Last visit was: 18 Nov 2025, 23:54 It is currently 18 Nov 2025, 23:54
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2   3   
555-605 Level|   Exponents|   Remainders|      
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,378
Own Kudos:
778,143
 [6]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,378
Kudos: 778,143
 [6]
When 51^25 is divided by 13, the remainder obtained is: 22 Jul 2013, 22:11
1
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
VeritasPrepKarishma
sonalmsingh47
The remainder should be 1

Actually the remainder is -1 which is the same as 12. Check out this post: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/05 ... ek-in-you/
Show more

Yes, the correct answer is A (12), not D (1).

Similar question to practice:
what-is-the-remainder-when-43-86-is-divided-by-134778.html
what-is-the-remainder-of-126493.html
what-is-the-remainder-when-32-32-32-is-divided-by-100316.html
what-is-the-remainder-when-18-22-10-is-divided-by-99724.html
what-is-the-remainder-when-333-222-is-divided-by-156379.html

Theory on remainders problems: remainders-144665.html

DS remainders problems to practice: search.php?search_id=tag&tag_id=198
PS remainders problems to practice: search.php?search_id=tag&tag_id=199

Hope it helps.
User avatar
kundankshrivastava
User avatar
Joined: 14 May 2014
Last visit: 18 Oct 2014
Posts: 35
Own Kudos:
170
 [5]
Given Kudos: 1
Posts: 35
Kudos: 170
 [5]
When 51^25 is divided by 13, the remainder obtained is: 19 May 2014, 23:55
1
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
Concept of negative remainder is very helpful in solving these type of questions.

IT simply says that if an expression is divided by x and remainder is -a ( where a is positive) then actual remainder is x-a.

For example if after dividing an expression with 8 we find a remainder of -1 then actual remainder will be 8-1 i.e. 7.

Now let's use this concept:

51^25 can be written as
(52-1)^25
all the terms of expansion of this will have 52 as a factor and hence would be divisible by 13 (52 = 13 X 4). Last term will be (-1)^25 i.e. -1
here remainder is negative while divided by 13. Hence using above concept remainder will be 13-1 i.e. 12.

Answer is A
User avatar
usre123
User avatar
Joined: 30 Mar 2013
Last visit: 25 Nov 2022
Posts: 74
Own Kudos:
224
Given Kudos: 197
Location: United States
Schools: Alberta '23 (WL)
GMAT 1: 760 Q50 V44
Schools: Alberta '23 (WL)
GMAT 1: 760 Q50 V44
Posts: 74
Kudos: 224
When 51^25 is divided by 13, the remainder obtained is: 01 Sep 2014, 13:36
Kudos
Add Kudos
Bookmarks
Bookmark this Post
The method used by Bunuel above is the best way to get to the answer. Some time back, I wrote a post detailing the method. Here is the link:

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

IMO,it's 5. Let me know please whether I'm correct!



Please let me know why I'm wrong:
why cant we create a cycle for 2, so at 2^83, units digit would be 8. Then we have 8/9, so since divisor is greater than the number being divided, our answer would be 8? Can you please explain what is wrong with my reasoning?

In the same way, the question 51^25, the cycle of 1 always gives the same answer, that is, one. so 1/13, then answer should be 1?

My concept was that when the divider is greater than the dividend, such as 3/13, then the remainder would be the dividend itself. (3)

Now if we have a negative number, say -3/13, then we can say the remainder will be 13-3= 10, is that correct? Is even half of what I said correct?
Thank you
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 18 Nov 2025
Posts: 16,267
Own Kudos:
76,984
 [3]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 76,984
 [3]
When 51^25 is divided by 13, the remainder obtained is: 02 Sep 2014, 01:45
1
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
usre123
The method used by Bunuel above is the best way to get to the answer. Some time back, I wrote a post detailing the method. Here is the link:

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

IMO,it's 5. Let me know please whether I'm correct!



