Develop Reference

how to make a script to print out all prime numbers in js

I want to know how I can improve my code by helping it find out what number is prime and what is not. I was thinking that I would divide a number by a number and then if it is a decimal number then it is prime,
I want it to have a loop to check every number 1 to 100 and see if it is a prime number
This is what I have so far:
for(let i = 1; i <= 100; i++) {
if(i == 1) {
}else if(i == 2) {
console.log(`${i} is a prime number`);
}else if(i >= 3){
x = i / 2;
tf = Number.isInteger(x);
if(tf == false && i >= 3) {
console.log(`${i} is a prime number`);
}
}
}
and so far it outputs 1 2 and all the odd numbers.

Create a function to test whether a number is prime or not (divisible only by 1 and itself). Then call this function inside the loop on each number.
function isPrimeNumber(no) {
if (no < 2) {
return false;
}
for (let i = 2; i < no; i++) {
if (no % i == 0) {
return false;
}
}
return true;
}
for (let i = 1; i <= 100; i++) {
if (isPrimeNumber(i)) {
console.log(i);
}
}

var numbers = new Array(101).fill(0).map((it, index) => index);
var halfWay = Math.floor(numbers.length / 2);
for (let i = 2; i <= halfWay; i++) {
if (!numbers[i]) continue;
for (let j = 2; j * i < numbers.length; j++) {
console.log(`${i} * ${j} = ${i * j}`);
numbers[j * i] = null;
}
}
console.log(numbers.filter(it => it));
Here is an attempt to mathematically find numbers between 1-100 that are primes.
Fill an array of numbers 0-100
For every number (starting at 2), multiply it by itself and all numbers after it, up to half of the array
For every computed number, remore it from the array, as it is not a prime
At the end, filter out all numbers that are null

As Taplar stated primes are numbers that only divide by the number itself and 1.
As far as improving your code. I would say you want to eliminate as many possible numbers with the fewest questions.
An example would be is the number even and not 2 if so it is not prime? The interesting part of this question you eliminate dividing by all even numbers as well. This instantly answers half of all possible numbers and halves the seek time with the ones you need to lookup.
So what would this look like?
function isPrime(num) {
// Check it the number is 1 or 2
if (num === 1 || num === 2) {
return true
}
// Check if the number is even
else if (num % 2 === 0) {
return false;
}
// Look it up
else {
// Skip 1 and 2 and start with 3 and skip all even numbers as they have already been checked
for (let i = 3; i <= num/2; i+=2) {
// If it divides correctly then it is not Prime
if (num % i === 0) {
return false
}
}
// Found no numbers that divide evenly it is Prime
return true
}
}
console.log('1:', isPrime(1))
console.log('2:', isPrime(2))
console.log('3:', isPrime(3))
console.log('4:', isPrime(4))
console.log('11:', isPrime(11))
console.log('12:', isPrime(12))
console.log('97:', isPrime(97))
console.log('99:', isPrime(99))
console.log('65727:', isPrime(65727))
console.log('65729:', isPrime(65729))

Check for perfect number and print out divisors?

My goal is to create a program that checks whether the user input is a perfect number or not. It has validation for the numbers entered. If the input IS a perfect number, I'd like to print out each of the divisors. I tried using this method:
{
for(int number=2; number <= 10000 ; number++)
perfect(number);
return 0;
}
void perfect(int number)
{
int total = 0;
for (int i = 1; i < number; i++)
{
if (number % i == 0)
total += i;
}
if (number == total)
{
for (int x = 1; x < number; x++)
{
if (number % x == 0)
cout << x << " + ";
}
cout << " = " << number << endl;
}
}
However, I was unable to get the desired effect. I am very new to javascript and am struggling with inserting code in the correct way. Does anyone have a suggestion for how I can get the desired effect? Here is the code I have already written:
function check_prime() {
var input = document.getElementById("enteredNumber").value;
var number = parseInt(input);
if (isNaN(number)) {
alert("Oops! Please enter a valid number.");
document.getElementById("enteredNumber").value="";
document.getElementById("result").innerHTML = "";
document.getElementById("enteredNumber").focus();
}
else if (input.length === 0) {
alert("Please enter a number.");
document.getElementById("enteredNumber").focus();
}
else if (!isNaN(number)) {
if (is_perfect(number)) {
document.getElementById("answer").innerHTML = "Congratulations! " + number + " is a perfect number." ;
}
else {
document.getElementById("answer").innerHTML = "I'm sorry. " + number + " is not a perfect number. Try Again.";
}
}
else {
document.getElementById("answer").innerHTML = "Please enter a number.";
}
}
function is_perfect(number)
{
var temp = 0;
for(var i=1;i<=number/2;i++)
{
if(number%i === 0)
{
temp += i;
}
}
if(temp === number)
{
return true;
}
else
{
return false;
}
}
function clear_textbox(){
document.getElementById("answer").innerHTML = "";
document.getElementById("enteredNumber").value="";
document.getElementById("enteredNumber").focus();
}

