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What is the remainder when 2^86 is divided by 9? [#permalink]
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16 Apr 2014, 00:19
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What is the remainder when 2^86 is divided by 9? (A) 1 (B) 2 (C) 3 (D) 4 (E) 8
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Last edited by Bunuel on 16 Apr 2014, 01:58, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.



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Re: Can anyone help me to solve this strategically? [#permalink]
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16 Apr 2014, 00:36
drkomal2000 wrote: Question 2: What is the remainder when 2^86 is divided by 9? (A) 1 (B) 2 (C) 3 (D) 4 (E) 8 I think it is D2^86 / 9 > 28 times (2^3) * (2^2) / 9 Remainder when 2^3 divided by 9 is 1 and remainder when 2^2 divided by 9 is 4 So we have that 1*1*1.......28 times * 4 / 9 > 4/9 > Remainder = 4 Rules for Posting  Please Read this Before Posting
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Re: What is the remainder when 2^86 is divided by 9? [#permalink]
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16 Apr 2014, 03:11
Sorry I don't quite get it. Are we trying to factorize by breaking down into factors with same power (in this case 2^2 / 3^2)?



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Re: What is the remainder when 2^86 is divided by 9? [#permalink]
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16 Apr 2014, 03:31
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Rule : Expression \(\frac{A*B*C*D*E}{K}\) will give the same remainder as the expression \(\frac{Ar*Br*Cr*Dr*Er}{K}\) where Ar, Br, Cr, Dr, Er are the remainders when divided by K individually. 2^86 / 9 can be simplified as \(\frac{2*2*2*2.......86 times}{9}\) 2*2*2 = 8. Remainder of 8/9 is 1. We can form 28 such groups of (2*2*2) and every group will give us the remainder as 1. After forming 28 groups we are left with 2*2 which when divided by 9 will give the remainder as 4 Now apply the rule cited above Remainder of 2^86/9 is the same as remainder of \(\frac{1*1*1*......28 times * 4}{9}\) > \(\frac{4}{9}\) > 4 Let me know if anything still unclear
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Re: What is the remainder when 2^86 is divided by 9? [#permalink]
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06 Jun 2014, 02:30
I tackled this problem as the following: First, notice that you are asked to deal with one digit number: 9; therefore the remainder is one digit number too(take a look at the answer choices). From this it follows that this question is about units digit of 2^86.



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Re: What is the remainder when 2^86 is divided by 9? [#permalink]
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08 Jun 2014, 02:56
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Narenn wrote: Rule : Expression \(\frac{A*B*C*D*E}{K}\) will give the same remainder as the expression \(\frac{Ar*Br*Cr*Dr*Er}{K}\) where Ar, Br, Cr, Dr, Er are the remainders when divided by K individually.
2^86 / 9 can be simplified as \(\frac{2*2*2*2.......86 times}{9}\)
2*2*2 = 8. Remainder of 8/9 is 1. We can form 28 such groups of (2*2*2) and every group will give us the remainder as 1. After forming 28 groups we are left with 2*2 which when divided by 9 will give the remainder as 4
Now apply the rule cited above
Remainder of 2^86/9 is the same as remainder of \(\frac{1*1*1*......28 times * 4}{9}\) > \(\frac{4}{9}\) > 4
Let me know if anything still unclear I'm sorry, but isn't 8/9 yield a remainder of 8?



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Re: What is the remainder when 2^86 is divided by 9? [#permalink]
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10 Jun 2014, 12:16
I am not sure if the way i approached it is correct. The cyclicity of 2 is 4. So 86/4 gives us a remainder of 2 which means the units digit of 2^86 will have a units digit of 4.



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Re: What is the remainder when 2^86 is divided by 9? [#permalink]
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10 Jun 2014, 12:43
amz14 wrote: I am not sure if the way i approached it is correct. The cyclicity of 2 is 4. So 86/4 gives us a remainder of 2 which means the units digit of 2^86 will have a units digit of 4. This approach is not correct, because cyclicity will only tell you about the last digit and not about the reminder. consider this 2^10 also ends with 4 i.e. 1024, but when it is divided by 9 the remainder is not 4 but 7.



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Re: What is the remainder when 2^86 is divided by 9? [#permalink]
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10 Jun 2014, 13:10
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I did it the following way: as 9*7 = 63, 64/63 will always give a remainder of 1. 64 is 2^6. So the new number becomes, [{2^(6*14)}*2^2]/9
=> {2^(6*14)}/9 will always give remainder 1. 2^2/9 will give e remainder of 4. So my answer will be D(4)



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Re: What is the remainder when 2^86 is divided by 9? [#permalink]
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14 Jun 2014, 18:46
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I used the Binomial Theory (which I admit I'm a little shaky on...)
1. We need to find a relationship between our dividend (2^86) and divisor (9)
2^86 = (2^2)*(2^84) = 4*(2^3)^28 = 4*(8)^28
We can expand this into binomial form as:
4*(91)^28
Every term will have a factor of 9 in it EXCEPT the last term (1)^28 which is just equal to 1. Don't forget to distribute the 4 out, and voila, we have our remainder of 4*(1) = 4 therefore answer D.



