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# When 51^25 is divided by 13, the remainder obtained is:

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When 51^25 is divided by 13, the remainder obtained is: [#permalink]  04 Apr 2012, 07:43
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When 51^25 is divided by 13, the remainder obtained is:

A. 12
B. 10
C. 2
D. 1
E. 0
[Reveal] Spoiler: OA
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Kudos [?]: 17571 [7] , given: 2183

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]  04 Apr 2012, 08:03
7
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Expert's post
LM wrote:
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.

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Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]  04 Apr 2012, 09:03
16
KUDOS
Expert's post
LM wrote:
When 51^25 is divided by 13, the remainder obtained is:

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

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:

http://www.veritasprep.com/blog/2011/05 ... ek-in-you/

Once you go through it, this question should be very easy for you.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Manager Joined: 28 Jul 2011 Posts: 213 Followers: 0 Kudos [?]: 20 [0], given: 13 Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink] 05 Apr 2012, 20:25 Bunuel wrote: LM wrote: 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. Hey Bunuel, I am trying to understand this concept what happens if 66^25 is divided by 13 - then how will the above method work? [65 + 1]^25 then will remainder be 1? Senior Manager Joined: 12 Mar 2012 Posts: 370 Concentration: Operations, Strategy Followers: 2 Kudos [?]: 84 [0], given: 31 Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink] 05 Apr 2012, 21:59 kuttingchai wrote: Bunuel wrote: LM wrote: 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. Hey Bunuel, I am trying to understand this concept what happens if 66^25 is divided by 13 - then how will the above method work? [65 + 1]^25 then will remainder be 1? the concept is when you binomially expand the (65+1)^25 all terms except last one that is (25c25)*(65^0)*(1^25) will be divisible by 13. the remainder for all terms will zero except the last one that is 1/13 for which the remainder is 1. hence the answer to your question will be 1. hope this helps..!! _________________ Practice Practice and practice...!! If my reply /analysis is helpful-->please press KUDOS If there's a loophole in my analysis--> suggest measures to make it airtight. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4028 Location: Pune, India Followers: 857 Kudos [?]: 3612 [1] , given: 144 Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink] 06 Apr 2012, 00:31 1 This post received KUDOS Expert's post kuttingchai wrote: Bunuel wrote: LM wrote: 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. Hey Bunuel, I am trying to understand this concept what happens if 66^25 is divided by 13 - then how will the above method work? [65 + 1]^25 then will remainder be 1? Check out the link I have given above. It discusses the use of Binomial Theorem to solve such questions. And yes, 66^25 is (65+1)^25 and since 65 is divisible by 13, the remainder will be 1 in this case. On the other hand, if you have 64^25, then you get (65 - 1)^25 so here, remainder will be (-1) i.e. 12 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Manager
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Kudos [?]: 20 [1] , given: 13

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]  06 Apr 2012, 04:51
1
KUDOS
Thank you kraizada84 & Karishma

The explaination in the below link helped to understand the "why part"...

http://www.veritasprep.com/blog/2011/05 ... ek-in-you/
Manager
Joined: 12 Oct 2011
Posts: 135
GMAT 1: 700 Q48 V37
GMAT 2: 720 Q48 V40
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Kudos [?]: 46 [0], given: 23

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]  07 Apr 2012, 15:07
VeritasPrepKarishma wrote:
LM wrote:
When 51^25 is divided by 13, the remainder obtained is:

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

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:

http://www.veritasprep.com/blog/2011/05 ... ek-in-you/

Once you go through it, this question should be very easy for you.

How is a test-taker, who has never heard of that concept, supposed to answer this question in 2 minutes though?
Intern
Joined: 12 Dec 2011
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Kudos [?]: 0 [0], given: 2

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]  15 Mar 2013, 18:27
Hello Experts,
I am not able to understand why the answer is 1 in the first case (65+1)/13 where as it is 12 in the second case (65-1)/13. Is it because of minus (-)?
If it is so, then the result should be (9-1)*4 in the second example in the link provided by you.

