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Logical operator with different types

I'm trying to solve the famous FizzBuzz quiz but I decided to use the logical operator or instead of else to provide fullback.
for (var num = 1; num <= 100; num++) {
var output;
if (num % 5 === 0 && num % 3 === 0) {
output = "FizzBuzz";
} else if (num % 5 === 0) {
output = "Buzz";
} else if (num % 3 === 0) {
output = "Fizz";
}
console.log(output || num);
}
This was supposed to print all the numbers from 1 to 100, with some exceptions. For numbers divisible by 3, print "Fizz" instead of the number, and for numbers divisible by 5, print "Buzz" instead and "FizzBuzz", for numbers that are divisible by both 3 and 5.
But it doesn't print any numbers.

The output declaration could be num like :
var output = num;
So you don"t have to use the || operator and just print the output directly :
console.log(output);
for (var num = 1; num <= 100; num++) {
var output = num;
if (num % 5 === 0 && num % 3 === 0) {
output = "FizzBuzz";
} else if (num % 5 === 0) {
output = "Buzz";
} else if (num % 3 === 0) {
output = "Fizz";
}
console.log(output);
}

I will say that Zakaria's answer is correct, but for exposure's sake, here is my answer
for (var i = 1; i <= 100; i++) {
var output = "";
if (!(i % 3)) output += "Fizz";
if (!(i % 5)) output += "Buzz";
console.log(output || i);
}
My logic here:
set the output value to be equal to "", which evaluates to a falsey value.
If a number is divisible by 3, then i % 3 will be 0, this is also a falsey value so we flip it by using the ! operator. Assume that i=9, then ! (i%3) = !(9%3) = !(0) = !(false) = true.
Therefore, if !(i%3) becomes true we append our empty string with "Fizz", then we use the same sort of logic for i%5, but instead appending "Buzz"
Note the order of these two if statements is important -- flip them around and you'll get BuzzFizz instead of FizzBuzz.
If output is not the empty string we set it to originally, output || i will return the value of output, giving us "Fizz", "Buzz", or "FizzBuzz" depending.
If ouput is empty, then output || i will return the value for i

Use let to fix the output's scope:
for (var num = 1; num <= 100; num++) {
let output;
if (num % 5 === 0 && num % 3 === 0) {
output = "FizzBuzz";
} else if (num % 5 === 0) {
output = "Buzz";
} else if (num % 3 === 0) {
output = "Fizz";
}
console.log(output || num);
}
Also, the || could be removed if you initialize output with num:
for (var num = 1; num <= 100; num++) {
let output = num;
if (num % 5 === 0 && num % 3 === 0) {
output = "FizzBuzz";
} else if (num % 5 === 0) {
output = "Buzz";
} else if (num % 3 === 0) {
output = "Fizz";
}
console.log(output);
}

Greatest Prime Factor

I'm trying to complete an algorithm challenge to find the largest prime factor of 600851475143. I'm not necessarily asking for the answer. Just trying to figure out why this code isn't working. Why does it return 'undefined' instead of a number?
let isPrime = n => {
let div = n - 1;
while (div > 1) {
if (n % div == 0) return false;
div--;
}
return true;
};
let primeFactor = x => {
for (let i = Math.floor(x / 2); i > 1; i--) {
if (x % i == 0 && isPrime(i) == true) {
return i;
}
}
};
console.log(primeFactor(35)); // 7
console.log(primeFactor(13195)); // 29
console.log(primeFactor(600851475143)); // undefined

