%I
%S 1,1,3,1,1,3,2,1,9,1,2,3,2,2,3,1,2,9,2,1,3,2,2,3,1,2,9,2,2,3,3,1,6,2,
%T 2,9,3,2,3,1,5,3,3,2,9,2,2,3,2,1,3,2,3,9,2,2,3,2,2,3,2,3,9,1,2,6,3,2,
%U 3,2,3,9,2,3,3,2,2,3,4,1,9,5,3,3,2,3,3,2,2,9,2,2,3,2,2,3,2,2,18,1,2,3,2,2,3
%N Smallest possible sum of the digits of a multiple of n.
%C a(n) is not bounded since a(10^n1)=9n. (Rustem Aidagulov)
%C In May 2002, this sequence (up to n=1000 with some useful remarks) was constructed by Pavel V. Phedotov. Some problems at the Second International Distant School Olympiad in Math "Third Millennium" (January 2002) asked to find a(n) for n = 5, 6, 7, 8, 9, 55, 66, 77, 88, 99, 555, 666, 777, 888, 999, and 2002^2002 .  Valery P. Phedotov (vphedotov(AT)narod.ru), May 05 2010
%H A.V.Izvalov, S.T.Kuznetsov, <a href="/A077196/b077196.txt">Table of n, a(n) for n = 1..56000</a>
%H Pavel V. Phedotov, <a href="http://digitsum.narod.ru">Sum of digits of a multiple of a given number</a>, May 2002. (in Russian)
%H Valery P. Phedotov, <a href="http://matholimp.livejournal.com/2009/12/14">Problems from 2002 Math Olympiad "Third Millennium"</a> (in Russian)
%F a(n) = A007953(A077194(n)).
%F a(2n)=a(n) and a(5n)=a(n) for any n. In particular, a(2^a*5^b) = a(1) = 1 where a or b are nonnegative integer.
%Y Cf. A077194, A077195.
%K base,nonn
%O 1,3
%A _Amarnath Murthy_, Nov 01 2002
%E More terms from _Sascha Kurz_, Feb 10 2003
%E Corrected and extended by _Max Alekseyev_, Feb 26 2009
