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Remainder problem [#permalink]
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01 Sep 2009, 23:52
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What is the remainder when 2^92 / 7 ?



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Re: Remainder problem [#permalink]
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02 Sep 2009, 00:51
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]
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02 Sep 2009, 10:26
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]
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02 Sep 2009, 10:35
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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]
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02 Sep 2009, 18:33
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]
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03 Sep 2009, 08:34
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]
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03 Sep 2009, 11:18
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]
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04 Sep 2009, 00:06
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]
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26 Sep 2009, 02:22
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]
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26 Sep 2009, 13:04
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 12^4 / 7 => R 2 2^5 / 7 => R 4
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Re: Remainder problem [#permalink]
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26 Sep 2009, 13:24
I too get the answer to be 4..



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Re: Remainder problem [#permalink]
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26 Sep 2009, 19:55
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?
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Re: Remainder problem [#permalink]
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27 Sep 2009, 01:41
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]
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28 Sep 2009, 03:16
anybody clarifying my concept here?
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Re: Remainder problem [#permalink]
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28 Sep 2009, 03:52
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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? 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]
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28 Sep 2009, 19:59
I am sorry i didnt understand Where did modulus come from?!
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Re: Remainder problem [#permalink]
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29 Sep 2009, 00:36
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]
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29 Sep 2009, 04:56
Yup, remainder is 4.



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Re: Remainder problem [#permalink]
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13 Nov 2009, 09:54
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]
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31 Jan 2010, 05:40
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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