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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, 09:47

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 _________________

“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.”

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

1

This post received KUDOS

Expert's post

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]
24 Jul 2013, 11:04

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 ??

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

Ah...okay I see that is not possible the way I have done it. It will have to be converted to (-1)^25 ie. -1 as remainder...since negative remainder so we have to add the divisior that is 13 to it and we get remainder as 12...

I guess than for Q3 ie 66^25/13 will have to be done the same way..why then do we get correct answer to Q2 and q4 by finding the cyclicity or the method used by you to solve.. _________________

“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.”

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]
17 May 2014, 00:05

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

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]
18 May 2014, 20:40

Expert's post

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

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

1

This post received KUDOS

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]
19 May 2014, 11:52

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

Hi Gaurav, If you find the remainder as -1, that means for some quotient q, 51^25 = 13*q - 1 = 13 * [(q-1) + 1] -1 = 13 *(q-1) + 13 -1 = 13*(q-1) + 12 We follow the convention of mentioning the remainder as positive so , we can work out backwards if we get negative remainder as solution.

Bunuel has tried to show us one way of translating the negative remainder as positive by using negative quotient. Hope it helps.

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]
19 May 2014, 22:55

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 x-a.

For example if after dividing an expression with 8 we find a remainder of -1 then actual remainder will be 8-1 i.e. 7.

Now let's use this concept:

51^25 can be written as (52-1)^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 13-1 i.e. 12.

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]
02 Jul 2014, 18:52

I think there is another easy way to tap the question. 51^1 = 51 51^2 = 2601 51^3 = 132651 51^4 = …201 The pattern is repeating: 51, 01, 51, 01 ..... The leftover is 51, divided by 13, =12

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, 13-1 = 12. Was that just a lucky guess on my part?

Re: When 51^25 is divided by 13, the remainder obtained is: [#permalink]
24 Aug 2014, 22:25

Expert's post

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, 13-1 = 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 13-1 = 12. The options will not have a negative remainder. They will always give you positive remainders only.

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