Loop through a list and select elements

Question

I am new to Mathematica and I have no idea how to make a basic for loop work. I want to loop through the elements of a list: Range[45, 200]. I want to select the elements (i) for which the following conditions holds: if i mod 3 and i mod 8 and i mod 12 == 1 and i mod 5 == 0 I want to print the element. How would I achieve this? I want something like this:

for i in range(45, 201):
   if i%3==1 and i%8==1 and i%12==1 and i%5==0:
      print i

Of course in Mathematica language. Sorry if this is a dumb question but the Mathematica documentation is pretty hard to understand. Thanks.

Answer 1

You don't need to loop through the elements. Have a look at Select:

Select[Range[45, 200],Mod[#, 3] == Mod[#, 8] == Mod[#, 12] == 1 && Mod[#, 5] == 0 &]

(*

145

*)

Answer 2

If you only want to print the result :

  If[
   Mod[#, 3] == Mod[#, 8] == Mod[#, 12] == 1 && Mod[#, 5] == 0, 
   Print[#]
   ] & /@ Range[45, 200];

the same code without the /@ which may be mysterious for beginners :

Map[
  If[
    Mod[#, 3] == Mod[#, 8] == Mod[#, 12] == 1 && Mod[#, 5] == 0, 
    Print[#]
    ] & ,
   Range[45, 200]
  ];

"only print the result" means.. that you can' t affect the result to a variable.

If you want to affect a variable see sOrce' s answer and add variable = at the beginning of his code

Answer 3

Defining a selector function sF:

  sF = Mod[#, 3] == Mod[#, 8] == Mod[#, 12] == 1 && Mod[#, 5] == 0 &;

you can use any of the following functions to select the elements of Range[45,200] that satisfy the criteria coded in sF:

  Select[#, sF] &@Range[45, 200] (*as in @sOrce's answer *)
  Pick[#, sF /@ #] &[Range[45, 200]]
  Cases[#, x_ /; sF[x]] &@Range[45, 200]
  Cases[#, x_?sF] &@Range[45, 200]  (* thanks: @m_goldberg *)
  # /. (x_ /; ! sF[x] :> (## &[])) & /@ Range[45, 200] (* thanks: Mr.Wizard *)
  If[sF[#], #, ## &[]] & /@ Range[45, 200] (* a variant of @andre's answer *)
  #[[SparseArray[Boole /@ sF /@ #]["NonzeroPositions"][[1]]]] &[Range[45, 200]]
  (* 145 *)

(Related Q/A:1)