Teacher Smolná prepared a test for her class in three versions to prevent students from copying. $V$ In each version, she specified three edges of a cuboid and asked the students to calculate its volume. However, she did not solve the problems in advance and did not realize that the result was the same in all three versions. She wrote the following edge lengths for the students: 12, 18, 20, 24, 30, 33, and 70, all in centimeters. Out of the nine edge lengths that Teacher Smolná specified, we have revealed only seven and have not disclosed which lengths belong to the same problem. Determine the two remaining edge lengths.

(L. Šimünek)

Hint. Factorize the given lengths into prime factors.
Let's factorize the lengths of all edges into prime factors:

$$
\begin{gathered}
12=2 \cdot 2 \cdot 3,18=2 \cdot 3 \cdot 3,20=2 \cdot 2 \cdot 5, \\
24=2 \cdot 2 \cdot 2 \cdot 3,30=2 \cdot 3 \cdot 5,33=3 \cdot 11,70=2 \cdot 5 \cdot 7
\end{gathered}
$$

In these products, the factors 7 and 11 are present, so the resulting volume must be a multiple of 77. The factors 7 and 11 are each present only once in the given lengths, specifically in the edges 33 and 70. Let's decide whether these lengths can belong to two different cuboids.

If the edges 33 and 70 belonged to different cuboids, the cuboid with the edge 33 would have to have another edge that is a multiple of 7, the cuboid with the edge 70 would have to have another edge that is a multiple of 11, and the last cuboid would have to have among its edges a multiple of 7 and a multiple of 11. We have just assumed the existence of at least three edges that are not listed in the problem, but only two are missing in the problem. This shows that the edges 33 and 70 belong to the same cuboid. Now let's determine the third edge of this cuboid.

Both lengths of the edges that are not listed in the problem must be multiples of 77 and belong to the remaining cuboids. Therefore, we need to look for the third edge of our cuboid among the given edges. Notice the edges 18 and 24. From the first, we deduce that the resulting volume is a multiple of nine (i.e., 3 3), and from the second, it is also a multiple of eight (i.e., 2 2 2 2). In the products corresponding to the edges 33 and 70, the factors 2 and 3 each appear only once. Therefore, the third edge of the considered cuboid must have the product \(2 \cdot 2 \cdot 3\) in its factorization. In the problem, we can choose either the edge 12 or 24.

First, let's consider the possibility that one of the cuboids has edges 33, 70, and 12, and thus the cuboids have a volume of \(2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5 \cdot 7 \cdot 11\). For the second cuboid, we choose the edge 24 and see that it cannot have any other edge with the factor 2 in its length. However, the remaining lengths in the problem all have the factor 2. Therefore, we must reject the possibility of a cuboid with edges 33, 70, and 12.

Now let's consider the possibility that one of the cuboids has edges 33, 70, and 24, and thus the cuboids have a volume of

$$
2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5 \cdot 7 \cdot 11
$$

The edges of the remaining two cuboids can be easily determined if we follow the fact that a cuboid must have exactly one factor 5 and exactly two factors 3 in the lengths of its edges: the second cuboid has edges

$$
30=2 \cdot 3 \cdot 5,12=2 \cdot 2 \cdot 3,154=2 \cdot 7 \cdot 11
$$

and the edges of the third cuboid are

$$
20=2 \cdot 2 \cdot 5,18=2 \cdot 3 \cdot 3,154=2 \cdot 7 \cdot 11
$$

The lengths of the two remaining edges that are not listed in the problem are both \(154 \text{ cm}\).