Based from what I understand, a positive integer $n$ is said to be Kaprekar if it goes back to itself after employing Kaprekar's Routine given below:

  1. Arrange the digits in descending and then in ascending order.

  2. Subtract the smaller number from the bigger number.

  3. Back at 1. and continue the process till we get $n$.

Example: 495 is a Kaprekar Number since $954-459=495$.

I noticed from this point that a palindrome of odd number of digits is not Kaprekar.

Example: Consider 272, by applying Kaprekar Routine we have: $722-227=495$ and surely after getting 495, we cant go back to 272. Thus 272 is not Kaprekar.

As a question, Is my observation always TRUE? How can I show it in general?

Thanks for your valuable comments and answers in advance.

  • I will write a proof for this by tomorrow (Or give a counterexample). Give me some time; I've have an exam tomorrow. – Kugelblitz Mar 9 '15 at 15:27
  • @Kugelblitz Thanks, will wait for your answer and God bless on your exam. – Jr Antalan Mar 9 '15 at 15:46
  • Thank you so much sir, it means a lot to me. – Kugelblitz Mar 9 '15 at 16:07
up vote 1 down vote accepted

It looks like the definition of a Kaprekar number is different from what you've understood sir... : http://en.wikipedia.org/wiki/Kaprekar_number

You wish to say 'Kaprekar's constants' arising out of the Kaprekar's routine.http://en.wikipedia.org/wiki/6174_(number)

I will update my post tomorrow after finding a proof/ counterexample for the conjecture you've made regarding the Kaprekar constants.

Edit: The Kaprekar Constant for three digit numbers seems to be 495 as evidenced by this, and palindromic numbers don't seem to be there...:

http://upload.wikimedia.org/wikipedia/commons/thumb/2/2d/KaprekarRoutineFlowGraph495.svg/1500px-KaprekarRoutineFlowGraph495.svg.png

So tomorrow I can surely come up with a formal proof. Good night or good day to you sir.


Edit: Alright. This is regarding the Kaprekar's constant. Kaprekar's number is something else entirely.

It seems that for the set of numbers containing 3 digits, 495 is the constant; that is, most numbers resolve to the number 495, except the following numbers

100, 101, 110, 111, 112, 121, 122, 211, 212, 221, 222, 223, 232, 233, 322, 323, 332, 333, 334, 343, 344, 433, 434, 443, 444, 445, 454, 455, 544, 545, 554, 555, 556, 565, 566, 655, 656, 665, 666, 667, 676, 677, 766, 767, 776, 777, 778, 787, 788, 877, 878

http://oeis.org/A090429 : This gives you the list for 3-digit numbers that do not resolve to 495 under the Kaprekar routine which I've copy pasted.

So we can conclude, that the only three digit number which results in itself after the Kaprekar Routine is 459. There is nothing to do with odd digit palindromes; from 100 to 999, only 459 follows the routine (i.e. not only do all 3-digit palindromes follow the routine to return to itself, no other number can do so).

With reference to five digit numbers, {(53955, 59994), (61974, 82962, 75933, 63954), (62964, 71973, 83952, 74943)} are possible cycles a number may end up in. So it's a collection of cycling constants in case of a five digit number.

For further amount of digits....not much research has been done yet.

Basically, there's no need for a proof as such for the three digit palindromes, because only 495 is the number which can come back to itself after the Routine (All three digit non-palindromes also do not follow the routine; they also end up in 495 like the palindromes, which then follows the routine).

But yes, one conjecture that can be made here, is that all non-repetitive odd digit palindromes (i.e. All digits are not equal to each and every other) eventually result in a Kaprekar constant, or a Kaprekar's cycling constants.

  • thank you again. Yeah that must be constant. – Jr Antalan Mar 9 '15 at 16:11
  • @JrAntalan I've updated my answer. – Kugelblitz Mar 9 '15 at 16:21
  • I will wait for your proof, thanks – Jr Antalan Mar 10 '15 at 2:53
  • @JrAntalan I've updated, but the answer isn't what I thought it would be :( Looks like there are no Kaprekar's constants besides 495 at all for 3 digit numbers! – Kugelblitz Mar 10 '15 at 10:31
  • I think we already have a proof. Thanks again. – Jr Antalan Mar 10 '15 at 13:18

All palindromic fixed points are binary and are from the same family, here are the first few palindromic fixed points.

$1001_2, 10101_2, 101101_2, 1011101_2, 10111101_2, 101111101_2, \dots$

There are no other palindromic Kaprekar Palindromes.

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