16. (5 points) Observing the subtraction equation shown in the figure, we find that the result 175 is the reverse order of the digits of the minuend 571. Therefore, after subtracting 396, the number of three-digit minuends that result in a number whose digits are in reverse order of the minuend is $\qquad$.
$$
\begin{array}{r}
571 \\
-396 \\
\hline 175
\end{array}
$$
【Solution】Let the hundreds, tens, and units digits of the minuend be: $a, b, c$; then we have: $\overline{\mathrm{ab}}$ $=\overline{\mathrm{cba}}$

Therefore, $100 a+10 b+c-396=100 c+10 b+a$,
Rearranging gives: $99 a-99 c=396$,
$$
a-c=396 \div 99 \text {, }
$$
$a-c=4$; and $a, c$ are natural numbers not greater than 9.
Thus: (1) $a=5, c=1, b=0 \sim 9$; a total of 10;
(2) $a=6, c=2, b=0 \sim 9$; a total of 10;
(3) $a=7, c=3, b=0 \sim 9$; a total of 10;
(4) $a=8, c=4, b=0 \sim 9$; a total of 10;
(5) $a=9, c=5, b=0 \sim 9$; a total of 10;

Therefore, in total: $10 \times 5=50$ (numbers).
Answer: There are 50 three-digit minuends whose result is the reverse of the digits of the minuend.
The answer is: 50.