+11 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! 
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...
+23246 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.
+11 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.
Hello,
I found this may help you:
https://www.quora.com/What-is-the-remainder-when-2019-2019-is-divided-by-2020 from Quora.
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.
+118609 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
so
\(\frac{n^n}{n+1}\quad\text{has a remainder of +1 when n is an even positive integer}\\ and\\ \frac{n^n}{n+1}\quad\text{has a remainder of n when n is an odd positive integer}\\ \text{which means}\\ \frac{2019^{2019}}{2020}\quad\text{has a remainder of 2019 }\\\)