The problem is not your algorithm it is perfectly valid, check the below slightly modified algorithm, all I've done is replaced your starting point Math.floor(x/2) with a parameter that you can choose:
let isPrime = n => {
let div = n - 1;
while (div > 1) {
if (n % div == 0) return false;
div--;
}
return true;
};
function primeFactor(x, n){
for (let i = n; i > 1; i--) {
if (x % i == 0 && isPrime(i) == true) {
return i;
}
}
}
console.log(primeFactor(35, 35));
console.log(primeFactor(13195, 13195));
console.log(primeFactor(600851475143, 100000))
Using the above you'll get an answer that proves your implementation works, but the loop is too big to do the entire thing(i.e. from Math.floor(600851475143/2)). Say your computer can do 500million loops per second, going through every one from 300,425,737,571 down to 1 would take 167 hours, even at 5 billion loops per second it would take 16 and a half hours. Your method is extremely inefficient but will return the correct answer. The reason you're not getting an answer on JSBin is more likely to do with browser/service limitations.
Spoilers on more efficient solution below
The following implementation uses a prime sieve(Sieve of Eratosthenes) in order to generate any list of primes requested and then checks if they fully factor into the given number, as long as you use a large enough list of primes, this will work exactly as intended. it should be noted that because it generates a large list of primes it can take some time if ran incorrectly, a single list of primes should be generated and used for all calls below, and the cached list of primes will pay off eventually by having to perform less calculations later on:
function genPrimes(n){
primes = new Uint32Array(n+1);
primes.fill(1)
for(var i = 2; i < Math.sqrt(n); i++){
if(primes[i]){
for(var j = 2*i; j < n; j+=i){
primes[j] = 0;
}
}
}
primeVals = []
for(var i = 2; i < primes.length; i++){
if(primes[i]){
primeVals.push(i);
}
}
return primeVals;
}
function primeFactor(x, primes){
var c = x < primes.length ? x : primes.length
for (var i = c; i > 1; i--) {
if(x % primes[i] == 0){
return primes[i];
}
}
}
primes = genPrimes(15487457);
console.log(primeFactor(35, primes));
console.log(primeFactor(13195, primes));
console.log(primeFactor(600851475143, primes));
console.log(primeFactor(30974914,primes));

let primeFactor = x => {
if (x === 1 || x === 2) {
return x;
}
while (x % 2 === 0) {
x /= 2;
}
if (x === 1) {
return 2;
}
let max = 0;
for (let i = 3; i <= Math.sqrt(x); i += 2) {
while (x % i === 0) {
x /= i;
max = Math.max(i, max);
}
}
if (x > 2) {
max = Math.max(x, max);
}
return max;
};
console.log(primeFactor(35));
console.log(primeFactor(13195));
console.log(primeFactor(27));
console.log(primeFactor(1024));
console.log(primeFactor(30974914));
console.log(primeFactor(600851475143));
Optimizations
Dividing the number by 2 until it's odd since no even number is prime.
The iteration increment is 2 rather than 1 to skip all even numbers.
The iteration stops at sqrt(x). The explanation for that is here.

Inserting into a number string

Have the function DashInsert(num) insert dashes ('-') between each two odd numbers in num. For example: if num is 454793 the output should be 4547-9-3. Don't count zero as an odd number.
Here is my code (not working). When I run it, I get the same response as an infinite loop where I have to kill the page but I can't see why. I know there are ways to do this by keeping it as a string but now I'm wondering why my way isn't working. Thanks...
function DashInsert(num) {
num = num.split("");
for (i = 1; i < num.length; i++) {
if (num[i - 1] % 2 != 0 && num[i] % 2 != 0) {
num.splice(i, 0, "-");
}
}
num = num.join("");
return num;
}

Using num.splice you are inserting new entries into the array, therefor increasing its length – and that makes the value of i “running behind” the increasing length of the array, so the break condition is never met.
And apart from that, on the next iteration after inserting a -, num[i-1] will be that - character, and therefor you are practically trying to check if '-' % 2 != 0 … that makes little sense as well.
So, when you insert a - into the array, you have to increase i by one as well – that will a) account for the length of the array having increased by one, and also it will check the next digit after the - on the next iteration:
function DashInsert(num) {
num = num.split("");
for (i = 1; i < num.length; i++) {
if (num[i - 1] % 2 != 0 && num[i] % 2 != 0) {
num.splice(i, 0, "-");
i++; // <- this is the IMPORTANT part!
}
}
num = num.join("");
return num;
}
alert(DashInsert("454793"));
http://jsfiddle.net/37wA9/

