Prove that for any n natural, the number \[ \sum \limits_{k=0}^{n} \binom{2n+1}{2k+1} 2^{3k} \] cannot be divided by $5$.
2011 Morocco TST
Day 1
For positive integers $m$ and $n$, find the smalles possible value of $|2011^m-45^n|$. (Swiss Mathematical Olympiad, Final round, problem 3)
For a given triangle $ ABC$, let $ X$ be a variable point on the line $ BC$ such that $ C$ lies between $ B$ and $ X$ and the incircles of the triangles $ ABX$ and $ ACX$ intersect at two distinct points $ P$ and $ Q.$ Prove that the line $ PQ$ passes through a point independent of $ X$.
Day 2
Find all pairs $(m,n)$ of nonnegative integers for which \[m^2 + 2 \cdot 3^n = m\left(2^{n+1} - 1\right).\] Proposed by Angelo Di Pasquale, Australia
Let $x_1, \ldots , x_{100}$ be nonnegative real numbers such that $x_i + x_{i+1} + x_{i+2} \leq 1$ for all $i = 1, \ldots , 100$ (we put $x_{101 } = x_1, x_{102} = x_2).$ Find the maximal possible value of the sum $S = \sum^{100}_{i=1} x_i x_{i+2}.$ Proposed by Sergei Berlov, Ilya Bogdanov, Russia
The vertices $X, Y , Z$ of an equilateral triangle $XYZ$ lie respectively on the sides $BC, CA, AB$ of an acute-angled triangle $ABC.$ Prove that the incenter of triangle $ABC$ lies inside triangle $XYZ.$ Proposed by Nikolay Beluhov, Bulgaria