Remove even digits from a number with only numerical operations, no list or strings

For this assignment, I am tasked to:

1. Input a number
2. Remove even digits from the given number without changing the order of the digits

For example: 123407 will print 137

If a number starts with 246 then it prints 0.

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I am not allowed to use strings, lists, functions, packages, recursion, …, only mathematical operations.

Here is my code so far:

POSITION = 1  # give the digit position of the  number 
OUTPUT = 0  # printing the output of the result 
enterNum = int(input('Enter a Number'))  # user inputs  a number 

while enterNum != 0:
    digit = enterNum % 10  # last digit of the number if number is odd 
    enterNum = enterNum / 10  # shift decimal places and remove the digit position 
    if digit % 2 == 1:
        OUTPUT += POSITION * digit  # if digit is odd, add it to the position 
    else:
        continue

print(OUTPUT)

I am quite lost on what to do. When the number is 123407, I can only go up to printing 7. I do not know how to store the number in the data type and combine the digits in the end to get 137. Furthermore, I have a hard time looping through the number. It just gives me 7.

>Solution :

Tested it out. Starting with your approach and just making a few modifications (see comments for details on modifications):

POSITION=0  # This starts at 0, this is because the power of 10 is 0 based (position 0 is the 1's place of the answer)
OUTPUT=0 
enterNum= int(input('Enter a Number'))

while enterNum!=0:
    digit=enterNum%10
    enterNum=enterNum/10 #This will work for python2 I think, but in python3 you will need enterNum//10
    if digit%2==1:
        OUTPUT+=(digit * 10**POSITION) # The digit needs to go in the current place value, if it is the 3rd digit, it needs to be multiplied by 100 (or 10^2).
        POSITION += 1 # Add 1 to the position since we found one (need to place the next OUTPUT digit at the next position)
    else:
        continue

print(OUTPUT)