Pick a four digit number, all digits different, such that when you add its reverse and divide it by 10, you get the number you started with.
For example: 1749+9471=11220, 11220/10 = 1122. Since 1749 is not equal to 1122, this is not the right number.
Let the four digit number be ABCD.
By the problem,
(ABCD + DCBA)/10 = ABCD
Or, DCBA= 9*ABCD.....(*)
Let A> = 2. If so,DCBA is a five digit number, a contradiction.
Thus A= 1
9*A is less than 10 (otherwise (ABCD)*9 would have 5 digits),
so A is 0 or 1. If A is 0 then so is D, since the last digit
of D*9 is A. So A is 1.
Considering the leftmost digits of the equation (*), we have:
1 Mod 10 = 9D, giving D Mod 10 = 9, so that D =9
Since A = 9D, it follows that there is no carryover to the leftmost column when one multiplies B by 9, so B is 0 or 1. Since A = 1, it follows that B = 0
The equation now reduces to 9C01 = 9*10C9
This gives (9C + 8) Mod 10 = 0, so that:
9C = 2 Mod 10
Or C = 8 Mod 10, so that: C = 8
Thus, ABCD = 1089
Consequently, the required number is 1089