The answer is surprising: For positive integers N meeting the conditions,
the sum of the digits of 9N is ALWAYS 9. So M = 1, always.
Solutions were
received by Ken Duisenberg, Stephen Meskin, John Guilford, and Joseph
DeVincentis.
Let N be written in digits as a_0, a_1, ..., a_n, b, where b > a_n and the digits are nondecreasing. Multiplying any single
digit X by 9 gives the 2-digit number [X − 1] [10 − X], where each term is a digit. So multiplying N by 9 finds the digits of 9N as follows (ignoring carrying for a moment):
The ones digits is 10 − b.
The tens digit is the sum of b − 1 and 10 − a_n.
The 100s digit is (a_n) − 1 + 10 − a_(n − 1), and so on down to the last digit being a_0 − 1. Summing the pairs yields that the final form is as follows, where we still have to deal with the carrying:
[a0 − 1] [9 + a1 − a0] [9 + a2 − a1]
...
[9 + a_n − a_(n − 1)] [9 + b − a_n] [10 − b]
The units digits causes no carry. Because b is larger than a_n, the final 10s digit is b − 1 − a_n and 1 is carried to the 100s digit. This forces a similar carry of a 1 to the 1000s digit. And so on.
So the final representation is
[a0] [a1 − a0] [a2 − a1]
...
[a_n − a_(n − 1)] [b − 1− a_n] [10 − b]
The digit sum telescopes to 10 − 1 = 9. So the sum of the digits of 9 N is always 9.