On the table are written three natural numbers $a, b, c$, for which the following holds:

- the greatest common divisor of numbers $a, b$ is 15,
- the greatest common divisor of numbers $b, c$ is 6,
- the product of numbers $b, c$ is 1800,
- the least common multiple of numbers $a, b$ is 3150.

What are these numbers?

(L. Šimunek)
We will gradually write down the prime factors of the factorization of numbers $a, b, c$ in the table.

According to the first condition, the greatest common divisor of numbers $a$ and $b$ is $15=3 \cdot 5$. This means that both in row $a$ and in row $b$ there must be factors 3 and 5, and no other factor can be in both rows at the same time. After applying the first and similar second conditions, the table looks like this:

| $a$ | $3 \cdot 5 \ldots$ |
| :--- | :--- |
| $b$ | $2 \cdot 3 \cdot 5 \ldots$ |
| $c$ | $2 \cdot 3 \ldots$ |

According to the third condition, $b \cdot c=1800=2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5 \cdot 5$. This means that in rows $b$ and $c$ there must be these 7 factors and no more. According to the fourth condition, the least common multiple of numbers $a$ and $b$ is $3150=2 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdot 7$. For rows $a$ and $b$, this means that:

- in one of them there must be exactly one factor 2 and in the other at most once (the same applies to factor 7),
- in one of them there must be exactly two factors 3 and in the other at most twice (the same applies to factor 5),
- no other factor can be there.

According to the third condition, we only need to add two factors, 5 and 2, to rows $b$ and $c$. The factor 5 cannot be in row $c$, because then numbers $b$ and $c$ would have a common divisor $2 \cdot 3 \cdot 5=30$, which contradicts the second condition. The factor 2 cannot be in row $b$, because that would contradict the fourth condition about the least common multiple of numbers $a$ and $b$. After this reasoning, we have rows $b$ and $c$ completely filled:

| $a$ | $3 \cdot 5 \ldots$ |
| :--- | :--- |
| $b$ | $2 \cdot 3 \cdot 5 \cdot 5$ |
| $c$ | $2 \cdot 2 \cdot 3$ |

According to the fourth condition, only factors $2,3,5,7$ can be in row $a$. We cannot add factors 2 and 5 to this row, because a common divisor of numbers $a$ and $b$ would then arise, contradicting the first condition. We must add factors 3 and 7 to row $a$ due to the fourth condition:

| $a$ | $3 \cdot 3 \cdot 5 \cdot 7=315$ |
| :--- | :--- |
| $b$ | $2 \cdot 3 \cdot 5 \cdot 5=150$ |
| $c$ | $2 \cdot 2 \cdot 3=12$ |

The unknowns $a, b, c$ are in order equal to the numbers $315,150,12$.