Thanks
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Kudos [?]: 3612 [2] , given: 144

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]  17 Mar 2013, 21:04
2
KUDOS
Expert's post
meenua wrote:
Hello Experts,
I am not able to understand why the answer is 1 in the first case (65+1)/13 where as it is 12 in the second case (65-1)/13. Is it because of minus (-)?
If it is so, then the result should be (9-1)*4 in the second example in the link provided by you.

Thanks

The remainder in the first case is 1 whereas the remainder in the second case is -1. Why? Because the last term of the binomial will be (-1)^25 in the second case. I strongly suggest that you check out this post to understand binomial:
http://www.veritasprep.com/blog/2011/05 ... ek-in-you/

How do we handle a negative remainder such as -1 since the options will only give us positive remainders? That is also explained in detail in the post but let me add a small explanation here as well:

When I divide 32 by 10, I get a remainder of 2 - you know that. Can I also say that the remainder can also be said to be -8 (if negative remainders were allowed)? It's something like this: Say I have $32 and 10 people in front of me. I can give each person$3 and I will be left with $2 (or I can say that my balance is +2). Or I can give each person$4 and I will be poorer by $8 (i.e. my balance will be -8. I would have given$8 from my own pocket).
So when I divide a number by 10, I can say that my remainder could be 2 or it could be -8.
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Karishma
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My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Intern Joined: 17 May 2013 Posts: 9 Followers: 0 Kudos [?]: 1 [0], given: 7 Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink] 23 May 2013, 22:23 Quite amazing as the question above entirely contradicts GMAC's and several other prep site's and material's assurance that for GMAT you need not have a real awesome Maths background. I even never heard of Binomial theorem in my entire life and gave 2 hours to understand it. Nothing clear for me. Can anyone elaborate the complex method of the specific theorem OR some other way to solve the problem. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4028 Location: Pune, India Followers: 857 Kudos [?]: 3612 [1] , given: 144 Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink] 23 May 2013, 23:02 1 This post received KUDOS Expert's post genuinebot85 wrote: Quite amazing as the question above entirely contradicts GMAC's and several other prep site's and material's assurance that for GMAT you need not have a real awesome Maths background. I even never heard of Binomial theorem in my entire life and gave 2 hours to understand it. Nothing clear for me. Can anyone elaborate the complex method of the specific theorem OR some other way to solve the problem. Here is the thing - GMAC and all test prep companies are absolutely correct. You don't need an awesome Math background. Also, such questions can be easily solved by figuring out the pattern - not this one since calculating powers of 51 (51^2, 51^3 ...) etc is quite painful - but such question with smaller numbers can be easily solved without using Binomial theorem. Then why do such questions appear in GMAT material? Because either some test prep companies go overboard in making tougher questions or these questions are picked from non-GMAT prep material. Then why are we discussing Binomial theorem here? Because it is a great way to solve all such questions (either small numbers or large) and it is not difficult to understand. If you understand what (a + b)^2 = a^2 + b^2 + 2ab means, you are halfway there already! That is what I have tried to do in this post: http://www.veritasprep.com/blog/2011/05 ... ek-in-you/ I have tried to explain Binomial in very layman terms and just enough to help you solve all such questions in under a minute. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]  23 May 2013, 23:20
Expert's post
genuinebot85 wrote:
Quite amazing as the question above entirely contradicts GMAC's and several other prep site's and material's assurance that for GMAT you need not have a real awesome Maths background.

I even never heard of Binomial theorem in my entire life and gave 2 hours to understand it.
Nothing clear for me. Can anyone elaborate the complex method of the specific theorem OR some other way to solve the problem.

Bunuel and Karishma have already shown the best method. However, you can do this problem in another way too.

The remainder when 51 is divided by 13 can be either (-1) or 12.
Assuming that you are not comfortable with negative remainders, we can then restate the problem as : Remainder of\frac{(51^{25})}{13}= Remainder of\frac{(12^{25})}{13}. Now notice that 12^2 = 144 and Remainder of\frac{144}{13}= 1. Thus, Remainder of\frac{(12^{25})}{13}= [(144)^{12}*12]/13 = Remainder of \frac{12}{13}= 12.

Assuming that you are comfortable with negative remainders, Remainder of \frac{(51^{25})}{13}= Remainder of\frac{(-1)^{25}}{13}= Remainder of -1/13 = 12.
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Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]  24 May 2013, 01:15
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Expert's post
vinaymimani wrote:
genuinebot85 wrote:
Quite amazing as the question above entirely contradicts GMAC's and several other prep site's and material's assurance that for GMAT you need not have a real awesome Maths background.

