Baltic Way 1997 · Problem 6

Solution: $(10, 10, 10)$ is the only such triple.

The equality immediately implies $a^{3} + 9 b^{2} + 9 c = 1990 \equiv 1 \pmod{9}$. Hence $a^{3} \equiv 1 \pmod{9}$ and $a \equiv 1 \pmod{3}$. Since $13^{3} = 2197 > 1990$ then the possible values for $a$ are $1, 4, 7, 10$.

On the other hand, if $a \leqslant 7$ then by $a \geqslant b \geqslant c$ we have

$$a^{3} + 9 b^{2} + 9 c^{2} \leqslant 7^{3} + 9 \cdot 7^{2} + 9 \cdot 7 = 847 < 1990$$

a contradiction. Hence $a = 10$ and $9 b^{2} + 9 c = 990$, whence by $c \leqslant b \leqslant 10$ we have $c = b = 10$.