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Intern  Joined: 22 Feb 2014
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What is the remainder when 2^86 is divided by 9?  [#permalink]

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42 00:00

Difficulty:   45% (medium)

Question Stats: 64% (01:30) correct 36% (01:53) wrong based on 548 sessions

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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

Originally posted by GGMAT760 on 16 Apr 2014, 00:19.
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: What is the remainder when 2^86 is divided by 9?  [#permalink]

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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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Intern  Joined: 06 Nov 2013
Posts: 3
Re: What is the remainder when 2^86 is divided by 9?  [#permalink]

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3
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.
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Re: Can anyone help me to solve this strategically?  [#permalink]

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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 D

2^86 / 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

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Manager  Joined: 04 Jan 2014
Posts: 93
Re: What is the remainder when 2^86 is divided by 9?  [#permalink]

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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)?
Intern  Joined: 06 Dec 2013
Posts: 5
Re: What is the remainder when 2^86 is divided by 9?  [#permalink]

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1
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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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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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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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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1
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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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?
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Re: What is the remainder when 2^86 is divided by 9?  [#permalink]

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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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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?

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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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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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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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*(9-1)^28

All the terms in (9-1)^28 will be multiples of 9 except the last one (-1)^28 = 28.

Thus, you need to find the remainder when 4*(9-1)^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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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

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

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1
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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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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gps5441 wrote:
8/9 remainder is 1? should it not be 8?

Yes, 8 divided by 9 yields the remainder of 8.
_________________ Re: What is the remainder when 2^86 is divided by 9?   [#permalink] 21 Aug 2017, 01:59

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