%I #10 Feb 27 2023 16:53:00
%S 1,20,102,1004,10005,100002,1000013,10000008,100000008,1000000010,
%T 10000000010,100000000008,1000000000012,10000000000004,
%U 100000000000005,1000000000000016,10000000000000016,100000000000000008,1000000000000000018
%N Least n-digit multiple of n whose digit permutations yield at least n distinct multiples of n, or 0 if no such number exists.
%C Conjecture: No term is zero. The proof should be simple.
%C There are no zeros for n<=500. - _Sean A. Irvine_, Apr 10 2013
%C The conjecture is true: There are 9*10^(n-1) n-digit numbers, meaning at least floor(9*10^(n-1)/n) n-digit multiples of n. There are binomial(n+9, 9) multisets of n digits. Thus, by the pigeonhole principle, one of these multisets contains at least ceiling(floor(9*10^(n-1)/n)/binomial(n+9, 9)) multiples of n, and this number is at least n whenever n > 3. - _James Rayman_, Feb 14 2023
%e a(4) = 1004; the multiples of 4 arising as a digit permutation are 1004, 1040, 1400, 4100.
%K base,easy,nonn
%O 1,2
%A _Amarnath Murthy_, Nov 07 2005
%E More terms from _Sean A. Irvine_, Apr 10 2013
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