I'd suggest revising your is_perfect() function to return an array of divisors if the number is perfect and null if the number is not perfect. Then the calling code has the divisors available for display when the input is a perfect number.
function is_perfect(number) {
var temp = 0;
var divisors = [];
for(var i=1;i<=number/2;i++) {
if (number%i === 0) {
divisors.push(i);
temp += i;
}
}
return temp === number ? divisors : null;
}
Then:
var divisors = is_perfect(number);
if (divisors) {
document.getElementById("answer").innerHTML = "Congratulations! " + number + " is a perfect number.";
// display the divisors somewhere; the alert is just for show
alert("Divisors: " + divisors.toString());
} else {
...
}
[Note: In an earlier version of this answer, I had initialized temp to 1 and divisors to [1] and had started the loop at 2, on the theory that 1 is always a divisor. Unfortunately, that's wrong, since 1 is not a proper divisor of 1. The revised version of is_perfect() now returns null for an argument of 1 instead of [1]. An alternative fix would have been to test explicitly for the case number === 1, but that's uglier (if perhaps a tiny bit more efficient, since it avoids one % evaluation).]

so I use 2^(n-1)*(2^n -1) formula (to generate a perfect number) and checking if last digit is 6 or 8 to check if x is perfect number.
Note: It's not perfect 100%
function pn(x) {
x = '' + x
for (var i = 0; i < Infinity; i++) {
perfnumgen = Math.pow(2, i - 1) * (Math.pow(2, i) - 1)
if (x === "" + perfnumgen && (perfnumgen % 10 === 8 || perfnumgen % 10 === 6))
return true
else if (perfnumgen > x)
return false
console.log("" + perfnumgen)
}
}

Adding sum in a for loop in javascript

I'm trying to recreate a simple project I have in my fundamentals class to javascript (from C++) but sum doesn't add every time the for loop runs. All other parts are ok but sum just lists the numbers in the order I put them in. Any help is appreciated
var num = prompt("Please enter an integer");
var lrg = num;
var sml = num;
var avg = num;
var sum = num;
var cnt = 10;
function runMath () {
for (i = 1; i < 10; i++) {
var num = prompt("Please enter an integer");
if (num > lrg) {
lrg = num;
} else {
lrg = lrg;
}
if (num < sml) {
sml = num;
} else {
sml = sml;
}
sum += num;
}
}
runMath();
avg = sum/cnt;

The problem is that prompt() returns a String, whereas you are expecting a number. You can turn this into a number in a few different ways:
parseInt("33") will return 33, instead of "33"
Likewise, shorthand would look like:
+prompt("33") will return 33, instead of "33"

All input is from the prompt() command is a string. You can convert it to an integer using parseInt(), but the user can enter something other than a number, so you will need to check if it isNaN() (is Not a Number), and deal with it differently if it is.
var num = prompt("Please enter an integer");
num = parseInt(num, 10)
if (isNaN(num)) {
alert ("That's not a number")
num = 0 // or do something else
}
Caution: typeof NaN will return "number", so you can't rely on that as a test (see NaN)
An explanation of the + in +prompt: Unary plus (+)

Find the largest prime factor with Javascript

Thanks for reading. Pretty new to Javascript and programming in general.
I'm looking for a way to return the largest prime factor of a given number. My first instinct was to work with a while loop that counts up and finds prime factors of the number, storing the factors in an array and resetting each time it finds one. This way the last item in the array should be the largest prime factor.
var primerizer = function(input){
var factors = [];
var numStorage = input
for (x=2; numStorage != 1; x++){ // counter stops when the divisor is equal to the last number in the
// array, meaning the input has been fully factorized
if (result === 0) { // check if the number is prime; if it is not prime
factors.push(x); // add the divisor to the array of prime numbers
numStorage = numStorage/x // divide the number being calculated by the divisor
x=2 // reset the divisor to 2 and continue
};
};
primeFactor = factors.pop();
return primeFactor;
}
document.write(primerizer(50))
This only returned 2, undefined, or nothing. My concern was that the stop condition for the for loop must be defined in terms of the same variable as the start condition, so I tried it with a while loop instead.
var primerizer = function(input){
var factors = [];
var numStorage = input
x=2
while (numStorage != 1){
var result = numStorage%x;
if (result === 0) {
factors.push(x);
numStorage = numStorage/x
x=2
}
else {
x = x+1
}
}
return factors.pop();
}
document.write(primerizer(50)
Same problem. Maybe there's a problem with my syntax that I'm overlooking? Any input is much appreciated.
Thank you.

