## Task A-4.4.

Determine all triples of natural numbers $(m, n, k)$ for which

$$
D(m, 20)=n, \quad D(n, 15)=k \quad \text { and } \quad D(m, k)=5
$$

where $D(a, b)$ is the greatest common divisor of the numbers $a$ and $b$.
## Solution.

From the second two equalities, we get $5|k| 15$, and from the first two equalities, we get $k|n| 20$.

The only multiple of the number 5 that is also a divisor of the numbers 15 and 20 is the number 5, so $k=5$.

Furthermore, $n$ must be 5, 10, or 20, since these are the only divisors of 20 that are divisible by 5.

If $n=5$, then the equalities become

$$
D(m, 20)=5, \quad D(5,15)=5, \quad \text { and } \quad D(m, 5)=5
$$

We see that $m$ must be a multiple of the number 5, but it cannot be even, as then the number $D(m, 20)$ would be divisible by 10. Therefore, $m$ can be any odd natural number divisible by 5.

If $n=10$, then the equalities become

$$
D(m, 20)=10, \quad D(10,15)=5, \quad \text { and } \quad D(m, 5)=5
$$

Similarly to the previous case, we conclude that $m$ can be any natural number divisible by 10, but not by 20.

If $n=20$, then the equalities become

$$
D(m, 20)=20, \quad D(20,15)=5, \quad \text { and } \quad D(20,5)=5
$$

Therefore, $m$ can be any natural number divisible by 20.

Thus, all triples $(m, n, k)$ are $(5a, 5, 5)$ for odd numbers $a$, $(10a, 10, 5)$ for odd numbers $a$, and $(20b, 20, 5)$ for all natural numbers $b$.

Note: The allowed numbers $m$ can also be described in words as in the official solution (a natural number divisible by 10 but not by 20, a natural number divisible by 5 but not by 10, and so on).