Suppose we have a collection of infinite stones. On the Cartesian plane, a stone is kept on $(0,0)$. In a move, we can remove a stone from $(m,n)$ (provided that it contains a stone) and keep two stones on $(m+1,n)$ and $(m,n+1)$ (provided that these two points don't have any stones). Prove that at any point in the whole process, there will always be a stone placed on some lattice point $(a,b)$, with $a+b\le 3$. (The point $(a,b)$ is not fixed)
Source: some TST.
A Beautiful Combi
- What is the value of the contour integral around Western Europe?
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
- Phlembac Adib Hasan
- Posts:1016
- Joined:Tue Nov 22, 2011 7:49 pm
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Re: A Beautiful Combi
My solution relies on a weighting invariance.
- What is the value of the contour integral around Western Europe?
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.