Kaprekar Process and Palindromes.Inclusion-exclusion principle problemsnumber of $n$-digit palindromesCommon factors for all palindromesProof of convergence of Kaprekar's ConstantCan anyone extend my findings for Langford Pairings?Expressions evaluating to palindromesPalindromes and LCMTheory of removing alternate digitsPalindromes in multiple basesUnexpected pattern in consecutive 7 digit “double” palindromes?

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Kaprekar Process and Palindromes.


Inclusion-exclusion principle problemsnumber of $n$-digit palindromesCommon factors for all palindromesProof of convergence of Kaprekar's ConstantCan anyone extend my findings for Langford Pairings?Expressions evaluating to palindromesPalindromes and LCMTheory of removing alternate digitsPalindromes in multiple basesUnexpected pattern in consecutive 7 digit “double” palindromes?













3












$begingroup$


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.










share|cite|improve this question











$endgroup$











  • $begingroup$
    I will write a proof for this by tomorrow (Or give a counterexample). Give me some time; I've have an exam tomorrow.
    $endgroup$
    – Kugelblitz
    Mar 9 '15 at 15:27











  • $begingroup$
    @Kugelblitz Thanks, will wait for your answer and God bless on your exam.
    $endgroup$
    – Jr Antalan
    Mar 9 '15 at 15:46










  • $begingroup$
    Thank you so much sir, it means a lot to me.
    $endgroup$
    – Kugelblitz
    Mar 9 '15 at 16:07










  • $begingroup$
    @JrAntalan I have posted all the solutions for you. My final year project is on the Kaprekar routine and after seeing this question I decided to find out the answer for you.
    $endgroup$
    – Ben Crossley
    18 hours ago















3












$begingroup$


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.










share|cite|improve this question











$endgroup$











  • $begingroup$
    I will write a proof for this by tomorrow (Or give a counterexample). Give me some time; I've have an exam tomorrow.
    $endgroup$
    – Kugelblitz
    Mar 9 '15 at 15:27











  • $begingroup$
    @Kugelblitz Thanks, will wait for your answer and God bless on your exam.
    $endgroup$
    – Jr Antalan
    Mar 9 '15 at 15:46










  • $begingroup$
    Thank you so much sir, it means a lot to me.
    $endgroup$
    – Kugelblitz
    Mar 9 '15 at 16:07










  • $begingroup$
    @JrAntalan I have posted all the solutions for you. My final year project is on the Kaprekar routine and after seeing this question I decided to find out the answer for you.
    $endgroup$
    – Ben Crossley
    18 hours ago













3












3








3


1



$begingroup$


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.










share|cite|improve this question











$endgroup$




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.







combinatorics elementary-number-theory decimal-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 18 hours ago









nbarto

14.1k32682




14.1k32682










asked Mar 9 '15 at 15:07









Jr AntalanJr Antalan

1,3071822




1,3071822











  • $begingroup$
    I will write a proof for this by tomorrow (Or give a counterexample). Give me some time; I've have an exam tomorrow.
    $endgroup$
    – Kugelblitz
    Mar 9 '15 at 15:27











  • $begingroup$
    @Kugelblitz Thanks, will wait for your answer and God bless on your exam.
    $endgroup$
    – Jr Antalan
    Mar 9 '15 at 15:46










  • $begingroup$
    Thank you so much sir, it means a lot to me.
    $endgroup$
    – Kugelblitz
    Mar 9 '15 at 16:07










  • $begingroup$
    @JrAntalan I have posted all the solutions for you. My final year project is on the Kaprekar routine and after seeing this question I decided to find out the answer for you.
    $endgroup$
    – Ben Crossley
    18 hours ago
















  • $begingroup$
    I will write a proof for this by tomorrow (Or give a counterexample). Give me some time; I've have an exam tomorrow.
    $endgroup$
    – Kugelblitz
    Mar 9 '15 at 15:27











  • $begingroup$
    @Kugelblitz Thanks, will wait for your answer and God bless on your exam.
    $endgroup$
    – Jr Antalan
    Mar 9 '15 at 15:46










