text: 
  """CountDiv Practice Problem
  by Codility
  
  Solution by B Hamilton 7/24/2018
  
  given three integers A, B and K, return the number of integers
  within the range [A..B] that are divisible by K, i.e.:
  { i : A ≤ i ≤ B, i mod K = 0 }
  
  For example, for A = 6, B = 11 and K = 2,
  function should return 3,
  because there are three numbers divisible by 2 within the range [6..11],
  namely 6, 8 and 10.
  
  Assume that:
  A and B are integers within the range [0..2,000,000,000];
  K is an integer within the range [1..2,000,000,000];
  A ≤ B.
  Complexity:
  
  expected worst-case time complexity is O(1);
  expected worst-case space complexity is O(1).
  
  """
  
  # you can write to stdout for debugging purposes, e.g.
  # print("this is a debug message")
  
  def solution(A, B, K):
      # write your code in Python 3.6
      # This is in the "prefix sum" section
      # But I don't think we need a prefix sum
      # We just need the values and
      # Intuition
  
      return ( (B - A)// K  + 1)
  
  
  """
  RESULT:
  50% of Task Score
  75% of Performance
  25% of Correctess
  Having difficulty with this one. Will need to research further!
  Failed on: A=B in {0,1}, K = 11 for A=B=1 got 1 expected 0
  - A=10, B = 10, K in {5, 7, 20} for 5 and 20 (got 1 expected 0
  )
  Failed o A=101, B = 123M, K = 10K  (Got 12346 expected 12345)
  """
  
  
  def solution2(A, B, K):
      # write your code in Python 3.6
      # B is inclusive.... so 0 to 10k is 10k+1
  
      # Floor returns the integer part of the division result
      # e.g. 9 // 2 = 4
  
      # Modulo returns remainder part of the division result
      # e.g. 9 % 2 = 1
  
      # edge cases: A, B small (1)...
      # edge case A,B same
      # edge case: B >> A  (0 to 10k, 1)
      # edge case B > A, Large K
  
      # the max(A%K,2) is saying it will take the highest
      # max(A%K,K) means "give the remainder of A%K,
      # but count 0 as a full K instead of 0
      return ( ((B-A) + max(A%K, K)) // K)
  
  """RESULT: Exactly the same. """
  