Last visit was: 21 Apr 2026, 19:17 It is currently 21 Apr 2026, 19:17
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
Noshad
Joined: 02 Nov 2018
Last visit: 30 Mar 2026
Posts: 207
Own Kudos:
1,029
 [60]
Given Kudos: 110
Status:Manager
Location: Bangladesh
Posts: 207
Kudos: 1,029
 [60]
2
Kudos
Add Kudos
58
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
CrackverbalGMAT
User avatar
Major Poster
Joined: 03 Oct 2013
Last visit: 21 Apr 2026
Posts: 4,846
Own Kudos:
9,180
 [29]
Given Kudos: 226
Affiliations: CrackVerbal
Location: India
Expert
Expert reply
Posts: 4,846
Kudos: 9,180
 [29]
8
Kudos
Add Kudos
21
Bookmarks
Bookmark this Post
avatar
fireagablast
Joined: 30 Jun 2019
Last visit: 17 Aug 2021
Posts: 260
Own Kudos:
129
 [6]
Given Kudos: 8
Posts: 260
Kudos: 129
 [6]
5
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
General Discussion
User avatar
felipet190
Joined: 18 Jun 2020
Last visit: 18 Aug 2021
Posts: 7
Own Kudos:
1
 [1]
Posts: 7
Kudos: 1
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
As i understand it, the remainder of products is the product of remainders. So, when 2 is divided by 13, the remainder is 2. (2 = 0*13 + 2).
This remainder of 2 is then multiplied. 2^1000 is a massive number. But this doesn't lead to the correct answer. Can someone please explain why this approach doesn't work?
User avatar
AkshayC95
Joined: 28 Jan 2018
Last visit: 23 Feb 2021
Posts: 75
Own Kudos:
87
 [1]
Given Kudos: 186
Location: India
Concentration: Finance, International Business
GPA: 3.5
Products:
Posts: 75
Kudos: 87
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
CrackVerbalGMAT
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.
User avatar
yashikaaggarwal
User avatar
Senior Moderator - Masters Forum
Joined: 19 Jan 2020
Last visit: 29 Mar 2026
Posts: 3,089
Own Kudos:
3,158
 [5]
Given Kudos: 1,510
Location: India
GPA: 4
WE:Analyst (Internet and New Media)
Posts: 3,089
Kudos: 3,158
 [5]
4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.

Posted from my mobile device
User avatar
AkshayC95
Joined: 28 Jan 2018
Last visit: 23 Feb 2021
Posts: 75
Own Kudos:
Given Kudos: 186
Location: India
Concentration: Finance, International Business
GPA: 3.5
Products:
Posts: 75
Kudos: 87
Kudos
Add Kudos
Bookmarks
Bookmark this Post
fireagablast
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?
User avatar
yashikaaggarwal
User avatar
Senior Moderator - Masters Forum
Joined: 19 Jan 2020
Last visit: 29 Mar 2026
Posts: 3,089
Own Kudos:
Given Kudos: 1,510
Location: India
GPA: 4
WE:Analyst (Internet and New Media)
Kudos
Add Kudos
Bookmarks
Bookmark this Post
AkshayChungade01
fireagablast
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

Posted from my mobile device
User avatar
yashikaaggarwal
User avatar
Senior Moderator - Masters Forum
Joined: 19 Jan 2020
Last visit: 29 Mar 2026
Posts: 3,089
Own Kudos:
Given Kudos: 1,510
Location: India
GPA: 4
WE:Analyst (Internet and New Media)
Kudos
Add Kudos
Bookmarks
Bookmark this Post
fireagablast
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.

Posted from my mobile device
User avatar
SameerGupta2001
Joined: 05 Apr 2020
Last visit: 13 Jun 2023
Posts: 15
Own Kudos:
42
 [3]
Given Kudos: 108
Posts: 15
Kudos: 42
 [3]
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
User avatar
Kinshook
User avatar
Major Poster
Joined: 03 Jun 2019
Last visit: 21 Apr 2026
Posts: 5,986
Own Kudos:
5,855
 [1]
Given Kudos: 163
Location: India
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
GMAT 1: 690 Q50 V34
Posts: 5,986
Kudos: 5,855
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
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
User avatar
vijayram24
Joined: 01 Sep 2020
Last visit: 29 Mar 2026
Posts: 38
Own Kudos:
Given Kudos: 209
Location: India
Concentration: Economics, Technology
Schools: ISB '26 NUS
GMAT 1: 660 Q48 V34
GPA: 8.3
WE:Engineering (Technology)
Products:
Schools: ISB '26 NUS
GMAT 1: 660 Q48 V34
Posts: 38
Kudos: 37
Kudos
Add Kudos
Bookmarks
Bookmark this Post
From Fermats Little Therem ,

Remainder of (a^p-a)/p =0 where p is prime and a is any number .
So Remainder(2^13-2)/13 =0 =>2^13/13 leaves remainder 2.

So 2^1000 =((2^13)^76)*(2^12)
=>(2^88)/13
=>Re((2^13)^6*(2^10))/13
=>Re(2^16/13)
=>Re(2^4/13)
=>3 is the remainder
User avatar
PSKhore
Joined: 28 Apr 2025
Last visit: 27 Feb 2026
Posts: 190
Own Kudos:
33
 [1]
Given Kudos: 112
Posts: 190
Kudos: 33
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Noshad
When the number \(2^{1000}\) is divided by 13, the remainder in the division is

(A) 1

(B) 2

(C) 3

(D) 7

(E) 11
Find the number when raised to some power of 2 that will be closest to 13
2^4 = 16...the remainder in this case is 3

(2^4)^250
=(3)^250

3^3 = 27 when divided by 13, it gives remainder as 1
Now the power 250 is not completely divisible by 3. The remaining power is 1 (249 is divisible by 3)
= (1)^249 X (3)^1
This will give us 3/13 and hence the remainder is 3
Moderators:
Math Expert
109728 posts
Tuck School Moderator
853 posts