Sum multiples of 3 and 5: Difference between revisions

// between first multiple and last

let sumMults = (n, factor) => {

 

return Math.floor(factor * n1 * (n1 + 1) / 2);

 

sum35 = (n) => sumMults(n, 3) + sumMults(n, 5) - sumMults(n, 15);