A detailed guide on Armstrong Numbers in Java. Learn what an Armstrong number is, how it works, and how to write efficient Java programs to check Armstrong numbers using loops and mathematical logic. Includes clear explanations, examples, and optimized Java code.

Armstrong Number in Java

Last updated: February 8, 2025

An Armstrong Number (also known as a Narcissistic Number) is a number that is equal to the sum of its own digits each raised to the power of the number of digits. In this blog, we will understand the concept and learn how to check Armstrong numbers using Java.

What is an Armstrong Number?

A number with n digits is an Armstrong number if:

sum = digit₁ⁿ + digit₂ⁿ + digit₃ⁿ + ... + digitₙⁿ

If the sum equals the original number, it is an Armstrong number.

Examples

153 → 1³ + 5³ + 3³ = 153
370 → 3³ + 7³ + 0³ = 370
9474 → 9⁴ + 4⁴ + 7⁴ + 4⁴ = 9474

Java Program to Check Armstrong Number

Below is a simple Java program using loops and math logic:


import java.util.Scanner;

public class ArmstrongNumber {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        System.out.print("Enter a number: ");
        int num = sc.nextInt();

        int original = num;
        int sum = 0;

        // count digits
        int digits = String.valueOf(num).length();

        while (num > 0) {
            int digit = num % 10;
            sum += Math.pow(digit, digits);
            num /= 10;
        }

        if (sum == original) {
            System.out.println(original + " is an Armstrong Number.");
        } else {
            System.out.println(original + " is NOT an Armstrong Number.");
        }
    }
}

Optimized Version

A slightly optimized solution that avoids converting the number to string:


public static boolean isArmstrong(int num) {
    int digits = 0, temp = num;

    // count digits
    while (temp > 0) {
        temp /= 10;
        digits++;
    }

    int sum = 0;
    temp = num;

    while (temp > 0) {
        int digit = temp % 10;
        sum += Math.pow(digit, digits);
        temp /= 10;
    }

    return sum == num;
}

Conclusion