pe.1.js

//jshint esnext:true
/*
Multiples of 3 and 5
Problem 1

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.
*/

// getMultiples :: Integer -> Integer -> [Integer]
const getMultiples = R.curry((max, mult) => R.compose(R.map(R.multiply(mult)),
                                                      R.range(1))(parseInt(max / mult) + 1))

// sumOfMultiples :: [Integer] -> Integer
const sumOfMultiples = R.compose(R.reduce(R.add, 0),
                 R.uniq,
                 R.apply(R.concat),
                 R.map(getMultiples(1000)));

console.log(sumOfMultiples([5,3]));