Answers and Solutions

Answer

#include<iostream.h>

#include<conio.h>

void main() {

int num, rem, sum=0, n;

clrscr();

cout<<"Enter a 3 digit number: ";

cin>>num;

n = num;

while(n) {

rem = n % 10;

sum = sum + (rem*rem*rem);

n = n / 10;

}

if(sum == num)

cout<<num<<" is an armstrong number.";

else

cout<<num<<" is not an armstrong number.";

getch();

}

Answer

class ArmstrongExample{     public static void main(String[] args)  {       int c=0,a,temp;       int n=153;//It is the number to check armstrong       temp=n;       while(n>0)       {       a=n%10;       n=n/10;       c=c+(a*a*a);       }       if(temp==c)       System.out.println("armstrong number");        else           System.out.println("Not armstrong number");       }   }  

Answer

Rectification in C++ Code

#include<iostream>

using namespace std;

void main()

{

int sum=0,temp,num,rem;

cout<<"Enter any 3 digit number: ";

cin>>num;

temp = num;

while(num > 0)

{

rem = num%10;

num = num / 10;

sum=sum+rem*rem*rem;

}

if(sum == temp)

{

cout<<temp<<" is an Armstrong number";

}

else

{

cout<<temp<<" is not an Armstrong number";

}

}

Answer

As you haven't mentioned your programming language platform,

Here it is in python, c++ and java

Python approach

print("Enter the number")

#get the number

number = int(input())

#store it in a temp

temp = int(number)

Sum = 0

#loop till the quotient is 0

while(temp != 0):

rem  = temp % 10 #find reminder

Sum  = Sum + (rem * rem * rem) #cube reminder and add it to the Sum

temp = temp / 10 #find quotient, if 0 then loop again

#if the entered number and the Sum value matches, it is an Armstrong number

if(number == Sum):

print ("Armstrong Number")

else:

print ("Not an Armstrong Number")

C++ Approach using Dev c++ compiler

#include<iostream>

using namespace std;

void main()

{

int sum=0,temp,num,rem;

cout<<"Enter any 3 digit number: ";

cin>>num;

temp = num;

while(num > 0)

{

r = num%10;

num = num / 10;

sum=sum+r*r*r;

}

if(sum == temp)

{

cout<<temp<<" is an Armstrong number";

}

else

{

cout<<temp<<" is not an Armstrong number";

}

Java Approach, core java using netbeans

import java.io.*;

class Armstrong_Check

{

public static void main(String args[]) throws Exception

{

int num, temp, sum = 0, rem = 0;

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

System.out.println("\nEnter an Integer:");

num = Integer.parseInt(br.readLine());

temp = num;

while(num > 0)

{

rem = num%10;

sum = sum + (rem * rem * rem);

num = num/10;

}

if(temp == sum)

{

System.out.println("\n" + temp + " is an Armstrong Number\n");

}

else

{

System.out.println("\n" + temp + " is Not an Armstrong Number\n");

}

}

}

Answer

#include <stdio.h> #include <math.h>

int main() {     int number, originalNumber, remainder, result = 0, n = 0 ;

    printf("Enter an integer: ");     scanf("%d", &number);

     originalNumber = number;          while (originalNumber != 0)     {         originalNumber /= 10;         ++n;     }

    originalNumber = number;

    while (originalNumber != 0)     {         remainder = originalNumber%10;         result += pow(remainder, n);         originalNumber /= 10;     }

    if(result == number)         printf("%d is an Armstrong number.", number);     else         printf("%d is not an Armstrong number.", number);

    return 0; }