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Re: What is the remainder when 2^86 is divided by 9? [#permalink]
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14 Jun 2014, 19:32
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dcbark01 wrote: I used the Binomial Theory (which I admit I'm a little shaky on...)
1. We need to find a relationship between our dividend (2^86) and divisor (9)
2^86 = (2^2)*(2^84) = 4*(2^3)^28 = 4*(8)^28
We can expand this into binomial form as:
4*(91)^28
Every term will have a factor of 9 in it EXCEPT the last term (1)^28 which is just equal to 1. Don't forget to distribute the 4 out, and voila, we have our remainder of 4*(1) = 4 therefore answer D. I really like this method, I have a question though, and my sincere apologies if its stupid. When you do this step :2^86 = (2^2)*(2^84) = 4*(2^3)^28 = 4*(8)^28, I see where the (2^2) * (2^84) is coming from but I get confused where 4*(2^3)^28 = 4*(8)^28 is coming from. Step by step where is 4*(2^3)^28 coming from? Is there distribution happening?



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Re: What is the remainder when 2^86 is divided by 9? [#permalink]
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14 Jun 2014, 19:37
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When a power of 2 is divided by 9, the remainder start repeating itsel after 2^6 means from 2^1 to 2^6 the remainder goes like, 2,4,8,7,5,1 and then from 2^7 it start repeating itself. so if we divide 86 by 6 it gives 2 as remainder that means only 2*2 left which give 4 as remainder for the question.



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Re: What is the remainder when 2^86 is divided by 9? [#permalink]
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14 Jun 2014, 19:38
sagnik242 wrote: dcbark01 wrote: I used the Binomial Theory (which I admit I'm a little shaky on...)
1. We need to find a relationship between our dividend (2^86) and divisor (9)
2^86 = (2^2)*(2^84) = 4*(2^3)^28 = 4*(8)^28
We can expand this into binomial form as:
4*(91)^28
Every term will have a factor of 9 in it EXCEPT the last term (1)^28 which is just equal to 1. Don't forget to distribute the 4 out, and voila, we have our remainder of 4*(1) = 4 therefore answer D. I really like this method, I have a question though, and my sincere apologies if its stupid. When you do this step :2^86 = (2^2)*(2^84) = 4*(2^3)^28 = 4*(8)^28, I see where the (2^2) * (2^84) is coming from but I get confused where 4*(2^3)^28 = 4*(8)^28 is coming from. Step by step where is 4*(2^3)^28 coming from? Is there distribution happening? Yes its distribution of powers. 3*28=84



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Re: What is the remainder when 2^86 is divided by 9? [#permalink]
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01 Nov 2014, 13:49
Can someone help me figure out the next step with the method I tried to use? I'm having a hard time understanding the other methods used in this thread.
I found that 2^86 would end with a 6, but I'm unsure what to do after that. If the question asked what the remainder would be if divided by 10, this would be an easier question.



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What is the remainder when 2^86 is divided by 9? [#permalink]
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24 Oct 2015, 11:48
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GGMAT760 wrote: What is the remainder when 2^86 is divided by 9?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 8 \(2^86\) has a unit digit 4, 4/86 has a remainder 4 Answer (D)
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What is the remainder when 2^86 is divided by 9? [#permalink]
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24 Oct 2015, 12:09
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BrainLab wrote: GGMAT760 wrote: What is the remainder when 2^86 is divided by 9?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 8 \(2^86\) has a unit digit 4, 4/86 has a remainder 4 Answer (D) Again, you are making the mistake of going for digits to come up with the answer. Best way will be 2^84 = 2^2*(2^3)^28 = 4*(8)^28 = 4*(91)^28 All the terms in (91)^28 will be multiples of 9 except the last one (1)^28 = 28. Thus, you need to find the remainder when 4*(91)^28 is divided by 9 or 4*(1)^28 is divided by 9 or 4 is divided by 9, giving you a remainder of 4. Hope this helps.



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Re: What is the remainder when 2^86 is divided by 9? [#permalink]
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24 Oct 2015, 21:05
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GGMAT760 wrote: What is the remainder when 2^86 is divided by 9?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 8 we know that when 64 is divided by 9 , remainder is 1 . 2^86 = 2^2 * 2^84 = 4 * (2^6)^14 = 4 * 64^14 Now , remember that if X= A*B then R(X/K) = ( R(A/K) *R(B/K) ) /K for eg: 65/9 remainder is 2 65= 13*5 R(13/9) * R(5/9) = 4*5 = 20 and R(20/9) = 2 Back to original question now  2^86 = 2^2 * 2^84 = 4 * (2^6)^14 = 4 * ( 64^14 ) 4/9 = 4 remainder 64/9 =1 remainder so 4 * ( 64^14 ) / 9 remainder = 4 AnswerD
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Re: What is the remainder when 2^86 is divided by 9? [#permalink]
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19 Jul 2017, 10:52
we know cyclicity of 2 is 4. Now on dividing 86/4 we get 2 remainder for which unit's place of 2^86 should be 2^2=4. Hence dividing unit's place digit by 9 leaves remainder as 4. Option D



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Re: What is the remainder when 2^86 is divided by 9? [#permalink]
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20 Aug 2017, 23:23
8/9 remainder is 1? should it not be 8? Narenn wrote: Rule : Expression \(\frac{A*B*C*D*E}{K}\) will give the same remainder as the expression \(\frac{Ar*Br*Cr*Dr*Er}{K}\) where Ar, Br, Cr, Dr, Er are the remainders when divided by K individually.
2^86 / 9 can be simplified as \(\frac{2*2*2*2.......86 times}{9}\)
2*2*2 = 8. Remainder of 8/9 is 1. We can form 28 such groups of (2*2*2) and every group will give us the remainder as 1. After forming 28 groups we are left with 2*2 which when divided by 9 will give the remainder as 4
Now apply the rule cited above
Remainder of 2^86/9 is the same as remainder of \(\frac{1*1*1*......28 times * 4}{9}\) > \(\frac{4}{9}\) > 4
Let me know if anything still unclear



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Re: What is the remainder when 2^86 is divided by 9? [#permalink]
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