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Least alternating multiple of alternators.
5

%I #12 Jan 14 2017 11:41:04

%S 1,2,3,4,5,6,7,8,9,10,121,12,52,14,30,16,34,18,38,0,21,418,23,72,25,

%T 52,27,56,29,30,341,32,165,34,70,36,74,38,78,0,41,210,43,616,45,92,47,

%U 96,49,50,561,52,212,54,165,56,456,58,236,0,61,434,63,256,65,858,67,272,69

%N Least alternating multiple of alternators.

%C An alternating integer is a positive integer for which, in base-10, the parity of its digits alternates. E.g., 121 is alternating because its consecutive digits are odd-even-odd, 1 being odd and 2 even. Of course, 1234567890 is also alternating. An alternator is a positive integer which has a multiple which is alternating.

%C For n congruent to 0 mod 20, a(n) is shown as zero to indicate that n is not an alternator.

%H 45th International Mathematical Olympiad (45th IMO), <a href="http://www.jstor.org/stable/30044168">Problem #6 and Solution</a>, Mathematics Magazine, 2005, Vol. 78, No. 3, pp. 247, 250-251.

%e a(11) = 121 because 121 is the least multiple of 11 which is alternating.

%Y Cf. A030141, A110303, A110305.

%K base,easy,nonn

%O 1,2

%A _Walter Nissen_, Jul 18 2005