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 number
temp=n;
while(n>0)
{
r=n%10;
sum=sum+(r*r*r);
n=n/10;
}
if(temp==sum)
printf("armstrong number ");
else
printf("not armstrong number");
return 0;armstrong number in c
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int cube(int a)
{
int c;
c = a*a*a;
return c;
}
int armnum(int *a)
{
int x = *a, n = 0, rem, r = 0;
while (x != 0) {
x /= 10;
n++;
}
x = *a;
while (x != 0) {
rem = x % 10;
r += cube(rem);
x /= 10;
}
if(r == *a){
return 1;
}
}
int main()
{
int a, y;
scanf("%d", &a);
y = armnum(&a);
if(y == 1){
printf("It is an Armstrong number.");
}
else{
printf("It is not an Armstrong number.");
}
}armstrong number
#include<stdio.h>
int main()
{
int num ,n,n1,c=0,mul=1,sum=0,r,f,i;
printf("enter any num: \n");
scanf("%d",&num);
n=num;
n1=num;
while(n!=0)
{
r=n%10;
c++;
n=n/10;
}
while (num!=0)
{
f=num%10;
mul=1;
for(i=1;i<=c;i++)
{
mul=mul*f;
}
sum=sum+mul;
num=num/10;
}
if(n1==sum)
printf("Armstrong Number");
else
printf("Not an Armstrong Number");
return 0;
}armstrong number function
def is_armstrong_number(number: int)-> bool:
arm = str(number)
lenght = len(arm)
sum = 0
for digit in arm:
sum += int(digit)**lenght
return sum == numberArmstrong 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;
}Copyright © 2021 Codeinu
Forgot your account's password or having trouble logging into your Account? Don't worry, we'll help you to get back your account. Enter your email address and we'll send you a recovery link to reset your password. If you are experiencing problems resetting your password contact us
