Question

Recursive Algorithms Quick Check 1 of 51 of 5 Items Question Consider the recursive function for the calculation of a factorial. def factorial(n): if n == 1: return n else: return n * factorial(n - 1) print("The factorial of", 7, "is", factorial(7)) How could the program be modified to print the product of all the even numbers less than 7? (1 point) Responses def factorial(n): if n <= 1: return 1 else: if n % 2 != 0: return n * factorial(n - 2) else: return factorial(n - 1) print("The answer is", factorial(7)) def factorial(n): if n <= 1: return 1 else: if n % 2 != 0: return n * factorial(n - 2) else: return factorial(n - 1) print("The answer is", factorial(7)) def factorial(n): if n == 1: return n else: return n * factorial(n - 2) print("The answer is", factorial(7)) def factorial(n): if n == 1: return n else: return n * factorial(n - 2) print("The answer is", factorial(7)) def factorial(n): if n == 0: return 1 else: if n % 2 == 0: return n * factorial(n - 2) else: return factorial(n - 1) print("The answer is", factorial(7)) def factorial(n): if n == 0: return 1 else: if n % 2 == 0: return n * factorial(n - 2) else: return factorial(n - 1) print("The answer is", factorial(7)) def factorial(n): if n <= 1: return n else: return n * factorial(n - 2) print("The answer is", factorial(7)) def factorial(n): if n <= 1: return n else: return n * factorial(n - 2) print("The answer is", factorial(7)) Skip to navigation page 9 of 9

Answer 1

To modify the program to compute the product of all even numbers less than 7, we want the function to return the product of all even numbers up to (but not including) 7.

Here's a step-by-step breakdown of what changes need to be made:

1. We want to handle the even numbers specifically, so we'll only include those numbers.
2. The base case should check when the number is less than or equal to 0.
3. If the number is even, multiply it by the result of the function called with the next even number down.
4. If the number is odd, just call the function with the next number down, effectively skipping it.

Given this explanation, the correct implementation of the function would look something like this:

def factorial(n):
    if n <= 0:
        return 1  # Base case: return 1 when n is less than or equal to 0
    else:
        if n % 2 == 0:  # Check if n is even
            return n * factorial(n - 2)  # Multiply n with the product of smaller even numbers
        else:
            return factorial(n - 1)  # If n is odd, skip it

print("The answer is", factorial(6))  # Call with 6 since we want evens less than 7

Among your provided options, the closest appropriate choice to achieve this goal is:

def factorial(n):
    if n == 0:
        return 1
    else:
        if n % 2 == 0:
            return n * factorial(n - 2)
        else:
            return factorial(n - 1)

print("The answer is", factorial(7))

This correctly uses the even check, operates recursively, and handles odd numbers appropriately. It should indeed compute the product of all even numbers less than 7 when called with factorial(6). So, the answer is:

def factorial(n): 
    if n == 0: 
        return 1 
    else: 
        if n % 2 == 0: 
            return n * factorial(n - 2) 
        else: 
            return factorial(n - 1) 

print("The answer is", factorial(7))