6174 is known as Kaprekar’s constant after the Indian mathematician D. R. Kaprekar. This number is notable for the following rule:
- Take any four-digit number, using at least two different digits (leading zeros are allowed).
- Arrange the digits in descending and then in ascending order to get two four-digit numbers, adding leading zeros if necessary.
- Subtract the smaller number from the bigger number.
- Go back to step 2 and repeat.
The above process, known as Kaprekar’s routine, will always reach its fixed point, 6174, in at most 7 iterations. Once 6174 is reached, the process will continue yielding 7641 – 1467 = 6174. For example, choose 3524:5432 – 2345 = 30878730 – 0378 = 83528532 – 2358 = 61747641 – 1467 = 6174
The only four-digit numbers for which Kaprekar’s routine does not reach 6174 are repdigits such as 1111, which give the result 0000 after a single iteration. All other four-digit numbers eventually reach 6174 if leading zeros are used to keep the number of digits at 4.
Finding the number of times using Python
Code language: Python (python)
# Taken from http://ao.gl/how-to-sort-an-integer-in-python def sort_asc(n): return int("".join(sorted([i for i in str(n)]))) # Taken from http://ao.gl/how-to-sort-an-integer-in-python def sort_desc(n): return int("".join(sorted([i for i in str(n)], reverse=True))) def kaprekar_times(n): # count how many times count = 0 # set the last answer to compare against last = n # loop forever while True: # get the smallest and biggest smaller = sort_asc(last) bigger = sort_desc(last) # get the answer answer = bigger-smaller # increment count += 1 # return the count if it's a match if answer==last: return count else: # otherwise continue last = answer # return if all else fails return count # Some test cases print(kaprekar_times(9272)) # 6 print(kaprekar_times(1263)) # 8 print(kaprekar_times(9820)) # 8 print(kaprekar_times(2489)) # 4