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.

This solves prettily easily as a substitution arithmetic problem:

   a  b  c  d
+ d  c  b  a
-------------
a b  c  d  0

Clearly, a must equal 1 because the sum must be less than 20,000.

Then d must equal 9 because a + d = 10.

And 10 + b = (a + d), or (a+d+1) if there is a carryover = 10 or 11.  Since b can't be 1 (a is already 1), then b = 0

Then c = 8, because c + 1 = b + d = 9

Therefore, the only solution is 1089.

Edited on October 23, 2004, 2:04 pm

Posted by Steve Herman on 2004-10-23 14:04:19