39. Find all three-digit numbers $x$, in the notation of which the digits do not repeat, such that the difference between this number and the number written with the same digits but in reverse order is also a three-digit number consisting of the same digits as the number $x$.
39. Let $x=100 a+10 b+c$. Then $(100 a+10 b+c)-(100 c+10 b+a)=$ $=99(a-c)$. The difference of digits $(a-c)$ can take values from 2 to 8. By assigning such values ( $a-c$ ), we get $198,297,396,495,594,693,792$. Let's consider all possible cases.
1) The desired number can consist of the digits 1,9 and 8. Then we have

$$
891-198=693,981-189=792,918-819=99
$$

None of these numbers satisfy the condition.

2) Suppose the desired number consists of the digits 2,9 and 7. Then we have

$792-297=495,972-279=693,927-729=198$.

None of these numbers satisfy the condition.

3) Suppose the desired number consists of the digits 3,9 and 6. Then we have

$693-396=297,963-369=594 ; 936-639=297$.

None of these numbers satisfy the condition.

4) Suppose the desired number consists of the digits 4, 9 and 5. Then we have

$594-495=99,954-459=495,945-549=396$.

The numbers 954 and 459 satisfy the condition.