Croatian Mathematical Society Competitions

a) We claim that Miljen can play so that he leaves some odd number $n$ and $2019-n$ ones on the board after each of his moves, i.e. before Rudi's move. In that case, after $1009$ of Miljen's moves only the number $2019$ is written on the board, so Rudi cannot make a move and Miljen wins.

In the beginning, the board can be described as above, with $n=1$. Rudi can erase two ones and write number $2$ on the board. In that case, Miljen can erase $n$ and $2$, which he can do since $n$ is odd, and write $n+2$ on the board, which leaves the board as described above.

In each subsequent turn, Rudi and Miljen can play as above and our claim holds. Alternatively, Rudi can erase $n$ and one of the ones, and write $n+1$. Note that $2019-n$ is even, so Rudi's move leaves an odd number of ones on the board. Miljen can now select $n+1$ and a one, and write $n+2$, which again only leaves some odd number and ones on the board, as before.

b) In this case, Miljen uses the same strategy as in a), except when making the final move. Namely, after each of Miljen's moves an odd number $n$ and $2020-n$ ones are written on the board. Note that now $2020-n$ is odd. It remains to describe Miljen's last, i.e. his $1009$th move.

If after Rudi's $1009$th move the numbers on the board are $2018$, $1$ and $1$, Miljen can erase the two ones, and replace them with $2$. Rudi cannot make a move after that, since $2018$ and $2$ are not relatively prime, and Miljen wins.

If after Rudi's $1009$th move the numbers on the board are $2017$, $2$ and $1$, Miljen can erase $2017$ and $1$, and replace them with $2018$, which again leaves Rudi with $2018$ and $2$, and Miljen wins.