A132378 - OEIS

%I #11 Sep 08 2024 22:31:57

%S 1,2,3,4,5,6,7,8,9,10,11,12,132,140,15,162,175,184,198,20,210,22,234,

%T 24,250,264,2702,280,2916,30,312,324,33,348,3510,36,378,384,396,40,

%U 412,420,432,44,4500,468,476,48,4932,50,510,520,5310,540,55,5610,5705,5800,5940

%N Smallest integer that starts with n and is divisible by every nonzero digit of n.

%C Note that a number that starts with n and is divisible by lcm(1, 2, ..., 9) = 2520 always satisfies the conditions. And there is such a number between n0000 and n2519. Therefore a(n) < 10000n + 2520. Hence to find a(n) one needs to check numbers n, n0..n9, n00..n99, n000..n999, n0000..n2519.

%H Robert Israel, <a href="/A132378/b132378.txt">Table of n, a(n) for n = 1..10000</a>

%H TopCoder problem <a href="http://www.topcoder.com/stat?c=problem_statement&amp;pm=8318">DivisibleByDigits</a>

%p f:= proc(n) local r, d, q ;

%p    r:= ilcm(op(convert(convert(n,base,10),set) minus {0,1}));

%p    if n mod r = 0 then return n fi;

%p    for d from 1 do

%p      q:= -(n*10^d) mod r;

%p      if q < 10^d then return n*10^d+q fi;

%p    od

%p end proc:

%p map(f, [$1..100]); # _Robert Israel_, Feb 01 2017

%K base,easy,nonn

%O 1,2

%A Mikhail Dvorkin (mikhail.dvorkin(AT)gmail.com), Nov 10 2007