This is pretty easy if you understood that we live in a world of base 10 and the principles of division with rest:
n = 781049;
Rest@Reverse[Flatten@Last@Reap@FixedPointList[(Sow[Mod[#, 10]]; Quotient[#, 10]) &, n]]
(* {7, 8, 1, 0, 4, 9} *)
or as J.M. suggested with one call to get the quotient and the remainder
With[{n = 781049}, Reverse[Reap[NestList[Block[{q, r},
{q, r} = QuotientRemainder[#, 10]; Sow[r]; q] &, n, IntegerLength[n]]][[-1, 1]]]]
If you really prefer shorter methods, than you could go with
IntegerDigits[n]
or maybe
ToExpression@StringCases[ToString[n], DigitCharacter]
if you like strings. Without giving explicit timings you can safely assume, that every function will be slower than IntegerDigits.
IntegerDigits. – b.gates.you.know.what Apr 10 '13 at 11:39ToString,CharactersandToExpression, in this order. – István Zachar Apr 10 '13 at 11:47IntegerDigitsis really the way to go about this. – Daniel Lichtblau Apr 10 '13 at 13:37