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

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?

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.

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

(-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: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/03 ... -the-gmat/

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.

Bunuel

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

51^25

here the last digit of 51 is 1, and the cyclicity of 1 is 1. So, the unit digit of 51^25 will be 1

thus, 1, when divided by 13 will produce a quotient "0" and will leave a remainder "1", and I am stumped here ...

please help me understand the problem ...

thanks in advance ...

You cannot do this way.

For example, 1 divided by 13 gives the remainder of 1 but 21 divided by 13 gives the remainder of 8. There are many different approaches and links to the underlying theory on previous 5 pages of the discussion.

We need to find When \(51^{25}\) is divided by 13, the remainder obtained is

Now, to solve this problem we will be using the concept of Binomial Theorem

We will try to split 51 into two numbers. One number will be a multiple of 13 and will be very close to 51 and other number will be a very small number.

\(51^{25}\) = \((52 - 1)^{25}\) (52 = 13*4)

Now, when we open this using Binomial Theorem then all terms apart from the last term will be a multiple of 52 => a multiple of 13 => Remainder by 13 will be 0

=> Question is reduced to what is the remainder when the last term ( 25C25 \((-1)^{25} * 52^0\) is divided by 13

=> Remainder when -1 is divided by 13
=> -1 + 13 = 12

So, Answer will be A
Hope it helps!

Watch the following video to Learn about Binomial Theorem and How to Solve Similar problems

Bunuel, could u clarify this? Thanks a lot.

The process for finding the remainder when dividing a negative integer by a positive integer follows the same principles as when dividing a positive integer by a positive integer.

For example:

• What is the remainder when dividing 21 by 6? We find the closest multiple of 6 that is less than 21, which is 18. Then, we calculate 21 - 18, yielding a remainder of 3.
• What is the remainder when dividing -23 by 7? Here, we find the closest multiple of 7 that is less than -23, which is -28. Then, we calculate -23 - (-28), resulting in a remainder of 5.

What about dividing -1 by 13? The closest multiple of 13 less than -1 is -13. So, the remainder is -1 - (-13) = 12.

Alternatively, consider this method:

• What is the remainder when dividing -23 by 7? Dividing 23 by 7 gives a remainder of 2. To find the remainder for -23 divided by 7, subtract this 2 from the divisor. Thus, the remainder when dividing -23 by 7 is 7 - 2 = 5.

Similarly, what is the remainder when dividing -1 by 13? Dividing 1 by 13 gives a remainder of 1. Therefore, the remainder when dividing -1 by 13 is 13 - 1 = 12.

Hope it helps.

\(51*51 = 2601\)

\(\frac{2601}{13} =\) Rem \(1\)

Now, \(51^{25} = 51^{12*2+1}\)

Thus, \(\frac{51^{25}}{13} = \frac{51^{12*2+1}}{13} = \frac{51^{24}}{13}*\frac{51}{13} \)

\(\frac{51^{24}}{13} =\) Rem \(1\), thus the question boils down to \(\frac{51}{13}\)


Calculate the rem of \(\frac{51}{13} = 13*3 + 12\), Rem is \(12\), Answer must be (A)