Project Euler Problem 1 - Multiples of 3 and 5
I think that any is your friend here.

aMultipleOf factors x = any (\f -&gt; x `rem` f == 0) factors
prob001 :: (Integral a) =&gt; [a] -&gt; [a] -&gt; a
prob001 nums factors = sum $ filter (aMultipleOf factors) nums