title: ISBN final digit finder
body: <p>I am working in python 3 and I am making a program that will take in a 10 digit ISBN Number and applying a method to it to find the 11th number.</p>
  
  <p>Here is my current code</p>
  
  <pre><code>ISBN=input('Please enter the 10 digit number: ')
  
  while len(ISBN)!= 10:
  
      print('Please make sure you have entered a number which is exactly 10 characters long.')
      ISBN=int(input('Please enter the 10 digit number: '))
      continue
  
  
  else:
  
      Digit1=int(ISBN[0])*11
      Digit2=int(ISBN[1])*10
      Digit3=int(ISBN[2])*9
      Digit4=int(ISBN[3])*8
      Digit5=int(ISBN[4])*7
      Digit6=int(ISBN[5])*6
      Digit7=int(ISBN[6])*5
      Digit8=int(ISBN[7])*4
      Digit9=int(ISBN[8])*3
      Digit10=int(ISBN[9])*2
      Sum=(Digit1+Digit2+Digit3+Digit4+Digit5+Digit6+Digit7+Digit8+Digit9+Digit10)
      Mod=Sum%11
      Digit11=11-Mod
      if Digit11==10:
         Digit11='X'
      ISBNNumber=str(ISBN)+str(Digit11)
      print('Your 11 digit ISBN Number is ' + ISBNNumber)
  </code></pre>
  
  <p>I want to create some kind of loop so that the number after "Digit" for the variable name increases starting from 1 (or zero if it makes life easier), the number in the square brackets increases starting from 0 and the multiplication number to decrease from 11 to 2.</p>
  
  <p>Is there any way of doing this code in a more efficient way?</p>