Even Fibonacci Numbers

To solve this coding challenge, we'll determine the sum of all natural numbers below 1000 that are multiples of either 3 or 5. By identifying and summing these multiples, we can find the required total. Let's walk through the solution step-by-step to understand the logic and breakdown the pseudocode reflecting the process.

Explanation

  1. Initialize a sum variable :
    • We'll need a variable to hold our accumulating sum. This ensures that every time we find a multiple of 3 or 5, we can add it to our running total.
  2. Loop through numbers from 0 to 999 :
    • We need to check every number below 1000 to see if it meets our criteria (i.e., being a multiple of either 3 or 5).
  3. Check if a number is a multiple of 3 or 5 :
    • Use the modulus operator to determine if a number is divisible by 3 or 5. If a number modulo 3 or 5 equals 0, it's a multiple of that number.
  4. Add valid multiples to the sum :
    • For every number that meets our criteria, we add it to our sum variable.
  5. Return the total sum :
    • After looping through all numbers below 1000 and accumulating valid multiples, the result stored in the sum variable is the desired total.

    Step-by-Step Explanation

  6. Initialization :
    • Start by setting up a sum variable (let's call it
      total_sum
      ).
  7. Iteration :
    • Use a loop (e.g.,
      for
      loop) to iterate through each number from 0 up to, but not including, 1000.
  8. Condition check :
    • Inside the loop, use an
      if
      statement to check if the current number is divisible by 3 or 5. This can be done using the condition
      number % 3 == 0
      or
      number % 5 == 0
      .
  9. Accumulate Sum :
    • If the condition is true (i.e., the number is a multiple of 3 or 5), add the number to
      total_sum
      .
  10. Output Result :
    • After completing the loop, the value of
      total_sum
      will be the answer.

Detailed Steps in Pseudocode

                                            
# Initialize sum variable
total_sum = 0

# Loop through all numbers below 1000
for number from 0 to 999:
    # Check if the number is a multiple of 3 or 5
    if number % 3 == 0 or number % 5 == 0:
        # Add the number to the sum
        total_sum = total_sum + number

# The total_sum now holds the sum of all multiples of 3 or 5 below 1000
return total_sum
  • Each part of the pseudocode corresponds to a specific part of the explanation to ensure clarity and understanding.
By following this step-by-step method, we can solve the given problem efficiently and understand clearly how we arrive at the solution. The key is to break the problem down into manageable parts – initializing, iterating, checking conditions, and accumulating results. This method ensures that we systematically solve the problem and have clear logic that can be easily translated into any programming language.