Can divisibility rules for digits be generalized to sum of digits












2












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Suppose that we are given a two digit number $AB$, where $A$ and $B$ represents the digits, i.e 21 would be A=2 , B=1. I wish to prove that the sum of $AB$ and $BA$ is always divisible by $11$.
My initial idée was to use the fact that if a number is divisible by $11$ then the sum of its digits with alternating sign is also divisible by 11. For example
$$1-2+3-3+2-1=0 $$
so $11$ divides $123321$. So my proof would then be to consider the two digit number $(A+B)(B+A)$ or $CC$ which clearly is divisible by $11$ by the above statement if $C$ is $1$ through $9$. However, I am having truble justifying the case were $A+B$ is greater than or equal to $10$ and it got me wondering if the more generel is true: Let $ABCD...$ be a $n-digit$ number, if $$A-B+C-... equiv 0 (mod 11)$$
then $$S=sum_{k=1}^{n}(A+B+C+...)10^{k} equiv0(mod 11)$$
I am not really familliar with the whole congruence thingy, so incase the above is trivial I would be greatful on some source which could aid the solving of the above . Any tips or suggestion are also very welcome






divisibility







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  • $begingroup$
    math.stackexchange.com/questions/328562/…
    $endgroup$
    – lab bhattacharjee
    2 days ago










  • $begingroup$
    Use pmod{11} to produce $pmod{11}$. So aequiv bpmod{11} produces $aequiv bpmod{11}$.
    $endgroup$
    – Arturo Magidin
    2 days ago
















2












$begingroup$


Suppose that we are given a two digit number $AB$, where $A$ and $B$ represents the digits, i.e 21 would be A=2 , B=1. I wish to prove that the sum of $AB$ and $BA$ is always divisible by $11$.
My initial idée was to use the fact that if a number is divisible by $11$ then the sum of its digits with alternating sign is also divisible by 11. For example
$$1-2+3-3+2-1=0 $$
so $11$ divides $123321$. So my proof would then be to consider the two digit number $(A+B)(B+A)$ or $CC$ which clearly is divisible by $11$ by the above statement if $C$ is $1$ through $9$. However, I am having truble justifying the case were $A+B$ is greater than or equal to $10$ and it got me wondering if the more generel is true: Let $ABCD...$ be a $n-digit$ number, if $$A-B+C-... equiv 0 (mod 11)$$
then $$S=sum_{k=1}^{n}(A+B+C+...)10^{k} equiv0(mod 11)$$
I am not really familliar with the whole congruence thingy, so incase the above is trivial I would be greatful on some source which could aid the solving of the above . Any tips or suggestion are also very welcome






divisibility







share|cite|improve this question











$endgroup$












  • $begingroup$
    math.stackexchange.com/questions/328562/…
    $endgroup$
    – lab bhattacharjee
    2 days ago










  • $begingroup$
    Use pmod{11} to produce $pmod{11}$. So aequiv bpmod{11} produces $aequiv bpmod{11}$.
    $endgroup$
    – Arturo Magidin
    2 days ago














2












2








2


1



$begingroup$


Suppose that we are given a two digit number $AB$, where $A$ and $B$ represents the digits, i.e 21 would be A=2 , B=1. I wish to prove that the sum of $AB$ and $BA$ is always divisible by $11$.
My initial idée was to use the fact that if a number is divisible by $11$ then the sum of its digits with alternating sign is also divisible by 11. For example
$$1-2+3-3+2-1=0 $$
so $11$ divides $123321$. So my proof would then be to consider the two digit number $(A+B)(B+A)$ or $CC$ which clearly is divisible by $11$ by the above statement if $C$ is $1$ through $9$. However, I am having truble justifying the case were $A+B$ is greater than or equal to $10$ and it got me wondering if the more generel is true: Let $ABCD...$ be a $n-digit$ number, if $$A-B+C-... equiv 0 (mod 11)$$
then $$S=sum_{k=1}^{n}(A+B+C+...)10^{k} equiv0(mod 11)$$
I am not really familliar with the whole congruence thingy, so incase the above is trivial I would be greatful on some source which could aid the solving of the above . Any tips or suggestion are also very welcome






divisibility







share|cite|improve this question











$endgroup$




Suppose that we are given a two digit number $AB$, where $A$ and $B$ represents the digits, i.e 21 would be A=2 , B=1. I wish to prove that the sum of $AB$ and $BA$ is always divisible by $11$.
My initial idée was to use the fact that if a number is divisible by $11$ then the sum of its digits with alternating sign is also divisible by 11. For example
$$1-2+3-3+2-1=0 $$
so $11$ divides $123321$. So my proof would then be to consider the two digit number $(A+B)(B+A)$ or $CC$ which clearly is divisible by $11$ by the above statement if $C$ is $1$ through $9$. However, I am having truble justifying the case were $A+B$ is greater than or equal to $10$ and it got me wondering if the more generel is true: Let $ABCD...$ be a $n-digit$ number, if $$A-B+C-... equiv 0 (mod 11)$$
then $$S=sum_{k=1}^{n}(A+B+C+...)10^{k} equiv0(mod 11)$$
I am not really familliar with the whole congruence thingy, so incase the above is trivial I would be greatful on some source which could aid the solving of the above . Any tips or suggestion are also very welcome






divisibility




divisibility






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share|cite|improve this question








edited 2 days ago







André Armatowski

















asked 2 days ago









André ArmatowskiAndré Armatowski

263




263












  • $begingroup$
    math.stackexchange.com/questions/328562/…
    $endgroup$
    – lab bhattacharjee
    2 days ago










