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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, 06:10
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Question Stats:

52% (01:41) correct 48% (01:31) wrong based on 232 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, 06:31
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1
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, 06:28
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2
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, 06:38
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, 10:18
1
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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, 09:03
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, 00:33
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, 09:26
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, 14:40
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, 11:12
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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What is the remainder when 2^83 is divided by 9?  [#permalink]

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10 Feb 2020, 11:54
Can anybody tell me why cyclity method does not work there as applied in other problems?
2 has four cycles: 2, 4, 8, 6
Then 2^83 should have 8 as the unit digit. So, why does 8 have not been the answer when divided by 9??
I completely understood the above solution. I just want to satisfy my curiosity.
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What is the remainder when 2^83 is divided by 9?  [#permalink]

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10 Feb 2020, 12:20
T1101 wrote:
What is the remainder when 2^83 is divided by 9?

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

Method 1

We know that, if $$N$$ leaves remainder $$r$$ when divided by $$D, N^k$$leaves remainder $$r^k$$ when divided by $$D$$ (provided $$r^k$$ is smaller than $$D$$); else you need to divided $$r^k$$ again by $$D$$ to get the actual remainder

We can observe that $$2^6 = 64$$ when divided by 9 leaves remainder 1

Thus, we express $$2^{83}$$ in terms of $$2^6$$:

$$2^{83} = (2^6)^{13} * 2^5$$

Since $$2^6$$ when divided by 9 leaves remainder 1, the remainder when $$(2^6)^{13}$$ is divided by 9 is $$1^{13} = 1$$

Thus, required remainder is simply the one when the left-over $$2^5 = 32$$ is divided by 9 => Remainder = 5

Method 2

Look for patterns in remainders for powers of 2 when divided by 9:

Remainder when $$2^1$$ is divided by 9 = 2
Remainder when $$2^2$$ is divided by 9 = 4
Remainder when $$2^3$$ is divided by 9 = 8
Remainder when $$2^4$$ is divided by 9 = 7
Remainder when $$2^5$$ is divided by 9 = 5
Remainder when $$2^6$$ is divided by 9 = 1
Remainder when $$2^7$$ is divided by 9 = 2 ... and the cycle repeats

Thus, the cycle is for 6 places

Thus, we express $$2^{83}$$ in terms of $$2^6$$:

$$2^{83} = (2^6)^{13} * 2^5 = 2^{78} * 2^5$$

The 13th time the cycle repeats, the remainder comes to 1. Now, the cycle will repeat but will stop at the 5th position since we have $$2^5$$ left-over

Thus, the remainder corresponding to the 5th position in the cycle is 5

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

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10 Feb 2020, 12:34
1
sajon wrote:
Can anybody tell me why cyclity method does not work there as applied in other problems?
2 has four cycles: 2, 4, 8, 6
Then 2^83 should have 8 as the unit digit. So, why does 8 have not been the answer when divided by 9??
I completely understood the above solution. I just want to satisfy my curiosity.

The cycle of 2 that you are referring to: 2, 4, 8, 6 is basically the cycle for the units digit

This is also the cycle for remainder when powers of 2 are DIVIDED BY 10

Check:
Unit digit of 2^1 = 2 <=> Remainder when 2^1 is divided by 10 is 2
Unit digit of 2^2 = 4 <=> Remainder when 2^2 is divided by 10 is 4
Unit digit of 2^3 = 8 <=> Remainder when 2^3 is divided by 10 is 8
Unit digit of 2^4 = 6 <=> Remainder when 2^4 is divided by 10 is 6

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

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10 Feb 2020, 13:28
2
sajon wrote:
Can anybody tell me why cyclity method does not work there as applied in other problems?
2 has four cycles: 2, 4, 8, 6
Then 2^83 should have 8 as the unit digit. So, why does 8 have not been the answer when divided by 9??
I completely understood the above solution. I just want to satisfy my curiosity.

Btw I am no expert.

The unit digit of the a number has no significance to the remainder (unless the divisor is 10).

I will explain it with a few examples .
(1) if your number is 8, then 8/9 = remainder 8
(2) if your number is 18, then 18/9 = remainder 0
(3) if your number is 118, then 118/9 = remainder 1
The unit digit is 8, but then the remainders are all different, so extending this logic you can't really say what you are suggesting.

I hope this "satisfies your curiosity"!
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Re: What is the remainder when 2^83 is divided by 9?  [#permalink]

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10 Feb 2020, 21:40
AnirudhaS wrote:
sajon wrote:
Can anybody tell me why cyclity method does not work there as applied in other problems?
2 has four cycles: 2, 4, 8, 6
Then 2^83 should have 8 as the unit digit. So, why does 8 have not been the answer when divided by 9??
I completely understood the above solution. I just want to satisfy my curiosity.

Btw I am no expert.

The unit digit of the a number has no significance to the remainder (unless the divisor is 10).

I will explain it with a few examples .
(1) if your number is 8, then 8/9 = remainder 8
(2) if your number is 18, then 18/9 = remainder 0
(3) if your number is 118, then 118/9 = remainder 1
The unit digit is 8, but then the remainders are all different, so extending this logic you can't really say what you are suggesting.

I hope this "satisfies your curiosity"!

Wow, now i got this point. Thanks a lot. Actually i solved few question using 10 and found cyclity method effective for all number. But, it is the only one that works.
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Re: What is the remainder when 2^83 is divided by 9?  [#permalink]

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10 Feb 2020, 21:47
sujoykrdatta wrote:
sajon wrote:
Can anybody tell me why cyclity method does not work there as applied in other problems?
2 has four cycles: 2, 4, 8, 6
Then 2^83 should have 8 as the unit digit. So, why does 8 have not been the answer when divided by 9??
I completely understood the above solution. I just want to satisfy my curiosity.

The cycle of 2 that you are referring to: 2, 4, 8, 6 is basically the cycle for the units digit

This is also the cycle for remainder when powers of 2 are DIVIDED BY 10

Check:
Unit digit of 2^1 = 2 <=> Remainder when 2^1 is divided by 10 is 2
Unit digit of 2^2 = 4 <=> Remainder when 2^2 is divided by 10 is 4
Unit digit of 2^3 = 8 <=> Remainder when 2^3 is divided by 10 is 8
Unit digit of 2^4 = 6 <=> Remainder when 2^4 is divided by 10 is 6

Hope this is clear now

Re: What is the remainder when 2^83 is divided by 9?   [#permalink] 10 Feb 2020, 21:47