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When 51^25 is divided by 13, the remainder obtained is:
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04 Dec 2016, 11:49
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: (Had to delete the veritas link since I have never posted before on GMAT Club)Once you go through it, this question should be very easy for you. I haven't been able to go through all the replies on this post here and I don't consider myself great at math, but wanted to look at an alternative way of doing this question and remaining within the domains of the GMAT topics (I'd think it would be harsh if the people there started checking us on advanced topics like Binomial Theorem). So let's do prime factorization. 51^25 = (3*17)^25 = (3^25)*(17^25) Using the Last Digit of a Power (pg 14 of GMAT Club Math Book for people unaware of this method), powers of 3 have cyclicality of 4, therefore last digit (3^25) = last digit(3^1) 17^25 would basically check cyclicality of the power of 7 (which is 4), i.e. last digit (17^25) = last digit (17^1). We can therefore simply write (3*17)^25 as (3*17) or 51. When you divide 51/13, you get 12 as the remainder. I can't judge if it is as quicker as the method above, but I definitely think it is simpler to understand.



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Re: When 51^25 is divided by 13, the remainder obtained is:
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04 Dec 2016, 20:13
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?



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Re: When 51^25 is divided by 13, the remainder obtained is:
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04 Dec 2016, 20:59
DonnieDrastic wrote: 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? 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.
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Re: When 51^25 is divided by 13, the remainder obtained is:
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04 Dec 2016, 22:05
omairanw wrote: 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: (Had to delete the veritas link since I have never posted before on GMAT Club)Once you go through it, this question should be very easy for you. I haven't been able to go through all the replies on this post here and I don't consider myself great at math, but wanted to look at an alternative way of doing this question and remaining within the domains of the GMAT topics (I'd think it would be harsh if the people there started checking us on advanced topics like Binomial Theorem). So let's do prime factorization. 51^25 = (3*17)^25 = (3^25)*(17^25) Using the Last Digit of a Power (pg 14 of GMAT Club Math Book for people unaware of this method), powers of 3 have cyclicality of 4, therefore last digit (3^25) = last digit(3^1) 17^25 would basically check cyclicality of the power of 7 (which is 4), i.e. last digit (17^25) = last digit (17^1). We can therefore simply write (3*17)^25 as (3*17) or 51. When you divide 51/13, you get 12 as the remainder. I can't judge if it is as quicker as the method above, but I definitely think it is simpler to understand. Cyclicity can help you find the remainder only in case of division by 2, 5 or 10. Check this post to know why: https://www.veritasprep.com/blog/2015/1 ... questions/The last digit will not decide the remainder in every case in case the divisor is other than 2, 5 or 10.
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Re: When 51^25 is divided by 13, the remainder obtained is:
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24 Jan 2017, 01:59
BunuelHi 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



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Re: When 51^25 is divided by 13, the remainder obtained is:
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24 Jan 2017, 02:42
Vaidya wrote: BunuelHi 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 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.
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Re: When 51^25 is divided by 13, the remainder obtained is:
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04 Sep 2017, 00:33
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 \(51^{25}=(521)^{25}\), 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 ...



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Re: When 51^25 is divided by 13, the remainder obtained is:
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04 Sep 2017, 00:36
gmatcracker2017 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 \(51^{25}=(521)^{25}\), 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 ... 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 5 pages of the discussion.
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Re: When 51^25 is divided by 13, the remainder obtained is:
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04 Sep 2017, 16:33
via fermat theorem a^(p1)= 1 mod p where a is an integer and p prime as 51 contains 13 it can be rewritten and simplified 51/13 remainder 12
12^(131) = 1 mod 13 12^25 = 12^(12*2+1)= 12



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Re: When 51^25 is divided by 13, the remainder obtained is:
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29 Sep 2017, 00:13
[i]
Hello Bunuel and Karishma,
Can anyone from you, please tell me the limitation of cyclic rule? So far i was using that rule blindly for all question. But it has not worked for this particular problem.[i]



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Re: When 51^25 is divided by 13, the remainder obtained is:
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29 Sep 2017, 01:12
goalMBA1990 wrote: [i]
Hello Bunuel and Karishma,
Can anyone from you, please tell me the limitation of cyclic rule? So far i was using that rule blindly for all question. But it has not worked for this particular problem.[i] Cyclicity helps you figure out the units digit in every case. Also, it has nothing to do with remainders. In some cases, the units digit helps you figure out the remainder, more specifically in case of division by 2, 5 or 10. https://www.veritasprep.com/blog/2015/1 ... questions/https://www.veritasprep.com/blog/2015/1 ... nspart2/
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Re: When 51^25 is divided by 13, the remainder obtained is:
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21 Feb 2018, 08:07
LM wrote: When 51^25 is divided by 13, the remainder obtained is:
A. 12 B. 10 C. 2 D. 1 E. 0 Remainder (A * B) = R(A)*R(B) Therefore 51^25/13 = [51*51*51*51....(25 times)] / 13 = [12*12*12.........(25 times)] / 13 (As 51/13 is 12) = [144*144..........(12 times) *12] / 13 ( Grouping two adjacent 12's together so finally one 12 will be left) = [1*1*1.............(12 times) *12] / 13 (144/13 is 1) = 12 / 13 = 12