  • $begingroup$
    Thank you so much sir, it means a lot to me.
    $endgroup$
    – Kugelblitz
    Mar 9 '15 at 16:07










  • $begingroup$
    @JrAntalan I have posted all the solutions for you. My final year project is on the Kaprekar routine and after seeing this question I decided to find out the answer for you.
    $endgroup$
    – Ben Crossley
    18 hours ago















$begingroup$
I will write a proof for this by tomorrow (Or give a counterexample). Give me some time; I've have an exam tomorrow.
$endgroup$
– Kugelblitz
Mar 9 '15 at 15:27





$begingroup$
I will write a proof for this by tomorrow (Or give a counterexample). Give me some time; I've have an exam tomorrow.
$endgroup$
– Kugelblitz
Mar 9 '15 at 15:27













$begingroup$
@Kugelblitz Thanks, will wait for your answer and God bless on your exam.
$endgroup$
– Jr Antalan
Mar 9 '15 at 15:46




$begingroup$
@Kugelblitz Thanks, will wait for your answer and God bless on your exam.
$endgroup$
– Jr Antalan
Mar 9 '15 at 15:46












$begingroup$
Thank you so much sir, it means a lot to me.
$endgroup$
– Kugelblitz
Mar 9 '15 at 16:07




$begingroup$
Thank you so much sir, it means a lot to me.
$endgroup$
– Kugelblitz
Mar 9 '15 at 16:07












$begingroup$
@JrAntalan I have posted all the solutions for you. My final year project is on the Kaprekar routine and after seeing this question I decided to find out the answer for you.
$endgroup$
– Ben Crossley
18 hours ago




$begingroup$
@JrAntalan I have posted all the solutions for you. My final year project is on the Kaprekar routine and after seeing this question I decided to find out the answer for you.
$endgroup$
– Ben Crossley
18 hours ago










2 Answers
2






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    thank you again. Yeah that must be constant.
    $endgroup$
    – Jr Antalan
    Mar 9 '15 at 16:11










  • $begingroup$
    @JrAntalan I've updated my answer.
    $endgroup$
    – Kugelblitz
    Mar 9 '15 at 16:21










  • $begingroup$
    I will wait for your proof, thanks
    $endgroup$
    – Jr Antalan
    Mar 10 '15 at 2:53










  • $begingroup$
    @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!
    $endgroup$
    – Kugelblitz
    Mar 10 '15 at 10:31










  • $begingroup$
    I think we already have a proof. Thanks again.
    $endgroup$
    – Jr Antalan
    Mar 10 '15 at 13:18


















0












$begingroup$

All palindromic fixed points:



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



$101101_2, 213312_4, 325523_6, 437734_8,549945, dots$



There are no other palindromic Kaprekar Palindromes.






share|cite|improve this answer











$endgroup$












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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      thank you again. Yeah that must be constant.
      $endgroup$
      – Jr Antalan
      Mar 9 '15 at 16:11










    • $begingroup$
      @JrAntalan I've updated my answer.
      $endgroup$
      – Kugelblitz
      Mar 9 '15 at 16:21










    • $begingroup$
      I will wait for your proof, thanks
      $endgroup$
      – Jr Antalan
      Mar 10 '15 at 2:53










    • $begingroup$
      @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!
      $endgroup$
      – Kugelblitz
      Mar 10 '15 at 10:31










    • $begingroup$
      I think we already have a proof. Thanks again.
      $endgroup$
      – Jr Antalan
      Mar 10 '15 at 13:18















    1












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      thank you again. Yeah that must be constant.
      $endgroup$
      – Jr Antalan
      Mar 9 '15 at 16:11










    • $begingroup$
      @JrAntalan I've updated my answer.
      $endgroup$
      – Kugelblitz
      Mar 9 '15 at 16:21










    • $begingroup$
      I will wait for your proof, thanks
      $endgroup$
      – Jr Antalan
      Mar 10 '15 at 2:53










    • $begingroup$
      @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!
      $endgroup$
      – Kugelblitz
      Mar 10 '15 at 10:31










    • $begingroup$
      I think we already have a proof. Thanks again.
      $endgroup$
      – Jr Antalan
      Mar 10 '15 at 13:18













