Project Euler: A (much) better way to solve problem #5?

Question

You probably know about project Euler question 5: get the smallest number divisble by all numbers 1 to 20.

The logic I applied was "start with the first number greater than the largest of the list(20) and also divisible by it which is 40" and stepsize of 20 (largest number)

I did this using list comprehension but it's pretty lame.

pe5 = head    [x|x<-[40,60..],x`mod`3==0,x`mod`4==0,x`mod`6==0,x`mod`7==0,x`mod`8==0,x`mod`9==0,x`mod`11==0,x`mod`12==0,x`mod`13==0,x`mod`14==0,x`mod`15==0,x`mod`16==0,x`mod`17==0,x`mod`18==0,x`mod`19==0] 

Can we do this better perhaps using zipWith and filter maybe?

Just to clarify, this is not a homework assignment. I'm doing this to wrap my brain around Haskell. (So far I'm losing!)

:Thanx all

I think this is a saner way (there may be thousand more better ways but this would suffice) to do it

listlcm'::(Integral a)=> [a] -> a
listlcm' [x] = x
listlcm' (x:xs) = lcm x (listlcm' xs)  

Answer 1

In this particular case, you can get it for free using foldl and lcm:

euler = foldl lcm 2 [3..20]

This gives me 232792560 instantaneously.

Answer 2

Since the spoiler has already been posted, I thought I'd explain how it works.

The smallest number divisible by two numbers is also known as the least common multiple of those numbers. There is a function for calculating this in the Prelude.

λ> lcm 10 12
60

Now, to extend this to multiple numbers we exploit the following property

lcm(a₁, ... a_n) = lcm(lcm(a₁, ... a_n-1), a_n)

In Haskell, f(f(... f(a₁, a₂), ...), a_n) can be written foldl1 f [a1, a2, ... an], so we can solve the problem with this simple one-liner:

λ> foldl1 lcm [1..20]
232792560

This finds the solution in a fraction of a second.