Palindrome vs Level of Palindrome












0












$begingroup$


The palindrome, example: $131$, $82728$, $55655$.



But from the palindrome maker algorithm
say:



If $17$ isn't palindrome you must additive by reverse of them



$33$ is say $P(1)$ palindrome



$38$ is say $P(2)$ semipalindrome of level $1$



Because:



$38+83=121$



$182$ is say $P(5)$ semipalindrome of level $4$



Because:



$182+281=463$



$463+364=827$



$827+728=1555$



$1555+5551=6666$




So the question is:



How many semipalindrome of level $1$ between $1-1000$ are?







number-theory palindrome







share|cite|improve this question











$endgroup$












  • $begingroup$
    Well, if there are no carries involved in the addition then you always get a palindrome on one iteration, as in $134+431=565$. That gives you a lot of them, but not all (as your $38+83$ example shows). I wouldn't think it would be too hard to count those examples as well...after all they must begin and end in $1$ after the sum.
    $endgroup$
    – lulu
    Dec 30 '18 at 15:01










  • $begingroup$
    Is $2$ a semipalindrome because $2+2=4$ and $4$ is a palindrome? How about $22$?
    $endgroup$
    – Ross Millikan
    Dec 30 '18 at 15:41
















0












$begingroup$


The palindrome, example: $131$, $82728$, $55655$.



But from the palindrome maker algorithm
say:



If $17$ isn't palindrome you must additive by reverse of them



$33$ is say $P(1)$ palindrome



$38$ is say $P(2)$ semipalindrome of level $1$



Because:



$38+83=121$



$182$ is say $P(5)$ semipalindrome of level $4$



Because:



$182+281=463$



$463+364=827$



$827+728=1555$



$1555+5551=6666$




So the question is:



How many semipalindrome of level $1$ between $1-1000$ are?







number-theory palindrome







share|cite|improve this question











$endgroup$












  • $begingroup$
    Well, if there are no carries involved in the addition then you always get a palindrome on one iteration, as in $134+431=565$. That gives you a lot of them, but not all (as your $38+83$ example shows). I wouldn't think it would be too hard to count those examples as well...after all they must begin and end in $1$ after the sum.
    $endgroup$
    – lulu
    Dec 30 '18 at 15:01










  • $begingroup$
    Is $2$ a semipalindrome because $2+2=4$ and $4$ is a palindrome? How about $22$?
    $endgroup$
    – Ross Millikan
    Dec 30 '18 at 15:41














0












0








0


1



$begingroup$


The palindrome, example: $131$, $82728$, $55655$.



But from the palindrome maker algorithm
say:



If $17$ isn't palindrome you must additive by reverse of them



$33$ is say $P(1)$ palindrome



$38$ is say $P(2)$ semipalindrome of level $1$



Because:



$38+83=121$



$182$ is say $P(5)$ semipalindrome of level $4$



Because:



$182+281=463$



$463+364=827$



$827+728=1555$



$1555+5551=6666$




So the question is:



How many semipalindrome of level $1$ between $1-1000$ are?







number-theory palindrome







share|cite|improve this question











$endgroup$




The palindrome, example: $131$, $82728$, $55655$.



But from the palindrome maker algorithm
say:



If $17$ isn't palindrome you must additive by reverse of them



$33$ is say $P(1)$ palindrome



$38$ is say $P(2)$ semipalindrome of level $1$



Because:



$38+83=121$



$182$ is say $P(5)$ semipalindrome of level $4$



Because:



$182+281=463$



$463+364=827$



$827+728=1555$



$1555+5551=6666$




So the question is:



How many semipalindrome of level $1$ between $1-1000$ are?







number-theory palindrome




number-theory palindrome






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 15:13









Zacky

7,89511061




7,89511061










asked Dec 30 '18 at 14:52









HeartHeart

30519




30519












  • $begingroup$
    Well, if there are no carries involved in the addition then you always get a palindrome on one iteration, as in $134+431=565$. That gives you a lot of them, but not all (as your $38+83$ example shows). I wouldn't think it would be too hard to count those examples as well...after all they must begin and end in $1$ after the sum.
    $endgroup$
    – lulu
    Dec 30 '18 at 15:01










  • $begingroup$
    Is $2$ a semipalindrome because $2+2=4$ and $4$ is a palindrome? How about $22$?
    $endgroup$
    – Ross Millikan
    Dec 30 '18 at 15:41


















  • $begingroup$
    Well, if there are no carries involved in the addition then you always get a palindrome on one iteration, as in $134+431=565$. That gives you a lot of them, but not all (as your $38+83$ example shows). I wouldn't think it would be too hard to count those examples as well...after all they must begin and end in $1$ after the sum.
    $endgroup$
    – lulu
    Dec 30 '18 at 15:01










