When I participated at IMO 1992 in Moscow, people were laughing that our year was so unlucky to have Moscow - but I am actually happy it was Moscow.

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:

If you can solve it, boast in the comments.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 \(f(m)\).

**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.

## snail feedback (6) :

You see, a nice task without real numbers. You want to learn them real numbers and the majority of people can't solve tasks with integers. What is leq?

I probably can't prove it, I am weak in algebra. :P

Disappointing to hear that your web page didn't get converted to LaTeX. Did you turn off JavaScript? Or what browser/OS do you use?

\leq appears as \(\leq\) and it means "

less than orequal to".Let f(m)=2^(2m) when m is a positive integer and f(m)=2^(-2m-1) when m is a negative integer.

Amsterdam dam in 2002 ?

really ? don't remember the LMoth face in 2002...nope

I use NoScript and my javascript was partialy disabled. Thanks.

This is a humble try and I could be wrong.

Apologies that I don't know how to use the Mathjax.

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The problem can be re-stated

Given k.f(m) = f(n)

prove than k is an element of Z+

----

f(m) - f(n) = j.f(m - n), where j is an element of Z.

j can be expressed as a function,

f(m) - f(n) = j(m, n).f(m - n), where j(m, n) is an element of Z.

For convenience, this can be re-stated,

f(n) - f(m) = [-j(m, n)].f(m - n)

f(n) - f(m) = p(m, n).f(m - n)

f(n) >= f(m) implies p(m, n) is an element of N.

----

k.f(m) - f(m) = p(m, n).f(m - n)

k = 1 + p(m, n).f(m - n) / f(m)

Therefore, k is an element of Z+ if f(m - n) | f(m)

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Let n = 0

f(0) - f(m) = p(m, 0).f(m)

This can be generalized for any n or m,

f(0) = f(x) [1 + p(x, 0)]

Therefore,

f(0) is the maximum and is always a factor of f(n) and f(m) including f(m - n) for any n or m

f is a set of positive factors.

----

f(0) | f(m)

f(0) | f(m - n)

f(0) | [f(m - n) - f(m)]

f(0) = y.[f(m - n) - f(m)], where y is an element of Z

f(0) / f(m) = y.[f(m - n) / f(m) - 1]

f(0) | f(m) implies f(m - n) | f(m)

k is an element of Z+

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