Please let me know why I'm wrong:
why cant we create a cycle for 2, so at 2^83, units digit would be 8. Then we have 8/9, so since divisor is greater than the number being divided, our answer would be 8? Can you please explain what is wrong with my reasoning?

In the same way, the question 51^25, the cycle of 1 always gives the same answer, that is, one. so 1/13, then answer should be 1?

My concept was that when the divider is greater than the dividend, such as 3/13, then the remainder would be the dividend itself. (3)

Now if we have a negative number, say -3/13, then we can say the remainder will be 13-3= 10, is that correct? Is even half of what I said correct?
Thank you
Show more


Responding to a pm:

When dividend is smaller than divisor, remainder is dividend (Correct)

Is \(2^{83}\) smaller than 9?

If you use cyclicity, you get that \(2^{83}\) is some huge number which "ends" with an 8. 8 is just the units digit of that number. \(2^{83}\) is NOT 8. It is actually equal to 9671406556917033397649408.
The dividend is not smaller than divisor. In fact, it is much much greater than the divisor.

Would you like to re-think now? Let me know if there are still some doubts.
User avatar
usre123
User avatar
Joined: 30 Mar 2013
Last visit: 25 Nov 2022
Posts: 74
Own Kudos:
224
Given Kudos: 197
Location: United States
Schools: Alberta '23 (WL)
GMAT 1: 760 Q50 V44
Schools: Alberta '23 (WL)
GMAT 1: 760 Q50 V44
Posts: 74
Kudos: 224
When 51^25 is divided by 13, the remainder obtained is: 23 Sep 2014, 06:41
Kudos
Add Kudos
Bookmarks
Bookmark this Post
VeritasPrepKarishma
vinaymimani
VeritasPrepKarishma


Actually, you are using Binomial too.

How do you explain Remainder of\(\frac{(51^{25})}{13}\)= Remainder of\(\frac{(12^{25})}{13}\)?

Binomial leads to this equality!

I don't think I am using Binomial. Maybe you are correct.

However, I would try to prove the above point, not using Binomial. Let me know if I went wrong somewhere.

--> \(51^{25} = 13Q_1 + R_1\)

and \(12^{25} = 13Q_2 + R_2\)

with all the understood notations.

I want to prove that \(R_1\) = \(R_2\)

I will assume they are equal and replace the value of \(R_1\) in the first equation by that of \(R_2\)

Thus, \(51^{25} = 13Q_1 +12^{25} - 13Q_2\)

or \(51^{25}-12^{25} = 13(Q_1-Q_2)\)

Thus, if I could show that \(51^{25}-12^{25}\) is divisible by 13, my assumption would be correct.

Now,\(x^n-a^n\), is always divisible by (x-a), x and a are integers, n is odd.

Thus,\(51^{25}-12^{25}\) is always divisible by (51-12) = 39 --> 13*3. Thus, it is divisible by 13.

In the spirit of a healthy discussion, I would like to point out that you can certainly prove that the two are equal since they ARE equal. It can be done in many ways. The point is that when you see \(51^{25}\), what makes you think of \(12^{25}\)?
You think of it because 51 = 39 + 12. You separate out the part which is divisible by 13 and take control of the rest. Why?

Because \((39 + 12)^{25}\), when divided by 13 will have the same remainder as \(12^{25}\) because every term in this expansion is divisible by 39 except the last term which is \(12^{25}\). You understand this intuitively and that is all binomial is about.
Show more



this is what I did, I went down, instead of going up. I mean I did (39+12)^ 25. and obviously 12^25 is going to be an issue when dividing by 13. But is it also ok to go one up? (52-1)^25. (i see that's what Bunuel has done).

