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

Question

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

FillFM · Accepted Answer

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

T1101 · Answer

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.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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

Abhishek009 · Answer

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

Princ · Answer

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.

Peddy · Answer

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!!

T1101 · Answer

Hi Peddy, hope i can help you! I have done this question with the same approach as described here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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?

Samine · Answer

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 =   / 9
(9-1)^27 will leave (-1)^27 = -1. When -1 / 9, the quotient is 1 and remainder is 8
Now this   / 9 = (4*8)/9 = 32 / 9 yield 5 as remainder
I think this explanation is more precisely and easier to understand

GMATestaker · Answer

Can anybody please provide a further step-by-step explanation ? Thanks

ScottTargetTestPrep · Answer

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.

Answer: D

sajon · Answer

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. 
Expert, please shade some light here.

sujoykrdatta · Answer

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

Answer D

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

Answer D

sujoykrdatta · Answer

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

AnirudhaS · Answer

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"!

sajon · Answer

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.

sajon · Answer

Thanks for your reply. now i got my point

Andrewcoleman · Answer

this is the easiest way to solve this problem.
the post attached will teach you pretty much everything you need to know to solve this type of problem

satya2029 · Answer

2^3*2^3(26)*2^2/(9) 
 (-1)*(-1)^(26)*(4) 
Remainder=-4 
since remainder is always positive
-4+9=5
D:)

minustark · Answer

For such questions the first step is to convert the base of the dividend (here its 2) into an integer that is 1 more or less than the divider(in this question its 9). So, we can write 2^83 as (2^3)^27*2^2, (as 8 is 1 less than 9).

8^27 *4. Now we will focus only on 8^27 part. Convert it into (9-1)^27. As per the binomial theory, all the elements of the expression except the last one will be divisible by 9. For example: (a+1)^3 = a^3 + 3a^2 + 3a +1...here first 3 values are divisible by 3, so we are left only with 1, which is the integer. Similarly, now we can write 8^27 as 9k - 1 (as the last element is (-1)^27= -1). So, 8^83 = 4* (9k -1) = 36 k - 4, which is 4 less than the integer divisible by 9, so the remainder will be 5. We can find out by pick-in integers for k. when k =1, it is 32, divided by 9 provides 5 as remainder. when k =2, 68 (9*7 + 5). So answer is D.

Kinshook · Answer

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

2^3mod9 = 8mod9 = - 1 
 2^{83}mod9 = 2^{3*27+2}mod9 = (-1)^{27} 4mod9 = - 4mod9 = 5mod9

IMO D

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?

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