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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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05 Sep 2018, 07:10
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Difficulty:

65% (hard)

Question Stats:

53% (01:45) correct 47% (01:23) wrong based on 179 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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Re: What is the remainder when 2^83 is divided by 9?  [#permalink]

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05 Sep 2018, 07:31
5
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..

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

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05 Sep 2018, 07:28
1
1
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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05 Sep 2018, 07:38
1
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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05 Sep 2018, 11:18
1
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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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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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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What is the remainder when 2^83 is divided by 9?  [#permalink]

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18 Oct 2018, 10:26
Hi ,T1101
Well, it seems strange for me when the remainder is arrived by 9-4=5
On the basis, I agree with your approach with 2^83 / 9 = [4*(9-1)^27] / 9
(9-1)^27 will leave (-1)^27 = -1. When -1 / 9, the quotient is 1 and remainder is 8
Now this [4*(9-1)^27] / 9 = (4*8)/9 = 32 / 9 yield 5 as remainder
I think this explanation is more precisely and easier to understand
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Re: What is the remainder when 2^83 is divided by 9?  [#permalink]

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25 Jul 2019, 15:40
Can anybody please provide a further step-by-step explanation ? Thanks
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Re: What is the remainder when 2^83 is divided by 9?  [#permalink]

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26 Jul 2019, 12:12
T1101 wrote:
What is the remainder when 2^83 is divided by 9?

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

Notice that 2^6 = 64 and 64/9 = 7 R 1. Now, notice that 2^83 = (2^6)^13 x 2^5 Since 2^6 has a remainder 1 when it’s divided by 9, (2^6)^13 will have a remainder of 1^13 = 1 when it’s divided by 9. Because of this, the remainder when 2^83 is divided by 9 will be same as the remainder when 2^5 is divided by 9. Since 2^5 = 32 and 32/9 = 3 R 5, the remainder is 5.

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Re: What is the remainder when 2^83 is divided by 9?   [#permalink] 26 Jul 2019, 12:12
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