But on your blog, Hamza asked the same question for remainder when 2^86/9
and he went one up, to 87. in that case we have 2*(2^3)^29... which means we can make it
2*(9-1)^29, so : -1^29 , which must be * by 2 as well, so that's -2. ..which gives us the wrong answer. Correct answer is:

2^ (84)(4)----- let's make it (2^3)^29----(9-1)^29..... -1^29 which is a -1. multiply by the 4 I ignored so I have -4. For negative numbers make it 9-4, so 5....
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 18 Nov 2025
Posts: 16,267
Own Kudos:
76,984
 [1]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 76,984
 [1]
When 51^25 is divided by 13, the remainder obtained is: 23 Sep 2014, 22:07
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
usre123

this is what I did, I went down, instead of going up. I mean I did (39+12)^ 25. and obviously 12^25 is going to be an issue when dividing by 13. But is it also ok to go one up? (52-1)^25. (i see that's what Bunuel has done).

But on your blog, Hamza asked the same question for remainder when 2^86/9
and he went one up, to 87. in that case we have 2*(2^3)^29... which means we can make it
2*(9-1)^29, so : -1^29 , which must be * by 2 as well, so that's -2. ..which gives us the wrong answer. Correct answer is:

2^ (84)(4)----- let's make it (2^3)^29----(9-1)^29..... -1^29 which is a -1. multiply by the 4 I ignored so I have -4. For negative numbers make it 9-4, so 5....
Show more

It doesn't matter whether you go up or down - as long as you don't change the question.

\(51^{25} = (52 - 1)^{25}\) - fine

But \(2^{86} \neq 2^{87}\)

\(2^{86} = 2^2 * 2^{84}\) - fine
or
\(2^{86} = \frac{2^{87}}{2}\) but this just complicates our question more since the denominator has changed now
or
\(2^{86} = 2^{87 - 1}\) Again, how do we handle such an exponent?
User avatar
GuptaDarsh
User avatar
Joined: 09 Jul 2014
Last visit: 08 Jun 2015
Posts: 11
Own Kudos:
5
Given Kudos: 63
Posts: 11
Kudos: 5
When 51^25 is divided by 13, the remainder obtained is: 13 Oct 2014, 06:34
Kudos
Add Kudos
Bookmarks
Bookmark this Post
VeritasPrepKarishma :
okay... got your point .... can you please explain when to use the concept of cyclicity and when to use the concept of binomial theoram ?
Why with the cyclicity of 1 this Question gives the remainder 1
:: 51^25 last digit will be 1^ 25 and when 1^ 25 is divided 13 it should give remainder 1 and not 12
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 18 Nov 2025
Posts: 16,267
Own Kudos:
76,984
 [3]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 76,984
 [3]
When 51^25 is divided by 13, the remainder obtained is: 13 Oct 2014, 21:01
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
GuptaDarsh
VeritasPrepKarishma :
okay... got your point .... can you please explain when to use the concept of cyclicity and when to use the concept of binomial theoram ?
Why with the cyclicity of 1 this Question gives the remainder 1
:: 51^25 last digit will be 1^ 25 and when 1^ 25 is divided 13 it should give remainder 1 and not 12
Show more

These are two different concepts:

Cyclicity only gives us the units digit of a number when you raise it to a power. Say, using cyclicity, I can say that \(51^{107}\) will end with a 1. It usually doesn't help us get the remainder when the number is divided by another number. Using cyclicity, we can say that 3^4 ends with 1. But what will be the remainder when 3^4 is divided by 7? We can't say. Will it be 1? Not necessary. Every number that ends with 1 doesn't give a remainder of 1 when divided by 7. 3^4 = 81 which when divided by 7 gives remainder 4. So cyclicity usually has nothing to do with remainders. It is useful for remainders only when we are talking about division by 10. Why? Because whenever you divide a number by 10, the remainder will always be the last digit of the number. Say, 81/10 remainder 1.
145/10 remainder is 5.
237/10 remainder is 7.
and so on...
So we can use cyclicity in this case to just get the last digit and that will be our remainder.

