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I need to do a code to verify whether or not the entered number is an Armstrong number, but my code does not work for every number.
Could anyone tell me what am I missing? Are there any other ways to do this?
Thank you!
let e, x, d = 0;
let b = prompt("Enter a number");
x=b;
while (x > 0) {
e = x % 10;
x = parseInt(x/10);
d = d + (e*e*e);
}
if (b==d)
alert("given number is an armstrong number");
else
alert("given number is not an armstrong number");<!DOCTYPE HTML>
<html>
<head>
<title>Armstrong</title>
</head>
<body>
</body>
</html>I need to do a code to verify whether or not the entered number is an Armstrong number, but my code does not work for every number.
Could anyone tell me what am I missing? Are there any other ways to do this?
Thank you!
let e, x, d = 0;
let b = prompt("Enter a number");
x=b;
while (x > 0) {
e = x % 10;
x = parseInt(x/10);
d = d + (e*e*e);
}
if (b==d)
alert("given number is an armstrong number");
else
alert("given number is not an armstrong number");<!DOCTYPE HTML>
<html>
<head>
<title>Armstrong</title>
</head>
<body>
</body>
</html>-
1
Your code does not initialize
z. The loop never runs. – Pointy Commented Aug 31, 2017 at 16:52 - An Armstrong number being a number that is equal to the sum of the cubes of each digit? (pages.mtu.edu/~shene/COURSES/cs201/NOTES/chap04/arms.html) I don't think that's what your algorithm does. – piisexactly3 Commented Aug 31, 2017 at 17:03
-
@Filburt where is
zinitialized to 0? The initial declaration statement only initializesd. – Pointy Commented Aug 31, 2017 at 17:05 -
Also, even if
zwere initialized, the loop doesn't change the value ofzanyway. – Pointy Commented Aug 31, 2017 at 17:06 -
@Filburt try it yourself. That declaration statement only initializes
d. Each variable in avardeclaration needs its own initialization expression in order to be initialized. – Pointy Commented Sep 1, 2017 at 12:07
13 Answers
Reset to default 2I think the way you pute the result is wrong. According to Wikipedia, an Armstrong number, also called narcissistic number, has the following property:
[An Armstrong number] is a number that is the sum of its own digits each raised to the power of the number of digits.
You can pute it like this:
var number = prompt("Enter a number");
var numberOfDigits = number.length;
var sum = 0;
for (i = 0; i < numberOfDigits; i++) {
sum += Math.pow(number.charAt(i), numberOfDigits);
}
if (sum == number) {
alert("The entered number is an Armstrong number.");
} else {
alert("The entered number is not an Armstrong number.");
}You can try this method, very easy to understand.
const armstrongNumber = (num) => {
const toArray = num.toString().split('').map(Number);
const newNum = toArray.map(a => {return a**3}).reduce((a, b) => a + b);
if(newNum == num){
console.log('This is an armstrong number');
}else{
console.log('This is not an armstrong number');
}
}
armstrongNumber(370);
//This is an armstrong number
Here is the solution to check Armstrong number without using Math Object.
function armstrongNum(number) {
const numberArr = String(number).split('');
const power = numberArr.length;
let TotalSum = numberArr.reduce((acc, cur) => {
return acc + (function(cur,power){
let curNum = Number(cur);
let product = 1;
while(power > 0) {
product *= curNum;
power --;
}
return product;
}(cur,power))
}, 0)
if (TotalSum === number) {
return true
}
return false
}
armstrongNum(153);
Here is an example of functional code for Armstrong Numbers.
<script>
function execute() {
var num1 = document.getElementById("Number1").value;
var num2 = document.getElementById("Number2").value;
var num3 = document.getElementById("Number3").value;
var concatNum = num1 + num2 + num3;
var num1Sqrt = num1 * num1 * num1;
var num2Sqrt = num2 * num2 * num2;
var num3Sqrt = num3 * num3 * num3;
if (num1Sqrt + num2Sqrt + num3Sqrt == concatNum) {
Answer.value = "The three integers you entered are Armstrong numbers.";
}
if (num1Sqrt + num2Sqrt + num3Sqrt != concatNum) {
Answer.value = "The three integers you entered are not Armstrong numbers.";
}
}
</script>
You first set your variables to what is entered, you then concatenate the string with the three integers.
