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.

Yes, you can use a program to "brute-force" test every four digit number, but it is better if you can find it and prove it without crunching all of those numbers.

First, as Lewis said, the first and last digits must add up to 10.

Second, since the first digit of the five digit number can only be 1, the first digit of the four digit number can only be 1.

So the number is of the form 1xx9

Third, since 1xx9 + 9xx1 = 1xx90 = 10(1xx9), then 9xx1 = 9(1xx9). Since 9xx1 is divisible by 9, and 1xx9 and 9xx1 have the same digits, then by the "casting out nines property" 1xx9 must be divisible by 9. And also, the two middle numers must add to either 8 or 17.

If I were inclined to crunch numbers, I would only need to crunch 1089, 1179, 1269, 1359, 1449, 1899, and 1989. The last three contain duplicated digits and are thus disqualified. So we have reduced the "pool" from 10!/6! = 5040 (Already reduced from 10,000 by the no duplicates rule) to 4.

But perhaps we can reduce it further. At this point it becomes necessary to identify the digits by unique variables. abcd, where a=1 and d=9.
we get:

1. 9+1=10
   
2. c+b+1=9 => c+b=8 (which confirms the earlier conclusion)
   
3. b+c=c => b=0 OR b+c=10+c =>b=10 (which is wrong)
   
4. 1+9=10+b => b=0

This gives us a=1, b=0, c=8, d=9 or 1089 as a unique solution.

Posted by TomM on 2003-07-06 13:02:15