Java Program to check Armstrong Number

Given a number x, determine whether the given number is Armstrong number or not. A positive integer of n digits is called an Armstrong number of order n (order is number of digits) if.

abcd... = pow(a,n) + pow(b,n) + pow(c,n) + pow(d,n) + .... 

Example:

Input : 153
Output : Yes
153 is an Armstrong number.
1*1*1 + 5*5*5 + 3*3*3 = 153

Input : 120
Output : No
120 is not a Armstrong number.
1*1*1 + 2*2*2 + 0*0*0 = 9

Input : 1253
Output : No
1253 is not a Armstrong Number
1*1*1*1 + 2*2*2*2 + 5*5*5*5 + 3*3*3*3 = 723

Input : 1634
Output : Yes
1*1*1*1 + 6*6*6*6 + 3*3*3*3 + 4*4*4*4 = 1634

Java

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// Java program to determine whether the number is
// Armstrong number or not
public class Armstrong
{
    /* Function to calculate x raised to the
       power y */
    int power(int x, long y)
    {
        if( y == 0)
            return 1;
        if (y%2 == 0)
            return power(x, y/2)*power(x, y/2);
        return x*power(x, y/2)*power(x, y/2);
    }
  
    /* Function to calculate order of the number */
    int order(int x)
    {
        int n = 0;
        while (x != 0)
        {
            n++;
            x = x/10;
        }
        return n;
    }
  
    // Function to check whether the given number is
    // Armstrong number or not
    boolean isArmstrong (int x)
    {
        // Calling order function
        int n = order(x);
        int temp=x, sum=0;
        while (temp!=0)
        {
            int r = temp%10;
            sum = sum + power(r,n);
            temp = temp/10;
        }
  
        // If satisfies Armstrong condition
        return (sum == x);
    }
  
    // Driver Program
    public static void main(String[] args)
    {
        Armstrong ob = new Armstrong();
        int x = 153;
        System.out.println(ob.isArmstrong(x));
        x = 1253;
        System.out.println(ob.isArmstrong(x));
    }
}

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