[bonanova]

I thought of a four-digit number, p = abcd; where each of a, b, c, d are in the set [0, 9].

Boy, was that fun.

But how interesting can a four-digit number be?

Soon I tired of that number; I wanted a new one.

So I made q = klmn, where k, l, m, n are the down-ordering of a, b, c, d.

But that looked too familiar, having the same digits and all, and so did r = nmlk.

So I subtracted the two: the smaller from the larger, and ended up with s = |q-r| = uvwx.

Great, I thought, and I started thinking about s.

But s looked even more familiar than q and r did.

No wonder: s turned out to be exactly equal to my first number, p.

If I give you the hint - if you think you need one - that b < c, can you tell me my original number p?

[Guest]

Just to clarify, by down-ordering you mean rearragning in descending order? so if, a<b<c<d then k=d, l=c, m=b and n=a??

[bonanova]

  DeeGee said:

Just to clarify, by down-ordering you mean rearragning in descending order? so if, a<b<c<d then k=d, l=c, m=b and n=a??

Yes.

If p were 4736 then q = 7643, r = 3467 and s = 4176.

[Guest]

  Reveal hidden contents

The number originally chosen was 6174 (abcd)

q = KLMN then is 7641

r = NMLK is 1467

s = q-r = 6174

[bonanova]

Kudos. That's correct.

The process of changing p = abcd into s = uvwx has an interesting property.

Start with any p [where abcd are not all the same] to get s = uvwx.

Continue the process, this time using s.

Repeated application of the process always leads to a final result of 6174.

Which gives rise to two questions:

1. How many repetitions are sufficient?
   
2. What do the initials D.R.K. have to do with any of this?
   

[Guest]

I like this problem, but I'm not sure how special we can say 6174 is. Surely, it simply has an order of 1 under this function.

Could we not say that the sequence 81,63,27,45,9 is just as special because it repeats under this function for 2 digit numbers?

Or that 495 is special because it also it also is order 1 for 3 digit numbers?

As for your questions, you are looking for:

  Reveal hidden contents

the Indian mathematician Kaprekar (Wikipedia entry)

but I can't give you an answer to the convergence rate to 6174. Though I can do 3-digits, I do not have the energy to try four:

  Reveal hidden contents

taking a number with digits a, b and c, arbitrarily such that a >=b >= c and a != b != c (that's a not equals sign!), then we can at least say a > c.

so,

.	a b c

.  - c b a

.---------

.	A B C


C = 10 + c - a (because a>c)

B = 10 + b - 1 - b = 9

A = a - 1 - c


leaving,


A + C = 9

B = 9


after only one iteration, which is a fairly small subset of 3-digit numbers.


And for every iteration except the first one, we know the figure must have a 9 (because B = 9), so that a = 9, and running this through gives


A = 8 - c

B = 9

C = c + 1


and so with each iteration, c increments by 1 until C = 5 and A = 4, a stable solution. Notice that b is irrelevant in convergence to 495.

Therefore, the longest sequence we could generate is (working backwards):

495

594

693

792

891

990

there is no 3 digit number that could generate the digits 9, 9, and 0 as it would require a - c = 10. Therefore, I propose that 5 steps is the longest it could take.

Unfortunately, because 6174 is not symetrical we do not have the advantage of 'b' being dropped and so I imagine that it becomes significantly hardert to establish the convergence for 4 digits.....I'd like to hear a solution though....

[Guest]

And having read this article, I'm intruiged by the table showing how many iterations it takes to get to 6174:

Iteration Frequency

0 1

1 356

2 519

3 2124

4 1124

5 1379

6 1508

7 1980

(aside: any reason why [table] tags don't seem to work?)

Why is there such a peak at 3 iterations? Is there a subset of 3-digit numbers that have a particularly easyt route to 6174?