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Re: When the number 2^1000 is divided by 13, the remainder in the division [#permalink]
CrackVerbalGMAT wrote:
This problem is not the most usual type of remainder questions, tested on the GMAT. What makes this question special is, that, although it appears difficult and intimidating at first look, it can actually be solved using some very fundamental concepts on Remainders.

In any remainder question, play smart by trying to express the dividend in the form of (divisor + 1) or (divisor – 1) (as far as you can) so that the remainder turns out to be 1. Apart from this, use the following shortcuts to find out remainders for large values:

Remainder \(\frac{( A+ B )}{K}\) = Remainder \(\frac{(A)}{K}\) + Remainder \(\frac{(B)}{K}\)

Remainder \(\frac{(A * B)}{K}\) = Remainder \(\frac{(A)}{K}\) * Remainder \(\frac{(B)}{K}\)

The above concepts can be extended to the sum or product of any number of values.

Also, when you write the dividend (which has a power) in the form of \((divisor + 1) ^{power}\), the remainder will always be 1. However, if you write the same dividend as \((divisor – 1) ^ {power}\), the remainder will be 1 if the power is even and will be -1 if the power is odd. Since the remainder cannot be negative, we add the divisor to the negative remainder to obtain the final remainder.

Let us now try and apply these concepts to the problem at hand.

By writing down the first few powers of 2, we observe that \(2^6\) i.e. 64 is the number we can write in the form of (13k -1).

Therefore, \(2^{1000}\) can be written as (\(2^{996}\)) * (\(2^4\)). So,

Remainder \(\frac{(2^{1000})}{13}\) = Remainder \(\frac{(2^{996})}{13}\) * Rem\(\frac{(2^4)}{13}\).

Let’s now calculate the individual remainders.

Remainder \(\frac{(2^{996})}{13}\):

\(2^{996}\) = \((2^6) ^{166}\) = \((13*5 – 1)^{166}\).
So, here, we have written the dividend, \(2^{996}\) in the form of \((13k – 1) ^{even power}\), where 13 is the divisor. As discussed earlier, since the power is even, the remainder here will be 1.


Remainder\(\frac{(2^4)}{13}\) = Remainder\(\frac{(16)}{13}\)= 3.

Therefore, final remainder = 1 * 3 = 3.

So, the correct answer option is C.

It’s important to not get flustered by the sheer magnitude of the dividend. Because, as we have demonstrated, if you know the concepts that we have discussed in this post, this question is actually very straight forward. The only effort you will have to put in is to find out the power of 2 which can be written in the form of (13k -1) or (13k + 1). Once that is established, the rest of the solution is simple.

Hope that helps!



But sir, If I divide 2^36 bye 13 I get 9 as remainder, It is not working according to the theory of getting 13k-1 as remainder at 2^6/13, If I divide 2^16 by 13 I get 3 as remainder. I don't get It can you please elaborate on it. Because 2^6 = 13k-1 or 13k+1 only seems to work on some powers and not on every even power. I will be very grateful of you to elaborate on the highlighted portion for clear understanding. Thank you very much in advance sir.
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Re: When the number 2^1000 is divided by 13, the remainder in the division [#permalink]
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felipet190 and AkshayChungade01 have you heard about Euler's theorem?
The formula is expressed as A^x/N {where A and N are co-prime}

First we calculate ∅(n) where we use n is prime factor
Let N = P^a+Q^b+R^c.......
So, ∅(n) = n(1-1/p)(1-1/q)(1-1/r)........

We have 2^1000/13 (where 2 & 13 are co prime and we have to find remainder)
N = 13 and 13 prime factors = 13^1
So, ∅(n) = 13(1-1/13)
∅(n) = 13(12/13)
∅(n) = 12

So formula will become
A^∅(n) mod n = 1
2^12 mod 13 = 1 (you can calculate this to verify)

So,
2^(12)x mod 13 = will also be 1 always.
The maximum no. Nearest to 1000 multiple of 12 = 996
2^(12)83 mod 13 = 1
2^996 mod 13 = 1
Remaining is 2^4
Now calculate 2^4/13 (remainder)
16/13 leaves remainder 3.
1 from 2^996 * 3 from 2^4 = 3 as remainder.