  • $begingroup$
    Is $2$ a semipalindrome because $2+2=4$ and $4$ is a palindrome? How about $22$?
    $endgroup$
    – Ross Millikan
    Dec 30 '18 at 15:41
















$begingroup$
Well, if there are no carries involved in the addition then you always get a palindrome on one iteration, as in $134+431=565$. That gives you a lot of them, but not all (as your $38+83$ example shows). I wouldn't think it would be too hard to count those examples as well...after all they must begin and end in $1$ after the sum.
$endgroup$
– lulu
Dec 30 '18 at 15:01




$begingroup$
Well, if there are no carries involved in the addition then you always get a palindrome on one iteration, as in $134+431=565$. That gives you a lot of them, but not all (as your $38+83$ example shows). I wouldn't think it would be too hard to count those examples as well...after all they must begin and end in $1$ after the sum.
$endgroup$
– lulu
Dec 30 '18 at 15:01












$begingroup$
Is $2$ a semipalindrome because $2+2=4$ and $4$ is a palindrome? How about $22$?
$endgroup$
– Ross Millikan
Dec 30 '18 at 15:41




$begingroup$
Is $2$ a semipalindrome because $2+2=4$ and $4$ is a palindrome? How about $22$?
$endgroup$
– Ross Millikan
Dec 30 '18 at 15:41










1 Answer
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$begingroup$

If you start with a two digit number $ab$ you get a palindrome in two cases:




  • if $a+b lt 10$ there will be no carry so you get $11(a+b)$


  • if you carry $1$ and the ones digit is $1$, so when $a+b=11$



If you start with a three digit number $abc$ you get a palindrome:




  • if there are no carries at all

  • if $a+c =11$ so the ones digit of the sum is $1$ and there is a carry, then $2b+1$ does not carry and matches the $1$ that $a+c$ gives in the hundreds, so $b=0$

  • if $a+c=11$ and $2b+1$ does carry there will be a $2$ in the hundreds, which $2b+1$ cannot match. This does not work.






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    $begingroup$

    If you start with a two digit number $ab$ you get a palindrome in two cases:




    • if $a+b lt 10$ there will be no carry so you get $11(a+b)$


    • if you carry $1$ and the ones digit is $1$, so when $a+b=11$



    If you start with a three digit number $abc$ you get a palindrome:




    • if there are no carries at all

    • if $a+c =11$ so the ones digit of the sum is $1$ and there is a carry, then $2b+1$ does not carry and matches the $1$ that $a+c$ gives in the hundreds, so $b=0$

    • if $a+c=11$ and $2b+1$ does carry there will be a $2$ in the hundreds, which $2b+1$ cannot match. This does not work.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      If you start with a two digit number $ab$ you get a palindrome in two cases:




      • if $a+b lt 10$ there will be no carry so you get $11(a+b)$


      • if you carry $1$ and the ones digit is $1$, so when $a+b=11$



      If you start with a three digit number $abc$ you get a palindrome:




      • if there are no carries at all

      • if $a+c =11$ so the ones digit of the sum is $1$ and there is a carry, then $2b+1$ does not carry and matches the $1$ that $a+c$ gives in the hundreds, so $b=0$

      • if $a+c=11$ and $2b+1$ does carry there will be a $2$ in the hundreds, which $2b+1$ cannot match. This does not work.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        If you start with a two digit number $ab$ you get a palindrome in two cases:




        • if $a+b lt 10$ there will be no carry so you get $11(a+b)$


        • if you carry $1$ and the ones digit is $1$, so when $a+b=11$



        If you start with a three digit number $abc$ you get a palindrome:




        • if there are no carries at all

        • if $a+c =11$ so the ones digit of the sum is $1$ and there is a carry, then $2b+1$ does not carry and matches the $1$ that $a+c$ gives in the hundreds, so $b=0$

        • if $a+c=11$ and $2b+1$ does carry there will be a $2$ in the hundreds, which $2b+1$ cannot match. This does not work.






        share|cite|improve this answer









        $endgroup$



        If you start with a two digit number $ab$ you get a palindrome in two cases:




        • if $a+b lt 10$ there will be no carry so you get $11(a+b)$


        • if you carry $1$ and the ones digit is $1$, so when $a+b=11$



        If you start with a three digit number $abc$ you get a palindrome:




        • if there are no carries at all

        • if $a+c =11$ so the ones digit of the sum is $1$ and there is a carry, then $2b+1$ does not carry and matches the $1$ that $a+c$ gives in the hundreds, so $b=0$

        • if $a+c=11$ and $2b+1$ does carry there will be a $2$ in the hundreds, which $2b+1$ cannot match. This does not work.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 30 '18 at 15:05









        Ross MillikanRoss Millikan

        300k24200375




        300k24200375






























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