29. (8th grade) When asked what his ticket number was, someone answered like this: "If all six two-digit numbers that can be formed from the digits of the number are added together, then half of the resulting sum will be exactly my ticket number." Determine the ticket number.
29. Let's denote the ticket number as: $N=100 x+10 y+\boldsymbol{z}$. According to the problem,

$$
\begin{gathered}
2 N=10 x+y+10 x+z+10 y+x+10 y+z+10 z+x+10 z+y= \\
=22 x+22 y+22 z, \text { i.e. } N=11(x+y+z)
\end{gathered}
$$

By equating the first and second results, we get the following equation: $100 x+10 y+z=11(x+y+z)$. The right side is divisible by 11, and the left side must also be divisible by 11. This is possible if $x+\boldsymbol{z}-\boldsymbol{y}=11 k$. Since $x+\boldsymbol{z}$ can take a value from 0 to 18, and $y$ from 0 to 9, the equation is possible only for $k=0, k=1$. Answer: $N=198$.