IMO Hk TST

No. We consider the numbers modulo $3$.
* If $3 \mid a, b$, then $3 \mid a + b$ and $3 \mid ab$. The number of multiples of $3$ remains unchanged.
* If $3 \mid a$ and $3 \nmid b$, then $3 \nmid a + b$ and $3 \mid ab$. The number of multiples of $3$ remains unchanged.
* If $a \equiv b \equiv \pm 1 \pmod{3}$, then $3 \nmid a + b$ and $3 \nmid ab$. The number of multiples of $3$ remains unchanged.
* If $a \equiv 1 \pmod{3}$ and $b \equiv 2 \pmod{3}$, then $3 \mid a + b$ and $ab \equiv 2 \pmod{3}$. The number of multiples of $3$ is increased by $1$, and there is a number which is congruent to $2$ modulo $3$.
Now, note that $3 \mid 21, 27, 180, 540$ and $64 \equiv 1 \pmod{3}$. If the $5$ numbers can appear on the board at the same time, there must be an instant when the number of multiples of $3$ changes from $3$ to $4$. This must be the last case in the above list, hence the $5$ numbers are now congruent to $0, 0, 0, 0, 2$ modulo $3$ respectively. Afterwards, no matter which $2$ numbers we choose, the new numbers are still congruent to $0, 0, 0, 0, 2$ modulo $3$ respectively. Therefore, it is impossible to generate $21$, $27$, $64$, $180$, $540$. This is a contradiction.