MikeMirzayanov's blog

By MikeMirzayanov, 3 weeks ago, In English,

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!

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3 weeks ago, # |
  Vote: I like it +68 Vote: I do not like it

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).

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3 weeks ago, # |
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3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

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

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3 weeks ago, # |
Rev. 2   Vote: I like it +29 Vote: I do not like it

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.

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3 weeks ago, # |
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Why areas are 3 in B? I used affine transformations in B, to solve

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    3 weeks ago, # ^ |
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    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.

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3 weeks ago, # |
  Vote: I like it +8 Vote: I do not like it

Well, I for got the +1 in Problem K :'(

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3 weeks ago, # |
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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

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    3 weeks ago, # ^ |
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    You have already got all answers in this tutorial... Any reasons to submit?

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      3 weeks ago, # ^ |
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      Just wanted to test ourselves, how good we are in maths, before seeing editorial.

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        3 weeks ago, # ^ |
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        Then write all answers on a paper, then compare it with the editorial.

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2 weeks ago, # |
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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.