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# How to check if a number is Armstrong in Java

Harsh Jain

In order to understand if a number is an Armstrong number in Python, one should be well versed with the following topics:

• Conditional statements
• Loops

### What is an Armstrong number?

A positive integer is called an Armstrong number of order n if it is equal to the sum of the powers of its own digits. For example:

abc…(up to n digits) = a^n + b^n +c^n ...(up to n digits)


In the case of the Armstrong number of 3 digits, it is equal to the sum of the cubes for each digit. For example, the number 407 = 4 x 4 x 4 + 0 x 0 x 0 + 7 x 7 x 7 ⇒ 407 is an Armstrong number.

### Code

Let’s take a look at the code below.

import java.util.Scanner;
import java.lang.Math;
class CheckArmstrongNumber
{
static boolean isArmstrong(int n)
{
int temp, digits=0, last=0, sum=0;
temp = n;
while(temp>0)
{
temp = temp/10;
digits++;
}
temp = n;
while(temp>0)
{
last = temp % 10;
sum +=  (Math.pow(last, digits));
temp = temp/10;
}
if(n == sum)
return true;
else
return false;
}

public static void main(String args[])
{
int num;
Scanner sc= new Scanner(System.in);
num = sc.nextInt();

if(isArmstrong(num))
System.out.print(num + " is Armstrong.");
else
System.out.print(num + " is not Armstrong.");
}
}  

Enter the input below

Check whether a number is Armstrong or not in Java

### Explanation

• In lines 1 and 2, we import the required header files.

• In lines 5 to 25, we create a function in which we run a loop to find the sum of the power of the order of each digit. We obtain the unit digit each time with the modulus operator %, since the remainder of the number when it is divided by 10 is the last digit of that number.

• In lines 15 to 20, we take the power of the order of the number with the pow() function.

• In lines 29 to 31, we read the input number from the user.

• In lines 33 to 37, we call the isArmstrong() function to check whether the number is Armstrong and print the statement accordingly.

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