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Remainder problem

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Remainder problem [#permalink] New post 01 Sep 2009, 22:52
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What is the remainder when 2^92 / 7 ?
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Re: Remainder problem [#permalink] New post 01 Sep 2009, 23:51
2^3=8=1 in mod 7
so
(2^3)^30=2^90=1 in mod 7
2^92=2^90 x 2^2 = 1 x 4 = 4 in mod 7

4
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Re: Remainder problem [#permalink] New post 02 Sep 2009, 09:26
yup..agree with maliyeci..

so we cannot use power cycle of 2 ??...then we should get remainder as 6 because the last digit in the power cycle of 2 is 6...or is mod the best approach..hmmm..need to think..
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Re: Remainder problem [#permalink] New post 02 Sep 2009, 09:35
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2/7 = 2
4/7 = 4
8/7 = 1
16/7 = 2
So 2, 4, 1 remainders repeat in cyclic order when the powers of 2 are divided by 7.
2^92/7
92/3 = 30 + 2/3
So we have to find the remainder for 2^2 divided by 7.
Answer is 4
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Re: Remainder problem [#permalink] New post 02 Sep 2009, 17:33
ya as Economist said the power cycle method cannot be used here.

Aleehsgonji method is very good one in such cases.
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Re: Remainder problem [#permalink] New post 03 Sep 2009, 07:34
maliyeci wrote:
2^3=8=1 in mod 7
so
(2^3)^30=2^90=1 in mod 7
2^92=2^90 x 2^2 = 1 x 4 = 4 in mod 7

4


Quick question : does that mean that (2^3)^30 has the same remainder that 2^3 when divided by 7, same as (2^3)^10, (2^3)^20, (2^3)^100.....?
Thx
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Re: Remainder problem [#permalink] New post 03 Sep 2009, 10:18
maliyeci wrote:
2^3=8=1 in mod 7
so
(2^3)^30=2^90=1 in mod 7
2^92=2^90 x 2^2 = 1 x 4 = 4 in mod 7

4


Maliyeci, I am really very slow in understanding this kind of problems.
Do I understand your thinking right?

2^3=8
8 divided by 7 gives remainder 1
Therefire, (2^30)^3 gives remainder 1 (by the way, what does "1 in mod 7" mean?)

2^2=4
4 is smaller than 7, hence no remainder is possible and we count 4 itself

As a result, (2^30)^3*2^2=1*4

This is just my guess...

Does anyone has a link to giudance on remainder problems, please?... Like MGMAT's ot anything.. I would be so very much grateful..
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Re: Remainder problem [#permalink] New post 03 Sep 2009, 23:06
CasperMonday wrote:
maliyeci wrote:
2^3=8=1 in mod 7
so
(2^3)^30=2^90=1 in mod 7
2^92=2^90 x 2^2 = 1 x 4 = 4 in mod 7

4


Maliyeci, I am really very slow in understanding this kind of problems.
Do I understand your thinking right?

2^3=8
8 divided by 7 gives remainder 1
Therefire, (2^30)^3 gives remainder 1 (by the way, what does "1 in mod 7" mean?)

2^2=4
4 is smaller than 7, hence no remainder is possible and we count 4 itself

As a result, (2^30)^3*2^2=1*4

This is just my guess...

Does anyone has a link to giudance on remainder problems, please?... Like MGMAT's ot anything.. I would be so very much grateful..

Exactly yes :)

defoue wrote:
Quick question : does that mean that (2^3)^30 has the same remainder that 2^3 when divided by 7, same as (2^3)^10, (2^3)^20, (2^3)^100.....?
Thx
maliyeci wrote:
2^3=8=1 in mod 7
so
(2^3)^30=2^90=1 in mod 7
2^92=2^90 x 2^2 = 1 x 4 = 4 in mod 7

4


Quick question : does that mean that (2^3)^30 has the same remainder that 2^3 when divided by 7, same as (2^3)^10, (2^3)^20, (2^3)^100.....?
Thx


Exactly yes too :)
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Re: Remainder problem [#permalink] New post 26 Sep 2009, 01:22
HUH?Guys..wait..im so confused..what?Why?how?Why is the power cycle of two not used her..we would get the remainder as 6..
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Re: Remainder problem [#permalink] New post 26 Sep 2009, 12:04
tejal777 wrote:
HUH?Guys..wait..im so confused..what?Why?how?Why is the power cycle of two not used her..we would get the remainder as 6..


the question doesnot ask us to check cyclicity of 2.....as we r interested in what will happen if divided by 7....hence cyclicity when divided by 7 to this series is 3...hence 92 / 3 ...remainder = 2...which correspondes to 4

cyclicity table for 2^x / 7
for x = [1,2,3,....]
2^1 / 7 => R 2
2^2 / 7 => R 4
2^3 / 7 => R 1

2^4 / 7 => R 2
2^5 / 7 => R 4
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Re: Remainder problem [#permalink] New post 26 Sep 2009, 12:24
I too get the answer to be 4..
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Re: Remainder problem [#permalink] New post 26 Sep 2009, 18:55
I am MAJORLY confused..okay lets start from the beginning..