Binomial helps you find the remainder when you divide a number with a power by any other given number.
avatar
intheend14
avatar
Joined: 12 Sep 2014
Last visit: 08 Sep 2019
Posts: 125
Own Kudos:
144
Given Kudos: 103
Concentration: Strategy, Leadership
Schools: Stanford '18 Haas '18 Wharton '18 Kellogg PT '18 Insead Sept '16
GMAT 1: 740 Q49 V41
GPA: 3.94
Schools: Stanford '18 Haas '18 Wharton '18 Kellogg PT '18 Insead Sept '16
GMAT 1: 740 Q49 V41
Posts: 125
Kudos: 144
When 51^25 is divided by 13, the remainder obtained is: 05 Nov 2014, 20:03
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I tried to base this on the cyclicity of exponents--is the following valid?

51^25=3^25*17^25. The cyclicity of both 3 and 17 are 4... so this is similar to 3^1*17^1=51.
51/13 leaves a remainder of 12... maybe I just got lucky?
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 18 Nov 2025
Posts: 16,267
Own Kudos:
76,984
 [2]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 76,984
 [2]
When 51^25 is divided by 13, the remainder obtained is: 10 Nov 2014, 22:38
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
intheend14
I tried to base this on the cyclicity of exponents--is the following valid?

51^25=3^25*17^25. The cyclicity of both 3 and 17 are 4... so this is similar to 3^1*17^1=51.
51/13 leaves a remainder of 12... maybe I just got lucky?
Show more

Cyclicity of last digit doesn't have anything to do with remainders. It is useful in remainders only when the last digit of the number is useful in finding the remainder e.g. when the divisor is 10.

You get the answer here because the cyclicity of the remainders of powers of 51 is 2. The remainder is 12 for every odd power and 1 for every even power. Hence remainder of 51^25 will be the same as remainder of 51^1.
User avatar
dabral
User avatar
Tutor
Joined: 19 Apr 2009
Last visit: 29 Nov 2024
Posts: 557
Own Kudos:
664
 [1]
Given Kudos: 19
Affiliations: GMATQuantum
Expert
Expert reply
Posts: 557
Kudos: 664
 [1]
When 51^25 is divided by 13, the remainder obtained is: 10 Nov 2014, 23:50
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Couple comments here:

First this question is not from GMAT Paper tests as quoted in the original post. I have scoured all the GMAT Paper tests and I have never seen this question. It is possible that I have missed it, but the odds are extremely low. I would love to see the original reference.

Second, the GMAT does not expect one to know the binomial theorem and this question is beyond the scope of the GMAT, at least question in the first post. I would love to hear a confirmation from someone else who has seen an official GMAT question like this on the exam. Sometimes I feel these discussions only end up creating unnecessary fear in students.

Dabral
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 18 Nov 2025
Posts: 16,267
Own Kudos:
76,984
 [2]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 76,984
 [2]
When 51^25 is divided by 13, the remainder obtained is: 11 Jan 2016, 23:39
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
naveenq

E.g. What is the remainder when 10^5 is divided by 13?

From binomial method, we can take (13-3)^5/13
Next, we're left with (-3)^5/13

Two things - 1st, if I go the long route and take patterns then
3^1 / 13 = R3
3^2 / 13 = R4
3^3 / 13 = R1
and the pattern repeats, in 3's.

so, now, (-3)^5, i know the answer is R4.

However, finding the pattern is time-consuming. Can you help me understand how I can use the binomial theorem for the negative base?
(-3)^5 / 13

THANK YOU!!
Show more

(-3)^5/13
Since the power is odd, the negative sign remains. Don't worry about it just yet.
-3^5/13

Note that 3^3 = 27 which is 1 more than a multiple of 13.

-27 * 9/13
-(26+1) * 9/13
Remainder is -9/13

So we have a remainder of -9. This means that remainder is actually 4.

