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I write a function to find the prime number , but it's not working because I'm not able to break the loop

function isPrime(n) {
if (n == 1 ) {
return (`${n} is niether a prime number nor composite number`)
}
else if( n < 1){
return (`${n} is not a prime number`)
}
else{
for (let i = 2; i < n; i++) {
if( n % 2 == 0){
return result = `${n} is not a prime number`
break;
}
else{
return result = `${n} is a prime number`
}
}
}
}
console.log(isPrime(15))
I write a function to find the prime number , but it's not working because I'm not able to break the loop

Although answer by Dave addresses the question I would suggest this stack overflow answer as an better alternative.
https://stackoverflow.com/a/38643868/10738743

There are two glitches preventing your function working as intended.
the loop which tests for primeness is repeating n%2 on each iteration, which will evaluate the same, regardless of the value of i in the loop. Instead, you should be testing to see whether n%i is true.
With that fixed, the final return (which only happens if all non-prime conditions have been rejected with earlier returns) should be outside of any block as it becomes the default return value.
your code functions fine with the minor changes suggested above incorporated:
function isPrime(n) {
if (n == 1 ) {
return (`${n} is niether a prime number nor composite number`);
}
else if( n < 1){
return (`${n} is not a prime number`)
}
else {
for (let i = 2; i < n; i++) {
if( n % i == 0){
return result = `${n} is not a prime number`
// break; // (return breaks anyway)
} // end if
}// end for i;
} // end else
return result = `${n} is a prime number`; // default after no earlier rejection
}
console.log(isPrime(15));

isPrime function logs 15 as true (15 is not a prime number)

function isPrime(num){
// loop numbers to see if any num is divisble by that number
if (num < 2){ return false
} else if (num === 2) {
return true
}
for (let i = 2; i < num; i++) {
if (num % i === 0) {
return false
} else {
return true
}
}
}
Tried changing the conditional of the for loop but still comes out as true.

Wait until the whole loop finishes before returning true, otherwise you only check the first iteration and make a decision right away whether the input is prime before checking all the rest of the numbers. Also surely you don't need to iterate the entire length of the number, Math.ceil(num/2) + 1 should suffice. This is because there's no point to ever check if anything larger than half of a number will factor into that number. (This will make a big impact on larger numbers)
function checkPrime(){
var num = document.getElementById("num").value;
var result = document.getElementById("result");
var arr=[];
if (num.length===0)
result.innerHTML = "Please, specify a number.";
else if (num<0)
result.innerHTML = "Please, specify a positive number.";
else
result.innerHTML = isPrime(num).toString();
return false;
}
function isPrime(num){
// loop numbers to see if any num is divisble by that number
if (num < 2)
return false;
else if (num === 2)
return true;
const max = Math.ceil(num/2) + 1;
for (let i = 2; i < max; i++) {
if (num % i === 0)
return false;
}
return true;
}
<form onsubmit=" return checkPrime()">
<label>Enter a number check prime</label>
<input type="number" id="num" name="num"><br>
<input type="submit" value="Submit" id="submit">
<div id="result"></div>
</form>

This means you are testing any number entered to return a bool (true/false) if it is a prime number.
I would modify your code a little bit to give this function a correct answer.
function isPrime(num){
// loop numbers to see if any num is divisble by that number
if (num < 2){
return false
}else if(num==2){
return true;
}else{
for (let i = 2; i < num; i++) {
if (num % i == 0) {
return false;
//break;
}
}
return true;
}
}

Printing 100 to 200, with three exceptions?

