Write a function, `persistence`, that takes in a positive parameter `num` and returns its multiplicative persistence, which is the number of times you must multiply the digits in `num` until you reach a single digit.

For example:

```python
 persistence(39) => 3  # Because 3*9 = 27, 2*7 = 14, 1*4=4
                       # and 4 has only one digit.
                 
 persistence(999) => 4 # Because 9*9*9 = 729, 7*2*9 = 126,
                       # 1*2*6 = 12, and finally 1*2 = 2.
                  
 persistence(4) => 0   # Because 4 is already a one-digit number.
```
```python
 persistence(39) # returns 3, because 3*9=27, 2*7=14, 1*4=4
                 # and 4 has only one digit
                  
 persistence(999) # returns 4, because 9*9*9=729, 7*2*9=126,
                  # 1*2*6=12, and finally 1*2=2

 persistence(4) # returns 0, because 4 is already a one-digit number
```

Difficulty: introductory

Starter code:

def persistence(n):
	

Solutions:
def persistence(n):
    n = str(n)
    count = 0
    while len(n) > 1:
        p = 1
        for i in n:
            p *= int(i)
        n = str(p)
        count += 1
    return count
    # your code


def persistence(n):
    if n < 10: return 0
    mult = 1
    while(n > 0):
        mult = n%10 * mult
        n = n//10
    return persistence(mult) + 1

def persistence(n):
    if str(n) == 1:
        return 0
    count = 0
    while len(str(n)) > 1:
        total = 1
        for i in str(n):
            total *= int(i)
        n = total
        count += 1
    return count

def multiply_digits(n):
    product = 1;
    while(n > 0):
        product *= n % 10;
        n //= 10;
        
    return product;

def persistence(n):
    count = 0;
    while(n > 9):
        n = multiply_digits(n);
        count += 1;
        
    return count;