For more on negative remainders, see: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/03 ... -the-gmat/
User avatar
DonnieDrastic
User avatar
Joined: 22 Jul 2016
Last visit: 28 Dec 2018
Posts: 9
Own Kudos:
3
Given Kudos: 2
Posts: 9
Kudos: 3
When 51^25 is divided by 13, the remainder obtained is: 04 Dec 2016, 20:13
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I don't get 12 when I plug (51^25)/13 in to a calculator to see what comes out. What am I misunderstanding about the question?
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 18 Nov 2025
Posts: 16,267
Own Kudos:
76,984
 [2]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 76,984
 [2]
When 51^25 is divided by 13, the remainder obtained is: 04 Dec 2016, 20:59
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
DonnieDrastic
I don't get 12 when I plug (51^25)/13 in to a calculator to see what comes out. What am I misunderstanding about the question?
Show more

There are two different ways of expressing the result of a division:

Say, I tell you the following: Divide 11 by 4. What do you get?

You could answer me with one of the following:

Case 1: You could say, “I get 2.75”

Case 2: You could say, “I get 2 as the quotient and 3 as the remainder.”

Either ways, you are correct. 11/4 = (2 ¾)

When you use the decimal form, you get a .75 which you add to 2 to give you 2.75. This .75 is nothing but the way you express the remainder 3. When you divide 11 by 4, 4 goes into 11 two times and then 3 is left over. When 4 goes into 3, you get 0.75 which is ¾. That is the reason why you can write 11/4 as (2 ¾) in mixed fractions.

The calculator gives you the result of case 1. Anyway, the calculator will give you an approximate value.
This question asks you for the remainder, as in case 2.
avatar
Vaidya
avatar
Joined: 02 Oct 2016
Last visit: 10 Sep 2025
Posts: 32
Own Kudos:
21
Given Kudos: 20
Location: India
Schools: HEC Dec '17 (D) IE (A) INSEAD Jan'23 intake
GMAT 1: 740 Q49 V42
Schools: HEC Dec '17 (D) IE (A) INSEAD Jan'23 intake
GMAT 1: 740 Q49 V42
Posts: 32
Kudos: 21
When 51^25 is divided by 13, the remainder obtained is: 24 Jan 2017, 01:59
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel

Hi Bunuel, I solved this using the remainder and cyclicity method and got the correct answer. Can you please tell me if this is just a coincidence?

By applying cyclicity of 4: 51^25 ~ 51^1
Therefore, 51/13 ---> Remainder = 12
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,378
Own Kudos:
778,143
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,378
Kudos: 778,143
When 51^25 is divided by 13, the remainder obtained is: 24 Jan 2017, 02:42
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Vaidya
Bunuel

Hi Bunuel, I solved this using the remainder and cyclicity method and got the correct answer. Can you please tell me if this is just a coincidence?

By applying cyclicity of 4: 51^25 ~ 51^1
Therefore, 51/13 ---> Remainder = 12
Show more

The question is how do you know the cycle? In fact the remainder repeats in blocks of two here:

The remainder of 51^1 divided by 13 is 12;
The remainder of 51^2 divided by 13 is 1;
The remainder of 51^3 divided by 13 is 12;
The remainder of 51^4 divided by 13 is 1;
The remainder of 51^5 divided by 13 is 12;
...
The remainder of 51^odd divided by 13 is 12;
The remainder of 51^even divided by 13 is 1.
User avatar
testcracker
User avatar
Joined: 24 Mar 2015
Last visit: 02 Dec 2024
Posts: 202
Own Kudos:
130
Given Kudos: 541
Status:love the club...
Posts: 202
Kudos: 130
When 51^25 is divided by 13, the remainder obtained is: 04 Sep 2017, 00:33
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
LM
When 51^25 is divided by 13, the remainder obtained is:

A. 12
B. 10
C. 2
D. 1
E. 0

\(51^{25}=(52-1)^{25}\), now if we expand this expression all terms but the last one will have \(52=13*4\) in them, thus will leave no remainder upon division by 13, the last term will be \((-1)^{25}=-1\). Thus the question becomes: what is the remainder upon division -1 by 13? The answer to this question is 12: \(-1=13*(-1)+12\).

