RRemainderrr

Number Theory Level 2

Find remainder when 1^2013+2^2013+3^2013............2012^2013 is divided by 2013.

The answer is 0.

4 solutions

Mayank Chaturvedi
May 17, 2014

a^n+b^n is divisible by a+b IF n IS ODD. This way, we can pair out the numbers: (1^2013+2012^2013)+(2^2013+2011^2013)+.........................(100^2013+1913^2013)..and so on. We can figure out by pairing that the REMAINDER IS 0.

Liked the way you approached, Mayank.

Just reshared it.

Ninad Akolekar - 6 years, 6 months ago

i to approached in the same way.

tushar agrawal - 7 years ago

I did a little different way but yes, must say keep it up, Mayank!

Kartik Sharma - 7 years ago

Hi! In What way did you approach this @Kartik Sharma

Krishna Ar - 7 years ago

As I said it's only a little different way and it states just why a^n + b^n/a+b gives remainder 0.

1^2013 = 1 mod 2013

2012^2013 = (-1)^2013 mod 2013

2^2013 = 2^2013 mod 2013

2011^2013 = (-2)^2013 mod 2013

and so on

Then, add everything and we will see that everything gets cut.

1^2013 + 2^2013....... 2013^2013 = 1 + 2^2013.... -2^2013 -1^2013 mod 2013

Therefore, we get 1^2013 + 2^2013....... 2013^2013 = 0 mod 2013

So, it is also the reason why this property goes true.

Kartik Sharma - 7 years ago

@Kartik Sharma – Ah! The long method. But, yes that's because I forgot that property that time. But I don't know I reached to its proof just by chance.

Kartik Sharma - 7 years ago

Thanks Kartik for my praise. But, really i wanna know the method by which you solved the question.

Mayank Chaturvedi - 7 years ago

Vishwesh Agrawal
Apr 13, 2015

Udsing fermat theorem each time we get 1 adding 2013 tms we get 0

Manas Verma
Jun 24, 2014

USE FERMETS THEORUM

Saurabh Mallik
Jun 14, 2014

We can solve this by $mod$ function.

$(1^{2013}+2^{2013}+3^{2013}+......+2012^{2013})$ $mod$ $2013=0$

Thus, the answer is $\boxed{0}$

Nice Solution -_-.

milind prabhu - 6 years, 7 months ago

$mod$ is \pmod{.....}

Adarsh Kumar - 6 years, 6 months ago

RRemainderrr

The answer is 0.

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