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 == 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 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 == 1:
return n
else:
return n * factorial(n - 2)
print("The answer is", factorial(7))

Answer 1

To modify the program for calculating the product of all even numbers less than 7, we need a function that effectively identifies and multiplies the even numbers (in this case, 2, 4, and 6) within the specified range.

The most suitable function from the given options is:

```python
def factorial(n):
if n <= 1:
return n
else:
return n * factorial(n - 2)
print("The answer is", factorial(7))
```

However, this function is not specifically designed for the even numbers; it will calculate a product starting at `n` and going down by 2 each time, but it does not check if `n` is even.

A more explicit and accurate function for calculating the product of all even numbers less than 7 would look like this:

```python
def product_of_evens(n):
if n <= 2:
return 1 # Base case: return 1 for even numbers less than 2
if n % 2 != 0: # ignore odd numbers
n -= 1
return n * product_of_evens(n - 2)

print("The product of all even numbers less than", 7, "is", product_of_evens(6)) # Starts with 6 (the highest even number less than 7)
```

However, since this option was not included, the closest option that gives the product of numbers in a similar pattern (though will not output just evens accurately) is the following:

```python
def factorial(n):
if n <= 1:
return n
else:
return n * factorial(n - 2)
print("The answer is", factorial(7))
```

But in practice, none of the provided options accurately provides the solution for the product of all even numbers less than 7. The ideal function should loop through even numbers or recursively calculate as shown above.