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?

Leming: Your numbers are incorrect - the numbers of 4s in the first number equals the number of 5s in the second.
Solution: Such numbers don't exist. Basing on what Charlie mentioned in the first post, there can be no carry for all digits. So let's assume we have those 2 numbers: a and b, let's pick first digit of a d_a and first digit of b d_b - they must sum up to 9, so digit d_b must exist in a on some other position and d_a must exist in a on some other position. Moreover they must be at the same index and sum up to 9. So let's cut these 2 digits from a and the same for b. Repeating this procedure we eventualy cut out even number of digits, so we end up with single digit in a and single digit in b (because the sum of a and b is odd). These 2 digits must sum up to 9 and be equal, which is impossible.

Posted by bubu on 2007-10-16 09:17:17