Estonian Mathematical Olympiad

Denote the value of the given expression by $k$. We know that $2024=2^3\cdot 11\cdot 23$ and $CC=C\cdot 11$. In a perfect square, all prime exponents are even. Thus one of the numbers $AB$, $BA$, $CC$ must be divisible by $23$. This cannot be $CC$, so let it be $AB$ without loss of generality (the other option is symmetrical). The two-digit multiples of $23$ are $23$, $46$, $69$ and $92$.

If $AB=23$, then $BA=32=2^5$, so $k=2^8\cdot 11^2\cdot 23^2\cdot C$. Then $k$ is a perfect square iff $C$ is a perfect square, meaning $C=1$, $C=4$ or $C=9$. The sum $A+B+C$ is $6$, $9$ or $14$ respectively.

If $AB=46$, then $BA=64=2^6$, so $k=2^{10}\cdot 11^2\cdot 23^2\cdot C$. Then $k$ is a perfect square iff $C$ is a perfect square, meaning $C=1$ or $C=9$ (the digit $4$ is already in use). The sum $A+B+C$ is $11$ or $19$ respectively.

If $AB=69$, then $BA=96=2^5\cdot 3$, so $k=2^8\cdot 3^2\cdot 11^2\cdot 23^2\cdot C$. Then $k$ is a perfect square iff $C$ is a perfect square, meaning $C=1$ or $C=4$ (the digit $9$ is already in use). The sum $A+B+C$ is $16$ or $19$ respectively.

If $AB=92$, then $BA=29$, so $k=2^5\cdot 11^2\cdot 23^2\cdot 29\cdot C$. This cannot be a perfect square, as $C$ cannot be divisible by $29$.

The sum $A+B+C$ can thus obtain $6$ different values ($19$ was present in two different cases).