[QUOTE="Cataclism"]

[QUOTE="SpinoRaptor24"]

Still doesn't explain where life originated from.

SpinoRaptor24

No, it doesn't. But that's not what we are discussing here now is it?

*shrugs*

Still feel like it is somewhat related to this discussion.

[QUOTE="entropyecho"]

In that case, consider this:

The digits of any four digit number can be arranged into a maximum number by putting the digits in descending order, and a minimum number by putting them in ascending order. So for four digits a,b,c,d where,

9 ≥ a ≥ b ≥ c ≥ d ≥ 0

and a, b, c, d are not all the same digit, the maximum number is abcd and the minimum is dcba.

We can calculate the result of Kaprekar's operation using the standard method of subtraction applied to each column of this problem, i.e.

abcd -dcba

---

ABCD

which gives the relations

D = 10 + d - a (as a >d) C = 10 + c - 1 - b = 9 + c - b (as b >c - 1) B = b - 1 - c (as b >c) A = a - d

for those numbers where a>b>c>d.

A number will be repeated under Kaprekar's operation if the resulting number ABCD can be written using the initial four digits a,b,c and d. So we can find the kernels of Kaprekar's operation by considering all the possible combinations of {a,b,c,d} and checking if they satisfy the relations above. Each of the 4! = 24 combinations gives a system of four simultaneous equations with four unknowns, so we should be able to solve this system for a, b, c and d.

It turns out that only one of these combinations has integer solutions that satisfy 9 ≥ a ≥ b ≥ c ≥ d ≥ 0. That combination is ABCD = bdac, and the solution to the simultaneous equations is a=7, b=6, c=4 and d=1. That is ABCD = 6174. There are no valid solutions to the simultaneous equations resulting from some of the digits in {a,b,c,d} being equal.

cjek

I don't 'get' British humour.