Adam has two cuboids with volumes of $12 \mathrm{~cm}^{3}$ and $30 \mathrm{~cm}^{3}$, and the dimensions of each are expressed in centimeters as different whole numbers. Adam found that the cuboids can be glued together so that the glued faces coincide, resulting in a new cuboid.

What dimensions could the new cuboid have? Determine all possibilities.

(E. Semerádová)
Possible dimensions of the cuboids can be determined using the factorization of the given volumes into the product of three different natural numbers:

$$
\begin{aligned}
12 & =1 \cdot 2 \cdot 6=1 \cdot 3 \cdot 4, \\
30=1 \cdot 2 \cdot 15 & =1 \cdot 3 \cdot 10=1 \cdot 5 \cdot 6=2 \cdot 3 \cdot 5 .
\end{aligned}
$$

In these factorizations, we look for pairs of numbers that represent the dimensions of the glued faces. Such pairs are exactly three and correspond to the following options:

a) For the common face $1 \mathrm{~cm} \times 2 \mathrm{~cm}$, the third dimension of the new cuboid is $6+15=21(\mathrm{~cm})$.

b) For the common face $1 \mathrm{~cm} \times 3 \mathrm{~cm}$, the third dimension of the new cuboid is $4+10=14(\mathrm{~cm})$.

c) For the common face $1 \mathrm{~cm} \times 6 \mathrm{~cm}$, the third dimension of the new cuboid is $2+5=7(\mathrm{~cm})$.

The new cuboid could have dimensions $1 \mathrm{~cm} \times 2 \mathrm{~cm} \times 21 \mathrm{~cm}$, $1 \mathrm{~cm} \times 3 \mathrm{~cm} \times 14 \mathrm{~cm}$, or $1 \mathrm{~cm} \times 6 \mathrm{~cm} \times 7 \mathrm{~cm}$.

Another solution. The sum of the volumes of both cuboids is $42 \mathrm{~cm}^{3}$. Just like the original cuboids, the dimensions of the new cuboid are expressed in centimeters and are whole numbers. Unlike the original cuboids, however, these dimensions do not have to be different. Possible dimensions of the new cuboid can be determined by factorizing its volume into the product of three natural numbers:

$$
42=1 \cdot 1 \cdot 42=1 \cdot 2 \cdot 21=1 \cdot 3 \cdot 14=1 \cdot 6 \cdot 7=2 \cdot 3 \cdot 7 \text {. }
$$

The pairs of numbers in these factorizations representing the common face must consist of different numbers, whose product must be a divisor of the number 12 (the volume of the smaller of the original cuboids). Such pairs are four, and they correspond to the following options:
a) For the common face $1 \mathrm{~cm} \times 2 \mathrm{~cm}$, the third dimension of the smaller, respectively larger, cuboid would be $12: 2=6(\mathrm{~cm})$, respectively $30: 2=15(\mathrm{~cm})$.

b) For the common face $1 \mathrm{~cm} \times 3 \mathrm{~cm}$, the third dimension of the smaller, respectively larger, cuboid would be $12: 3=4(\mathrm{~cm})$, respectively $30: 3=10(\mathrm{~cm})$.

c) For the common face $1 \mathrm{~cm} \times 6 \mathrm{~cm}$, the third dimension of the smaller, respectively larger, cuboid would be $12: 6=2(\mathrm{~cm})$, respectively $30: 6=5(\mathrm{~cm})$.

d) For the common face $2 \mathrm{~cm} \times 3 \mathrm{~cm}$, the third dimension of the smaller, respectively larger, cuboid would be $12: 6=2(\mathrm{~cm})$, respectively $30: 6=5(\mathrm{~cm})$.

In the case of d), the smaller of the original cuboids would not have different side lengths, the other options meet all the requirements. The new cuboid could have dimensions $1 \mathrm{~cm} \times 2 \mathrm{~cm} \times 21 \mathrm{~cm}$, $1 \mathrm{~cm} \times 3 \mathrm{~cm} \times 14 \mathrm{~cm}$, or $1 \mathrm{~cm} \times 6 \mathrm{~cm} \times 7 \mathrm{~cm}$.

Grading suggestion. 2 points for determining all factorizations; 3 points for discussion and selection of suitable options; 1 point for the conclusion.