## Task B-1.2.

Determine the periodic numbers $0 . \dot{x}, 0 . \dot{x} \dot{y}$ and $0 . \dot{x} y \dot{z}$ for which

$$
0 . \dot{x}+0 . \dot{x} \dot{y}+0 . \dot{x} y \dot{z}=\frac{445}{333}
$$

where $x, y, z$ (not necessarily distinct) are digits.
## Solution.

We represent the given periodic numbers as fractions:

$$
\begin{aligned}
0 . \dot{x} & =\frac{x}{9} \\
0 . \dot{x} \dot{y} & =\frac{10 x+y}{99} \\
0 . \dot{x} y \dot{z} & =\frac{100 x+10 y+z}{999}
\end{aligned}
$$

After substituting into the given equation, we get

$$
\frac{x}{9}+\frac{10 x+y}{99}+\frac{100 x+10 y+z}{999}=\frac{445}{333}
$$

Now we multiply by the least common denominator:

$$
x \cdot 37 \cdot 3 \cdot 11+(10 x+y) \cdot 37 \cdot 3+(100 x+10 y+z) \cdot 11=445 \cdot 3 \cdot 11
$$

Since all addends, except the second one, contain the factor 11 and are divisible by 11, the only possibility is that the number $10 x+y$ is divisible by 11, meaning that the digits $x$ and $y$ are equal.

Then the previous equation becomes

$$
\begin{aligned}
& x \cdot 37 \cdot 3 \cdot 11+(11 x) \cdot 37 \cdot 3+(110 x+z) \cdot 11=445 \cdot 3 \cdot 11 \\
& x \cdot 37 \cdot 3+x \cdot 37 \cdot 3+(110 x+z)=445 \cdot 3 \\
& 332 x+z=1335 \\
& z=1335-332 x
\end{aligned}
$$

Since $z$ is a digit, we have:

$$
\begin{aligned}
& 0 \leqslant 1335-332 x \leqslant 9 \\
\Rightarrow \quad & 3.99<x<4.1 \\
\Rightarrow \quad & x=4=y \\
& z=1335-332 \cdot 4=7
\end{aligned}
$$

The numbers are $\frac{4}{9}=0 . \dot{4}, \frac{44}{99}=0 . \dot{4} \dot{4}, \frac{447}{999}=0 . \dot{4} 4 \dot{7}$.