I need to prompt the user to enter a series of numbers, or the word "quit".
Then, If the user enters a number, add the new number to a running total.
And, If the user enters the word "quit" the loop should stop execution.
I cant figure what should I do here. Im a beginner. i don't know how to make it work when user enter a word quit
let total = 0;
let number = 0;
do {
total += number;
number = parseInt(prompt('Enter a number: '));
if (number >= 0) {
document.writeln("<p>You entered " + number + "!</p>");
} else {
if (isNaN(number)) {
number = parseInt(prompt('Enter a number: '));
document.writeln("<p>Try again. We are looking for a number!</p>");
}
}
} while (number >= 0)
document.writeln("<p>The total is " + total + "</p>")
Use break to stop the javascript loop:
let total = 0;
let number = 0;
do {
total += number;
text = prompt('Enter a number: ');
if (text == "quit") {
break;
}
number = parseInt(text);
if (number >= 0) {
document.writeln("<p>You entered " + number + "!</p>");
} else {
if (isNaN(number)) {
number = parseInt(prompt('Enter a number: '));
document.writeln("<p>Try again. We are looking for a number!</p>");
}
}
} while(number >= 0)
document.writeln("<p>The total is " + total + "</p>")
Related
So I need a tiny bit of help with this code, Some background information: The user inputs a number, the code takes the number and outputs various combinations of numbers that multiply to it.
For example:
Input: 7
Output: (1,7)(7,1).
*But what really happens:
*
Input: 7
Output: (7,1)
I want my code to reverse the numbers as well, so it makes can look like it has two combinations
var input= parseInt(prompt("Please enter a number larger than 1"));
var arr = [];
if(input <= 1) {
console.log("Goodbye!")
}
while(input > 0) {
var arr = [];
var input = parseInt(prompt("Please enter a number larger than 1"));
for (var i = 0; i < input; ++input) {
var r = ((input / i) % 1 === 0) ? (input / i) : Infinity
if(isFinite(r)) {
arr.unshift(r + ", " + i)
}
}
console.log("The multiplicative combination(s) are: " + "(" + arr.join("), (") + "). ");
}
My code just need this tiny bit of problem fixed and the rest will be fine!
Your code has 2 infinite loop because you never change i and always increase input.
also in this line for (var i = 0; i < input; ++input) you never let i to be equal to the input so in your example (input=7) you can not have (7,1) as one of your answers. I think this is what you looking for:
var input = 1;
while(input > 0) {
input = parseInt(prompt("Please enter a number larger than 1"));
if(input > 1) {
var arr = [];
for (var i = 0; i <= input; ++i) {
var r = ((input / i) % 1 === 0) ? (input / i) : Infinity
if(isFinite(r)) {
arr.unshift(r + ", " + i)
}
}
console.log("The multiplicative combination(s) are: " + "(" + arr.join("), (") + "). ");
continue;
}
else{
console.log("Goodbye!");
break;
}
}
I need my code to only accept whole numbers, no decimals and should prompt an error when a decimal gets entered. I don't want a new function , I'm hoping I can just add lines to my function but I don't know what I need to add.
function number_function() {
number = parseInt(prompt('Enter a positive integer:'));
if (number < 0) {
alert('Error! Factorial for negative number does not exist. But I will show you the positive number');
number = number * -1;
let n = 1;
for (i = 1; i <= number; i++) {
n *= i;
}
alert("The factorial of " + number + " is " + n + ".");
} else if (number === 0) {
alert("Please enter a number greater than 0");
} else {
let n = 1;
for (i = 1; i <= number; i++) {
n *= i;
}
alert("The factorial of " + number + " is " + n + ".");
}
}
number_function();
You can do this to check if the number has decimals
const val = 10.7 % 1;
if (val !== 0) {
console.log('has decimals');
} else {
console.log('has no decimal');
}
JavaScript provides the built-in function Number.isInteger(n)
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)
}
}
Here's the problem:
Create a sum program that calculates the sum of all odd numbers between 1 and the number entered by the user. For example if the user enters in the number 7 the program would calculate 1 + 3 + 5 + 7. The total and the expression should be displayed on the document. The answer would be 16.
My code so far
//declare the variables
var sternum = prompt("enter a number");
var tantalum = 1;
var increase = 1;
var expression = "+";
//finding the sum
document.write(" the sum of all numbers are: ");
do {
if(sternum % 2 == 0) {
}
else{
document.write(increase + expression);
increase = increase + 1;
tantalum = tantalum + increase;
}
}while(increase < sternum);
document.write(sternum + " = " + tantalum);
You have created an infinite loop. Make sure you increment increase every iteration:
var sternum = prompt("enter a number");
var tantalum = 0;
var increase = 1;
var expression = "+";
//finding the sum
document.write(" the sum of all numbers are: ");
do {
if(increase % 2 == 0) {
}
else{
document.write(increase + expression);
tantalum = tantalum + increase;
}
increase = increase + 1;
}while(increase <= sternum);
document.write(" = " + tantalum);
To make it more efficient you could change increase = increase + 1; to increase = increase + 2;. No need to process even numbers. Also tantalum should be set to 0 to start.
Question, how do I make it so when an user-inputted number is negative, it isn't added to the total variable that will be ouput?
Code below!
function lab10logicInLoopsPart1()
{
lCounter = 1;
var total = 0;
userNumber = 0;
while(lCounter < 6) {
lCounter++;
userNumber = prompt("Enter a number.");
total += +userNumber;
document.write("Entered number was: " + userNumber + "\n");
}
document.write("\nTotal: " + total);
}
After prompting the user to enter a number, just check it's not a negative with an if statement:
if(userNumber >=0)
{ //do your stuff here}