You then square the three integers and set each value to a variable. You then have a basic check for equality, if the three squared values add up to your concatenated string then you have an Armstrong number.
This is how i solved mine:
function armstrong(num) {
var digits = num.toString().split('')
var realDigits = num
var a = 0
for (let i = 0; i < digits.length; i++){
digits[i] = Math.pow(digits[i], digits.length)
a += digits[i]
}
if (a == realDigits) {
console.log("Number is armstrong")
} else if (a != realDigits) {
console.log("Number is not armstrong")
}
}
armstrong(371)
//feel free to pass any value here
You can copy/paste and run this code at https://www.typescriptlang/play/
I hope this helps someone.
This works as well..
function isArmstrong (n) {
const res = parseInt(n, 10) === String(n)
.split('')
.reduce((sum, n) => parseInt(sum, 10) + n ** 3, 0);
console.log(n, 'is', res, 'Armstrong number')
return res
}
isArmstrong(153)Here is another way to solve it.
let c = 153
let sum = 0;
let d = c.toString().split('');
console.log(d)
for(let i = 0; i<d.length; i++) {
sum = sum + Math.pow(d[i], 3)
}
if(sum == c) {
console.log("armstrong")
}
else {
console.log("not a armstrong")
}
Correct way to find Armstrong
var e, x, d = 0, size;
var b = prompt("Enter a number");
b=parseInt(b);
x=b;
size = x.toString().length;
while (x > 0) {
e = x % 10;
d = d + Math.pow(e,size);
x = parseInt(x/10);
}
//This is I solved without function
let num = prompt();
let num1 = num;
let sum = 0;
while(num > 0){
rem = num % 10;
sum = sum + Math.pow(rem, num1.length);
num = parseInt (num /10);
}
if (sum == num1) console.log("Armstrong");
else console.log("not Armstrong");number is an Armstrong number or not.
let inputvalue=371
let spiltValue=inputvalue.toString().split('')
let output=0
spiltValue.map((item)=>output+=item**spiltValue.length)
alert(`${inputvalue} ${inputvalue==output? "is armstrong number" : "is not armstrong number"}`);In order to get a Narcissistic/Armstrong number, you need to take the length of the number as
nfor taking the power for summing the value.
Here's another solution that works with an input >= 3 digits
(With Math.pow() each character is added as a number to the power of the array size)
const isArmstrongNumber = (n) => {
//converting to an array of digits and getting the array length
const arr = [...`${n}`].map(Number);
let power = arr.length;
const newNum = arr
.map((a) => {
return Math.pow(parseInt(a), power);
})
.reduce((a, b) => a + b);
return newNum == n;
};
console.log("Is it Armstrong? " + isArmstrongNumber(370));
console.log("Is it Armstrong? " + isArmstrongNumber(123));
console.log("Is it Armstrong? " + isArmstrongNumber(1634));
May be this help you output without using any function.
let armstrongNumber = '153';
armstrongNumber = isNaN(armstrongNumber) ? armstrongNumber : armstrongNumber.toString();
let sumationValue = 0;
let result = (function () {
for (var digit of armstrongNumber) {
sumationValue += parseInt(digit) ** armstrongNumber.length;
}
return sumationValue;
})();
if (result == armstrongNumber) {
console.log(armstrongNumber + ' is an Armstrong Number');
} else {
console.log(armstrongNumber + ' is not an Armstrong Number');
}You can try for This, May this help for you:
<!DOCTYPE HTML>
<html>
<head>
<title>Armstrong</title>
</head>
<body>
</body>
</html>
<script>
var z,e,x,d=0;
var b=prompt("Enter a number");
x=parseInt(b);
while(x>0) //Here is the mistake
{
e=x%10;
x=parseInt(x/10);
d=d+(e*e*e);
console.log("d: "+d+" e: "+e);
}
if(parseInt(b)==d){
alert("given no is amstrong number");
}
else {
alert("given no is not an amstrong number");
}
</script>
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