    1












    1








    1





    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$



    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Oct 11 '17 at 18:06

























    answered Mar 9 '15 at 16:09









    KugelblitzKugelblitz

    3,3301449




    3,3301449











    • $begingroup$
      thank you again. Yeah that must be constant.
      $endgroup$
      – Jr Antalan
      Mar 9 '15 at 16:11










    • $begingroup$
      @JrAntalan I've updated my answer.
      $endgroup$
      – Kugelblitz
      Mar 9 '15 at 16:21










    • $begingroup$
      I will wait for your proof, thanks
      $endgroup$
      – Jr Antalan
      Mar 10 '15 at 2:53










    • $begingroup$
      @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!
      $endgroup$
      – Kugelblitz
      Mar 10 '15 at 10:31










    • $begingroup$
      I think we already have a proof. Thanks again.
      $endgroup$
      – Jr Antalan
      Mar 10 '15 at 13:18
















    • $begingroup$
      thank you again. Yeah that must be constant.
      $endgroup$
      – Jr Antalan
      Mar 9 '15 at 16:11










    • $begingroup$
      @JrAntalan I've updated my answer.
      $endgroup$
      – Kugelblitz
      Mar 9 '15 at 16:21










    • $begingroup$
      I will wait for your proof, thanks
      $endgroup$
      – Jr Antalan
      Mar 10 '15 at 2:53










    • $begingroup$
      @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!
      $endgroup$
      – Kugelblitz
      Mar 10 '15 at 10:31










    • $begingroup$
      I think we already have a proof. Thanks again.
      $endgroup$
      – Jr Antalan
      Mar 10 '15 at 13:18















    $begingroup$
    thank you again. Yeah that must be constant.
    $endgroup$
    – Jr Antalan
    Mar 9 '15 at 16:11




    $begingroup$
    thank you again. Yeah that must be constant.
    $endgroup$
    – Jr Antalan
    Mar 9 '15 at 16:11












    $begingroup$
    @JrAntalan I've updated my answer.
    $endgroup$
    – Kugelblitz
    Mar 9 '15 at 16:21




    $begingroup$
    @JrAntalan I've updated my answer.
    $endgroup$
    – Kugelblitz
    Mar 9 '15 at 16:21












    $begingroup$
    I will wait for your proof, thanks
    $endgroup$
    – Jr Antalan
    Mar 10 '15 at 2:53




    $begingroup$
    I will wait for your proof, thanks
    $endgroup$
    – Jr Antalan
    Mar 10 '15 at 2:53












    $begingroup$
    @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!
    $endgroup$
    – Kugelblitz
    Mar 10 '15 at 10:31




    $begingroup$
    @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!
    $endgroup$
    – Kugelblitz
    Mar 10 '15 at 10:31












    $begingroup$
    I think we already have a proof. Thanks again.
    $endgroup$
    – Jr Antalan
    Mar 10 '15 at 13:18




    $begingroup$
    I think we already have a proof. Thanks again.
    $endgroup$
    – Jr Antalan
    Mar 10 '15 at 13:18











    0












    $begingroup$

    All palindromic fixed points:



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



    $101101_2, 213312_4, 325523_6, 437734_8,549945, dots$



    There are no other palindromic Kaprekar Palindromes.






    share|cite|improve this answer











    $endgroup$

















      0












      $begingroup$

      All palindromic fixed points:



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



      $101101_2, 213312_4, 325523_6, 437734_8,549945, dots$



      There are no other palindromic Kaprekar Palindromes.






      share|cite|improve this answer











      $endgroup$















        0












        0








        0





        $begingroup$

        All palindromic fixed points:



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



        $101101_2, 213312_4, 325523_6, 437734_8,549945, dots$



        There are no other palindromic Kaprekar Palindromes.






        share|cite|improve this answer











        $endgroup$



        All palindromic fixed points:



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



        $101101_2, 213312_4, 325523_6, 437734_8,549945, dots$



        There are no other palindromic Kaprekar Palindromes.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 18 hours ago

























        answered Dec 6 '18 at 7:33









        Ben CrossleyBen Crossley

        904418




        904418



























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