7 give a remainder of 4.


\(3^4=81=11\cdot{7}+4\). \(7\) is prime, but the remainder is neither \(3\), nor a prime.[/quote]


Yes.. I was referring to 3.. and not the remainder 4..
But I got your point.. thanks..
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Re: What is the remainder when you divide 2^200 by 7?  [#permalink]

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New post 26 Oct 2012, 11:34
Another approach is the Theorem of Remainder
2^200 = 4*(2^3)^66;7 = 2^3 -1
Theorem of Remainder of f(X)/X-a is f(a)
2^200 = 4*f(X)/X-a X=2^3 a = 1 ==> Remainder of f(X)/X-a = 1
Remainder of 2^200 divided by7 is 4*1 = 4
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Re: What is the remainder when you divide 2^200 by 7?  [#permalink]

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New post 08 Aug 2013, 23:39
Easiest and the smallest possible solution ever:

Keep your mind open while dealing with such question.It is not that you have to be math pro for that.

REM(2^200/7) [ REM(x/y) means remainder when x is divided by y]

We know that : Rem when 8 is divided by 7 is '1'.

Also by powers of 2 we can reach to 8.Using this concept:

[ (2^3)^198 * 2^2] /7

[ (8)^198 * 2^2] /7

Since REM(8/7) =1

We are left with REM(4/7) = 4
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Re: Remainder problem.  [#permalink]

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New post 09 Aug 2013, 02:56
VeritasPrepKarishma wrote:
g3kr wrote:
I am confused . Please help!

What is the remainder when you divide 2^200 by 7?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

my approach :

2^x has a cyclicity of 4
Therefore, Rem(200/4) = 0

Rem(2^0/7) =1

Am i missing something here?

OA is D


I think you are getting confused between cyclicity of last digit and cyclicity of remainders.

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64

If you see, the last digits are 2, 4, 8, 6 i.e. cyclicity of 4.

On the other hand,

2^1/7 Rem = 2
2^2/7 Rem = 4
2^3/7 Rem = 1
2^4/7 Rem = 2
2^5 / 7 Rem = 4
2^6/7 Rem = 1

Here the cyclicity is 3.
\(2^{198}\) will give a remainder of 1. \(2^{200}\) gives a remainder of 4.

Or, you can easily use binomial theorem here.
\(\frac{2^{200}}{7} = 2*2*\frac{2^{198}}{7} = 4*\frac{8^{66}}{7} = 4*\frac{(7 + 1)^{66}}{7}\)

Remainder must be 4. (Check out this link: http://www.veritasprep.com/blog/2011/05 ... ek-in-you/)



...
2^200 = (2^3)^66 × 2^2 = (7+1)^66 × 4
1^66 × 4 = 4 (Answer)

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Re: What is the remainder when you divide 2^200 by 7?  [#permalink]

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