1. A three-digit number is 29 times the sum of its digits. Then this three-digit number is $\qquad$
Let the three-digit number be $\overline{a b c}$. Then
$$
\begin{array}{l}
100 a+10 b+c=29(a+b+c) \\
\Rightarrow 71 a=19 b+28 c \\
\Rightarrow b=4 a-2 c+\frac{5}{19}(2 c-a) .
\end{array}
$$

Since $a$, $b$, and $c$ are positive integers, 19 divides $(2 c - a)$.
And $-9 \leqslant 2 c-a \leqslant 17$, then
$$
2 c-a=0 \Rightarrow a=2 c, b=6 c \text {. }
$$

From $0<b<10$, we know $c=1$.
Thus, $a=2, b=6$.