How to check if all the digits of a number are odd in python?

Question

I was told to solve a problem in which I would have to find out the number of 4-digit numbers which are all composed of odd digits. I tried the following python code:

Answer 1

You could try something like this

def isAllOdd(num):
  if num < 10: return num % 2 == 1;
  return isAllOdd(num % 10) and isAllOdd(int(num / 10))

Answer 2

I believe something like this would work if you're looking to model what you already have, but I echo the comment by yotommy that some intuitive multiplication would do the trick.

 for a in range(1111,10000):
       allOdd = True
       for b in str(a):
           if int(b) % 2 == 0:
               allOdd = False
       if(allOdd):
          new_list.append(a)

Answer 3

for i in range(1000,3001):
  s=str(i)
  if (int(s[0])%2==0 and int(s[1])%2==0 and int(s[2])%2==0 and int(s[3])%2==0):
      print(i,end=",")

Answer 4

A 4 digit number that is composed only of odd digits can only use the digits 1, 3, 5, 7 and 9. That gives you 5 to the power 4 is 625 different numbers. That didn't require trying them all out.

You can still do that of course, using itertools.product() for example:

from itertools import product

print sum(1 for combo in product('13579', repeat=4))

because product('13579', repeat=4) will produce all possible combinations of 4 characters from the string of odd digits.

Your code needs to test if all digits are odd; break out early if any digit is not odd:

new_list =[] # A list which holds the numbers
for a in range(1111,10000):
    for b in str(a):
        if int(b) % 2 == 0:
            break
    else:
        # only executed if the loop wasn't broken out of
        new_list.append(a)

You could use the all() function with a generator expression for that test too:

new_list = []
for a in range(1111,10000):
    if all(int(b) % 2 == 1 for b in str(a)):
        new_list.append(a)

which then can be collapsed into a list comprehension:

new_list = [a for a in range(1111,10000) if all(int(b) % 2 == 1 for b in str(a))]