Hans Havermann:

It is 72, I think, for x = 125.

Yes, you are right. It seems likely that 72 should suffice for all integers x. One may extend the question to arbitrary nonzero real numbers x; write down the decimal expansions of x, 2x, 3x, 4x, ..., kx. We need a convention for the digit 0, but let us count any occurrence of 0 after the leading nonzero digit of the number, including the case that the expansion is terminating. (When x is an integer, we need not bother with 0, as it always appears as early as in 10x.) Note that the smallest possible k in the "real" case is an upper bound on the smallest possible k in the "integer" case. I claim that k = 81 in the "real" case. Note that this k is required for x = 1/9, because ax contains no 9 for a<81. To prove the claim, first note that it suffices to look at numbers x between 1/10 and 1. We divide into three cases, depending on how x relates to 1/9 and 10/81. Case 1: 1/10 <= x < 1/9. Then we may pick k = 10, because ax contains the digit a for a<10, and x contains a (non-leading) 0. Case 2: 1/9 < x < 10/81. By an argument similar to the one of Michael Kleber, all nonzero digits have appeared at least once as a leading digit as soon as kx is at least 9. Moreover, 9x contains a 0, because 1 < 9x < 10/9. Picking k minimal, we get (k-1)x < 9, and so k < 9/x + 1 < 82; hence we may pick k = 81. Case 3: 10/81 <= x < 1. Again, all nonzero digits have appeared as a leading digit as soon as kx is at least 9. To make sure we also find a 0, pick k such that kx is at least 10. For the smallest such k, we have that kx contains a 0, because 10 <= kx < 11. This time, we get (k-1)x < 10, and so k < 10/x + 1 <= 82; hence again we may pick k = 81. I hope I got things right. Jakob