by January 12, 2012on
This is my start of hopefully a series of posts on solving the problems of Project Euler. So let’s start with the first problem:
The problem’s description is as following:
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
I am pretty sure there are more elegant solutions to this fairly simple problem, but this is my first shot:
solution :: Int solution = sum multiples where multiples = [x | x <- [1..999], x `mod` 3 == 0 || x `mod` 5 == 0]