Answer: A.
Show more


hi

51^25

here the last digit of 51 is 1, and the cyclicity of 1 is 1. So, the unit digit of 51^25 will be 1

thus, 1, when divided by 13 will produce a quotient "0" and will leave a remainder "1", and I am stumped here ... :(

please help me understand the problem ...

thanks in advance ...
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,378
Own Kudos:
778,143
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,378
Kudos: 778,143
When 51^25 is divided by 13, the remainder obtained is: 04 Sep 2017, 00:36
Kudos
Add Kudos
Bookmarks
Bookmark this Post
gmatcracker2017
Bunuel
LM
When 51^25 is divided by 13, the remainder obtained is:

A. 12
B. 10
C. 2
D. 1
E. 0

\(51^{25}=(52-1)^{25}\), now if we expand this expression all terms but the last one will have \(52=13*4\) in them, thus will leave no remainder upon division by 13, the last term will be \((-1)^{25}=-1\). Thus the question becomes: what is the remainder upon division -1 by 13? The answer to this question is 12: \(-1=13*(-1)+12\).

Answer: A.


hi

51^25

here the last digit of 51 is 1, and the cyclicity of 1 is 1. So, the unit digit of 51^25 will be 1

thus, 1, when divided by 13 will produce a quotient "0" and will leave a remainder "1", and I am stumped here ... :(

please help me understand the problem ...

thanks in advance ...
Show more

You cannot do this way.

For example, 1 divided by 13 gives the remainder of 1 but 21 divided by 13 gives the remainder of 8. There are many different approaches and links to the underlying theory on previous pages of the discussion.
User avatar
BrushMyQuant
User avatar
Joined: 05 Apr 2011
Last visit: 21 Oct 2025
Posts: 2,284
Own Kudos:
2,552
 [1]
Given Kudos: 100
Status:Tutor - BrushMyQuant
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE:Information Technology (Computer Software)
Expert
Expert reply
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
Posts: 2,284
Kudos: 2,552
 [1]
When 51^25 is divided by 13, the remainder obtained is: 19 Oct 2022, 11:56
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
We need to find When \(51^{25}\) is divided by 13, the remainder obtained is

Now, to solve this problem we will be using the concept of Binomial Theorem

We will try to split 51 into two numbers. One number will be a multiple of 13 and will be very close to 51 and other number will be a very small number.

\(51^{25}\) = \((52 - 1)^{25}\) (52 = 13*4)

Now, when we open this using Binomial Theorem then all terms apart from the last term will be a multiple of 52 => a multiple of 13 => Remainder by 13 will be 0

=> Question is reduced to what is the remainder when the last term ( 25C25 \((-1)^{25} * 52^0\) is divided by 13

=> Remainder when -1 is divided by 13
=> -1 + 13 = 12

So, Answer will be A
Hope it helps!

Watch the following video to Learn about Binomial Theorem and How to Solve Similar problems

User avatar
Yes2GMAT
User avatar
Joined: 30 Dec 2020
Last visit: 29 Jun 2024
Posts: 38
Own Kudos:
18
Given Kudos: 175
Posts: 38
Kudos: 18
When 51^25 is divided by 13, the remainder obtained is: 26 Jan 2024, 04:55
Kudos
Add Kudos
Bookmarks
Bookmark this Post
gaurav1418z
Bunuel,

How did you get 12 as remainder when -1 is divided by 13, i didnt quite understand how did you substitute quotient and the remainder in the formula below

-1 = 13(-1) + 12
Show more

Bunuel, could u clarify this? Thanks a lot.
   1   2   3   
Moderators:
Bunuel
Math Expert
105378 posts
Krunaal
Tuck School Moderator
805 posts