Once you insert a dash -, the if statement is checking this '-'%2 != 0 which is always true and thus inserts another dash, ad infinitum.
Here's one way to do it with replace using a regex and function:
function DashInsert(n) {
var f = function(m,i,s) { return m&s[i+1]&1 ? m+'-' : m; };
return String(n).replace(/\d/g,f);
}
DashInsert(454793) // "4547-9-3"

When you are adding a dash, this dash will be processed as a number on the next iteration. You need to forward one step.
function DashInsert(num) {
var num = num.split("");
for (var i = 1; i < num.length; i++) {
if ((num[i - 1] % 2 != 0) && (num[i] % 2 != 0)) {
num.splice(i, 0, "-");
i++; // This is the only thing that needs changing
}
}
num = num.join("");
return num;
}

It's because there are cases when you use the % operator on dash '-' itself, e.g. right after you splice a dash into the array.
You can correct this behavior by using a clone array.
function DashInsert(num) {
num = num.split("");
var clone = num.slice(0);
var offset = 0;
for (i = 1; i < num.length; i++) {
if (num[i - 1] % 2 != 0 && num[i] % 2 != 0) {
clone.splice(i + offset, 0, "-");
offset++;
}
}
return clone.join("");
}
alert(DashInsert("45739"));
Output: 45-7-3-9
Demo: http://jsfiddle.net/262Bf/

To complement the great answers already given, I would like to share an alternative implementation, that doesn't modify arrays in-place:
function DashInsert(num) {
var characters = num.split("");
var numbers = characters.map(function(chr) {
return parseInt(chr, 10);
});
var withDashes = numbers.reduce(function(result, current) {
var lastNumber = result[result.length - 1];
if(lastNumber == null || current % 2 === 0 || lastNumber % 2 === 0) {
return result.concat(current);
} else {
return result.concat("-", current);
}
}, []);
return withDashes.join("");
}
It's longer, but IMHO reveals the intention better, and avoids the original issue.

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)))

Count the prime numbers from 2 to 100 with simpler code than this

It has to be with just functions, variables, loops, etc (Basic stuff). I'm having trouble coming up with the code from scratch from what I've I learned so far(Should be able to do it). Makes me really mad :/. If you could give me step by step to make sure I understand I'd really really appreciated. Thanks a bunch in advanced.
How could I get the same result with a simpler code than this one:
var primes=4;
for (var counter = 2; counter <= 100; counter = counter + 1)
{
var isPrime = 0;
if(isPrime === 0){
if(counter === 2){console.log(counter);}
else if(counter === 3){console.log(counter);}
else if(counter === 5){console.log(counter);}
else if(counter === 7){console.log(counter);}
else if(counter % 2 === 0){isPrime=0;}
else if(counter % 3 === 0){isPrime=0;}
else if(counter % 5 === 0){isPrime=0;}
else if(counter % 7 === 0){isPrime=0;}
else {
console.log(counter);
primes = primes + 1;
}
}
}
console.log("Counted: "+primes+" primes");

I'm feeling naughty today so:
function printPrimesBetweenTwoAndOneHundredSimply(){
var primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97],
i,
arrayLength = primes.length;
for(i = 0; i < arrayLength; i++){
console.log(primes[i]);
}
console.log("Counted: " + arrayLength + " primes");
}

