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

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01 Sep 2014, 13:21

WoundedTiger 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 ??

Consider another Q2 question 2^86/9 ? Cyclicity of 2 is 4 so dividing the power of 2 i.e 86/4 we get remainder 4 {cyclicity 2,4,8,6}

While in this Question we can get remainder straight away

Similarly another Question Q3 " What is remainder when 66^25/13 ?" Discussed in this thread

Now again cyclicity of 6 is 1 and hence remainder should be 6. However the answer seems to be 1 to this question.

Take another Q4 " what is remainder when 2^83 is divided by 9" taken from Karishma's blog on veritas prep Again Cyclicity of 2 is 4 so 2^ 83 will end up in 8 and hence remainder should be 8

Alternatively using other method discussed in Quant Forum 2^83-----> 2^3*2^80------> 2^3*4^40------> 2^3*8^20

Consider 8^20 can written as (9-1)^20 so remainder will be (-1)^20/9 = 1 and multiplying this by 8 we get remainder 8.Is this solution correct ??

If so can you elaborate that why in Q2 & Q4 we are getting the right answer but not in Q1 (original Q) and Q3?

Many Thanks!!!

Mridul

As for the pink part, I simply don't understand how you incorporated 9 in this....I understand the cycle well enough, so reminder would be 4. Then we divide 4/9? Since dividend is smaller than the divisor, answer should be 4?

Also, isn't this a hard (perhaps a late 600s) questions? Why is it categorized as a sub 600?

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

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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 13-3= 10, is that correct? Is even half of what I said correct? Thank you

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?

This is all good. There is nothing wrong in the 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?

Nope...remember the remainder can never be negative. Whenever you have a negative remainder, add the divider..

So why the remainder is negative..The questions What is the remainder when (51)^25/13.... You can write the expression as as (52-1)^25/13...Now barring the last term in this expression (When expanded using Binomial theroum) will not be divisble by13... so remainder will be (-1)^25 =-1...Remainder is negative so add divider and we get -1+13=12---> remainder

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)

This is correct.

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

This is correct.
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

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
_________________

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

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01 Sep 2014, 23:43

[quote="WoundedTiger"]

Quote from VeritasprepKarishma"

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: "

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?

The answer given by Karishma is a 5, not a 8

This is all good. There is nothing wrong in the 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?

Nope...remember the remainder can never be negative. Whenever you have a negative remainder, add the divider..

So why the remainder is negative..The questions What is the remainder when (51)^25/13.... You can write the expression as as (52-1)^25/13...Now barring the last term in this expression (When expanded using Binomial theroum) will not be divisble by13... so remainder will be (-1)^25 =-1...Remainder is negative so add divider and we get -1+13=12---> remainder

Thank you. I understand binomial theorem (Atleast I think I do, read Karishma's wonderful blog post)

So my only question now is, why should I use binomial (or split up 51 into 52-1) when I can, as in the question before it, just look at the cycle of units digit, which is 1, and 1^25 is still 1, and so the remainder is 1? Why should I apply binomial here, when I didn't have to for the 2^83 question. Is it because the number being raised to the power of 25 a double digit, and not a single one?

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

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02 Sep 2014, 00:08

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?

The answer given by Karishma is a 5, not a 8

I missed the actual question. Ans is indeed 5...

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?

Nope...remember the remainder can never be negative. Whenever you have a negative remainder, add the divider..

So why the remainder is negative..The questions What is the remainder when (51)^25/13.... You can write the expression as as (52-1)^25/13...Now barring the last term in this expression (When expanded using Binomial theroum) will not be divisble by13... so remainder will be (-1)^25 =-1...Remainder is negative so add divider and we get -1+13=12---> remainder

Thank you. I understand binomial theorem (Atleast I think I do, read Karishma's wonderful blog post)

So my only question now is, why should I use binomial (or split up 51 into 52-1) when I can, as in the question before it, just look at the cycle of units digit, which is 1, and 1^25 is still 1, and so the remainder is 1? Why should I apply binomial here, when I didn't have to for the 2^83 question. Is it because the number being raised to the power of 25 a double digit, and not a single one?

No...The problem here is last digit is of the type (-1)^odd Integer=-1...

If the the last term of the Bionomial expression was of the type (-1)^even number then you can say Remainder to be 1..

Let's apply Binomial theorum and say what we get...

The question is remainder when 2^83/9...

It can be re-written as 2^2*2^81/9----> 2^2*8^27/9...Keep 2^2 aside for a moment and see what we have

8^27----> \((9-1)^{27}/9\).....If you expand this expression, we the last term will be of the type (-1)^27=-1 and remainder will be \(2^2*-1/9\) or -4/9...Since remainder is negative we add divider and get -4+9=5...

Ans is 5.
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

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]

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

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.

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....

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]

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11 Oct 2014, 03:03

VeritasPrepKarishma i 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

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

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11 Oct 2014, 03:33

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

\(51^{25}=(52-1)^{25}\), 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 Bunuel .. I am confused about When to use the concept of cyclicity and when to use the concept of binomial theoram ? in 51^25 -- the last digit will be the same as 1^25 .As cyclicity of 1 is 1, the last digit of 1^25 will be 1. and 1/13 gives remainder 1. Why is this incorrect ? Please explain

VeritasPrepKarishma i 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.
_________________

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

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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

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.
_________________

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?

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.
_________________

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.

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

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01 Jan 2015, 10:12

I used another method to find the result. It would be great if someone told everyone if this is a valid method to use.

So, I decided to use the "find the last digit of a power" to solve it. 51^25: First, I would have to find the cyclisity of 1. So, raise 1 to its powers, such as 1^1, 1^2, 1^3. In this case, it is always one. So, the last digit of 51^25 would be 1.

Then, I just chose 51^1/25 and ended up with a remainder of 12. If you only test the first 2 powers of 51 (namely 51^1 and 51^2) you see that the last digit is always 1. So, this is how I checked, just in case this was the wrong way to do it. Then, I solved it in a few seconds.

I would like to also post this link:http://totalgadha.com/mod/forum/discuss.php?d=4405 , which I found later, that deals with How to find last two digits of a number.

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

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01 Mar 2015, 08:42

LM wrote:

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

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

i am doing anything new than what have already been done by Masters and Gurus, just writing it here on the forum to reinforce the idea into my mind. (52-1)^25 = (52-1)Y where Y=(52-1)^24 when we divide (52-1)Y by 13 we get to same question again -1Y/13 = -(52-1)^24 , if we repeat the same step as above, we will get ^23 , ^22.....^1 . so question really is if (52-1)/13 what is the remainder . . . 12 is the answer .
_________________

Thanks, Lucky

_______________________________________________________ Kindly press the to appreciate my post !!

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

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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?