A308237 Numbers m not ending with 0 that contain a digit, other than the leftmost digit, that can be removed such that the resulting number d divides m.

11, 12, 13, 14, 15, 16, 17, 18, 19, 22, 24, 26, 28, 33, 36, 39, 44, 48, 55, 66, 77, 88, 99, 105, 108, 121, 132, 135, 143, 154, 165, 176, 187, 192, 195, 198, 225, 231, 242, 253, 264, 275, 286, 297, 315, 341, 352, 363, 374, 385, 396, 405, 451, 462, 473, 484, 495, 561, 572, 583, 594, 671, 682, 693

Offset: 1

Author

Bernard Schott, May 16 2019

Keywords

• nonn
• base
• fini

Comments

When m is a term, then, necessarily, the digit that is removed is the second from the left.

This sequence is finite with 95 integers and the greatest term is 180625. The number of terms with respectively 2, 3, 4, 5, 6 digits is 23, 44, 10, 17, 1.

The obtained quotients m/d belong to: { 6, 7, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19 } (all proofs in Diophante link).

Examples

			264 is a term because 264/24 = 11.
34875 is a term because 34875/3875 = 9.
		

Links

• Marius A. Burtea, Table of n, a(n) for n = 1..95
• Diophante, A333. Un chiffre à la trappe, Oct. 2011 (in French).

Crossrefs

Cf. A034837, A178157.

Programs

MATLAB

m=1;
for u=10:700 digit=dec2base(u,10)-'0';
   if digit(length(digit))~=0 aa=str2num(strrep(num2str(digit), ' ', ''));
      digit(2)=[]; a=str2num(strrep(num2str(digit), ' ', ''));
if mod(aa,a)==0 sol(m)=u;  m=m+1;  end; end; end;
sol % Marius A. Burtea, May 16 2019

Mathematica

Select[Range[700], With[{m = #}, And[Mod[#, 10] != 0, AnyTrue[FromDigits@ Delete[IntegerDigits[m], #] & /@ Range[2, IntegerLength@ m], Mod[m, #] == 0 &]]] &] (* Michael De Vlieger, Jun 09 2019 *)

PARI

isok(m) = {if (m % 10, my(d=digits(m)); for (k=2, #d, mk = fromdigits(vector(#d-1, i, if (iMichel Marcus, Jun 21 2019