$10$ -by- $10$ grid game?

Suppose that the grid is filled so that touching cells contain coprime integers.

Since all even numbers share the common factor $2$ , it is easy to see that there can be no more than five even numbers in each adjacent pair of rows. The left-hand picture shows a possible choice of five cells that could be even. Similarly, since the numbers $3$ and $9$ share the common factor $3$ , there can be no more than five $3$ s and $9$ s together in each adjacent pair of rows. Thus there can be no more than $25$ even numbers in the whole grid. Similarly the number of $3$ s and $9$ s together in the whole grid cannot exceed $25$ . Thus at least $50$ of the numbers in the grid must be one of $1$ , $5$ or $7$ . Since $3 \times 16 = 48$ , we deduce that at least one of $1$ , $5$ or $7$ must occur at least $17$ times. Thus we deduce that $a \ge 16$ . It remains to show that $a=16$ .

The middle picture shows an arrangement of $25$ even numbers (ten $2$ s, ten $4$ s and five $8$ s), thirteen $3$ s and twelve $9$ s which is acceptable. The right-hand picture adds seventeen $1$ s, seventeen $5$ s and sixteen $7$ s to give a winning grid. Thus the players could win with $a=17$ , and hence $a=16$ .

Divide the grid into 25 of 2x2 quadrants.

Inside each of these quadrant, there exist no more than 1 number divisible by 2, or by 3. Thus, there can be at most 25 of: "2, 4, 6, 8, 10", as each quadrant can only contain 1 of those numbers Similarly there can exist at most 25 of "3, 9"

Thus, the remaining 50 is divided between "1, 5, 7". This means we need at least 17 repeating numbers (50/3 rounded up).

Finally, we need to prove that by allowing 17 repeats, it is possible to fill the grid, which can be done easily.