Perhaps I'm missing something, but it seems to me that the answer to this modified question must be 1. Take the decimal log of x, and add (say) 17 to it, to get a number n, and then set N = 10^n. Now without a doubt, the smallest multiple of x exceeding 1023456789N begins with the digits 1023456789, and hence that single multiple has all ten digits in it. On Fri, Dec 16, 2011 at 12:34 PM, Michael Reid <reid@gauss.math.ucf.edu>wrote:
I give you an arbitrary positive integer x > 0.
You write down x, 2x, 3x, 4x, ..., kx in decimal.
How large much k be to guarantee that every digit appears somewhere in your list, regardless of x?
What is the smallest k for which there are k positive integers m_1 , m_2 , ... , m_k such that, for every positive integer x , all ten decimal digits occur on the list m_1 x , m_2 x , ... , m_k x ?
(The solution of the original problem shows that such k exist.)
Michael Reid
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