Write a program using integers user_num and x as input, and output user_num divided by x three times. Ex: If the input is: 2000 2 Then the output is: 1000 500 250

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```python # Define a function to perform the division operation def divide_user_num_by_x(user_num, x): # Initialize an empty list to store the results results = [] # Loop to perform the division three times for _ in range(3): # Divide user_num by x and store the result user_num = user_num // x # Append the result to the list results.append(user_num) # Return the list of results return results ``` ```python # Main function that tests the divide_user_num_by_x function def main(): # Test the divide_user_num_by_x function # Use case 1 assert divide_user_num_by_x(2000, 2) == [1000, 500, 250] # Use case 2 assert divide_user_num_by_x(1500, 3) == [500, 166, 55] # Call the main function to execute the tests main() ``` ∻Explanation∻ ⚹The provided code defines a function that takes two integers as input and performs integer division of the first number by the second number three times, each time updating the first number with the result of the division. The results are stored in a list and returned. The main function tests this behavior with two use cases.⚹ ∻Step-by-step Instruction∻ ‖ ▪ ⋮ Define the main function required here: perform integer division of user_num by x three times and store the results. ‖ ‖ ▪ ⋮ Use a loop to perform the division three times and update user_num with the result after each division. ‖ ‖ ▪ ⋮ Store each result in a list to keep track of the division results. ‖ ‖ ▪ ⋮ Define concrete use cases and testing cases for the function: when user calls divide_user_num_by_x(2000, 2), it should return [1000, 500, 250]. ‖ ∻Time Complexity∻ ⚹The time complexity is O(1) because the loop runs a constant three times, regardless of the input size.⚹ ∻Space Complexity∻ ⚹The space complexity is O(1) as well, since the output list always contains three integers, which does not change with the size of the input.⚹

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