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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, 03:22
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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, 19:42.
Last edited by Bunuel on 17 Oct 2012, 03:22, 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, 20:46
3
5
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, 20:18
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, 20:11
2
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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Re: Remainder problem.  [#permalink]

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New post 17 Oct 2012, 00:17
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, 00:29
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, 02:48
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, 02:29.
Last edited by EvaJager on 17 Oct 2012, 02:48, edited 1 time in total.
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Re: Remainder problem.  [#permalink]

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New post 17 Oct 2012, 02:43
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, 03:01
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, 03:09
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, 10:34
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, 22:39
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, 01:56
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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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Re: What is the remainder when you divide 2^200 by 7?  [#permalink]

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New post 24 May 2020, 08:00
VeritasKarishma 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/)


VeritasKarishma : why this question can't be solved with cyclicity approach, here is similar question thread i am copying, and is solved by cylicity. why e need to bring remainder cylicity.

https://gmatclub.com/forum/what-is-the- ... l#p2066643
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Re: What is the remainder when you divide 2^200 by 7?  [#permalink]

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New post 25 May 2020, 03:07
1
rishab0507 wrote:
VeritasKarishma 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/)


VeritasKarishma : why this question can't be solved with cyclicity approach, here is similar question thread i am copying, and is solved by cylicity. why e need to bring remainder cylicity.

https://gmatclub.com/forum/what-is-the- ... l#p2066643



Cyclicity tells you the units digit of exponential expressions. Now think about this - if you know the units digit of a number, can you say what the remainder is upon division by say 7?

e.g. What is the remainder when 792947 is divided by 7? Would you say the remainder here is 0? It is not.
The remainder depends on what the actual number is, not just the units digit.

Only in case of division by 2, 5 or 10 does the units digit give us the remainder.

Check out these two posts:
https://www.veritasprep.com/blog/2015/1 ... questions/
https://www.veritasprep.com/blog/2015/1 ... ns-part-2/
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Re: What is the remainder when you divide 2^200 by 7?  [#permalink]

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New post 25 May 2020, 11:12
VeritasKarishma wrote:
rishab0507 wrote:
VeritasKarishma 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/)


VeritasKarishma : why this question can't be solved with cyclicity approach, here is similar question thread i am copying, and is solved by cylicity. why e need to bring remainder cylicity.

https://gmatclub.com/forum/what-is-the- ... l#p2066643



Cyclicity tells you the units digit of exponential expressions. Now think about this - if you know the units digit of a number, can you say what the remainder is upon division by say 7?

e.g. What is the remainder when 792947 is divided by 7? Would you say the remainder here is 0? It is not.
The remainder depends on what the actual number is, not just the units digit.

Only in case of division by 2, 5 or 10 does the units digit give us the remainder.

Check out these two posts:
https://www.veritasprep.com/blog/2015/1 ... questions/
https://www.veritasprep.com/blog/2015/1 ... ns-part-2/



thanks, It was indeed a good learning. I never paid attention i was solving only 2,5, 10 divisor problems .I think for other divisors i need to follow split method, not cyclicity to find remainders
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Re: What is the remainder when you divide 2^200 by 7?   [#permalink] 25 May 2020, 11:12

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

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