Let s = 666 or any other prefix, and n >= 1, define k(s, n) = the smallest k so that kn starts with s in base 10. First to show k <= 10s. Let d be the number of digits of n, that is, 10^(d-1) <= n < 10^d. Let j = ceil(s*10^d/n), that is, s*10^d/n <= j < s*10^d/n + 1 This gives s*10^d <= jn < s*10^d + n < s*10^d + 10^d This means that in base 10, jn starts with s followed by d digits. By the definition of k, k <= j <= s*10^d/n But 10^(d-1) <= n so 10^d/n <= 10 giving k <= 10s. We knew 10s was a bound on n, this proves it. For a given s, let m(s) largest value of k(s, n) over all n. I am convinced that m(s) = ceil(10 s^2/(s+1)). For n >= 9, this gives m(s) = 10s-9, and in particular m(666) = 6651. The smallest n that produces k(s,n) = m(s) seems to be of the form ceil(10^j*(s+1)/s) for some j >= 0. The value of j depends on s, but seems to wiggle around, I haven't investigated closely. For s = 666, j = 6, giving n = 1001502 which produces k = m(s) = 6651. Now let's talk substrings instead of prefixes. Defined k'(s, n) = the smallest k so that kn has substring s in base 10. Clearly a prefix is a substring, so k'(s, n) <= k(s, n). If we define m'(s) analogously to m(s), we then have m'(s) <= m(s). By computer search, I find m'(s) = m(s) for s = 1, 4, 11, and 12. However, it looks like m'(s) < m(s) often enough, e.g, it appears that m'(7) = 35 < m(7) = 62. It appears that m'(666) = 4263 < m(666) = 6651 ----- Original Message ----- From: "Michael Reid" <reid@gauss.math.ucf.edu> To: <math-fun@mailman.xmission.com> Sent: Thursday, October 02, 2008 5:36 PM Subject: Re: [math-fun] Beastly number problem
i blathered:
A beastly number (A051003) is a number with substring 666.
It is easy enough to prove that for each positive integer n, there is an integer k such that kn is beastly. Let f(n) be the smallest such k.
Show that f(n) is bounded over all integers n. What is the largest possible value of f(n)?
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to get the largest value of f(n) that occurs, i found it easier to consider final digits and split it into numerous cases; perhaps there is an easier calculation using initial digits.
or maybe not.
let g(n) denote the smallest positive integer k , such that kn begins with the digits 666 . what is the largest value of the function g(n) ?
mike
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