IMO 2021 Day 2 solutions and discussion

Alternative problem 5 solution:
In the case N=3 , k clearly exists,
Induction step by condraddiction: If N is odd we can assume k doesn't exist -> if we substitute 1 and his surrounding numbers by the larger one of those numbers we should have still a cycle of size N-2 where k doesn't exist.
This proves than for all N k exist and we don't even need to swap anything.

how did you do it?

Ta giả định trường hợp lí tưởng nhất rằng với mọi phương trình ẩn m thì vô nghiệm với bậc chẵn và luôn có cùng 1 nghiệm m=mo với bậc lẻ, hơn nữa B1 là con B2, B2 là con B3,...Ta có B1 chứa 1 phần tử là m, B2 chứa 2 phần tử là m; m^2-m (do m^2-m khác m theo giả định),B3 chứa 2 phần tử là m;m^2-m (vì m^3-m^2=m theo giả định),...Cứ như vậy thì rõ ràng A có ít nhất m/2 phần tử.

God help me when i get to these levels of dimension breaking mathematical problems i am fricking crying right now well rip to my brain i just busted a gasket

I think there is some mistakes in ptoblem 5

Have you solved on your own.

nice challenges :-)

Thank you for this beautiful work

For problem 6, Since for m>=2, m+m^2+....+m^{m-1}<m^m, B_m cannot be a subset of the union of B_1 to B_{m-1}, therefore there is an element in B_m not in any of B_1 to B_{m-1}, repeat this argument for B_{m-1} and you get an element in B_{m-1} not in any of previous subsets.... and so you can conclude that n>=m-1.

Thanks for ur videos
I've uploaded all imo problems of 2021 and 2020 and I will upload the past years soon
I've also uploaded national olympiads of USA,UK,Switzerland,Belgium,Luxembourg,Netherlanda,Indonesia,....

Excellent!

Appreciate your time. For the last problem, it seems, the product of c-vector and 0/1-matrix, gives much fewer # of combinations than (m(m-1)+1)^n, as about half of the c's are multiplied by zeros in the matrix. Thus, n has to be much greater than m/2, in order to match the RHS of {1..m^m}. In fact, to minimize n, it may be argued that, m^m has to be a member of A, as it is not obtainable by adding the rest of {m^k}; the same applies to m^(m-1), and then m^(m-2)… Therefore, it seems n>=m, rather than >=m/2. After all, the sum of {1~10} gives only 55 values, not 2^10-1..

Can you suggest some combinatorics resources from basics to a good level

In general, how long do you spend to solve a hard problem ?
Which one is you preference when practicing ?
1) Try to solve problems by yourself
2) Solve and read solution as many as possible

This is merely beautiful, applauses for your work

What do you think of the day 2 problems? :) Solutions are relatively short!