# Problem 6. (3 points)

Natural numbers $x, y, z$ are such that $\operatorname{GCD}(\operatorname{LCM}(x, y), z) \cdot \operatorname{LCM}(\operatorname{GCD}(x, y), z)=1400$.

What is the greatest value that $\operatorname{GCD}(\operatorname{LCM}(x, y), z)$ can take?

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# Answer: 10

## Solution:

Notice 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 $\operatorname{LCM}(\operatorname{GCD}(x, y), z)$ is divisible by $\operatorname{GCD}(\operatorname{LCM}(x, y), z)$.

$1400=2^{3} \cdot 5^{2} \cdot 7$ and in the second factor, each prime number appears 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 $x=y=10, z=140$.