I am trying to write a program that prints the numbers from 100 to 200, with three exceptions:
If the number is a multiple of 3, the string "yes" should be returned instead of the number.
If the number is a multiple of 4, the string "yes and yes" instead of the number should be returned.
If the number is a multiple of both 3 and 4, the string "yes, yes and yes" instead of the number.
I am new to JavaScript so I try to do this step by step.
I wrote this code to print the numbers from 100 to 200:
function hundredTwoHundred() {
result = [];
for (let i = 100; i <= 200; i++) {
result.push(i);
}
return result;
}
console.log(hundredTwoHundred());
Then I tried to use else/if for the exceptions:
function hundredTwoHundred() {
result = [];
for (let i = 100; i <= 200; i++) {
if (i % 3 == 0) {
console.log("yes");
} else if (i % 4 == 0) {
console.log("yes and yes")
} else if (i % 3 == 0 && i % 4 == 0) {
console.log("yes, yes and yes");
} else {
result.push(i)
}
}
return result;
}
console.log(hundredTwoHundred());
The code of course, does not work. I have tried moving result.push(i) around, but I don't want to just mindlessly move things around, without knowing the reasoning behind it.
How do I use conditional operators to find these exceptions? What am I doing wrong?
Thank you.

You need to test if the number is (divisible by 3 and divisible by 4) before checking whether it's (individually) divisible by 3 or 4, otherwise the first condition if (i % 3 == 0) will evaluate to true and you'll get yes rather than yes, yes and yes. You should also push to the result in the conditionals rather than console.logging in the conditionals, since you want to create an array of numbers and yeses and then console.log the whole constructed array afterwards.
Also make sure to declare the result with const (or var, for ES5) - it's not good to implicitly create global variables.
Also, although it doesn't matter in this case, when comparing, it's good to rely on === by default rather than == - best to only use == when you deliberately want to rely on implicit type coercion, which can result in confusing behavior.
function hundredTwoHundred() {
const result = [];
for (let i = 100; i <= 200; i++) {
if (i % 3 === 0 && i % 4 === 0) {
result.push("yes, yes and yes");
} else if (i % 3 === 0) {
result.push("yes");
} else if (i % 4 === 0) {
result.push("yes and yes")
} else {
result.push(i)
}
}
return result;
}
console.log(hundredTwoHundred());

If a number is a multiple of 3 and 4, then it is a multiple of 12. I’d also use a switch statement, so you can rewrite as follow:
for (let i = 100; i <= 200; i = i + 1) {
switch (0) {
case i % 12: console.log('yes, yes and yes'); break;
case i % 4: console.log('yes and yes'); break;
case i % 3: console.log('yes'); break;
default: console.log(i);
}
}
If you want it as an array:
// Fill an array with numbers from 100 to 200
const arr = Array(101).fill().map((_, i) => i + 100);
// Map it to numbers and strings
const hundredTwoHundred = arr.map(i => {
switch (0) {
case i % 12: return 'yes, yes and yes';
case i % 4: return 'yes and yes';
case i % 3: return 'yes';
default: return i
}
});
// Print it:
console.log(hundredTwoHundred);

When you have a complex set of conditions you need to be careful with the order in which you evaluate them.
function logExceptions(start, end) {
var divisibleByThree, divisibleByFour;
for (var i = start; i <= end; ++i) {
divisibleByThree = i % 3 == 0;
divisibleByFour = i % 4 == 0;
if (divisibleByThree && divisibleByFour) {
console.log("yes, yes and yes");
}
else if (divisibleByThree) {
console.log("yes");
}
else if (divisibleByFour) {
console.log("yes and yes");
}
else {
console.log(i);
}
}
}
logExceptions(100, 200);
If you want to save the result in an array and only later print it:
function logExceptions(start, end) {
var result = [];
var divisibleByThree, divisibleByFour;
for (var i = start; i <= end; ++i) {
divisibleByThree = i % 3 == 0;
divisibleByFour = i % 4 == 0;
if (divisibleByThree && divisibleByFour) {
result.push("yes, yes and yes");
}
else if (divisibleByThree) {
result.push("yes");
}
else if (divisibleByFour) {
result.push("yes and yes");
}
else {
result.push(i);
}
}
return result;
}
console.log(logExceptions(100, 200).toString());

Getting Boolean values if a number is a prime number in JavaScript

I am trying to solve this problem where I need to print true or false on the console depending on the number (true if a number is Prime and false if it is not). I seem to have the solution but I know there is a mistake somewhere because I get different answers if I change the boolean values of the 2 isPrime variables.
Here is the question:
Return true if num is prime, otherwise return false.
Hint: a prime number is only evenly divisible by itself and 1
Hint 2: you can solve this using a for loop.
Note: 0 and 1 are NOT considered prime numbers
Input Example:
1
7
11
15
20
Output Example:
false
true
true
false
false
My code:
function isPrime(num){
let isPrime = '';
for(let i = 2; i <= Math.sqrt(num); i++){
if(num % i === 0){
isPrime = false;
} else {
isPrime = true;
}
}
return isPrime;
}
isPrime(11);

