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Solution 6 to 12
Solution 7

Armstrong Numbers in a Range

Armstrong numbers are numbers that are equal to the sum of their own digits each raised to the power of the number of digits. For example, 153 is an Armstrong number because 1^3 + 5^3 + 3^3 = 153 . In this example, we will find all Armstrong numbers within a given range. The program prompts the user to enter a lower and upper bound for the range and then identifies and displays all Armstrong numbers within that range.

Code Breakdown

Algorithm

  1. Start
  2. Initialize variables: i, n, d, lb, ub, p, s, x, t.
  3. Prompt the user to enter the lower bound lb and upper bound ub.
  4. Loop through each number x from lb to ub:
    1. Count Digits:
      • Initialize n to count the number of digits in x.
      • For each digit of x, increase n by 1.
    2. Calculate Armstrong Sum:
      • Initialize s to 0.
      • For each digit of x:
        • Calculate p as the digit raised to the power of n.
        • Add p to s.
    3. Check Armstrong Number:
      • If s equals x, then x is an Armstrong number. Print x.
  5. End

Code Explanation

Example Flowchart

                                  Start
                                    |
                                    V
                       Input lowerBound, upperBound
                                    |
                                    V
                  For each number from lowerBound to upperBound
                                    |
                                    V
           Calculate the number of digits (numDigits) in current number
                                    |
                                    V
                  Initialize sum of powered digits (sum = 0)
                                    |
                                    V
                      +---------------------------+
                      | For each digit in number: |
                      | ------------------------- |
                      | Extract last digit:       |
                      | digit = number % 10       |
                      | Raise digit to power:     |
                      | power = digit^numDigits   |
                      | Add power to sum:         |
                      | sum += power              |
                      | Remove last digit:        |
                      | number /= 10              |
                      +---------------------------+
                                    |
                                    V
            Is sum of powered digits equal to the original number?
                     /                             \
                    /                               \
                  Yes                               No
                   |                                 |
                   V                                 V
      Print number as Armstrong number      Continue to next number
                   |                                 |
                   V                                 |
                  End                                |
                                                     |
                                                     V
                                                    End
Solution 6Solution 8