We need to determine a one-digit, a two-digit, and a three-digit natural number based on the following:

The two-digit number is even, the other two are odd.

If we multiply the numbers, the end of the product is 68, before which the sum of the three numbers stands, which is a square number.

If the hundreds place digit of the three-digit number is written in front of the two-digit number, then the new three-digit number and the remaining two-digit number are both square numbers.
We need to determine a one-digit, a two-digit, and a three-digit natural number based on the following:

(1) The two-digit number is even, the other two are odd.

(2) If we multiply the numbers, the product ends in 68, and before that, the sum of the three numbers stands, which is a perfect square.

(3) If the hundreds place digit of the three-digit number is moved to the front of the two-digit number, then the new three-digit number and the remaining two-digit number are both perfect squares.

Solution. Let the numbers in decimal notation be $A, B C, D E F$, so $A, B, D$ are none of them 0.

From (2), the last digit of the product $A \cdot C \cdot F$ is 8, which means it is not divisible by 5, and none of the factors are. Therefore, according to (1), the even digit $C$ is not 0, the odd digits $A$ and $F$ are not 5, and $A$ is one of 1, 3, 7, and 9.

$C$ and $F$ are further restricted by the fact that they are perfect square endings according to (3). Therefore, $C$ can only be 4 or 6, and $F$ can only be 1 or 9.

If $C=4$, then according to (2) and (1), the ending of the product $A \cdot F$ is 7, which is only satisfied by $F=1$ and $A=7$, or $F=9$ and $A=3$. However, the former set of values is excluded because, according to (2), the sum $s$ of the numbers must also be a perfect square, and thus cannot end in 2, which is derived from $A+C+F=7+4+1$. The set of values $C=4, F=9, A=3$ results in a sum ending in 6, which is still to be investigated.

Similarly, if $C=6$, the ending of $A \cdot F$ is 3, and with $F=1$ and $A=3$, $3+6+1$ can give a possible 0 at the end of the square, but with $F=9$ and $A=7$, the forbidden 2 would appear again, which is not allowed.

In both promising cases, $A=3$, and from the addition scheme of the three numbers, we know so far:

![](https://cdn.mathpix.com/cropped/2024_05_02_75f46131de191abcb2ddg-1.jpg?height=179&width=802&top_left_y=886&top_left_x=651)

In the b) scheme, the second last digit of $s$ is also 0, so in such a solution, only $B+E=9$ is possible, and in $s$ the hundreds place is $D+1$. Furthermore, $s100 \cdot D E F \\
A \cdot B C=3(10 B+4)>100, \quad B \geqq 3
\end{gathered}
$$

so $B \neq 2$. The value $B=8$ is also excluded, as it implies $D \geqq 4$, because in this case the value of the product given in two ways

$$
\begin{gathered}
3 \cdot 84 \cdot D E 9=25200 D+2520 E+2268 \\
100 s+68=10000 D+1000 E+9668
\end{gathered}
$$

differs by

$$
15200 D+1520 E-7400>7600(D-1)>0
$$

For the remaining $D B C=144$, i.e., $B=4, D=1$, from the equation

$$
A \cdot B C \cdot D E F=3 \cdot 44 \cdot 1 E 9=100(3+44+10 E+109)+68
$$

we get $E=4$. Thus, we have found the solution to the problem, because $E=4$ is a valid digit, and with it, $E F=49$ is a perfect square, and the sum of the three numbers, 196, is also a perfect square.

Therefore, the numbers we are looking for are: 3, 44, and 149; there is no other solution to the problem.

Remark. The fact that $E F$ is a perfect square was only used in the digit $F$, and similarly, in the a) scheme, only the last digit of $s$ was considered, which is why a final check was necessary.