Kaprekar's Constant

My professor showed us this in class today:

1. Take any 4 digit number with at least two different digits:

2. Arrange the numbers in ascending and descending order and subtract them

3. Use step two on the subtracted number

Within 7 times, you'll come up with the number 6174 and it will stop there.

For example, I'll pick 5218

8521 - 1258 = 7263

7632 - 2367 = 5265

6552 - 2556 = 3996

9963 - 3699 = 6264

6642 - 2466 = 4176

7641 - 1467 = 6174

I tried this on at least ten different numbers and they all turned out to 6174 in at least 7 steps.

Can anyone find a theorem or explanation for how it turns out to be 6174?

Sounds like hw ;)

Sounds like hw ;)

entropyecho

I ended up with 900 on the fourth go :?

9889 - 9889 = 0000

:(

EDIT: Ohh..I read the instruction wrong. Yeah I heard of this constant too, there are tons of cool formulas out there though.

I ended up with 900 on the fourth go :?

Big_Bad_Sad

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 (write it vertically, align the columns)

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.

Edit: Damn html

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.

entropyecho

[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.