# Problem 7. (4 points)

Natural numbers $a, b, c$ are such that $\operatorname{GCD}(\operatorname{LCM}(a, b), c) \cdot \operatorname{LCM}(\operatorname{GCD}(a, b), c)=200$.

What is the greatest value that $\operatorname{GCD}(\operatorname{LCM}(a, b), c)$ can take?
Answer: 10

## Solution:

Note that $\operatorname{LCM}(\operatorname{GCD}(x, y), z)$ is divisible by $z$, and $z$ is divisible by $\operatorname{GCD}(\operatorname{LCM}(x, y), z)$, so $\operatorname{LCM}(\operatorname{GCD}(x, y)$ is divisible by $\operatorname{GCD}(\operatorname{LCM}(x, y), z)$.
$200=2^{3} \cdot 5^{2}$ and in the second factor, each prime number enters with an exponent no less than in the first. Therefore, the maximum possible value of $\operatorname{GCD}(\operatorname{LCM}(x, y), z)$ is $2 \cdot 5=10$. This value is achieved when $x=y=10, z=20$.