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I would like to produce elements (preferably by Table or For-If commands) that they belong to specific sets. For example I want to write positive integers from 1 to 100 such that they are $2 \pmod 3$, $3,\, 5$ or $6 \pmod 7$ or $7,\,8$ or $10 \pmod {11}$.

I wrote the following code but it was not correct.

A1={2};
A2={3,5,6};
A3={7,8,10};

Union[Flatten[Table[If[(Element[Mod[i,3],A1]
||Element[Mod[i,7],A2]||Element[Mod[i, 11], A3]), i, 0], {i, 1, 100}]]]
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3 Answers 3

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You have a series of conditions you want to be true, so use Select,

A1 = {2};
A2 = {3, 5, 6};
A3 = {7, 8, 10}; Select[Range[100], 
 MemberQ[A1, Mod[#, 3]] || MemberQ[A2, Mod[#, 7]] || 
   MemberQ[A3, Mod[#, 11]] &]

(* {2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 17, 18, 19, 20, 21, \
23, 24, 26, 27, 29, 30, 31, 32, 33, 34, 35, 38, 40, 41, 43, 44, 45, \
47, 48, 50, 51, 52, 53, 54, 55, 56, 59, 61, 62, 63, 65, 66, 68, 69, \
71, 73, 74, 75, 76, 77, 80, 82, 83, 84, 85, 86, 87, 89, 90, 92, 94, \
95, 96, 97, 98} *)
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3
  • $\begingroup$ You know, if you keep this up, it will force me one step closer to being off the first page of users. Thanks. :) $\endgroup$
    – rcollyer
    Apr 15, 2016 at 19:09
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    $\begingroup$ @rcollyer Once I start at alpha in June I'm sure I won't be on here as much lol $\endgroup$
    – Jason B.
    Apr 15, 2016 at 20:06
  • $\begingroup$ I'm "vaguely" familiar with that. $\endgroup$
    – rcollyer
    Apr 15, 2016 at 20:11
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Using MemberQ I wrote the following code and it worked:

A1={2};
A2={3,5,6};
A3={7,8,10}
Union[Flatten[Table[If[(MemberQ[A1,Mod[i,3]]==True
||MemberQ[A2,Mod[i,7]]==True||MemberQ[A3,Mod[i,11]]==True),i,0], 
{i,1,100}]]]
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3
  • $\begingroup$ Here you have Mod[i,1] which is just 0 for every number. Also, your result has a 0, but you only want numbers from 1 to 100 $\endgroup$
    – Jason B.
    Apr 15, 2016 at 13:34
  • $\begingroup$ @JasonB, thanks for comment, it was 11 not 1, I corrected it. $\endgroup$
    – asad
    Apr 15, 2016 at 17:14
  • $\begingroup$ you don't need MemberQ[...]==True; just MemberQ[...] works fine. $\endgroup$
    – kglr
    Apr 21, 2016 at 0:31
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s1 = Union@Flatten[Position[Mod[Range[100], #], Alternatives @@ #2] & @@@ 
        {{3, A1}, {7, A2}, {11, A3}}];

s2 = Select[Range[100], MemberQ[A1, Mod[#, 3]] || MemberQ[A2, Mod[#, 7]] || 
 MemberQ[A3, Mod[#, 11]] &]; (* from Jason's answer *)

s1 == s2

True

Although speed is not the issue in this Q/A, surprisingly, for large lists, this is faster than the approach using Select or Pick:

Union@Flatten[Position[Mod[Range[100000], #], Alternatives @@ #2] & @@@
      {{3, A1}, {7, A2}, {11, A3}}, 2]; // AbsoluteTiming // First

0.128634

Pick[Range[100000],  MemberQ[A1, Mod[#, 3]] || MemberQ[A2, Mod[#, 7]] || 
       MemberQ[A3, Mod[#, 11]] & /@ Range[100000]]; // AbsoluteTiming // First

0.326279

Select[Range[100000], MemberQ[A1, Mod[#, 3]] || MemberQ[A2, Mod[#, 7]] || 
      MemberQ[A3, Mod[#, 11]] &]; // AbsoluteTiming // First

0.312955

And corrected version of OP's Table approach is somewhere in between:

Table[If[MemberQ[A1, Mod[i, 3]] || MemberQ[A2, Mod[i, 7]] || 
        MemberQ[A3, Mod[i, 11]], i, ## &[]], {i, 1, 100000}]; // AbsoluteTiming // First

0.288671

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