  • $begingroup$
    Use pmod{11} to produce $pmod{11}$. So aequiv bpmod{11} produces $aequiv bpmod{11}$.
    $endgroup$
    – Arturo Magidin
    2 days ago


















  • $begingroup$
    math.stackexchange.com/questions/328562/…
    $endgroup$
    – lab bhattacharjee
    2 days ago










  • $begingroup$
    Use pmod{11} to produce $pmod{11}$. So aequiv bpmod{11} produces $aequiv bpmod{11}$.
    $endgroup$
    – Arturo Magidin
    2 days ago
















$begingroup$
math.stackexchange.com/questions/328562/…
$endgroup$
– lab bhattacharjee
2 days ago




$begingroup$
math.stackexchange.com/questions/328562/…
$endgroup$
– lab bhattacharjee
2 days ago












$begingroup$
Use pmod{11} to produce $pmod{11}$. So aequiv bpmod{11} produces $aequiv bpmod{11}$.
$endgroup$
– Arturo Magidin
2 days ago




$begingroup$
Use pmod{11} to produce $pmod{11}$. So aequiv bpmod{11} produces $aequiv bpmod{11}$.
$endgroup$
– Arturo Magidin
2 days ago










3 Answers
3






active

oldest

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2












$begingroup$

More generally, recall that the radix $rm,b,$ digit string $rm d_n cdots d_1 d_0 $ denotes a polynomial expression $rm P(b) = d_n b^n +:cdots: + d_1 b + d_0,, $ where $rm P(x) = d_n x^n +cdots+ d_1 x + d_0., $ Recall the reversed (digits) polynomial is $rm {bf tilde {rm P}}(x) = x^n P(1/x).,$ If $rm:n:$ is odd the Polynomial Congruence Rule yields $$rm: mod b!+!1: color{#c00}{bequiv -1} Rightarrow {bf tilde {rm P}}(b) = color{#c00}b^n P(1/color{#c00}b) equiv (color{#c00}{-1})^n P(color{#c00}{-1})equiv {-}P(-1),:$$ therefore we conclude that $rm P(b) + {bf tilde {rm P}}(b)equiv P(-1)-P(-1)equiv 0.,$ OP is case $rm,b=10, n=1$.



Remark $ $ Essentially we have twice applied the radix $rm,b,$ analog of casting out elevens (the analog of casting out nines).






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    11












    $begingroup$

    It's simpler than you are making it...and no congruences are needed:



    We have $$overline {AB}=10A+B quad &quad overline {BA}=10B+A$$



    It follows that $$overline {AB}+overline {BA}=11times (A+B)$$ and we are done.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Very clean, totally escaped me!
      $endgroup$
      – André Armatowski
      2 days ago



















    2












    $begingroup$

    You can easily push through what you were trying though. Say your numbers are $AB$ and $BA$. If $A+Bgt 10$, then write it as $A+B=10+c$; note that $0leq cleq 8$, because two digits cannot add to $19$. That means that when you do the carry, the second digit is $(c+1)$, and so $AB+BA$ will be a three digit number: $1$, then $c+1$, and then $c$. At this point, your test gives you $1-(c+1)+c = 0$, so you get a multiple of $11$.






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

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      active

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      active

      oldest

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      2












      $begingroup$

      More generally, recall that the radix $rm,b,$ digit string $rm d_n cdots d_1 d_0 $ denotes a polynomial expression $rm P(b) = d_n b^n +:cdots: + d_1 b + d_0,, $ where $rm P(x) = d_n x^n +cdots+ d_1 x + d_0., $ Recall the reversed (digits) polynomial is $rm {bf tilde {rm P}}(x) = x^n P(1/x).,$ If $rm:n:$ is odd the Polynomial Congruence Rule yields $$rm: mod b!+!1: color{#c00}{bequiv -1} Rightarrow {bf tilde {rm P}}(b) = color{#c00}b^n P(1/color{#c00}b) equiv (color{#c00}{-1})^n P(color{#c00}{-1})equiv {-}P(-1),:$$ therefore we conclude that $rm P(b) + {bf tilde {rm P}}(b)equiv P(-1)-P(-1)equiv 0.,$ OP is case $rm,b=10, n=1$.



      Remark $ $ Essentially we have twice applied the radix $rm,b,$ analog of casting out elevens (the analog of casting out nines).






