Do the three empty fields in the image belong to such natural numbers so that the product of the three numbers on each side of the triangle is the same.

![](https://cdn.mathpix.com/cropped/2024_04_17_cd0342d3a611cc15345bg-1.jpg?height=297&width=346&top_left_y=622&top_left_x=889)

What is the smallest and what is the largest number that can be written in the gray-shaded field under this condition?
Given numbers on the left side of the triangle give the product

$$
16 \cdot 42=2^{5} \cdot 3 \cdot 7
$$

and the given numbers on the right side give the product

$$
42 \cdot 72=2^{4} \cdot 3^{3} \cdot 7 \text {. }
$$

For the product of all numbers on the left side to be the same as on the right, the number in the lower left corner must contain the factor $3^{2}$ in its factorization, and the number in the lower right corner must contain the factor 2. The number in the lower left corner can be expressed as $3^{2} \cdot c$, where $c$ represents any natural number. In the lower right corner, it must then be $2 \cdot c$, and the product on both the left and right sides is $2^{5} \cdot 3^{3} \cdot 7 \cdot c$.

![](https://cdn.mathpix.com/cropped/2024_04_17_cd0342d3a611cc15345bg-1.jpg?height=414&width=454&top_left_y=1643&top_left_x=835)

If the number in the gray field is denoted by $x$, then the condition from the problem statement means

$$
3^{2} \cdot c \cdot x \cdot 2 \cdot c=2^{5} \cdot 3^{3} \cdot 7 \cdot c
$$

from which, after rearrangement, we can express

$$
x=\frac{2^{4} \cdot 3 \cdot 7}{c}=\frac{336}{c}
$$

The value of $x$ is the smallest when $c$ is the largest divisor of the number in the numerator, i.e., $c=336$ and $x=1$. The value of $x$ is the largest when $c$ is the smallest positive divisor of the number in the numerator, i.e., $c=1$ and $x=336$.