You are very close. I will build on top of your solution. You need to return false as soon as you detect that the number is not prime. This avoids us making additional redundant calculations.
You also did not take into account the first hint.
Hint - Note: 0 and 1 are NOT considered prime numbers
For this you can simply return false if number is 0 or 1.
function isPrime(num) {
if (num === 0 || num === 1) return false;
for (let i = 2; i <= Math.sqrt(num); i++) {
if (num % i === 0) return false;
}
return true;
}
Also in your code you are using a variable isPrime to keep track of boolean values while initialising it with an empty string. This is not correct. Javascript allows this because it's weakly typed, but this is misleading. So in future if you define a variable for boolean value don't initialise it with a string ;)
To address your comment further. If you do want to stick with having a flag in your code, then you can initialise isPrime with true instead of an empty string.
That way you can get rid of the else part in the for loop, making the code shorter. I used code provided by #Commercial Suicide (by the way he did say that there are more elegant solutions) as an example:
function isPrime(num) {
let flag = true;
let isPrime = true;
if (num === 0 || num === 1) return false;
if (num === 2) return true;
for (let i = 2; i <= Math.sqrt(num); i++) {
if (flag) {
if(num % i === 0) {
isPrime = false;
flag = false;
}
}
}
return isPrime;
}
const numbers = [1, 7, 11, 15, 20];
const booleans = [false, true, true, false, false];
numbers.forEach((item, i) => {
if (isPrime(item) === booleans[i]) {
console.log("CORRECT");
} else {
console.log("WRONG");
}
})
You can then go a step further and remove the flag variable.
function isPrime(num) {
let isPrime = true;
if (num === 0 || num === 1) return false;
if (num === 2) return true;
for (let i = 2; i <= Math.sqrt(num); i++) {
if (isPrime) {
if(num % i === 0) {
isPrime = false;
} else {
isPrime = true;
}
}
}
return isPrime;
}
But now it should become even more clear that those flags are actually redundant, as I pointed out initially. Simply return false as soon as you detect that the number is not prime to avoid further unnecessary loop runs. From my experience in javascript over the years I can see that we are heading towards functional programming mindset. It helps to avoid variable mutations generally.

It's because your isPrime variable can be overriding many times, simple flag (for example) will prevent this issue. I have also added check for 0, 1 and 2. There are more elegant ways to do that, but I wanted to keep your logic, here is an example with some tests:
function isPrime(num) {
let flag = true;
let isPrime = '';
if (num === 0 || num === 1) return false;
if (num === 2) return true;
for (let i = 2; i <= Math.sqrt(num); i++) {
if (flag) {
if(num % i === 0) {
isPrime = false;
flag = false;
} else {
isPrime = true;
}
}
}
return isPrime;
}
const numbers = [1, 7, 11, 15, 20];
const booleans = [false, true, true, false, false];
numbers.forEach((item, i) => {
if (isPrime(item) === booleans[i]) {
console.log("CORRECT");
} else {
console.log("WRONG");
}
})

FizzBuzz program (details given) in Javascript [closed]

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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")
}
}

for (let i = 1; i <= 100; i++) {
let out = '';
if (i % 3 === 0) out += 'Fizz';
if (i % 5 === 0) out += 'Buzz';
console.log(out || i);
}

/*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”*/
var str="",x,y,a;
for (a=1;a<=100;a++)
{
x = a%3 ==0;
y = a%5 ==0;
if(x)
{
str+="fizz"
}
if (y)
{
str+="buzz"
}
if (!(x||y))
{
str+=a;
}
str+="\n"
}
console.log(str);
Your functions return falsy values no matter what, but will print anyway. No need to make this overly complicated.
fiddle: http://jsfiddle.net/ben336/7c9KN/

