Algorithm


  1. Start
  2. Accept the lower and upper limits from the user.
  3. Loop through each number in the given range (from lower limit to upper limit). a. Initialize sum to 0. b. Calculate the number of digits in the current number. c. Temporarily store the current number in a variable. d. Repeat until the temporary number becomes 0: i. Extract the last digit. ii. Raise the extracted digit to the power of the number of digits. iii. Add the result to the sum. iv. Remove the last digit from the temporary number. e. If the sum is equal to the original number, print it as an Armstrong number.
  4. End

 

Code Examples

#1 Code Example- Armstrong Numbers Between Two Integers

Code - Java Programming

class Main {
  public static void main(String[] args) {

    int low = 999, high = 99999;

    for(int number = low + 1; number  <  high; ++number) {
      int digits = 0;
      int result = 0;
      int originalNumber = number;

      // number of digits calculation
      while (originalNumber != 0) {
        originalNumber /= 10;
        ++digits;
      }

      originalNumber = number;

      // result contains sum of nth power of its digits
      while (originalNumber != 0) {
        int remainder = originalNumber % 10;
        result += Math.pow(remainder, digits);
        originalNumber /= 10;
      }

      if (result == number) {
        System.out.print(number + " ");
      }
    }
  }
}
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Output

x
+
cmd
1634 8208 9474 54748 92727 93084
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Demonstration


Java Programing Example to Display Armstrong Number Between Two Intervals-DevsEnv