Romanian Mathematical Olympiad - District Round

If exactly 16 robots were sold each month, the number of the robots sold would have been $12 \cdot 16 = 192$. The difference of 43 robots comes from the months where 20 robots were sold (4 more each month) and from the months where 25 robots were sold (9 more each month). Denote by $a$ and $b$ the number of months where 20 robots were sold and where 25 robots were sold, respectively. We have $4a + 9b = 43$,

therefore $b$ is an odd number which does not exceed 4. We obtain that $b = 3$, therefore exactly 20 robots were sold in $a = 4$ months.

Alternative solution.
Let $x, y, z$ be the number of months in which 16, 20, and 25 robots were sold, respectively. We have $16x + 20y + 25z = 235$, thus $x = 5(47 - 3x - 4y - 5z)$, therefore $5 \mid x$, consequently $x \in \{5, 10\}$. If $x = 10$, we obtain $4y + 5z = 15$ and $y + z = 12 - 10 = 2$, which leads to $z = 7 > y + z = 2$, false. If $x = 5$, we have $4y + 5z = 31$ and $y + z = 12 - 5 = 7$ and we obtain $z = 3$, therefore exactly 20 robots were sold in $y = 7 - z = 4$ months.