Was fooling around with FizzBuzz and JavaScript as comparison to C#.
Here's my version, heavily influenced by more rigid languages:
function FizzBuzz(aTarget) {
for (var i = 1; i <= aTarget; i++) {
var result = "";
if (i%3 === 0) result += "Fizz";
if (i%5 === 0) result += "Buzz";
if (result.length ===0) result = i;
console.log(result);
}
}
I like the structure and ease of read.
Now, what Trevor Dixon cleverly did is relay on the false-y values of the language (false , null , undefined , '' (the empty string) , 0 and NaN (Not a Number)) to shorten the code.
Now, the if (result.length ===0) result = i; line is redundant and the code will look like:
function FizzBuzz(aTarget) {
for (var i = 1; i <= aTarget; i++) {
var result = "";
if (i%3 === 0) result += "Fizz";
if (i%5 === 0) result += "Buzz";
console.log(result || i);
}
}
Here we relay on the || operator to say : "if result is false, print the iteration value (i)". Cool trick, and I guess I need to play more with JavaScript in order to assimilate this logic.
You can see other examples (from GitHub) that will range from things like :
for (var i=1; i <= 20; i++)
{
if (i % 15 == 0)
console.log("FizzBuzz");
else if (i % 3 == 0)
console.log("Fizz");
else if (i % 5 == 0)
console.log("Buzz");
else
console.log(i);
}
No variables here, and just check for division by 15,3 & 5 (my above one only divides by 3 & 5, but has an extra variable, so I guess it's down to microbenchmarking for those who care, or style preferences).
To:
for(i=0;i<100;)console.log((++i%3?'':'Fizz')+(i%5?'':'Buzz')||i)
Which does it all in on line, relaying on the fact that 0 is a false value, so you can use that for the if-else shorthanded version (? :), in addition to the || trick we've seen before.
Here's a more readable version of the above, with some variables:
for (var i = 1; i <= 100; i++) {
var f = i % 3 == 0, b = i % 5 == 0;
console.log(f ? b ? "FizzBuzz" : "Fizz" : b ? "Buzz" : i);
}
All in all, you can do it in different ways, and I hope you picked up some nifty tips for use in JavaScript :)

.fizz and .buzz could be CSS classes, no? In which case:
var n = 0;
var b = document.querySelector("output");
window.setInterval(function () {
n++;
b.classList[n%3 ? "remove" : "add"]("fizz");
b.classList[n%5 ? "remove" : "add"]("buzz");
b.textContent = n;
}, 500);
output.fizz:after {
content: " fizz";
color:red;
}
output.buzz:after {
content: " buzz";
color:blue;
}
output.fizz.buzz:after {
content: " fizzbuzz";
color:magenta;
}
<output>0</output>

With ternary operator it is much simple:
for (var i = 0; i <= 100; i++) {
str = (i % 5 == 0 && i % 3 == 0) ? "FizzBuzz" : (i % 3 == 0 ? "Fizz" : (i % 5 == 0) ? "Buzz" : i);
console.log(str);
}

for(i = 1; i < 101; i++) {
if(i % 3 === 0) {
if(i % 5 === 0) {
console.log("FizzBuzz");
}
else {
console.log("Fizz");
}
}
else if(i % 5 === 0) {
console.log("Buzz");
}
else {
console.log(i)
}
}

In your by3 and by5 functions, you implicitly return undefined if it is applicable and false if it's not applicable, but your if statement is testing as if it returned true or false. Return true explicitly if it is applicable so your if statement picks it up.

As an ES6 generator: http://www.es6fiddle.net/i9lhnt2v/
function* FizzBuzz() {
let index = 0;
while (true) {
let value = ''; index++;
if (index % 3 === 0) value += 'Fizz';
if (index % 5 === 0) value += 'Buzz';
yield value || index;
}
}
let fb = FizzBuzz();
for (let index = 0; index < 100; index++) {
console.log(fb.next().value);
}

