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What is the remainder when 2^86 is divided by 9?

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Intern
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Joined: 21 Jul 2017
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Re: What is the remainder when 2^86 is divided by 9? [#permalink]

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New post 21 Aug 2017, 03:18
ronr34 wrote:
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?


How do you get the number 28? I'm unclear on that.
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Re: What is the remainder when 2^86 is divided by 9? [#permalink]

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New post 21 Aug 2017, 23:07
In all the remainder of a Number with power problems, the easiest way to solve the question is by finding the cyclicity of remainders.
Remainder of 2/9 = 2
Remainder of 2^2/9 = 4
Remainder of 2^3/9 =8
Remainder of 2^4/9 =7
Remainder of 2^5/9 =5
Remainder of 2^6/9 =1
Remainder of 2^7/9 =2
Remainder of 2^8/9 =4

The remainder start repeating after 2^6 so the cyclicity is 6.
So 2^86 =2^(84+2)will have the same remainder as 2^2 when divided by 9. So the remainder is 4.
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What is the remainder when 2^86 is divided by 9? [#permalink]

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New post 04 Sep 2017, 16:17
We can also use a fermat's little theorem with a small customization as 9 is not a prime number
9 = 3 * 3
so 2^86 = 2^43 * 2^43
according to fermat's theorem
a^(p-1) = 1 mod p where a integer and p prime
so (2^2)/3=1 mod 3

2^43 * 2^43 = (2^42+1)*(2^42+1) (as we have two 3s) = so 2 * 2 = 4
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Re: What is the remainder when 2^86 is divided by 9? [#permalink]

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New post 19 Sep 2017, 04:50
Rem|2^86/9|

Now Rem|(2^3)/9| = -1

Rem|2^86/9| = Rem|(2^84 x 2^2)/9|

= Rem|[(2^3)^28 x 2^2]/9|

= (-1)^28 x 4

= 4

Final Remainder is 4
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Re: What is the remainder when 2^86 is divided by 9? [#permalink]

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New post 19 Sep 2017, 05:36
remainder when 2^86 is divided by 9
= 2^2(43)/9
= 4^43/9
= -5^43/9

now 5 has cyclicity of either 0 or 5, when power is odd , last digit will be 5
Therefore remainder=-5
since it is negative we need to add it to the divisor
therefore remainder= 9-5=4
Re: What is the remainder when 2^86 is divided by 9?   [#permalink] 19 Sep 2017, 05:36

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