The number N is obtained by rearranging the digits of a positive decimal whole number M.
Can M+N be equal to 99.....9, where the digit 9 is repeated precisely 2007 times?
The rightmost digit of M plus the rightmost digit of N can't exceed 18, so these two digits must add up to 9, with no carry. The same applies to the digits to the left of these. So there is no carry taking place in any of the place positions.
Thus M and N each have 2007 digits an odd number.
Another way of showing that is that the leftmost digit of the total is not a 1, also leading to M and N each having 2007 digits.
Edited on October 15, 2007, 2:49 pm

Posted by Charlie
on 20071015 14:45:05 