armstrong numbers
int main(){
int a;
cin >> a;
int check = a;
int sum = 0;
while(a!=0){
int rem = a%10;
sum += pow(rem,3); //include cmath
a/=10;
}
if(sum==check){
cout << "armstrong";
} else {
cout << "not armstrong";
}
return 0;
}Armstrong numbers
Numbers -- Armstrong numbers
Write a method that can check if a number is Armstrong number
Solution:
public static boolean ArmStrongNumber (int num) {
int a = 0, b =0, c= num;
while(num>0){
a=num%10;
num/=10;
b=b+(a*a*a);
}
if(c==b) {
return true;
}
return false;
}How do you calculate Armstrong number in maths?
import java.util.Scanner;
import java.lang.Math;
public class one //ArmstsrongNumberExample
{
//function to check if the number is Armstrong or not
static boolean isArmstrong(int n)
{
int temp, digits=0, last=0, sum=0;
//assigning n into a temp variable
temp=n;
//loop execute until the condition becomes false
while(temp>0)
{
temp = temp/10;
digits++;
}
temp = n;
while(temp>0)
{
//determines the last digit from the number
last = temp % 10;
//calculates the power of a number up to digit times and add the resultant to the sum variable
sum += (Math.pow(last, digits));
//removes the last digit
temp = temp/10;
}
//compares the sum with n
if(n==sum)
//returns if sum and n are equal
return true;
//returns false if sum and n are not equal
else return false;
}
//driver code
public static void main(String args[])
{
int num;
Scanner sc= new Scanner(System.in);
System.out.print("Enter the limit: ");
//reads the limit from the user
num=sc.nextInt();
System.out.println("Armstrong Number up to "+ num + " are: ");
for(int i=0; i<=num; i++)
//function calling
if(isArmstrong(i))
//prints the armstrong numbers
System.out.print(i+ ", ");
}
}Copyright © 2021 Codeinu
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