A245680 Numbers x whose digits can be permuted to produce a multiple of x.

1035, 1089, 1359, 1386, 1782, 2178, 2475, 10035, 10089, 10350, 10449, 10890, 10899, 10989, 11688, 11883, 12375, 12903, 13029, 13359, 13449, 13590, 13599, 13659, 13860, 13986, 14085, 14247, 14724, 14859, 15192, 16782, 17604, 17802, 17820, 17832, 17982, 18027

Offset: 1

Author

Paolo P. Lava, Jul 29 2014

Keywords

• nonn
• base

Comments

A008919 is a subset of this sequence.

Every element of the sequence is divisible by 3. - Emmanuel Vantieghem, Oct 27 2015

It is an obvious fact that if a(n) is the n-th term of the sequence, then a(n)*(10^k) is also a member of the sequence for all k > 0. - Altug Alkan, Nov 01 2015

Examples

			A permutation of 1782 is 7128 and 7128 / 1782 = 4.
A permutation of 11688 is 81816 and 81816 / 11688 = 7.
		

Links

• Paolo P. Lava, Table of n, a(n) for n = 1..100

Crossrefs

Cf. A008919, A096092, A096093, A245682.

Programs

Maple

P:=proc(q) local a, b, c, i, j, k, n, t; for n from 1 to q do a:=n; b:=[];
while a>0 do b:=[a mod 10, op(b)]; a:=trunc(a/10); od;
t:=0; for i from 2 to 9 do a:=i*n; c:=[];
while a>0 do c:=[a mod 10, op(c)]; a:=trunc(a/10); od;
if sort(b)=sort(c) then print(n); break; fi; od; od; end: P(10^6);
# Alternative:
N:= 100: # to get the first N entries
count:= 0:
for x from 10 while count < N do
  M:= 10^(ilog10(x)+1)-1;
  L:= sort(convert(x,base,10));
  for i from 2 to floor(M/x) do
    Lp:= sort(convert(i*x,base,10));
    if Lp = L then
      count:= count+1;
      A[count]:= x;
      break;
    fi
   od
od:
seq(A[i],i=1..count); # Robert Israel, Jul 29 2014

Mathematica

fQ[n_] := AnyTrue[Rest[FromDigits /@ Permutations[IntegerDigits@ n]], Divisible[#, n] &]; Select[Range@ 20000, fQ] (* Michael De Vlieger, Oct 27 2015, Version 10 *)

PARI

for(n=1,10^8,d=vecsort(digits(n));p=0;for(k=2,9,dd=vecsort(digits(n*k));if(d==dd,p++;break));if(p>0,print1(n,", "))) \\ quicker program Derek Orr, Jul 29 2014

Python

import itertools
from itertools import permutations
for n in range(1,10**5):
  plist = list(permutations(str(n)))
  for i in plist:
    num = ''
    for j in range(len(i)):
      num += i[j]
    if int(num)%n==0 and int(num)/n > 1:
      print(n,end=', ') # Derek Orr, Jul 29 2014