For three unknown natural numbers, it holds that:

- the greatest common divisor of the first and second is 8,
- the greatest common divisor of the second and third is 2,
- the greatest common divisor of the first and third is 6,
- the least common multiple of all three numbers is 1680,
- the largest of the numbers is greater than 100, but not greater than 200,
- one of the numbers is the fourth power of an integer.

Which numbers are these? Determine all possibilities.

(E. Semerádová)
From the first three conditions, it follows that the first number is a multiple of 24 (to be divisible by 8 and 6), the second number is a multiple of 8, and the third number is a multiple of 6. Since the mentioned divisors are the largest possible, the unknown numbers must be of the form

$$
24a, \quad 8b, \quad 6c
$$

where $a, b, c$ are pairwise coprime numbers.

The least common multiple of such a triplet of numbers is $24 \cdot a \cdot b \cdot c$, which, according to the problem, is equal to $1680 = 24 \cdot 2 \cdot 5 \cdot 7$. For one of the numbers to be a fourth power of an integer, it must be the second number, and this is the case when $b=2$ (the first and third numbers are divisible by three, but the least common multiple of all three numbers is not divisible by $3^4 = 81$). Therefore, the unknown numbers are of the form

$$
24a, \quad 16, \quad 6c
$$

where the numbers $a, c$ are 5, 7, or 1, 35 (in any order).

To ensure that none of the numbers is greater than 200, neither $a$ nor $c$ can be 35. Thus, there are two possibilities: either $a=5$ and $c=7$, or $a=7$ and $c=5$. Both cases satisfy the last required condition, i.e., that the largest of the sought numbers is greater than 100. The unknown triplet of numbers is one of the following

$$
\begin{array}{lll}
120, & 16, & 42, \\
168, & 16, & 30.
\end{array}
$$

Evaluation. 2 points for deriving (1); 2 points for factoring the number 1680 and deriving (2); 2 points for analyzing the possibilities and the result.