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CGMO-2012 (China Girls Math Olympiad 2012) Problem 8

Find the number of integers k in the set \{0, 1, 2,\cdots, 2012\}   such that  \binom{2012}{k} is a multiple of 2012


CGMO-2012 (China Girls Math Olympiad 2012) Problem 4

There is a stone at each vertex of a given regular 13-gon, and the color of each stone is black or white. Prove that we may exchange the position of two stones such that the coloring of these stones are symmetric with respect to some symmetric axis of the 13-gon

Functional equations challenge 1

This is a challenging problem !!

Find all continuous  functions  f: R \longrightarrow R    such that for all x\inR:

f(1 - x) = 1 - f(f(x))

Solution : ( first try it , if not then you can read )

1) replacing the equation by a simpler one :   g(g(x))=-g(-x)
Let   g(x)=\frac 12-f(\frac 12-x)  and so  f(x)=\frac 12-g(\frac 12-x) and the equation becomes -g(x-\frac 12)=g(g(\frac 12-x)) and so :

g(x) is a continuous function such that g(g(x))=-g(-x)

2) g(x)=x \forall x\in g(\mathbb R)

Let A=g(\mathbb R)
From 1) above, we get that a\in A\implies -a\in A and so, since g(x) is continuous :
Either A=\mathbb R, either A=[-u,u] for some u\ge 0.

g(g(x))=-g(-x) \implies g(g(g(g(x))))=-g(-g(g(x)))=-g(g(-x))=g(x) and so g(g(g(x)))=x \forall x\in A

Let then g_r(x) : A\to A the restriction of g(x) to A :
g(g(g(x)))=x \forall x\in A \implies g_r(g_r(g_r(x)))=x \forall x\in A and so g_r(x) is a bijection, so is monotonous
g(g(x))=-g(-x) \implies g_r(g_r(x))=-g_r(-x) implies then -g_r(-x) is increasing and so g_r(x) is increasing.

So, g_r(a)> a \implies g_r(g_r(a))>g_r(a)>a and so g_r(g_r(g_r(a)))>g_r(a)>a which is impossible and so g_r(a)\le a
Same, g_r(a)< a \implies g_r(g_r(a))<g_r(a)<a and so g_r(g_r(g_r(a)))<g_r(a)<a which is impossible and so g_r(a)\ge a

So g_r(x)=x

3) general solution of g(g(x))=-g(-x) with g(x) continuous

Then the original equation g(g(x))=-g(-x) becomes g(x)=-g(-x) (since g(g(x))=g_r(g(x))=g(x)) and so :

The general continuous solutions of g(g(x))=-g(-x) are any odd continuous function g(x) such that g(x)=x \forall x\in g(\mathbb R)
A simple way to build them is :
3.1) If g(\mathbb R)=\mathbb R : g(x)=x \forall x

3.2) If g(\mathbb R)=[-u,u] for some u\ge 0 :
Let any continuous fonction h(x) from [u,+\infty)\to[-u,+u] such that h(u)=u and define g(x) as :
\forall x\in(-\infty,-u) : g(x)=-h(-x)
\forall x\in[-u,+u] : g(x)=x
\forall x\in(u,+\infty) : g(x)=h(x)

Notice that u=0 gives the solution g(x)=0

4) general solution of f(1-x)=1-f(f(x)) with f(x) continuous
The general solution of original equation is f(x)=\frac 12-g(\frac 12-x)) where g(x) is any function described in 3) above

5) examples of solutions
g(x)=0 gives the solution f(x)=\frac 12

g(x)=x gives the solution f(x)=x

u=1 and h(x)=\sin(\frac{\pi}2x) gives the solution :
If |x-\frac 12|<1 : f(x)=x
If |x-\frac 12|\ge 1 : f(x)=\frac 12+\sin(\frac{\pi}2x-\frac{\pi}4)

u=2 and h(x)=4|2\{\frac{x-2}8\}-1|-2 gives the solution : f(x)=\frac 52-4|2\{\frac{-2x-3}{16}\}-1| :)

Notice that the only differentiable solutions are f(x)=\frac 12 and f(x)=x