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.

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.
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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

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 ...

via fermat theorem
a^(p-1)= 1 mod p where a is an integer and p prime
as 51 contains 13 it can be rewritten and simplified 51/13 remainder 12

12^(13-1) = 1 mod 13
12^25 = 12^(12*2+1)= 12

Cyclicity helps you figure out the units digit in every case. Also, it has nothing to do with remainders.

In some cases, the units digit helps you figure out the remainder, more specifically in case of division by 2, 5 or 10.


https://www.veritasprep.com/blog/2015/1 ... questions/
https://www.veritasprep.com/blog/2015/1 ... ns-part-2/
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Mam, why cyclicity method does not work here. I marked answer 1 thinking 1/13 remainder "1"
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You are not dividing 1 by 13 here. You are dividing a number that ends in 1 by 13 here. We don't know what that number exactly is.
Will every number ending in 1 when divided by 13 give 1 as remainder? Think about it:
11/13, remainder 11
21/13, remainder 8
31/13, remainder 5
and so on...
Cyclicity does not help get the remainder in most cases. It does so in very few cases. They are discussed in this post: https://www.veritasprep.com/blog/2015/1 ... questions/
https://www.veritasprep.com/blog/2015/1 ... ns-part-2/
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I was having hard time understanding why remainder is not simply -1 as -1/13 means that we could have simply done it as: 13(0)+(-1).

Thereafter I myself realised that remainder can never be less than zero and hence it has been multiplied by minimum possible integer to equate further i.e. 1.

Incase I am wrong, please correct. If not, this might be useful for somebody who might be having trouble understanding the same.

Thanks

Bunuel can you please explain the last part i.e "\(-1=13*(-1)+12\)". I am not able to understand why did u subtract?

Thanks