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?
(In reply to
solution by bubu)
As each place position is independent, for every 0 in M there must be a 9 in N; for every 1, an 8; for every 2, a 7; etc.
But M and N use the same digits: there must be the same number of 0's as 9's in this pool of digits, and the same number of 1's as 8's, and the same number of 2's as 7's; 3's as 6's and 4's as 5's.
That implies there is an even number of digits. But we need an odd number of digits. Thus it's impossible.

Posted by Charlie
on 20071016 09:35:30 