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persistent_bugger.py
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persistent_bugger.py
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#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.
# 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
def persistence(n):
# your code
aux = str(n)
digit = False
numero = 1
cont = 0
if len(aux) > 1:
while digit == False:
for i in range(len(aux)):
numero *= int(aux[i])
cont += 1
if numero//10 < 1:
digit = True
aux = str(numero)
numero = 1
return cont
#persistence