## SUBJECT 1.

If $x$ is the number of subsets of the set $A=\{\overline{a b c} \mid \sqrt{\overline{0, a(b c)}+\overline{0, b(c a)}+\overline{0, c(a b)}} \in \mathrm{N}$ and $\mathrm{a}>\mathrm{b}>\mathrm{c}>0\}$ and $y=\left(\left|3^{51}-2^{85}\right|+3^{2015}: 81^{491}\right):\left(-4^{41}\right)+\sqrt{1296}+\sqrt{(8-5 \sqrt{3})^{2}}-\sqrt{75}+\sqrt{2^{8}}$

Calculate the geometric mean of the numbers $x$ and $y$.
## Subject 1.

$$
\sqrt{\frac{\overline{a b c}-a+\overline{b c a}-b+\overline{c a b}-c}{990}}=\sqrt{\frac{110 a+110 b+110 c}{990}}=\sqrt{\frac{a+b+c}{9}} \in \mathbb{N} \text { so } a+b+c=9 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \ldots \mathbf{2}
$$

Since $a>b>c>0$, it follows that $A=\{621 ; 531 ; 432\}$, so $x=2^{\text{card } A}=2^{3}=8 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ 1p

Considering that $3^{51}=\left(3^{3}\right)^{17}=27^{17} ; 2^{85}=\left(2^{5}\right)^{17}=32^{17}=>3^{51}$

![](https://cdn.mathpix.com/cropped/2024_06_07_053b2b9fe3ee34c188bfg-2.jpg?height=68&width=1840&top_left_y=931&top_left_x=88)

$m_{g}=\sqrt{x \cdot y}=\sqrt{8 \cdot 36}=12 \sqrt{2} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .1 p$