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To me, the answer for K seems ambiguous as there can be considered to be 2018 sequences covering the whole circle. Hence, 2018*2019/2 is the answer that I was getting (even though I did not participate).

1164R — Divisible by 83 , Anser is 0, 0 mod 83 = ???? :)))

n is natural, 0 is x_0, 0 index is not natural

ok . thanks :3 <3

I think the answer to problem K should be $$$\frac{2018*2017}{2}+2018=2037171$$$

The whole circle may be counted 2018 times (because a sequence is an enumerated collection of numbers and there are 2018 possible starting points)

It can be seen that these 2018 sequences are pairwise distinct.

Why areas are 3 in B? I used affine transformations in B, to solve

The area of triangle ABP will be equal to 3/(1+3) of the area of the triangle ABC as P divides AC as 3:1. The area of triangle AMP will be equal to 1/(1+3) of the area of the triangle ABP as M divides AB as 1:3. Therfore, area of triangle AMP will be equal to 3/4 * 1/4 = 3/16 of the area of ABC.

Well, I for got the

+1in Problem K :'(Is there a way to submit my answers now ? I couldn't take part of the contest and I would like to submit it ... Thanks :3

You have already got all answers in this tutorial... Any reasons to submit?

Just wanted to test ourselves, how good we are in maths, before seeing editorial.

Then write all answers on a paper, then compare it with the editorial.

Yeah, but that doesn't still solved his problem, he still wanted to submit. Isn't it?

In some of these problems, you could take one particular instance and work out the answer for that, since you know there is only 1 answer to the problem (because of how the system works).

For example, for the 2018 integers written on a circle with sum 1. You can assume 2017 of them are positive and one of them is the negative sum of all the others +1. It's now easy to see that all sequences that do not contain the negative number are positive.