For any natural number, let's call the numbers formed from its digits and have the same "digit" arrangement with the initial number as the "partial numbers". For example, the partial numbers of $149$ are ${1, 4, 9, 14,19, 49, 149},$ and the partial numbers of $313$ are ${3, 1, 31,33, 13, 313}.$ Find all natural numbers whose partial numbers are all prime. Justify your opinion.

Let $I$ be the incenter in the acute triangle $ABC.$ Rays $BI$ and $CI$ intersect the circumcircle of triangle $ABC$ at points $S$ and $T,$ respectively. The segment $ST$ intersects the sides $AB$ and $AC$ at points $K$ and $L,$ respectively. Prove that $AKIL$ is a rhombus.

Find all the real roots of the system of equations: $$ \begin{cases} x^3+y^3=19 \\ x^2+y^2+5x+5y+xy=12 \end{cases} $$

Solve the following diophantine equation in the set of nonnegative integers: $11^{a}5^{b}-3^{c}2^{d}=1$.

Baklavas with nuts are laid out on the table in a row at the Nowruz celebration. Kosa and Kechel saw this and decided to play a game. Kosa eats one baklava from either the beginning or the end of the row in each move. Kechel either doesn't touch anything in each move or chooses the baklava he wants and just eats the nut on it. They agree that the first Kosa will start the game and make $20$ moves in each step, and the Kechel will only make $1$ move in each step. If the last baklava eaten by the Kosa is a nut, he wins the game. It is given that the number of baklavas is a multiple of $20.$ $A)$ If the number of baklavas is $400,$ prove that Kosa will win the game regardless of which strategy Kechel chooses. $B)$ Is it always true that no matter how many baklavas there are and what strategy Kechel chooses, Kosa will always win the game?