persistent_bugger.py

# 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:
# 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.


def persistence(n):
    print('init', n)
    res = n
    count = 0
    
    while res >= 10:
        coll = []
        temp = 1
        
        for digit in str(res):
            coll.append (int(digit))
            
        for num in coll:
            temp = temp*num
            
        count += 1
        res = temp
    return count