FizzBuzz program (details given) in Javascript

Question

Can someone please correct this code of mine for FizzBuzz? There seems to be a small mistake. This code below prints all the numbers instead of printing only numbers that are not divisible by 3 or 5.

Write a program that prints the numbers from 1 to 100. But for multiples of three, print "Fizz" instead of the number, and for the multiples of five, print "Buzz". For numbers which are multiples of both three and five, print "FizzBuzz".

function isDivisible(numa, num) {
  if (numa % num == 0) {
    return true;
  } else {
    return false;
  }
};

function by3(num) {
  if (isDivisible(num, 3)) {
    console.log("Fizz");
  } else {
    return false;
  }
};

function by5(num) {
  if (isDivisible(num, 5)) {
    console.log("Buzz");
  } else {
    return false;
  }
};

for (var a=1; a<=100; a++) {
  if (by3(a)) {
    by3(a);
    if (by5(a)) {
      by5(a);
      console.log("\n");
    } else {
      console.log("\n");
    }
  } else if (by5(a)) {
    by5(a);
    console.log("\n");
  } else {
    console.log(a+"\n")
  }
}

Answer 1

for (i=1; i<=100; i++) {
  output = "";
  if (i%5==0) output = "buzz"; 
  if (i%3==0) output = "fizz" + output;
  if (output=="") output = i;
  console.log(output);
}

Answer 2

check this out!

function fizzBuzz(){
    for(var i=1; i<=100; i++){
        if(i % 3 ===0 && i % 5===0){
            console.log(i+' fizzBuzz');
        } else if(i % 3 ===0){
            console.log(i+' fizz');
        } else if(i % 5 ===0){
            console.log(i+' buzz');
        } else {
            console.log(i);
        }
    } 
}fizzBuzz();

Answer 3

Functional style! JSBin Demo

// create a iterable array with a length of 100
// and map every value to a random number from 1 to a 100
var series = Array.apply(null, Array(100)).map(function() {
  return Math.round(Math.random() * 100) + 1;
});

// define the fizzbuzz function which takes an interger as input
// it evaluates the case expressions similar to Haskell's guards
var fizzbuzz = function (item) {
  switch (true) {
    case item % 15 === 0:
      console.log('fizzbuzz');
      break;
    case item % 3 === 0:
      console.log('fizz');
      break;
    case item % 5 === 0:
      console.log('buzz');
      break;
    default:
      console.log(item);
      break;
  }
};

// map the series values to the fizzbuzz function
series.map(fizzbuzz);

Answer 4

Just want to share my way to solve this

for (i = 1; i <= 100; i++){
  if (i % 3 === 0 && i % 5 === 0) {
    console.log('fizzBuzz');
  } else if (i % 3 === 0) {
    console.log('fizz');
  } else if (i % 5 === 0){
    console.log('buzz');
  } else {
    console.log(i);
  }
}

Answer 5

In case someone is looking for other solutions: This one is a pure, recursive, and reusable function with optionally customizable parameter values:

const fizzBuzz = (from = 1, till = 100, ruleMap = {
  3: "Fizz",
  5: "Buzz",
}) => from > till || console.log(
  Object.keys(ruleMap)
    .filter(number => from % number === 0)
    .map(number => ruleMap[number]).join("") || from
) || fizzBuzz(from + 1, till, ruleMap);

// Usage:
fizzBuzz(/*Default values*/);

• The from > till is the anchor to break the recursion. Since it returns false until from is higher than till, it goes to the next statement (console.log):
• Object.keys returns an array of object properties in the given ruleMap which are 3 and 5 by default in our case.
  • Then, it iterates through the numbers and returns only those which are divisible by the from (0 as rest).
  • Then, it iterates through the filtered numbers and outputs the saying according to the rule.
  • If, however, the filter method returned an empty array ([], no results found), it outputs just the current from value because the join method at the end finally returns just an empty string ("") which is a falsy value.
• Since console.log always returns undefined, it goes to the next statement and calls itself again incrementing the from value by 1.

Answer 6

var limit = prompt("Enter the number limit");
var n = parseInt(limit);
var series = 0;
for(i=1;i<n;i++){    
  series = series+" " +check();
}

function check() {
  var result;
  if (i%3==0 && i%5==0) {  // check whether the number is divisible by both 3 and 5
    result = "fizzbuzz "; // if so, return fizzbuzz
    return result;
  }
  else if (i%3==0) {  // check whether the number is divisible by 3 
    result = "fizz ";  // if so, return fizz
    return result;
  }
  else if (i%5==0) {  // check whether the number is divisible by 5
    result = "buzz ";  // if so, return buzz
    return result;
  }
  else return i;  // if all the above conditions fail, then return the number as it is
}

alert(series);