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Re: When 51^25 is divided by 13, the remainder obtained is:
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29 Mar 2018, 23:02
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 ... ekinyou/Once you go through it, this question should be very easy for you. Mam, why cyclicity method does not work here. I marked answer 1 thinking 1/13 remainder "1"
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Re: When 51^25 is divided by 13, the remainder obtained is:
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30 Mar 2018, 02:09
disharupani wrote: Hey, I didn't understand:
When 1 is divided by 13 how can the reminder be 12? If I substitute in the equation 1=13q+r, so how can we arrive at r=12? hi I will try to help you okay, think about simple division, when you divide 5 by 2, you get 2 as quotient and 1 as remainder, because 4 is the greatest integer less than 5 now, on the same line of reasoning, when you divide 1 by 13, 13 is the greatest integer less than 1, and thus 12 is the remainder hope this helps and is clear! thanks and cheers!



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Re: When 51^25 is divided by 13, the remainder obtained is:
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30 Mar 2018, 02:26
QZ wrote: 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 ... ekinyou/Once you go through it, this question should be very easy for you. Mam, why cyclicity method does not work here. I marked answer 1 thinking 1/13 remainder "1" You are not dividing 1 by 13 here. You are dividing a number that ends in 1 by 13 here. We don't know what that number exactly is. Will every number ending in 1 when divided by 13 give 1 as remainder? Think about it: 11/13, remainder 11 21/13, remainder 8 31/13, remainder 5 and so on... Cyclicity does not help get the remainder in most cases. It does so in very few cases. They are discussed in this post: https://www.veritasprep.com/blog/2015/1 ... questions/https://www.veritasprep.com/blog/2015/1 ... nspart2/
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Re: When 51^25 is divided by 13, the remainder obtained is:
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08 Apr 2018, 06:58
5125=(52−1)255125=(52−1)25 , now if we expand this expression all terms but the last one will have 52=13∗452=13∗4 in them, thus will leave no remainder upon division by 13, the last term will be (−1)25=−1(−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)+12−1=13∗(−1)+12.
Answer: A.



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Re: When 51^25 is divided by 13, the remainder obtained is:
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26 May 2018, 04:06
I was having hard time understanding why remainder is not simply 1 as 1/13 means that we could have simply done it as: 13(0)+(1).
Thereafter I myself realised that remainder can never be less than zero and hence it has been multiplied by minimum possible integer to equate further i.e. 1.
Incase I am wrong, please correct. If not, this might be useful for somebody who might be having trouble understanding the same.
Thanks



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Re: When 51^25 is divided by 13, the remainder obtained is:
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06 Jul 2019, 17:56
I reasoned through it without actually knowing binomial theorem... it's important to understand it, though, as it will save tremendous amounts of time.
What I did: 1) 13 goes into 51 three times (Q1) with remainder 12. 2) Now we have 12^25 remainder from the initial division, obviously 13 goes into this extra amounts. 3) 12 is 1 less than 13, which means that when we divide 12^25 by 13 we have a deficit of 1 every time, 25 times. 4) That means we get some # (Q2) that's divisible by 13  25. But, 13 goes into 25 once, so that means (Q2+1) with remainder 12.
So, at the end of it all we had two numbers divisible by 13 (Q1 & Q2) with 25 leftover deficit, which we subtracted from the number to find that the remainder is 12.



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Re: When 51^25 is divided by 13, the remainder obtained is:
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13 Jul 2019, 06:29
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 \(51^{25}=(521)^{25}\), 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. Bunuel can you please explain the last part i.e "\(1=13*(1)+12\)". I am not able to understand why did u subtract? Thanks



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Re: When 51^25 is divided by 13, the remainder obtained is:
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13 Jul 2019, 07:02
Anoushka1995 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 \(51^{25}=(521)^{25}\), 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. Bunuel can you please explain the last part i.e "\(1=13*(1)+12\)". I am not able to understand why did u subtract? Thanks This is expanded several times on previous pages. There are many other useful staff there too. So, worth reading the whole thread.
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