url: https://javaistic.vercel.app/blog

title: Java Program to Check Armstrong Number

shortTitle: Check Armstrong Number

description: Learn how to check if a number is an Armstrong number in Java using digit extraction and power calculation with multiple approaches and examples.

An Armstrong number (also called a narcissistic number) is a number that equals the sum of its digits each raised to the power of the number of digits. For example:

- 153 = 1³ + 5³ + 3³ = 1 + 125 + 27 = 153 (3 digits)

- 9474 = 9⁴ + 4⁴ + 7⁴ + 4⁴ = 6561 + 256 + 2401 + 256 = 9474 (4 digits)

Before you start, you may want to review:

- [Java Variables and Literals](/docs/variables-and-literals)

- [Java Operators](/docs/operators)

- [Java Basic Input and Output](/docs/basic-input-output)

This method extracts digits and calculates the sum of powers.

class Main {

public static void main(String[] args) {

int num = 153; // change value to test

if (isArmstrong(num)) {

System.out.println(num + " is an Armstrong number.");

} else {

System.out.println(num + " is not an Armstrong number.");

}

}

static boolean isArmstrong(int num) {

int original = num;

int digits = countDigits(num);

int sum = 0;

while (num > 0) {

int digit = num % 10;

sum += Math.pow(digit, digits);

num /= 10;

}

return original == sum;

}

static int countDigits(int num) {

if (num == 0) return 1;

int count = 0;

while (num > 0) {

count++;

num /= 10;

}

return count;

}

}

Avoid floating-point operations by using integer multiplication.

class Main {

public static void main(String[] args) {

int num = 9474; // change value to test

if (isArmstrongOptimized(num)) {

System.out.println(num + " is an Armstrong number.");

} else {

System.out.println(num + " is not an Armstrong number.");

}

}

static boolean isArmstrongOptimized(int num) {

int original = num;

int digits = countDigits(num);

int sum = 0;

while (num > 0) {

int digit = num % 10;

sum += power(digit, digits);

num /= 10;

}

return original == sum;

}

static int power(int base, int exp) {

int result = 1;

for (int i = 0; i < exp; i++) {

result *= base;

}

return result;

}

static int countDigits(int num) {

if (num == 0) return 1;

int count = 0;

while (num > 0) {

count++;

num /= 10;

}

return count;

}

}

Program to find all Armstrong numbers within a given range.

import java.util.ArrayList;

import java.util.List;

class Main {

public static void main(String[] args) {

int start = 1;

int end = 10000;

System.out.println("Armstrong numbers between " + start + " and " + end + ":");

List<Integer> armstrongNumbers = findArmstrongInRange(start, end);

for (int num : armstrongNumbers) {

System.out.println(num + " (digits: " + countDigits(num) + ")");

}

System.out.println("\nTotal Armstrong numbers found: " + armstrongNumbers.size());

}

static List<Integer> findArmstrongInRange(int start, int end) {

List<Integer> result = new ArrayList<>();

for (int i = start; i <= end; i++) {

if (isArmstrong(i)) {

result.add(i);

}

}

return result;

}

static boolean isArmstrong(int num) {

int original = num;

int digits = countDigits(num);

int sum = 0;

while (num > 0) {

int digit = num % 10;

sum += power(digit, digits);

num /= 10;

}

return original == sum;

}

static int power(int base, int exp) {

int result = 1;

for (int i = 0; i < exp; i++) {

result *= base;

}

return result;

}

static int countDigits(int num) {

if (num == 0) return 1;

int count = 0;

while (num > 0) {

count++;

num /= 10;

}

return count;

}

}

- **Common Armstrong numbers**: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 371, 407, 1634, 8208, 9474

- **Performance**: The optimized version using integer multiplication is faster than Math.pow()

- **Large numbers**: For numbers with many digits, consider using `long` or `BigInteger` to avoid overflow

- **Time complexity**: O(d) where d is the number of digits

- **Space complexity**: O(1) for the basic check, O(n) for range finding