First, you really don't need to use === for this. The standard == will suffice. Second, you can make all of those lines that are the same thing, except for a single digit into one line:
var primes=4;
for (var counter = 2; counter <= 100; counter = counter + 1)
{
var isPrime = 0;
if(isPrime == 0){
if(counter == 2 || counter == 3 ||counter == 5 || counter == 7)console.log(counter);
else if(counter % 2 == 0 || counter % 3 == 0 || counter % 5 == 0 || counter % 7 == 0)isPrime=0;
else {
console.log(counter);
primes = primes + 1;
}
}
}
console.log("Counted: "+primes+" primes");
You'll also notice that the {} was removed on a few lines. This is because a single line of code following an if (among others) is always considered nested.
Next, we can change your primes = primes + 1; to this: primes++; which just tells primes to increment itself by one. The same can be done for your counter. We also know that isPrime equals "0" because you set it to that a second ago, so we no longer need that if statement:
var primes=4;
for (var counter = 2; counter <= 100; counter++)
{
var isPrime = 0;
if(counter == 2 || counter == 3 ||counter == 5 || counter == 7)console.log(counter);
else if(counter % 2 == 0 || counter % 3 == 0 || counter % 5 == 0 || counter % 7 == 0)isPrime=0;
else {
console.log(counter);
primes++;
}
}
console.log("Counted: "+primes+" primes");
Next, we can do a negative check (!= instead of ==) on the values and combine your else if with your else. Since we're doing a negative check (for this case) we have to switch the ORs (||) to ANDs (&&):
var primes=4;
for (var counter = 2; counter <= 100; counter++)
{
if(counter == 2 || counter == 3 ||counter == 5 || counter == 7)console.log(counter);
else if(counter % 2 != 0 && counter % 3 != 0 && counter % 5 != 0 && counter % 7 != 0) {
console.log(counter);
primes++;
}
}
console.log("Counted: "+primes+" primes");
There are many other ways to write it, but I felt it more beneficial to you to use what you started with and shorten it from there.

This finds all prime numbers between 2 and 100:
var primes=0;
var isprime = true;
for (var counter = 2; counter <= 100; counter = counter + 1)
{
// For now, we believe that it is a prime
isprime = true;
var limit = Math.round(Math.sqrt(counter)); // See comment from #AresAvatar, below
// We try to find a number between 2 and limit that gives us a reminder of 0
for (var mod = 2; mod <= limit; mod++) {
// If we find one, we know it's not a prime
if (counter%mod == 0) {
isprime = false;
break; // Break out of the inner for loop
}
}
if (isprime) {
console.log(counter, limit);
primes = primes + 1;
}
}
console.log("Counted: "+primes+" primes");

Of course "simpler" is not defined. :-)
The following can be made much shorter, but has a bit of robustness built in.
// Get primes from 0 to n.
// n must be < 2^53
function primeSieve(n) {
n = Number(n);
if (n > Math.pow(2, 53) || isNaN(n) || n < 1) {
return;
}
var primes = [];
var notPrimes = {};
var num;
for (var i=2; i<n; i++) {
for (var j=2; j<n/2; j++) {
num = i*j;
notPrimes[num] = num;
}
if (!(i in notPrimes)) {
primes.push(i);
}
}
return primes
}
So if you want less code:
// Get primes from 0 to n.
// n must be < 2^53
function primeSieve2(n) {
var primes = [], notPrimes = {};
for (var i=2; i<n; i++) {
for (var j=2; j<n/2; j++)
notPrimes[i*j] = i*j;
i in notPrimes? 0 : primes.push(i);
}
return primes
}

This is pretty simple (I mean, short):
console.log(2); console.log(3);
var m5=25, m7=49, i=5, d=2, c=2;
for( ; i<100; i+=d, d=6-d )
{
if( i!=m5 && i!=m7) { console.log(i); c+=1; }
if( m5 <= i ) m5+=10;
if( m7 <= i ) m7+=14;
}
c
It is an "unrolled" sieve of Eratosthenes with 2-3-wheel factorization.
On the one hand, we enumerate all the numbers below 100 (and bigger than 3) that are coprime with 2 and 3, as partial sums of 5+2+4+2+4+... , so that there are no multiples of 2 and 3 among thus enumerated numbers.
On the other hand, we discard from among them all the multiples of 5 and 7, by enumerating them as partial sums of 25+10+10+10+... and 49+14+14+14+... . Multiples of 2 and 3 are not there in the first place, and the first multiple of 11 that needs to be discarded by the sieve of Eratosthenes is 121.

This will be simpler for your Question....
for (int i = 2; i < 100; i++) {
int j = 0;
for (j = 2; j < i; j++)
if ((i % j) == 0)break;
if (i == j)
System.out.print(" " + i);}