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Java Program to Check Armstrong Number

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:

1) Basic Armstrong Check

This method extracts digits and calculates the sum of powers.

Example

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;
  }
}

Output (for num = 153)

153 is an Armstrong number.

2) Optimized Version (Without Math.pow)

Avoid floating-point operations by using integer multiplication.

Example

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;
  }
}

Output (for num = 9474)

9474 is an Armstrong number.

3) Find Armstrong Numbers in Range

Program to find all Armstrong numbers within a given range.

Example

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;
  }
}

Sample Output

Armstrong numbers between 1 and 10000:
1 (digits: 1)
2 (digits: 1)
3 (digits: 1)
4 (digits: 1)
5 (digits: 1)
6 (digits: 1)
7 (digits: 1)
8 (digits: 1)
9 (digits: 1)
153 (digits: 3)
371 (digits: 3)
407 (digits: 3)
1634 (digits: 4)
8208 (digits: 4)
9474 (digits: 4)

Total Armstrong numbers found: 15

Notes & Tips

  • 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