getting only the odd digits in a number with recursion

Question

So my problem is this i have an number like 123 and as the tilte sugests i want the result to be 13.

The problem is that firstly with the method im using im going to get

Answer 1

You could do it as follows -

def apenas_digitos_impares(n):
    if n == 0:
        return 0
    elif (n%10)%2 == 0:
        return apenas_digitos_impares(n//10)
    elif (n%10)%2 == 1:
        # Include the digit and recurse for remaining...
        return (n%10) + 10*apenas_digitos_impares(n//10)
        
print(apenas_digitos_impares(123))

OUTPUT :

13

The only change that your code needed was in the last line of the function.

• You will just include the odd digit(done by n%10) and,

• move on(or recurse) to check for remaining digits. You need to multiply next digit by 10, so - 10*apenas_digitos_impares(n//10)