jmc

36, 64, and 81 are all perfect squares. That means they can all be written as $1\cdot a^2+0\cdot a^1+0\cdot a^0=100_a$, where $a$ is the square root of each number. So, all three numbers can be represented by the digits $\boxed{100}$ when converted to other bases. To see that no other digits work, note that only the bases 4, 5, and 6 use three digits to represent the number 36. (This follows from $b^2\le 36<b^3$, which expresses the condition that 36 has 3 digits in base $b$). The representations are $100_6$, $121_5$ and $210_4$, only one of which fits the form $△\square \square$.