Find the real numbers $a$, $b$, $c$ and $d$ that satisfy the following equations: $$\left \{\begin{matrix} a\cdot b+c+d & = & 6, \\ b\cdot c+d+a & = & 2, \\ c\cdot d+a+b & = & 5, \\ d\cdot a+b+c & = & 3. \end{matrix}\right .$$

Consider a square $8 \times 8$ board with its $64$ cells initially white. Determine the minimum number of colors needed to color the cells (each one with only one color) in such a way that if four cells on the board can be covered by an $L$-shaped tile as shown in the figure, then the four cells are of different colors. [asy][asy] size(1.5cm); draw((0,1)--(1,1)--(1,2)--(0,2)--(0,1)--(0,0)--(1,0)--(2,0)--(2,1)--(1,1)--(1,0)--(1,1)--(1,2)--(1,3)--(0,3)--(0,2)); [/asy][/asy] Note: The $L$-shaped tile can be rotated or flipped.

Let $n$ be a positive integer. Determine the maximum number of positive integers less than or equal to $n^2$ that can be colored red in such a way that if $a$ and $b$ are red, with $a \neq b$, then $a \cdot b$ is not red.

On a table, there are $10\,000$ matches, two of which are inside a box. Ana and Beto take turns playing the following game. On each turn, a player adds to the box a number of matches equal to a proper divisor of the current number of matches in the box. The game ends when, for the first time, there are more than $2024$ matches in the box and the person who played the last turn is the winner. If Ana starts the game, determine who has a winning strategy.

In triangle $ABC$, let $A'$, $B'$ and $C'$ be points on the sides $BC$, $CA$ and $AB$, respectively, such that$$\frac{BA'}{A'C}=\frac{CB'}{B'A}=\frac{AC'}{C'B}.$$The line parallel to $B'C'$ passing through $A'$ intersects line $AC$ at $P$ and line $AB$ at $Q$. Prove that$$\frac{PQ}{B'C'} \geqslant 2.$$

An equilateral triangle with integer side length $n$ is subdivided into smaller equilateral triangles of side length $1$ by drawing lines parallel to its sides, as shown in the figure for $n = 4$. [asy][asy] size(5cm); // Function to draw an equilateral triangle with subdivisions and mark vertices void drawTriangleWithDots(pair A, pair B, pair C, int n) { real step = 1.0 / n; // Draw horizontal lines for (int i = 0; i <= n; ++i) { pair start = A + i * step * (C - A); pair end = start + i * step * (B - C); draw(start -- end, gray(0.5)); } // Draw left-leaning diagonal lines for (int i = 0; i <= n; ++i) { pair start = A + i * step * (B - A); pair end = start + (n - i) * step * (C - A); draw(start -- end, gray(0.5)); } // Draw right-leaning diagonal lines for (int i = 0; i <= n; ++i) { pair start = B + i * step * (C - B); pair end = start + (n - i) * step * (A - B); draw(start -- end, gray(0.5)); } // Mark dots at all vertices for (int i = 0; i <= n; ++i) { for (int j = 0; j <= i; ++j) { pair vertex = A + i * step * (C - A) + j * step * (B - C); dot(vertex, black); } } // Draw the outer triangle draw(A -- B -- C -- cycle, black+linewidth(1)); } // Main triangle vertices pair A = (0, 0); pair B = (4, 0); pair C = (2, 3.464); // Height = sqrt(3)/2 * side length // Subdivisions int n = 4; // Draw the subdivided equilateral triangle with dots drawTriangleWithDots(A, B, C, n); [/asy][/asy] Consider the set $A$ consisting of all points that are vertices of any of these smaller triangles. A subtriangle is defined as any equilateral triangle whose three vertices belong to the set $A$ and whose three sides lie along the lines of the initial subdivision. We wish to color all points in $A$ either red or blue such that no subtriangle has all three vertices of the same color. Let $C(n)$ denote the number of such valid colorings for each positive integer $n$. Calculate, in terms of $n$, the value of $C(n)$.