Solution to:
Palindrome Puzzle
Because the reverse of the starting number is greater than the starting number itself,
the first digit of the starting number must be less than the last digit.
Therefore, the starting number must be at least 102.
Secondly, we know that after two summations, the result has still only three digits.
abc
cba +

def
fed +

ghi
We know that def is not a palindrome.
Therefore, d differs from f.
This is only possible if d=f+1
(d can only be one greater than f, because b is at most 9).
Since abc is at least 102, def is at least 403, so d+f will be at least 7.
Since ghi is still a 3digit number but not a palindrome,
i can be at most 8, so d+f can be at most 8.
Since d=f+1, d+f can only be 7, from which we conclude that a=1 and c=2.
Now we have:
1b2
2b1 +

4e3
To make the first digit of 4e3 a 4, b must be 5, 6, 7, 8, or 9.
Now calculate the sum of 4e3 and 3e4:
4e3
3e4 +

8h7
Because the first digit of the sum must be 8,
e must be at least 5.
Therefore, the only remaining candidates for b are 8 (8+8=16)
and 9 (9+9=18).
Now it can easily be found that b must be 9 and the starting number we are looking for is 192:
192
291 + (291 is greater than 192)

483
384 +

867 (still a 3digit number)
768 +

1635
5361 +

6996 (the 4digit palindrome)
back to the puzzle
