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

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

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New post 11 Sep 2018, 18:31
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


Sorry, but may I ask what is the formula for the equation, I have to solve a question similiar to that, but my question is 2^63 divided by 5.
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Re: What is the remainder when 2^86 is divided by 9?  [#permalink]

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New post 30 Sep 2018, 06:49
GGMAT760 wrote:
What is the remainder when 2^86 is divided by 9?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 8


I don't think that such questions will appear on the real test. Sorry, Veritis Prep be realistic.
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Re: What is the remainder when 2^86 is divided by 9?  [#permalink]

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New post 30 Sep 2018, 07:27
My approach was to check the ending numbers of powers of 2.

2^1 ends w. 2
2^2 ends w. 4
2^3 ends w.8
2^4 ends w.6
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2^5 ends w.2
2^6 ends w.4

So we see there is a 4 pair cycle that repeats. If 2^4 ends with 8, 2^40 also ends with 8 and 2^80 too. Thus 2^84 ends with 8. 2^85 with 6 and 2^86 with 2 (beginning of new 4 pair group).

Sounds logic to you guys?
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What is the remainder when 2^86 is divided by 9?  [#permalink]

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New post 03 Oct 2018, 21:19
alitariquet wrote:
I don't think that such questions will appear on the real test. Sorry, Veritis Prep be realistic.

alitariquet , I would hate for you and others who may believe the assertion above to dismiss this topic based on the belief that the topic will not be tested, only to be confronted with exactly that topic on the real test.

You may well decide that your time is better spent on other topics. Totally understandable. :-)

This similar question is from GMAT PREP 2

This similar question is from Magoosh.

This similar question is from MGMAT
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What is the remainder when 2^86 is divided by 9?  [#permalink]

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New post 03 Oct 2018, 21:21
GGMAT760 wrote:
What is the remainder when 2^86 is divided by 9?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 8

Weido wrote:
My approach was to check the ending numbers of powers of 2.

2^1 ends w. 2
2^2 ends w. 4
2^3 ends w.8
2^4 ends w.6
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2^5 ends w.2
2^6 ends w.4

So we see there is a 4 pair cycle that repeats. If 2^4 ends with 8, 2^40 also ends with 8 and 2^80 too. Thus 2^84 ends with 8. 2^85 with 6 and 2^86 with 2 (beginning of new 4 pair group).

Sounds logic to you guys?

Weido - Sounds logical but does not always work. :-)

In fact, using units digits cyclicity approach for remainders works every time only when the divisor is 2, 5, or 10.

Otherwise, take your good instinct one step further. Find the cyclicity of remainders.

See my post below. As mentioned, good instincts. :-)
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What is the remainder when 2^86 is divided by 9?  [#permalink]

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New post 03 Oct 2018, 21:22
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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

Find the cyclicity of the remainders.

The units digit cyclicity approach works every time only when the divisor is 2, 5, or 10

People who used units digits only in this question got lucky. ;)

Here is one problem in which units digit cyclicity does NOT give the right answer.

Cyclicity of remainders
Increase 2, power by power. Divide each number by 9 and find the remainder. Continue until you see a repeating pattern.

\(2^1=2\), and \(\frac{2}{9}=0,\) Remainder 2
\(2^2=4\), and \(\frac{4}{9}=0\) Remainder 4
\(2^3=8\), and \(\frac{8}{9}=0\) Remainder 8
\(2^4=16\), and \(\frac{16}{9}=1\) Remainder 7
\(2^5=32\), and \(\frac{32}{9}=3\) Remainder 5
\(2^6=64\), and \(\frac{64}{9}=7\) Remainder 1
\(2^7=128\), and \(\frac{128}{9}=14\) Remainder 2
\(2^8=256\), and \(\frac{256}{9}=28\) Remainder 4
\(2^9=512\), and \(\frac{512}{9}=56\) Remainder 8

The remainder pattern started to repeat after \(2^6\): (2, 4, 8, 7, 5, 1)

So the cyclicity of remainders is 6

Divide the exponent, 86, by cyclicity of 6.
If there is a remainder, that remainder number is the power of 2 that yields our answer.

(Example: exponent/cyclicity leaves a remainder of 5? We use 2\(^5\) to find our answer.)

If 86/6 leaves no remainder, then we find our answer at the end of the cycle, at 2\(^6.\)
\(\frac{86}{6}=14\) + remainder of 2, therefore:

\(2^{86}\), when divided by 9, will have the same remainder as \(2^2\)*

From above, when \(2^2\) is divided by 9,
the remainder is 4

Answer D

*Divide the exponent by the cycle number to figure out where, in the cycle of 6, the exponent would fall.
(6*14)=84, so 2\(^{84}\) would "hit" the cycle in the same position as 2\(^6\) (84 is a multiple of 6).
Start the cycle of 6 over.
2\(^{85}\) = same remainder as 2\(^1\), and
2\(^{86}\) = same remainder as 2\(^2\)
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What is the remainder when 2^86 is divided by 9? &nbs [#permalink] 03 Oct 2018, 21:22

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