Codeacademy sprang a FizzBuzz on me tonight. I had a vague memory that it was "a thing" so I did this. Not the best way, perhaps, but different from the above:
var data = {
Fizz:3,
Buzz:5
};
for (var i=1;i<=100;i++) {
var value = '';
for (var k in data) {
value += i%data[k]?'':k;
}
console.log(value?value:i);
}
It relies on data rather than code. I think that if there is an advantage to this approach, it is that you can go FizzBuzzBing 3 5 7 or further without adding additional logic, provided that you assign the object elements in the order your rules specify. For example:
var data = {
Fizz:3,
Buzz:5,
Bing:7,
Boom:11,
Zing:13
};
for (var i=1;i<=1000;i++) {
var value = '';
for (var k in data) {
value += i%data[k]?'':k;
}
console.log(value?value:i);
}

This is what I wrote:
for (var num = 1; num<101; num = num + 1) {
if (num % 5 == 0 && num % 3 == 0) {
console.log("FizzBuzz");
}
else if (num % 5 == 0) {
console.log("Buzz");
}
else if (num % 3 == 0) {
console.log("Fizz");
}
else {
console.log(num);
}
}

for (var 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);
}
One of the easiest way to FizzBuzz.
Multiple of 3 and 5, at the same time, means multiple of 15.
Second version:
for (var i = 1; i <= 100; i++) {
if (i % 15 === 0) console.log("FizzBuzz");
else if (i%3 === 0) console.log("Fizz");
else if (i%5 === 0) console.log("Buzz");
else console.log(i);
}

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.

A Functional version of FizzBuzz
const dot = (a,b) => x => a(b(x));
const id = x => x;
function fizzbuzz(n){
const f = (N, m) => n % N ? id : x => _ => m + x('');
return dot(f(3, 'fizz'), f(5, 'buzz')) (id) (n);
}
for more options in the above replace dot with dots as below
const dots = (...a) => f0 => a.reduceRight((acc, f) => f(acc), f0);
function fizzbuzz(n){
const f = (N, m) => n % N ? id : x => _ => m + x('');
return dots(f(3, 'fizz'), f(5, 'buzz'), f(7, 'bam')) (id) (n);
}
Reference: FizzBuzz in Haskell by Embedding a Domain-Specific Language
by Maciej Piro ́g

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);
}

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);

Another solution, avoiding excess divisions and eliminating excess spaces between "Fizz" and "Buzz":
var num = 1;
var FIZZ = 3; // why not make this easily modded?
var BUZZ = 5; // ditto
var UPTO = 100; // ditto
// and easily extended to other effervescent sounds
while (num < UPTO)
{
var flag = false;
if (num % FIZZ == 0) { document.write ("Fizz"); flag = true; }
if (num % BUZZ == 0) { document.write ("Buzz"); flag = true; }
if (flag == false) { document.write (num); }
document.write ("<br>");
num += 1;
}
If you're using using jscript/jsc/.net, use Console.Write(). If you're using using Node.js, use process.stdout.write(). Unfortunately, console.log() appends newlines and ignores backspaces, so it's unusable for this purpose. You could also probably append to a string and print it. (I'm a complete n00b, but I think (ok, hope) I've been reasonably thorough.)
"Whaddya think, sirs?"

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();

Slightly different implementation.
You can put your own argument into the function. Can be non-sequential numbers like [0, 3, 10, 1, 4]. The default set is only from 1-15.
function fizzbuzz (set) {
var set = set ? set : [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
var isValidSet = set.map((element) => {if (typeof element !== 'number') {return false} else return true}).indexOf(false) === -1 ? true : false
var gotFizz = (n) => {if (n % 3 === 0) {return true} else return false}
var gotBuzz = (n) => {if (n % 5 === 0) {return true} else return false}
if (!Array.isArray(set)) return new Error('First argument must an array with "Number" elements')
if (!isValidSet) return new Error('The elements of the first argument must all be "Numbers"')
set.forEach((n) => {
if (gotFizz(n) && gotBuzz(n)) return console.log('fizzbuzz')
if (gotFizz(n)) return console.log('fizz')
if (gotBuzz(n)) return console.log('buzz')
else return console.log(n)
})
}

var num = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20];
var runLoop = function() {
for (var i = 1; i<=num.length; i++) {
if (i % 5 === 0 && i % 3 === 0) {
console.log("FizzBuzz");
}
else if (i % 5 === 0) {
console.log("Buzz");
}
else if (i % 3 === 0) {
console.log("Fizz");
}
else {
console.log(i);
}
}
};
runLoop();

