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Re: Remainder problem [#permalink]
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..

Re: Remainder problem [#permalink]
02 Sep 2009, 09:35

1

This post received KUDOS

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

Re: Remainder problem [#permalink]
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 _________________

Bhushan S. If you like my post....Consider it for Kudos

Re: Remainder problem [#permalink]
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? _________________

Re: Remainder problem [#permalink]
28 Sep 2009, 02:52

1

This post received KUDOS

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

Re: Remainder problem [#permalink]
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)

Re: Remainder problem [#permalink]
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 _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

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Re: Remainder problem
[#permalink]
31 Jan 2010, 04:40

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...