Adam has a secret natural number $x$ which Eve is trying to discover. At each stage Eve may only ask questions of the form "is $x+n$ a prime number?" for some natural number $n$ of her choice. Prove that Eve may discover $x$ using finitely many questions.

For a natural number $ n, $ a string $ s $ of $ n $ binary digits and a natural number $ k\le n, $ define an $ n,s,k$ [i]-block[/i] as a string of $ k $ consecutive elements from $ s. $ We say that two $ n,s,k\text{-blocks} , $ namely, $ a_1a_2\ldots a_k,b_1b_2\ldots b_k, $ are [i]incompatible[/i] if there exists an $ i\in\{1,2,\ldots ,k\} $ such that $ a_i\neq b_i. $ Also, for two natural numbers $ r\le n, l, $ we say that $ s $ is $ r,l $ [i]-typed[/i] if there are, at most, $ l $ pairwise incompatible $ n,s,r\text{-blocks} . $ Let be a $ 3,7\text{-typed} $ string $ t $ consisting of $ 10000 $ binary digits. Determine the maximum number $ M $ that satisfies the condition that $ t $ is $ 10,M\text{-typed} . $ [i]Cătălin Gherghe[/i]

Gandalf (the wizard) and Bilbo (the assistant) are presenting a magic trick to Nitzan (the audience). While Gandalf leaves the room, Nitzan chooses a number $1\leq x\leq 2^{2022}$ and shows it to Bilbo. Now bilbo writes on the board a long row of $N$ digits, each of which is $0$ or $1$. After this Nitzan can, if he wishes, switch the order of two consecutive digits in the row, but only once. Then Gandalf returns to the room, looks at the row, and guesses the number $x$. Can Bilbo and Gandalf come up with a strategy that allows Gandalf to guess $x$ correctly no matter how Nitzan acts, if [b]a)[/b] $N=2500$? [b]b)[/b] $N=2030$? [b]c)[/b] $N=2040$?

A "Hishgad" lottery ticket contains the numbers $1$ to $mn$, arranged in some order in a table with $n$ rows and $m$ columns. It is known that the numbers in each row increase from left to right and the numbers in each column increase from top to bottom. An example for $n=3$ and $m=4$: [asy] size(3cm); Label[][] numbers = {{"$1$", "$2$", "$3$", "$9$"}, {"$4$", "$6$", "$7$", "$10$"}, {"$5$", "$8$", "$11$", "$12$"}}; for (int i=0; i<5;++i) { draw((i,0)--(i,3)); } for (int i=0; i<4;++i) { draw((0,i)--(4,i)); } for (int i=0; i<4;++i){ for (int j=0; j<3;++j){ label(numbers[2-j][i], (i+0.5, j+0.5)); }} [/asy] When the ticket is bought the numbers are hidden, and one must "scratch" the ticket to reveal them. How many cells does it always suffice to reveal in order to determine the whole table with certainty?