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Re: When 51^25 is divided by 13, the remainder obtained is:
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17 May 2014, 01:05
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



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Re: When 51^25 is divided by 13, the remainder obtained is:
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18 May 2014, 21:40
gaurav1418z wrote: 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 Check this post for negative remainders: http://www.veritasprep.com/blog/2014/03 ... thegmat/
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Re: When 51^25 is divided by 13, the remainder obtained is:
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19 May 2014, 23:55
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 xa.
For example if after dividing an expression with 8 we find a remainder of 1 then actual remainder will be 81 i.e. 7.
Now let's use this concept:
51^25 can be written as (521)^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 131 i.e. 12.
Answer is A



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When 51^25 is divided by 13, the remainder obtained is:
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23 Aug 2014, 13:19
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. Hi Karishma & Bunuel, I'm failing to see the direct connection with your blog post(binomial) and the method that bunuel posted. In Binomial, you're trying to find an exponent that you can raise the base to, which will be close to the divisor. In this case, we just multiplied 13 by 4, and are not raising it to anything. 1)That being said, even if I make the leap and see the connection here, I'm having a hard time noticing why the remainder is 12 and not 1. 2) = (1)^25 = 1. If that choice was given in the answer choice, alongside 13, would 1 be the answer? 3) Additionally, I tried the approach of 51^any power will yield a units digit of 1. Since we are dividing by 13, and the units will always be 1, 131 = 12. Was that just a lucky guess on my part?



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Re: When 51^25 is divided by 13, the remainder obtained is:
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24 Aug 2014, 23:25
russ9 wrote: Hi Karishma & Bunuel,
I'm failing to see the direct connection with your blog post(binomial) and the method that bunuel posted.
In Binomial, you're trying to find an exponent that you can raise the base to, which will be close to the divisor. In this case, we just multiplied 13 by 4, and are not raising it to anything.
1)That being said, even if I make the leap and see the connection here, I'm having a hard time noticing why the remainder is 12 and not 1.
2) = (1)^25 = 1. If that choice was given in the answer choice, alongside 13, would 1 be the answer?
3) Additionally, I tried the approach of 51^any power will yield a units digit of 1. Since we are dividing by 13, and the units will always be 1, 131 = 12. Was that just a lucky guess on my part? In the method shown by Bunuel, \(51^{25}\) is split into \((52  1)^{25}\) and binomial is applied on it to find that when it is divided by 13, the remainder will be 1. My post explains you why the remainder will be 1 when \((52  1)^{25}\) is divided by 13. But note that remainder is a positive concept. So saying that remainder is 1 is same as saying that remainder is 131 = 12. The options will not have a negative remainder. They will always give you positive remainders only. This post gives you a discussion on negative remainders: http://www.veritasprep.com/blog/2014/03 ... thegmat/
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When 51^25 is divided by 13, the remainder obtained is:
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01 Sep 2014, 13:36
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 133= 10, is that correct? Is even half of what I said correct? Thank you



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Re: When 51^25 is divided by 13, the remainder obtained is:
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01 Sep 2014, 21:44
bagdbmba wrote: Could you please let me know what'll be the answer of this question given at the end of your blogpost :
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 91=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) ?
This is correct. \(8^{28} = (9  1)^{28}\) When you expand \((9  1)^{28}\) you get \(9^{28} + 28*9^{27}(1)^1 + ....+ (1)^{28} = 9^{28} + 28*9^{27}(1)^1 + ....+1\) Every term has 9 as a factor except the last term which is 1 so keep it separately. Now you can take out 9 common from all terms except the last term and put whatever is leftover as m to get 9m + 1
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Re: When 51^25 is divided by 13, the remainder obtained is:
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02 Sep 2014, 01:45
usre123 wrote: 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 133= 10, is that correct? Is even half of what I said correct? Thank you 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 rethink now? Let me know if there are still some doubts.
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Re: When 51^25 is divided by 13, the remainder obtained is:
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23 Sep 2014, 06:41
VeritasPrepKarishma wrote: 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 \(51^{25}12^{25} = 13(Q_1Q_2)\) Thus, if I could show that \(51^{25}12^{25}\) is divisible by 13, my assumption would be correct. Now,\(x^na^n\), is always divisible by (xa), x and a are integers, n is odd. Thus,\(51^{25}12^{25}\) is always divisible by (5112) = 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. 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? (521)^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*(91)^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(91)^29..... 1^29 which is a 1. multiply by the 4 I ignored so I have 4. For negative numbers make it 94, so 5....



