This year, kids compete in Amsterdam which I visited in 1997 and 2002, to make you start to think about functions with period 5 which you will actually find helpful momentarily. ;-)
Terence Tao will probably ask his readers to solve the most difficult problem, Q6. Phil Gibbs is modest so he chose Q1, maybe the easiest one.
As you know, I am even more modest, so let's start with a special problem for Miss USA participants of the IMO 2011:
A nice solution. Well, it was too easy. Here is a well-known problem for the climate alarmists:
You may prefer some problems involving smiling faces:
Via Tetrahedral where you may find additional jokes.
But back to IMO 2011.
Here are all the six problems and those from the previous years. Let me choose Q5 which looks easy:
Q5: Let \(f\) be a function from the set of integers to the set of positive integers. Suppose that, for any two integers \(m\) and \(n\), the difference \(f(m)-f(n)\) is divisible by \(f(m-n)\). Prove that, for all integers \(m\) and \(n\) with \(f(m)\leq f(n)\), the number \(f(n)\) is divisible by \(f(m)\).If you can solve it, boast in the comments.
Pilsner soccer rules
Czech top soccer league's champion, Victoria Pilsner, had defeated Pyunik Yerevan in the Champions League when it played in Armenia, 4-to-0. But what about the second match? Some people were afraid. I wasn't. A mile away from my home, on the "Pilsner construction site", our team smashed Yerevan 5-to-1, to make the aggregate score 9-to-1.