16. (6 points) If the two-digit number $\overline{a b}$ is 24 more than $\overline{c d}$, and the three-digit number $\overline{1 a b}$ is 15 more than $\overline{c d 1}$, then $\overline{a d}=$
【Analysis】Through analysis: from
$$
\begin{array}{r}
c d 1 \\
+\quad 15 \\
\hline 1 a b
\end{array}
$$

we get $b=1+5=6$, substituting $b=6$:
$$
\begin{array}{r}
c d \\
+24 \\
\hline a b
\end{array}
$$

we get $d=6-4=2$, substituting $b=6, d=2$ into $\overline{c d 1}+15=\overline{1 \mathrm{ab}}$, we get $a=2+1=3, c=1$, so $\overline{\mathrm{ab}}=36$, and the solution can be derived accordingly.

【Solution】Solution: Because:
$$
\begin{array}{r}
c d 1 \\
+\quad 15 \\
\hline 1 a b
\end{array}
$$

so $b=1+5=6$,
substituting $b=6$:
$$
\begin{array}{r}
c d \\
+\quad 24 \\
\hline a b
\end{array}
$$

we get $d=6-4=2$,
substituting $b=6, d=2$ into $\overline{c d 1}+15=\overline{1 \mathrm{ab}}$
we get: $a=2+1=3, c=1$
so: $\overline{\mathrm{ad}}=32$
Therefore, the answer is: 32.