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

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"!

\(2^6 = 64\)

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

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

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?

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.

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

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

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

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 ,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

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

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.

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.

Thanks for your reply. now i got my point

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

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

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.

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