I even never heard of Binomial theorem in my entire life and gave 2 hours to understand it.
Nothing clear for me. Can anyone elaborate the complex method of the specific theorem OR some other way to solve the problem.

Bunuel and Karishma have already shown the best method. However, you can do this problem in another way too.

The remainder when 51 is divided by 13 can be either (-1) or 12.
Assuming that you are not comfortable with negative remainders, we can then restate the problem as : Remainder of\frac{(51^{25})}{13}= Remainder of\frac{(12^{25})}{13}. Now notice that 12^2 = 144 and Remainder of\frac{144}{13}= 1. Thus, Remainder of\frac{(12^{25})}{13}= [(144)^{12}*12]/13 = Remainder of \frac{12}{13}= 12.

Assuming that you are comfortable with negative remainders, Remainder of \frac{(51^{25})}{13}= Remainder of\frac{(-1)^{25}}{13}= Remainder of -1/13 = 12.

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!
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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 619 Followers: 32 Kudos [?]: 468 [0], given: 133 Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink] 24 May 2013, 01:57 Expert's post VeritasPrepKarishma wrote: 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. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4028 Location: Pune, India Followers: 857 Kudos [?]: 3612 [1] , given: 144 Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink] 25 May 2013, 02:07 1 This post received KUDOS Expert's post vinaymimani wrote: VeritasPrepKarishma wrote: 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]  25 May 2013, 21:43
Expert's post
VeritasPrepKarishma wrote:
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?

Yes,I get what you are saying.Moreover, in the same spirit, I would just like to add ,that what made me think about 12^{25}, when I saw 51^{25} is this :

51mod13 = (17*3)mod13 = 17mod13*3mod13 = 4mod13*3mod13 = 12mod13.IMO, this doesn't have any undercurrents of Binomial Theorem.

And Yes you can also think of it as 51mod13 = (39+12)mod13 = 39mod13+12mod13 = 0+12mod13.I guess, this is more in line with what you pointed out and does have a flavor of Binomial Theorem.

Note: It is beyond the scope of GMAT and this discussion but there is still atleast one more way of doing this problem, and I believe that doesn't involve Binomial Theorem.Nonetheless, I appreciate the discussion we had. Let me know if I missed out on anything.
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Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]  26 May 2013, 03:30
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This is one of the best questions and best replies..really liked it..Although i took around an hour to understand binomial theorem in Karishma's blog
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Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]  16 Jun 2013, 04:46
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For this question we can get a pattern and solve this

51 when divided by 13 is 12
51^2 2601 divided by 13 is 1

for 51^3 remainder is 12

51^4 remainder is 1

The pattern for this question is {12,1}

PS you don't have to test the values for 3 and 4 just multiply the remainder values and divide by 13 for example

for the remainder of 51^3 multiply the remainder of 51^2 which is 12 and 51^1 which is 1 and divide that by 13

so when we divide 25/ 2( which is the pattern) we get remainder 1 so the final remainder will be 12
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Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]  21 Jul 2013, 15:09
Expert's post
VeritasPrepKarishma wrote:
LM wrote:
When 51^25 is divided by 13, the remainder obtained is:

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

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:

http://www.veritasprep.com/blog/2011/05 ... ek-in-you/

Once you go through it, this question should be very easy for you.

Hi VeritasPrepKarishma,
This blog-post is just awesome..Thanks! +1 from me.

Could you please let me know what'll be the answer of this question given at the end of your blog-post :

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

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

2^83 = (2^2) * (2^81) = (4) * (2^3)^27 = 4* (8^27)

Now, 8^27= (9 – 1)^27
So,(-1)^27 =-1 => remainder 9-1=8

So,we're left out with 4 and remainder for 2^83 divided by 9 is 5.

P.S: can you please clarify that how 8^28 = 9m + 1 (where m is some positive integer) ?
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Re: When 51^25 is divided by 13, the remainder obtained is:   [#permalink] 21 Jul 2013, 15:09
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# When 51^25 is divided by 13, the remainder obtained is:

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