Project Euler problem #1 solved in APL

euler1.apl

⍝ This is my very first APL program,
⍝ made from scratch using an APL primer at 
⍝ http://aplwiki.com/LearnApl/BuiltInFunctions
⍝ Evaluated using http://tryapl.org/
⍝ Yields the sum of all multiples of 3 or 5 below 1000.
n←⍳999⋄a←+/n×0=3|n⋄b←+/n×0=5|n⋄c←+/n×0=15|n⋄a+b-c

⍝ Second version using a function definition..
f←{+/⍺×0=⍵|⍺}⋄n←⍳999⋄a←n f 3⋄b←n f 5⋄c←n f 15⋄a+b-c

⍝ Third version using a closure, kind of..
n←⍳999⋄f←{+/n×0=⍵|n}⋄a←f 3⋄b←f 5⋄c←f 15⋄a+b-c

⍝ ..or a bit simpler without the temporary variables..
n←⍳999⋄f←{+/n×0=⍵|n}⋄(f 3)+(f 5)-(f 15)

⍝ ..or maybe this is better?!
n←⍳999⋄f←{+/n×0=⍵|n}⋄(+/f¨ 3 5) - f 15

⍝ Inlining function is also possible :)
⍝ Doesn't make much sense in this case though.
n←⍳999⋄(+/{+/n×0=⍵|n}¨ 3 5) - {+/n×0=⍵|n} 15

⍝ A variation using indexing.., 
⍝ and now inlining the func makes more sense.
n←⍳999⋄m←{+/n×0=⍵|n}¨ 3 5 15⋄m[1]+m[2]-m[3]

⍝ Slightly more dense version of the latter..
n←⍳999⋄m←{+/n×0=⍵|n}¨ 3 5 15⋄(+/2↑m)-m[3]

⍝ Programming APL is actually quite fun \o/