Additional solution for Project Euler: Smallest multiple

Contributors
#1

What is your hint or solution suggestion?

This can be done using ES6. Use the reduce() method to find the GCD of all numbers in the sequence. The LCM can then be found as n!/gcd using the reducer.

Solution 1 
function smallestMult(n) {

  const gcd = (a, b) => a ? gcd(b % a, a) : b;

  const lcm = (a, b) => a * b / gcd(a, b);

  return [...Array(n+1).keys()].slice(1).reduce(lcm);
}

Challenge: Project Euler: Problem 5: Smallest multiple

Link to the challenge:

#2