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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=52-1 ---> 51^25 = (52-1)^25

By binomial theorem, (a+b)^n = a^n*b^0+a^(n-1)*b^1+ a^(n-2)*b^2......a^(2)*b^(n-2)+a^1*b^(n-1)+a^0*b^n

Thus for (52-1)^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 13-1= 12 is divided by 13, giving you 12 as the remainder.

This is a method that can be applied to all such questions.

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

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11 Jan 2016, 05:53

VeritasPrepKarishma 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?

Hi Karishma and Bunuel,

Your guide on Binomial Theorem is AMAZING!! I wish i had gone through this before taking my GMAT for the first time. I have one question for you which I can't seem to solve. this is in regards to base of powers that are negative - and how to compute them. Can you provide a general rule for this?

E.g. What is the remainder when 10^5 is divided by 13?

From binomial method, we can take (13-3)^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 time-consuming. Can you help me understand how I can use the binomial theorem for the negative base? (-3)^5 / 13

Thank you for the quick response! That makes sense - essentially, in these questions, I need to find a way to get to +/- 1 from the divisor itself, or from a multiple of that divisor. Even if that means following binomial theorem more than once, like in the example above.

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

Show Tags

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.

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.
_________________

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.

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

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19 Dec 2016, 05:28

Hi,

I went through the solution, I went through the Blog and I went through some other discussions as well. But I am unable to understand the Remainder Concept.

You guys are saying that -1/13 gives the remainder as 12.

I just don't understand how this works.

I'll give you my chain of thoughts and how I am working things out:

Hi 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.
_________________

Hi 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.

Correct me if I'm wrong. I believe that it is not necessary to know the exact cyclicity of any number, since a cyclicity of 2 will also follow the cylicity of 4, ie, in this case, it does not matter if the cyclicity is 2 or 4 since they will be equal anyway. And this applies to any number which follows a cyclicity of 2. 2nd and 4th WILL be equal.

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

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22 Jun 2017, 08:03

Hi, 51^ 25 will have unit digit 1 and 1 divided by 13 will have remainder 1. So answer should be D. Now any power of 1 will always have unit digit one.

Please correct me where I am making mistake?

Is it not correct that any power of 1 will always have unit digit one?

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

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22 Jun 2017, 08:44

goalMBA1990 wrote:

Hi, 51^ 25 will have unit digit 1 and 1 divided by 13 will have remainder 1. So answer should be D. Now any power of 1 will always have unit digit one.

Please correct me where I am making mistake?

Is it not correct that any power of 1 will always have unit digit one?

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

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22 Jun 2017, 10:49

gmathopeful19 wrote:

goalMBA1990 wrote:

Hi, 51^ 25 will have unit digit 1 and 1 divided by 13 will have remainder 1. So answer should be D. Now any power of 1 will always have unit digit one.

Please correct me where I am making mistake?

Is it not correct that any power of 1 will always have unit digit one?

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

Show Tags

22 Jun 2017, 11:15

goalMBA1990 wrote:

Hi, 51^ 25 will have unit digit 1 and 1 divided by 13 will have remainder 1. So answer should be D. Now any power of 1 will always have unit digit one.

Please correct me where I am making mistake?

Is it not correct that any power of 1 will always have unit digit one?

Thanks.

Hi Bunuel, Can you please explain where am I making mistake?

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