MikeMirzayanov's blog

By MikeMirzayanov, 2 weeks ago, ,

Many thanks to problem authors — Tech Scouts instructors. Please, review the author's solutions. They are beautiful and short. Our community has many to learn from mathematicians!

•
• +49
•

 » 2 weeks ago, # |   +68 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).
 » 2 weeks ago, # |   0
 » 2 weeks ago, # |   0 1164R — Divisible by 83 , Anser is 0, 0 mod 83 = ???? :)))
•  » » 2 weeks ago, # ^ |   0 n is natural, 0 is x_0, 0 index is not natural
•  » » » 2 weeks ago, # ^ |   0 ok . thanks :3 <3
 » 2 weeks ago, # | ← Rev. 2 →   +29 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.
 » 2 weeks ago, # |   +1 Why areas are 3 in B? I used affine transformations in B, to solve
•  » » 2 weeks ago, # ^ |   +3 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.
 » 2 weeks ago, # |   +8 Well, I for got the +1 in Problem K :'(
 » 2 weeks ago, # |   +18 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
•  » » 2 weeks ago, # ^ |   +20 You have already got all answers in this tutorial... Any reasons to submit?
•  » » » 2 weeks ago, # ^ |   -10 Just wanted to test ourselves, how good we are in maths, before seeing editorial.
•  » » » » 2 weeks ago, # ^ |   +1 Then write all answers on a paper, then compare it with the editorial.
•  » » » » » 2 weeks ago, # ^ |   -20 Yeah, but that doesn't still solved his problem, he still wanted to submit. Isn't it?
 » 2 weeks ago, # |   +2 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.