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);
}
}

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);

Thats How i did it :
Not the best code but that did the trick
var numbers = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20];
for(var i = 0 ; i <= 19 ; i++){
var fizz = numbers[i] % 3 === 0;
var buzz = numbers[i] % 5 === 0;
var fizzBuzz = numbers[i] % 5 === 0 && numbers[i] % 3 === 0;
if(fizzBuzz){
console.log("FizzBuzz");
} else if(fizz){
console.log("Fizz");
} else if(buzz){
console.log("Buzz");
} else {
console.log(numbers[i]);
}
}

As much as this is easy logic it can be a daunting task for beginners. Below is my solution to the FizzBuzz problem:
let i = 1;
while(i<=100){
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);
}
i++;
}

considering performance and readability, please find my take on this problem
way 1: instead of doing a math modules operation in an if loop, which results in performing 3 times taking it a step above reduces the overhead
function fizzBuzz(n) {
let count =0;
let x = 0;
let y = 0;
while(n!==count)
{
count++;
x = count%3;
y = count%5;
if(x === 0 && y ===0)
{
console.log("fizzbuzz");
}
else if(x === 0)
{
console.log("fizz");
}
else if(y === 0)
{
console.log("buzz");
}
else
{
console.log(count);
}
}
}
fizzBuzz(15);
way 2: condensing the solution
function fizzBuzz(n) {
let x = 0;
let y = 0;
for (var i = 1; i <= n; i++) {
var result = "";
x = i%3;
y = i%5;
if (x === 0 && y === 0) result += "fizzbuzz";
else if (x === 0) result += "fizz";
else if (y === 0) result += "buzz";
console.log(result || i);
}
}
fizzBuzz(5)

Here's my favorite solution. Succinct, functional & fast.
const oneToOneHundred = Array.from({ length: 100 }, (_, i) => i + 1);
const fizzBuzz = (n) => {
if (n % 15 === 0) return 'FizzBuzz';
if (n % 3 === 0) return 'Fizz';
if (n % 5 === 0) return 'Buzz';
return n;
};
console.log(oneToOneHundred.map((i) => fizzBuzz(i)).join('\n'));

function fizzBuzz(n) {
for (let i = 1; i < n + 1; i++) {
if (i % 15 == 0) {
console.log("fizzbuzz");
} else if (i % 3 == 0) {
console.log("fizz");
} else if (i % 5 == 0) {
console.log("buzz");
} else {
console.log(i);
}
}
}
fizzBuzz(15);

Different functional style -- naive
fbRule = function(x,y,f,b,z){return function(z){return (z % (x*y) == 0 ? f+b: (z % x == 0 ? f : (z % y == 0 ? b: z))) }}
range = function(n){return Array.apply(null, Array(n)).map(function (_, i) {return i+1;});}
range(100).map(fbRule(3,5, "fizz", "buzz"))
or, to incorporate structures as in above example: ie [[3, "fizz"],[5, "buzz"], ...]
fbRule = function(fbArr,z){
return function(z){
var ed = fbArr.reduce(function(sum, unit){return z%unit[0] === 0 ? sum.concat(unit[1]) : sum }, [] )
return ed.length>0 ? ed.join("") : z
}
}
range = function(n){return Array.apply(null, Array(n)).map(function (_, i) {return i+1;});}
range(100).map(fbRule([[3, "fizz"],[5, "buzz"]]))
OR, use ramda [from https://codereview.stackexchange.com/questions/108449/fizzbuzz-in-javascript-using-ramda ]
var divisibleBy = R.curry(R.compose(R.equals(0), R.flip(R.modulo)))
var fizzbuzz = R.map(R.cond([
[R.both(divisibleBy(3), divisibleBy(5)), R.always('FizzBuzz')],
[divisibleBy(3), R.aklways('Fizz')],
[divisibleBy(5), R.always('Buzz')],
[R.T, R.identity]
]));
console.log(fizzbuzz(R.range(1,101)))