Let me know if you don't get any step.

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Re: When the number 2^1000 is divided by 13, the remainder in the division [#permalink]
2
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not sure if this is correct approach but
cyclicity of 2
2
4
8
6

2^1000 lands on 4. So Dividing 2^1000 by 13 should be the same remainder as with 2^4
2^4 = 16

16/13 = 1 reminder 3
Answer is 3
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Re: When the number 2^1000 is divided by 13, the remainder in the division [#permalink]
fireagablast wrote:
not sure if this is correct approach but
cyclicity of 2
2
4
8
6

2^1000 lands on 4. So Dividing 2^1000 by 13 should be the same remainder as with 2^4
2^4 = 16

16/13 = 1 reminder 3
Answer is 3


This is what exactly was there in my mind but After dividing 1000 by 4, I got 250 and I ended up dividing 250 by 4 and made a mistake. Thanks mate.
2
4
8
6

yashikaaggarwal what are your thoughts?
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Re: When the number 2^1000 is divided by 13, the remainder in the division [#permalink]
AkshayChungade01 wrote:
fireagablast wrote:
not sure if this is correct approach but
cyclicity of 2
2
4
8
6

2^1000 lands on 4. So Dividing 2^1000 by 13 should be the same remainder as with 2^4
2^4 = 16

16/13 = 1 reminder 3
Answer is 3


This is what exactly was there in my mind but After dividing 1000 by 4, I got 250 and I ended up dividing 250 by 4 and made a mistake. Thanks mate.
2
4
8
6

yashikaaggarwal what are your thoughts?

We can't use cyclicity here. Because we have to check both 2 as well as 13 cyclicity at the same time which is a time taking process.. Easy way is to use Eulers theorem and solve for ∅
Here we have ∅ as 12
So we will divide 1000 by 12.
Whose maximum value under 1000 is 996
2^4 is hence divided by 13.
Giving away 3

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When the number 2^1000 is divided by 13, the remainder in the division [#permalink]
fireagablast wrote:
not sure if this is correct approach but
cyclicity of 2
2
4
8
6

2^1000 lands on 4. So Dividing 2^1000 by 13 should be the same remainder as with 2^4
2^4 = 16

16/13 = 1 reminder 3
Answer is 3

It's coincidental, you can't depict all values using this.
1000 is an easy number. We could have some complicated number. And then it will all mess up. ∅ is calculated because of that complex number. Let's say if ∅ was 101 you can't use cyclicity, here 12 is a multiple of four and can be used as cyclicity number among number. So it's better to use Eulers theorem or any other for these specific methods.

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Re: When the number 2^1000 is divided by 13, the remainder in the division [#permalink]
3
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2^1000=(2^4)^250
=16^250
=16*16*16.......upto 250 terms
Every time 250 is divided by 13 it leaves a remainder we get
=3*3*3..............upto 250 terms= 3^250
Since in these there can be a factor of 13 ,
we know that 3^3=27( which when divided by 1 gives remainder 1)

so, 3^250= (3^249)*3
=(3^3)^83*3
=(27)^83*3
when divide by 13 we get
=(1)^83*3=3
Hence remainder we got is 3
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Re: When the number 2^1000 is divided by 13, the remainder in the division [#permalink]
Asked: When the number \(2^{1000}\) is divided by 13, the remainder in the division is

\(2^6 = 64 = 13*5 - 1\)
\(1000 = 6*166 + 4\)
\(2^1000 = 2^6*166 * 2^4\)

The remainder when 2^1000 is divided by 13 = (-1)^166 * (16-13) = 3

IMO C
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Re: When the number 2^1000 is divided by 13, the remainder in the division [#permalink]
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Re: When the number 2^1000 is divided by 13, the remainder in the division [#permalink]
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