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Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]
04 Apr 2012, 08:03

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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.

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]
17 Mar 2013, 21:04

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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.

Please help me understand this concept. 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. _________________

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]
06 Apr 2012, 00:31

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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 _________________

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]
23 May 2013, 23:02

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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. _________________

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]
23 May 2013, 23:20

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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. _________________

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]
24 May 2013, 01:15

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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! _________________

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]
25 May 2013, 02:07

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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. _________________

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 _________________

“Confidence comes not from always being right but from not fearing to be wrong.”

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) ?

Yes, 5 is correct. Good job. You get 4*(9 - 1)^{27} which gives a remainder of 4(-1)^{27} = -4 When divisor is 9, a remainder of -4 is equivalent to a remainder of 5.

"P.S: can you please clarify that how 8^28 = 9m + 1 (where m is some positive integer) "

8^{28} = ( 9 - 1)^{28} = [9 + (-1)]^{28} When you expand it, you get 9^{28} + 28*9^{27}*(-1) + ....... (-1)^{28} = 9^{28} - 28*9^{27} + ... + 1 All terms in the expansion will have 9 as a factor except the last term (which is 1). In the expansion, some terms will be positive and some will be negative. Notice that overall the sum must be positive because the sum is equal to 8^{28} which is positive. when you add/subtract multiples of 9, you will get a multiple of 9 as the answer.

So expansion becomes 9m + 1 (which is equal to 8^{28} so m must be positive) _________________

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]
24 Jul 2013, 09:50

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mridulparashar1 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.

Hello Bunuel,

I understood the solution to this Question quite well. But I need your inputs on the following. Can this Question be done by first finding the cyclicity of 51^25 which will be 1^ 25 and when 1^ 25 is divided 13 should give remainder 1 and not 12. What am I missing here ??

Please elaborate on red part. What does this mean? How are you going to do that? ... _________________

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]
19 May 2014, 09:37

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While solving this question, thought process should be like this 2^3 = 8 which can be written as (9-1) Since power of 2 i.e. 83 is not a multiple of 3 we have to write 2^83 as (2^2)(2^81) because 81 is divisible by 3. Now , 2^81 can be written as (2^3)^27 or (8)^27 or (9-1)^27. In the expansion of (9-1)^27 all the terms will be divisible by 9, except last term i.e. (-1)^27 = -1 Since the second last term will be 9, if we combined last two term, we will get 9-1=8 . This 8 will be remainder as all other terms will have a multiple of 9. We can write the number as 9k+8. Multiply this with 2^2 i.e. 4. we will get 36k+32 or 36k + 27 +5 First two terms are divisible by 9 and this will give 5 as a remainder. Hence Answer is 5 _________________

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]
02 Sep 2014, 00:45

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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 13-3= 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 re-think now? Let me know if there are still some doubts. _________________

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]
23 Sep 2014, 21:07

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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? (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....

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? _________________

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]
13 Oct 2014, 20:01

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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. _________________

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?

gmatclubot

Re: When 51^25 is divided by 13, the remainder obtained is:
[#permalink]
05 Apr 2012, 20:25