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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
_________________

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.

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

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.

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.

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*(9-1)^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.

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*(9-1)^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?

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.

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*(9-1)^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?

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.

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

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

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