Create a divisors :: Int -> [Int] function that returns the divisors of x.
I've successfully made a divides function that returns True if y is a divisor of x.
divides x y | mod x y == 0 = True
| mod x y /= 0 = False
I've tried to use it to filter numbers from [1..n] , but can't exactly get a grasp on how the filter function works. Can anyone set me in the right direction?
divisors n = filter divides n [1..n]
You are in the right track. Only that your problem is not exactly how filter works, but how Haskell works.
divisors n = filter (divides n) [1..n]
The above will do the trick. See, filter takes two arguments, so does divisors. But you are giving it three arguments at filter divides n [1..n].
Btw,
divides x y | mod x y == 0 = True
| mod x y /= 0 = False
is semantically equivalent to
divides x y = mod x y == 0
and operationally it doesn't repeat the mod calculation and the test mod x y /= 0.
filter takes as its second argument a list to filter - here your [1..n] - and as first argument a function which takes one of those list elements as input and returns a boolean.
So think about what this function would be, in terms of your pre-defined function divides. Since, as you have it, divides x y actually means "y is a divisor of x", you can easily express this with a lambda function:
divisors n = filter (\m -> divides n m) [1..n]
which can be further simplified by "eta-reduction" (or coloquially, "cancelling the m") to
divisors n = filter (divides n) [1..n]