Certainly k=100 suffices: (a) The number 10x is one digit longer than x, which means that adding x can change the lead digit by at most 1. (b) Both 10x and 100x start with the same digit as x. (c) Together these imply that by the time you've made it to 100x, you've cycled through every lead digit. Note that all of (a), (b), and (c) are base-independent statements :-). I suspect David's k=45 (for base 10) is the correct answer, but don't immediately see how to pull the upper bound that low. --Michael On Thu, Dec 15, 2011 at 8:47 AM, <dwcantrell@comcast.net> wrote:
Considering that x could be 2, we can conclude that k must be at least 45. I don't know whether k be larger than 45 or not.
David Cantrell
----- Original Message -----
From: "David Wilson" <davidwwilson@comcast.net> To: "Math Fun" <math-fun@mailman.xmission.com> Sent: Thursday, December 15, 2011 12:48:22 PM Subject: [math-fun] Puzzler
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?
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