Original source: http://mathforum.org/library/drmath/view/54205.html

i found the answer on the above website

This struck my mind as soon as i read the post bcz its a kind of weird one cant forgot easily

so, here we go lets guess the 5 digit number is abcde and lets calculate the possibility

(9 here because we need to find the one which get reversed when multiplied by 9)

9*(10000*a+1000*b+100*c+10*d+e) = 10000*e+1000*d+100*c+10*b+a

```
89999*a + 8990*b + 800*c = 910*d + 9991*e.
```

This implies that a = 1 and e = 9, by looking at last digits, as

before, and then

```
8990*b + 800*c = 910*d + 80
899*b + 80*c = 91*d + 8
```

This implies by looking at sizes that either b = 1 (so d = 1 and 899 +

80*c = 99, which has no solutions) or b = 0 (so 91*d - 80*c = 8, which

has the unique solution d = 8, c = 9). Thus there is just one 5-digit

solution 10989.