Let $$x$$ be a five-digit number. Let $$y$$ be the four-digit number formed from $$x$$ by deleting its middle digit. Find all such $$x$$ when $$y$$ divides $$x$$ completely.

Write $$x = \overline{abcde}$$ and $$y = \overline{abde}$$, and let $$z = 10d + e$$, so $$x = 10(1000a + 100b) + 100c + z$$ and $$y = 1000a + 100b + z$$. Then $$x - 10y = 100c - 9z$$, which must be divisible by $$y$$. Since $$0 \le c \le 9$$ and $$0 \le z \le 99$$, we have $$-891 \le 100c - 9z \le 900$$, while $$y \ge 1000$$, so the only possible multiple is 0, giving $$100c = 9z$$. Thus $$c$$ is a multiple of 9; $$c = 9$$ would need $$z = 100$$, impossible, so $$c = 0$$ and $$z = 0$$, i.e. $$c = d = e = 0$$. Therefore $$x = \overline{ab000}$$, that is $$x = 10000, 11000, 12000, \ldots, 99000$$, which is 90 numbers.