How to maintain the value of the variable in the field (number). Python

[nishika]

I've solved the task. It needs to find a multiplicative stable, which is the number of times you have to multiply the numbers until you reach the only number. The number is 39. 3*9 = 27. 2*7 = 14. Then 1*4 = 4. And stop. 3. Only one agrarian(number) is given. Here's my code:

from functools import reduce
amount - total length of blocks
67 - number
['6','7']
def rec(number, *args):

summ = args

amount = 0

t1 = tuple(map(int, list(str(number))))

l1 = len(t1)

amount += l1

lst = list(summ)

lst.append(1)

if amount > 1:

result1 = reduce(lambda x, y: x * y, t1)

rec(result1, lst)

else:

return sum(list(args))

[shizuka]

There's no other option with a double class. The significance of the variable is maintained through the positioning argument (the)y in this example:

def rec(x, y=0): 
    f = lambda i: i if i<10 else f(i//10) * (i%10)
    return y if x<10 else rec(f(x), y+1)
rec(39)  # 3