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What is the remainder when 2^83 is divided by 9?

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What is the remainder when 2^83 is divided by 9?  [#permalink]

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New post 05 Sep 2018, 07:10
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A
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C
D
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Difficulty:

  55% (hard)

Question Stats:

47% (01:20) correct 53% (00:43) wrong based on 81 sessions

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What is the remainder when 2^83 is divided by 9?

(A) -4
(B) 0
(C) 4
(D) 5
(E) 8

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What is the remainder when 2^83 is divided by 9?  [#permalink]

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New post 05 Sep 2018, 07:28
2^83/9
=(2^2*2^81)/9
= (4*(2^3)^27)/9
= (4*8^27)/9
= (4*(9-1)^27)/9

Leaving us with a negative remainder of -4, since remainder can't be negative; the remainder is 9-4=5!

For further explanation check out: https://www.veritasprep.com/blog/2011/0 ... ek-in-you/

Similar question with better explanation:
https://gmatclub.com/forum/what-is-the- ... 70273.html
https://gmatclub.com/forum/what-is-the- ... 07423.html
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Re: What is the remainder when 2^83 is divided by 9?  [#permalink]

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New post 05 Sep 2018, 07:31
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What is the remainder when 2^83 is divided by 9?

When a power of 2 is divided by 9, the remainder start repeating itsel after 2^6:

2^1 => remainder = 2
2^2 => remainder = 4
2^3 => remainder = 8
2^4 => remainder = 7
2^5 => remainder = 5
2^6 => remainder = 1

2^7 => remainder = 2

....

83/6 gives 5 as remainder => 2^83 and 2^5 have the same remainder when divided by 9..

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Re: What is the remainder when 2^83 is divided by 9?  [#permalink]

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New post 05 Sep 2018, 07:38
T1101 wrote:
What is the remainder when 2^83 is divided by 9?

(A) -4
(B) 0
(C) 4
(D) 5
(E) 8


\(2^6 = 64\)

\(\frac{2^6}{9} = Remainder \ 1\)

Now, \(2^{83} = 2^{6*13 + 5}\)

\(2^{78}\) will give a remainder of \(1\) when divided by \(9\) and \(2^5 = 32\) when divided by \(9\) will give a remainder of \(5\), Thus Answer must be (D) 5
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What is the remainder when 2^83 is divided by 9?  [#permalink]

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New post 05 Sep 2018, 11:18
T1101 wrote:
What is the remainder when 2^83 is divided by 9?

(A) -4
(B) 0
(C) 4
(D) 5
(E) 8


OA : D

\(2^{83}= 2^2*(2^3)^{27}=2^2*(9-1)^{27}\)

Remainder of \((9-1)^{27}\), when divided by \(9\) would be \(8 \quad (9-1)\).

So \((9-1)^{27}\) can be written as \(9m+8\), where \(m\) is a positive integer.

\(2^2*(9-1)^{27}=2^2*(9m+8)=4*9m+32\).

\(4*9m+32\) or \(32\) divided by \(9\) will leave \(5\) as remainder.
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Re: What is the remainder when 2^83 is divided by 9?  [#permalink]

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New post 15 Sep 2018, 10:03
Hi,

I tried to apply a formula which goes this way:
2*2*2 = 8. Remainder of 8/9 is 1. We can form 27 such groups of (2*2*2) and every group will give us the remainder as 1. After forming 27 groups we are left with 2*2 which when divided by 9 will give the remainder as 4.

I don't understand why this formula doesn't yield the right result. Can anybody help? Many thanks in advance!!
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Re: What is the remainder when 2^83 is divided by 9?  [#permalink]

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New post 19 Sep 2018, 01:33
Peddy wrote:
Hi,

I tried to apply a formula which goes this way:
2*2*2 = 8. Remainder of 8/9 is 1. We can form 27 such groups of (2*2*2) and every group will give us the remainder as 1. After forming 27 groups we are left with 2*2 which when divided by 9 will give the remainder as 4.

I don't understand why this formula doesn't yield the right result. Can anybody help? Many thanks in advance!!


Hi Peddy,

hope i can help you! I have done this question with the same approach as described here: https://www.veritasprep.com/blog/2011/0 ... ek-in-you/

\(\frac{{2}^{83}}{9}\)

=\(\frac{(2^2*{2}^{81})}{9}\)

=\(\frac{4*({2}^{3})^{27}}{9}\)

=\(\frac{4*({8})^{27}}{9}\)

=\(\frac{4*({9-1})^{27}}{9}\)

So \(4*9^{27}\) will yield a reminder of 0.
And \(4*(-1)^{27}\)= -4. But a reminder can not be negative! Therefore 4 is not the reminder... The reminder is arrived by 9-4=5

Does it help?
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Re: What is the remainder when 2^83 is divided by 9? &nbs [#permalink] 19 Sep 2018, 01:33
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