## Task 3 - 100723

Determine all possibilities to specify a natural number $t$ and a digit $\star$ such that the following equation holds: $9(230+t)^{2}=492 \star 04$.
Assuming the equation has a solution, then both sides must be divisible by 9. Since $4+9+2+0+4=19$, it follows that $\star=8$. Therefore, the number on the right side of the given equation can only be 492804.

From the equation, we further get $(230+t)^{2}=492804 \div 9=54756$, and thus we obtain: $230+t$ is a natural number, not less than 230, and such that its square is 54756.

This implies that for $t$, only the value 4 is possible. Because 54756 ends in 6, $t$ can only end in 4 or 6.

If $t>4$, then $(230+t)^{2}>234^{2}=54756$. Therefore, only $t=4$ is possible.

As the verification shows, $t=4 ; \star=8$ is the solution to the given equation, and it is the only one.