The shortest answer I've found is this:
function largestPrimeFactor(n){
var i=2;
while (i<=n){
if (n%i == 0){
n/=i;
}else{
i++;
}
}
console.log(i);
}
var a = **TYPE YOUR NUMBER HERE**;
largestPrimeFactor(a)

You can try with this
var x = 1, div = 0, primes = [];
while(primes.length != 10001) {
x++;
for(var i = 2; i < x && !div; i++) if(!(x % i)) div++;
if(!div) primes.push(x); else div = 0;
}
console.log(primes[primes.length-1]);
or this: (This solution uses more of your memory)
var dont = [], max = 2000000, primes = [];
for (var i = 2; i <= max; i++) {
if (!dont[i]) {
primes.push(i);
for (var j = i; j <= max; j += i) dont[j] = true;
}
}
console.log(primes);

here is my own solution.
//function
function largestPrimeFactor (num) {
//initialize the variable that will represent the divider
let i = 2;
//initialize the variable that will represent the quotient
let numQuot = num;
//array that will keep all the dividers
let primeFactors = [];
//execute until the quotient is equal to 1
while(numQuot != 1) {
/*check if the division between the number and the divider has no reminder, if yes then do the division keeping the quotient in numQuot, the divider in primeFactors and proceed to restart the divider to 2, if not then increment i by one an check again the condition.*/
if(numQuot % i == 0){
numQuot /= i;
primeFactors.push(i);
i = 2;
} else {
i++;
}
}
/*initialize the variable that will represent the biggest prime factor. biggest is equal to the last position of the array, that is the biggest prime factor (we have to subtract 1 of .length in order to obtain the index of the last item)*/
let biggest = primeFactors[primeFactors.length - 1];
//write the resutl
console.log(biggest);
}
//calling the function
largestPrimeFactor(100);

<script>
function LPrimeFactor() {
var x = function (input) {
var factors = [];
var numStorage = input;
x = 2;
while (numStorage != 1) {
var result = numStorage % x;
if (result === 0) {
factors.push(x);
numStorage = numStorage / x;
x = 2;
}
else {
x = x + 1;
}
}
return factors.pop();
}
document.write(x(50));
}
</script>
<input type="button" onclick="LPrimeFactor();" />
Here is an example i tried with your code

Here is the solution I used that should work in theory... except for one small problem. At a certain size number (which you can change in the code) it crashes the browser due to making it too busy.
https://github.com/gordondavidescu/project-euler/blob/master/problem%203%20(Javascript)
Adding the code inline:
<p id="demo">
</p>
<script>
function isPrime(value) {
for(var i = 2; i < value; i++) {
if(value % i === 0) {
return false;
}
}
return value > 1;
}
function biggestPrime(){
var biggest = 1;
for(var i = 600851470000; i < 600851475143; i++){
if (isPrime(i) != false)
{
biggest = i;
}
document.getElementById("demo").innerHTML = biggest;
}
}
biggestPrime();
</script>
</p>

<script>
//Finds largest prime factor
find = 2165415 ; // Number to test!
var prime = 0;
loop1:
for (i = 2; i < find; i++){
prime = 0;
if (find%i == 0){
document.write(find/i);
for (j = 2; j < (find / i); j++){
if ((find / i )%j == 0){
document.write(" divides by "+j+"<br>");
prime = prime + 1;
break;
}
}
if (prime == 0){
document.write("<br>",find/i, "- Largest Prime Factor")
prime = 1;
break;
}
}
}
if (prime==0)
document.write("No prime factors ",find," is prime!")

Program that uses While and For loops to find all Prime Numbers between an upper Bound

I'm having some trouble with the for loop. I have the while loop working properly, but I need help with the nested loop. The concept of loops is fairly new to me and I don't quite understand what I've done here in the code. Any help would be greatly appreciated.
var number = parseInt(window.prompt("Enter number: ", ""));
var divisor;
var check;
var prime = true;
var num;
document.write("The factors of ", number, " are: </br>");
divisor = 1;
while (divisor <= number) {
check = number % divisor;
if (check == 0) {
document.write(divisor, " ");
if ((divisor != 1) && (divisor != number)) {
prime = false;
}
}
divisor = divisor + 1;
}
if (prime == true) {
document.write("<br>The number is prime");
} else {
document.write("<br> The number is composite");
}

Something to get you excited. This program is using for and while to check prime numbers. The for loop is used as an if-check but this will work. You should also try using functions to minimize your code.
var number = parseInt(window.prompt("Enter number: ", ""));
var result = isPrime(number);
function isPrime(number) {
var start = 2;
//USING WHILE
while (start <= Math.sqrt(number)) {
if (number % start++ < 1) return false;
}
return number > 1;
}
//USING FOR
for (;result;){
document.write("<br>The number is prime");
}
if(!result) {
document.write("<br>The number is NOT prime");
}

As a fresh man,I think may you need a break in the while loop,beacause you just check if it is prime once.Just like that:
if((divisor != 1) && (divisor != number)){
prime= false;
break;
}