Armstrong number in c | c++ and Java

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What is an armstrong number?

Armstrong number is a number in which the sum of cubes of digits of that number is equals to that number itself. Example: 0, 1, 153, 370, 371, 407.

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How to check a number is armstrong or not?

• First split the digits of that number.
• Cube the digits.
• Sum up the cubes of the digits.
• If the sum is equal to that number then it is an armstrong number.

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Number	Digits	Sum of its cubes
0	0	0*0*0=0
1	1	1*1*1=1
153	1 , 5, 3	(1*1*1)+(5*5*5)+(3*3*3)=153
370	3,7,0	(3*3*3)+(7*7*7)+(0*0*0)=370
371	3,7,1	(3*3*3)+(7*7*7)+(1*1*1)=371
407	4,0,7	(4*4*4)+(0*0*0)+(7*7*7)=407

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How to extract a digit from a number in a program?

The method for extracting a digit is simply as follows:

• Copy the number to a temporary variable.
• Find the digit which is in unit’s place of that number i.e. the remainder after dividing by 10. For this % operator can be used.
• Then divide the temporary number by 10.
• perform the digit finding operation and dividing by 10 until the number is greater than 0.

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Armstrong number in c

Note: follow-up comment lines.

#include<stdio.h>
int main(int argc, char const *argv[])
{
	int number;
	//read a number
	printf("\nEnter a number to check Armstrong number");
	scanf("%d",&number);
	//find the sum of its cubes
	int sum_of_digit_cube=0,temp=number;
	while( temp>0 )
	{
		int digit=temp%10;
		sum_of_digit_cube+=(digit*digit*digit);
		temp=temp/10;
	}
	if( sum_of_digit_cube == number)
	{
		printf("The number %d is an Armstrong number",number);
	}
	else
	{
		printf("The number %d is not Armstrong number",number);
	}
	return 0;
}

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Armstrong number in c++

Note: follow-up comment lines.

#include<iostream>
#include<cmath>
using namespace std;
int main(int argc, char const *argv[])
{
	int number;
	//read a number
	cout<<"\nEnter a number to check Armstrong number";
	cin>>number;
	//find the sum of its cubes
	int sum_of_digit_cube=0,temp=number;
	while( temp>0 )
	{
		int digit=temp%10;
		sum_of_digit_cube+=pow(digit,3);
		temp=temp/10;
	}
	if( sum_of_digit_cube == number)
	{
		cout<<"The number "<<number<<" is an Armstrong number";
	}
	else
	{
		cout<<"The number "<<number<<" is not Armstrong number";
	}
	return 0;
}

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Armstrong number in java

Note: follow-up comment lines.

import java.io.*;
import java.util.Scanner;

public class HelloWorld {
//finds sum of the cubes of the digit
	static int sumCubes(int n) {
		int rv = 0;
		int temp = n;
		int digit;
		while (temp > 0)
                {
			digit = temp % 10;
			rv += Math.pow(digit, 3);
			temp = temp / 10;
		}

		return rv;
	}

	public static void main(String[] args) {
		//read a number
		  Scanner stdin=new Scanner(System.in);
		  System.out.println("Enter a number to check Armstrong Number "); int
		  num=stdin.nextInt(); 
                  if( num==sumCubes(num) )
		  System.out.println("It is Armstrong Number ");
                  else
		  System.out.println("It is not Armstrong Number ");
                  stdin.close();
	}

}

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Print all armstrong numbers below 1000 in c

Note: follow-up comment lines.

#include<stdio.h>
int main(int argc, char const *argv[])
{
	printf("Armstrong numbers below 1000\n");
	for( int i=0 ; i<1000 ; i++ )
	{
		int number=i;
		int sum_of_digit_cube=0;
		while( number>0 )
		{
		  int digit=number%10;
		  sum_of_digit_cube+=digit*digit*digit;
		  number=number/10;
		}
		if( sum_of_digit_cube==i )
		{
			printf( "%d\n",i );
		}
	}
	return 0;
}

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Print all armstrong numbers below 1000 in c++

Note: follow-up comment lines.

#include<iostream>
#include<cmath>
using namespace std;
int main(int argc, char const *argv[])
{
	cout<<"Armstrong numbers below 1000 in c++\n";
	for( int i=0 ; i<1000 ; i++ )
	{
		int number=i;
		int sum_of_digit_cube=0;
		while( number>0 )
		{
		  int digit=number%10;
		  sum_of_digit_cube+=pow(digit,3);
		  number=number/10;
		}
		if( sum_of_digit_cube==i )
		{
			cout<<i<<endl;
		}
	}

	return 0;
}

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Print all armstrong numbers below 1000 in java

Note: follow-up comment lines.

import java.io.*;
public class HelloWorld {
	static int sumCubes(int n) {
		int rv = 0;
		int temp = n;
		int digit;
		while (temp > 0) {
			digit = temp % 10;
			rv += Math.pow(digit, 3);
			temp = temp / 10;
		}

		return rv;
	}

	public static void main(String[] args) {
		System.out.println("Armstrong Numbers below 1000 in java ");
		 for( int i=0 ;i<10000000 ;i++ )
		 {
			 if( i==sumCubes(i) )
				 System.out.println(i);
		 }	 
	}
}

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