In each of the unit squares of a $10\times 10$ checkerboard, a positive integer not exceeding $10$ is written. Any two numbers that appear in adjacent or diagonally adjacent squares of the board are relatively prime. Prove that some number appears at least $17$ times.
Solution from book: In any $2\times 2$ square, only one of the four numbers can be divisible by $2$, and only one can be divisible by $3$. Tiling the board by $2\times 2$ squares, we deduce that at most $25$ numbers are divisible by $2$ and at most $25$ numbers are divisible by $3$. There are at least $50$ remaining numbers that are not divisible by $2$ or $3$, and thus must equal one of the numbers $1,5,$ or $7$. By the pigeonhole principle, one of these numbers appears at least $17$ times.
Source: St. Petersburgh Math Olympiad, 2001.
1) The first sentence of solutions seems to me very strange. What if we consider square with elements $1,3,5,7$ then we see that there is no element which is divisible by $2$. Can anyone clarify this sentence, please?
2) How to prove that in any $2\times 2$ square there is element divisible by $2$ and $3$. It seems correct to me but I cannot provide rigorous proof.
P.S. It would be interesting to look at another approach.
Question 1: Read "only one" to be "at most one" in the solution.
Question 2: There is no need to have any element divisible by 2 or 3 in any $2\times 2$ square, merely at most one.