With the exception of N=0, the proposition is true.
Not only that, but the 'two distinct digits' can be sharpened to 0 and 9:
1. Assume that N divides 10^n M times, then N*M=10^n, for some M. Multiplying this by 9 produces 9 followed by a number of zeroes.
2. Assume that N is odd, is not divisible by 5, and does not divide 10^n. Then M divides 10^n-1 (this is, after all, the basis of the decimal system) so that 10*N*M= 9999...90. In this case M is the periodic part of the decimal expansion of 1/N, see Sloane A036275.
3. Assume that N is even, or divisible by 5, and does not divide 10^n:
(a) Extract all powers of 2 and 5 to produce an odd number, and derive its periodic part, M1.
(b) Since matched powers of 2 and 5 just produce terminal zeroes, they can be ignored. For each unmatched power of 2, multiply M1 by 5; for each unmatched power of 5, multiply M1 by 2, to produce M2. In this case M is M1*M2. The result comprises a series of 9's followed by a series of zeroes, 999...000.
4. Since 0*M = 0, only 1 digit is needed to represent it.
A nice problem, but not D4 as we know it, Jim.
Edited on January 17, 2015, 9:05 pm
Posted by broll
on 2015-01-17 20:58:17