Task B-1.2. The sum of the digits of the natural number $x$ is $y$, and the sum of the digits of the number $y$ is $z$. Determine all numbers $x$ for which

$$
x+y+z=60
$$
## Solution.

From $x+y+z=60$ it follows that $x$ is a two-digit number.

We write $x=10a+b$, where $a$ and $b$ are digits and $a \neq 0$.

Then $y=a+b$, the sum of the digits of the number $x$.

If $a+b \leq 9$, then $z=a+b$ and from $x+y+z=60$ we get

$$
\begin{aligned}
& 10a+b+2(a+b)=60 \\
& 12a+3b=60 \\
& 4a+b=20 \\
& b=20-4a=4(5-a)
\end{aligned}
$$

Since $a$ and $b$ are digits whose sum is less than or equal to 9, we get $b=0, a=5$ or $b=4$, $a=4$.

If $a+b \geq 10$, then $z=a+b-9$, and from $x+y+z=60$ we get

$$
\begin{aligned}
& 10a+b+2(a+b)-9=60 \\
& 12a+3b=69 \\
& 4a+b=23 \\
& 4a=23-b
\end{aligned}
$$

If $b=3$, then $a=5$ and $a+b \geq 10$, is a solution.

Therefore, $x \in\{44,47,50\}$.