## Introduction

**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

To learn how to sort an integer in python, check out this tutorial on how to sort an integer.

```
# Taken from http://ataiva.com/how-to-sort-an-integer-in-python
def sort_asc(n):
return int("".join(sorted([i for i in str(n)])))
# Taken from http://ataiva.com/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
```