So I was doing some math, and I came across this question.


What is the remainder when \(2019^{2019}\) is divided by \(2020\)?


So, I thought, well, I definitely can't do that in my head! cheeky I did find a way to get the answer, but I was wondering if there was a simpler way to obtain the answer?


This is how I did it. By the way, I put the expression equal to its remainder.


\(1^1/2=1 \)

\( 2^2/3=2 \)

\( 3^3/4=3 \)

\( 4^4/5=1\)

\( 5^5/6=5\)


Then I saw a pattern (kind of). The answer is in fact \(2019\), so I was correct in my reasoning, but I was wondering if anyone had a different approach to the question. 


Oh, and you are not allowed to use a calculator.


Edit: Realized that 256 divided by 5 has a remainder of 1...angryblush

 Jan 23, 2020
edited by IMadeYouReadThis  Jan 23, 2020

How did you get the remainder of 44 when divided by 5 to be 4?

44 = 256

Divide 256 by 5 and you get a quotient of 51 and a remainder of 1.

 Jan 23, 2020

Oh! Thanks for catching that error!

IMadeYouReadThis  Jan 23, 2020

Well, my pattern seems to work, but it kind of broke at 5. Will it break at all multiples of 5? Or is there something else?


Based on what I am seeing, it seems as though all times a multiple of 5 appears, then it has a remainder of 1.

 Jan 23, 2020


I found this may help you: 

https://www.quora.com/What-is-the-remainder-when-2019-2019-is-divided-by-2020 from Quora. 

 Jan 23, 2020

By the way, your pattern works! If you looked at the link, there is a solution similar to yours.

Although If it asked for proof (In competitions etc..) using mods is a must and not just patterns. 

 Jan 23, 2020

Consider the sequence you discovered.

You found that  


\(1^1\;\mod2 = 1\;or \;-1\\ 2^2\;\mod3=+1\\ 3^3\;\mod4=-1\\ 4^4\;\mod5=+1\)


NOTE: If a number (mod n) = -1 Then the number divided by n will have a raimainder of  n-1

            If you want me to discuss this more I can.


So it looks like 

 \(n^n\mod(n+1)=+1 \text{ when n is even}\\ n^n\mod(n+1)=-1 \text{ when n is odd}\\\)

So can I prove that his is always the case??

\(n\mod(n+1)=-1\\ so\\ n^n\mod(n+1)\equiv (-1)^{n+1} \)

-1 raised to an even power will be +1     

-1 raised to an odd power will be -1



\(\frac{n^n}{n+1}\quad\text{has a remainder of +1 when n is an even positive integer}\\ and\\ \frac{n^n}{n+1}\quad\text{has a remainder of n when n is an odd positive integer}\\ \text{which means}\\ \frac{2019^{2019}}{2020}\quad\text{has a remainder of 2019 }\\\)

 Jan 23, 2020

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