Minimal readable Fizz Buzz test solution in JavaScript

fizzbuzz.js

// This function isolates the logic
// Its param is number `i`
// It returns 'Fizz', 'Bazz', 'FizzBuzz' or the number
// (This implementation takes advatage of some JavaScript specific features)
const toFizzBuzz = (i) =>
  // We construct a string

  // First part is 'Fizz' if the number is divisible by 3
  // and an empty string otherwise
  // `i%3` is 0 iff `i` is divisible by 3
  // 0 is a falsy value
  // condition ? truthy (= here other than 0) : falsy (= here 0)
  (i%3 ? '' : 'Fizz')

  // Second part is 'Buzz' if the number is divisible by 5
  // and an empty string otherwise
  + (i%5 ? '' : 'Buzz')

  // The constructed string is either
  // 1) Non-empty: The number is divisible by 3, 5, or both.
  //    We want to return the string
  // 2) Empty: The number is not divisible by 3 or 5
  //    We want to return the number `i`

  // The logical OR (||) operator
  // returns the expresion before `||` if this expression is truthy
  // otherwise returns the expression after `||`.
  // Empty string is falsy value, non-empty string is truthy
  || i

// Create an array of 100 items
Array(100)
  // Convert to a dense array so we can iterate it with `map` and `forEach`
  .fill()
  // The array is now a hundred times `undefined`,
  // Turn it into a sequence 0..99 by taking the position of an element `i`
  // Also shift by one so the result is 1..100
  .map((_, i) => i + 1)
  // Transform to FizzBuzz
  .map(toFizzBuzz)
  // Write each item to the output
  .forEach((i) => console.log(i))