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Re: When 51^25 is divided by 13, the remainder obtained is:
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23 Sep 2014, 22:07
usre123 wrote: 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? (521)^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*(91)^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(91)^29..... 1^29 which is a 1. multiply by the 4 I ignored so I have 4. For negative numbers make it 94, so 5.... 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?
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Re: When 51^25 is divided by 13, the remainder obtained is:
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11 Oct 2014, 03:03
VeritasPrepKarishmai checked your post .. it makes the concept very clear .. thanks . however i have one doubt : in the qstn , what is the remainder of 2^83 / 9. rather than considering (2^3)^27 * 2^2 /9 , can we consider (2^3)^28 * 2^1 /9



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Re: When 51^25 is divided by 13, the remainder obtained is:
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12 Oct 2014, 20:49
GuptaDarsh wrote: VeritasPrepKarishmai checked your post .. it makes the concept very clear .. thanks . however i have one doubt : in the qstn , what is the remainder of 2^83 / 9. rather than considering (2^3)^27 * 2^2 /9 , can we consider (2^3)^28 * 2^1 /9 If you do that, how will you handle the extra 2 you get in the denominator from \(2^{1}\)? Usually, we try not to change the divisor. It brings in its own complications.
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Re: When 51^25 is divided by 13, the remainder obtained is:
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13 Oct 2014, 06:34
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



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Re: When 51^25 is divided by 13, the remainder obtained is:
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13 Oct 2014, 21:01
GuptaDarsh wrote: 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 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.
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Re: When 51^25 is divided by 13, the remainder obtained is:
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05 Nov 2014, 20:03
I tried to base this on the cyclicity of exponentsis 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?



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Re: When 51^25 is divided by 13, the remainder obtained is:
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10 Nov 2014, 22:38
intheend14 wrote: I tried to base this on the cyclicity of exponentsis 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? 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.
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Re: When 51^25 is divided by 13, the remainder obtained is:
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10 Nov 2014, 23:50
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



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Re: When 51^25 is divided by 13, the remainder obtained is:
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29 Oct 2015, 18:02
How about this: 51 is 3*17, or 3*(13 + 4) if I distribute the 25 to each, I get (3^25)(13 + 4)^25 The units digit of 3^25 will be 3 and since we don't care about that 13, we only need to look at the 4 (I'm shaky on the binomial theorem, but this seems to be the pattern). Well, the units digit of 4^25 is 4, so I multiplied the units digit of each and got 12.
Can someone tell me if my logic makes sense and can be applied to other problems of this nature?
Happy Studying!



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Re: When 51^25 is divided by 13, the remainder obtained is:
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29 Oct 2015, 18:16
atd86 wrote: How about this: 51 is 3*17, or 3*(13 + 4) if I distribute the 25 to each, I get (3^25)(13 + 4)^25 The units digit of 3^25 will be 3 and since we don't care about that 13, we only need to look at the 4 (I'm shaky on the binomial theorem, but this seems to be the pattern). Well, the units digit of 4^25 is 4, so I multiplied the units digit of each and got 12.
Can someone tell me if my logic makes sense and can be applied to other problems of this nature?
Happy Studying! I dont think this method will work on other questions. All these questions can be worked in the following manner: For finding remainder of 51^25 when divided by 13, make sure to express 51 as a multiple of 13 \(\pm\) 1 > 51=521 > 51^25 = (521)^25 By binomial theorem, (a+b)^n = a^n*b^0+a^(n1)*b^1+ a^(n2)*b^2......a^(2)*b^(n2)+a^1*b^(n1)+a^0*b^n Thus for (521)^25 , all terms except the last (1)^25 will be multiples of 52 > multiple of 13 and hence we only need to care about (1)^25 = remainder when 1 is divided by 13 is = remainder when 131= 12 is divided by 13, giving you 12 as the remainder. This is a method that can be applied to all such questions. Hope this helps.



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Re: When 51^25 is divided by 13, the remainder obtained is:
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11 Jan 2016, 23:39
naveenq wrote: E.g. What is the remainder when 10^5 is divided by 13?
From binomial method, we can take (133)^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 timeconsuming. Can you help me understand how I can use the binomial theorem for the negative base? (3)^5 / 13
THANK YOU!! (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: http://www.veritasprep.com/blog/2014/03 ... thegmat/
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Re: When 51^25 is divided by 13, the remainder obtained is:
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