I had come across this question:
What is the remainder: 7^548/10
We know 7 has cycles of 4: 7,9,3,1,7,9,3,...
So,548=136 x 4 +4.Therefore remainder is 1.Correct ans.

Amother question:
What is the remainder: 7^131/5
Again the cycles theory,and we get the remainder as 3.Correct ans.

Now,coming back to our question:
2^92/7

Cycles theory of 2: 2,4,8,6,2,4,8...
In this way the remainder should come 6.

So the primary question is why is the third question different from the first two??Why are we not proceeding in the same way? :shock:
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Re: Remainder problem [#permalink] New post 27 Sep 2009, 00:41
7 = 2^3 - 1.

Expressing, 92 in terms of 3 we get 2^92 = 2^90 * 2^2

2^90 mod 7 = 1

So we are left with 2^2 = 4.

So I wud go with option 4
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Re: Remainder problem [#permalink] New post 28 Sep 2009, 02:16
anybody clarifying my concept here?
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Re: Remainder problem [#permalink] New post 28 Sep 2009, 02:52
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tejal777 wrote:
I am MAJORLY confused..okay lets start from the beginning..

I had come across this question:
What is the remainder: 7^548/10
We know 7 has cycles of 4: 7,9,3,1,7,9,3,...
So,548=136 x 4 +4.Therefore remainder is 1.Correct ans.

Amother question:
What is the remainder: 7^131/5
Again the cycles theory,and we get the remainder as 3.Correct ans.

Now,coming back to our question:
2^92/7

Cycles theory of 2: 2,4,8,6,2,4,8...
In this way the remainder should come 6.

So the primary question is why is the third question different from the first two??Why are we not proceeding in the same way? :shock:

You were really confused :D
You wrote there the cycles of 2 but wrote in the modulus of 10. Be aware that the answer is for modulus 7. Then cycles become

2, 4, 1, 2, 4, 1 ;)
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Re: Remainder problem [#permalink] New post 28 Sep 2009, 18:59
I am sorry i didnt understand :(
Where did modulus come from?! :oops:
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Re: Remainder problem [#permalink] New post 28 Sep 2009, 23:36
question asks us that what the remainder is when 2^n divided by 7. This means it asks us the results of 2^n's in modulus 7 or mod 7.
So lets dig this example.
2^1=2 it is 2 in mod 7 and 2 in mod 10 (i.e. when divided by 7 the remainder is 2, when divided by 10 the remainder is 2)
2^2=4 it is 4 in mod 7 and 4 in mod 10 (i.e. when divided by 7 the remainder is 4, when divided by 10 the remainder is 4)
2^3=8 it is 1 in mod 7 and 8 in mod 10 (i.e. when divided by 7 the remainder is 1, when divided by 10 the remainder is 8)
2^4=16 it is 2 in mod 7 and 6 in mod 10 (i.e. when divided by 7 the remainder is 2, when divided by 10 the remainder is 6)

I hope you got it ;)
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Re: Remainder problem [#permalink] New post 29 Sep 2009, 03:56
Yup, remainder is 4.
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Re: Remainder problem [#permalink] New post 13 Nov 2009, 08:54
Woah, some of these topics are really complicated with cycles and moduluses...

I really don't think any of that is necessary:

First we do need to find the pattern:

2^1 / 7 = R2
2^2 / 7 = R4
2^3 / 7 = R1
2^4 / 7 = R2
2^5 / 7 = R4
2^6 / 7 = R1

So it is pretty safe to say that every third term the pattern of 2,4,1 repeats itself.

Seeeing that the problem is 2^92nd power, lets divide 92 by 3 and find the remainder:

R = 2

So the remainder will be the second term in the pattern of 2,4,1

ANSWER: R = 4
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Re: Remainder problem [#permalink] New post 31 Jan 2010, 04:40
tkarthi4u wrote:
What is the remainder when 2^92 / 7 ?


2^92 = 2^90 * 2^2 = 4 * 2 ^ (3*30) = 4 * 8^30

Therefore (2^92) MOD 7 = (4 * 8^30) MOD 7 = 4 MOD 7 * (8 MOD 7)^30 = 4 * 1^30 = 4
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Re: Remainder problem   [#permalink] 31 Jan 2010, 04:40
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