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        More generally, recall that the radix $rm,b,$ digit string $rm d_n cdots d_1 d_0 $ denotes a polynomial expression $rm P(b) = d_n b^n +:cdots: + d_1 b + d_0,, $ where $rm P(x) = d_n x^n +cdots+ d_1 x + d_0., $ Recall the reversed (digits) polynomial is $rm {bf tilde {rm P}}(x) = x^n P(1/x).,$ If $rm:n:$ is odd the Polynomial Congruence Rule yields $$rm: mod b!+!1: color{#c00}{bequiv -1} Rightarrow {bf tilde {rm P}}(b) = color{#c00}b^n P(1/color{#c00}b) equiv (color{#c00}{-1})^n P(color{#c00}{-1})equiv {-}P(-1),:$$ therefore we conclude that $rm P(b) + {bf tilde {rm P}}(b)equiv P(-1)-P(-1)equiv 0.,$ OP is case $rm,b=10, n=1$.



        Remark $ $ Essentially we have twice applied the radix $rm,b,$ analog of casting out elevens (the analog of casting out nines).






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          More generally, recall that the radix $rm,b,$ digit string $rm d_n cdots d_1 d_0 $ denotes a polynomial expression $rm P(b) = d_n b^n +:cdots: + d_1 b + d_0,, $ where $rm P(x) = d_n x^n +cdots+ d_1 x + d_0., $ Recall the reversed (digits) polynomial is $rm {bf tilde {rm P}}(x) = x^n P(1/x).,$ If $rm:n:$ is odd the Polynomial Congruence Rule yields $$rm: mod b!+!1: color{#c00}{bequiv -1} Rightarrow {bf tilde {rm P}}(b) = color{#c00}b^n P(1/color{#c00}b) equiv (color{#c00}{-1})^n P(color{#c00}{-1})equiv {-}P(-1),:$$ therefore we conclude that $rm P(b) + {bf tilde {rm P}}(b)equiv P(-1)-P(-1)equiv 0.,$ OP is case $rm,b=10, n=1$.



          Remark $ $ Essentially we have twice applied the radix $rm,b,$ analog of casting out elevens (the analog of casting out nines).






          share|cite|improve this answer











          $endgroup$



          More generally, recall that the radix $rm,b,$ digit string $rm d_n cdots d_1 d_0 $ denotes a polynomial expression $rm P(b) = d_n b^n +:cdots: + d_1 b + d_0,, $ where $rm P(x) = d_n x^n +cdots+ d_1 x + d_0., $ Recall the reversed (digits) polynomial is $rm {bf tilde {rm P}}(x) = x^n P(1/x).,$ If $rm:n:$ is odd the Polynomial Congruence Rule yields $$rm: mod b!+!1: color{#c00}{bequiv -1} Rightarrow {bf tilde {rm P}}(b) = color{#c00}b^n P(1/color{#c00}b) equiv (color{#c00}{-1})^n P(color{#c00}{-1})equiv {-}P(-1),:$$ therefore we conclude that $rm P(b) + {bf tilde {rm P}}(b)equiv P(-1)-P(-1)equiv 0.,$ OP is case $rm,b=10, n=1$.



          Remark $ $ Essentially we have twice applied the radix $rm,b,$ analog of casting out elevens (the analog of casting out nines).







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday

























          answered 2 days ago









          Bill DubuqueBill Dubuque

          214k29196655




          214k29196655























              11












              $begingroup$

              It's simpler than you are making it...and no congruences are needed:



              We have $$overline {AB}=10A+B quad &quad overline {BA}=10B+A$$



              It follows that $$overline {AB}+overline {BA}=11times (A+B)$$ and we are done.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Very clean, totally escaped me!
                $endgroup$
                – André Armatowski
                2 days ago
















              11












              $begingroup$

              It's simpler than you are making it...and no congruences are needed:



              We have $$overline {AB}=10A+B quad &quad overline {BA}=10B+A$$



              It follows that $$overline {AB}+overline {BA}=11times (A+B)$$ and we are done.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Very clean, totally escaped me!
                $endgroup$
                – André Armatowski
                2 days ago














              11












              11








              11





              $begingroup$

              It's simpler than you are making it...and no congruences are needed:



              We have $$overline {AB}=10A+B quad &quad overline {BA}=10B+A$$



              It follows that $$overline {AB}+overline {BA}=11times (A+B)$$ and we are done.






              share|cite|improve this answer











              $endgroup$



              It's simpler than you are making it...and no congruences are needed:



              We have $$overline {AB}=10A+B quad &quad overline {BA}=10B+A$$



              It follows that $$overline {AB}+overline {BA}=11times (A+B)$$ and we are done.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 2 days ago

























              answered 2 days ago









              lulululu

              43.6k25081




              43.6k25081












              • $begingroup$
                Very clean, totally escaped me!
                $endgroup$
                – André Armatowski
                2 days ago


















              • $begingroup$
                Very clean, totally escaped me!
                $endgroup$
                – André Armatowski
                2 days ago
















              $begingroup$
              Very clean, totally escaped me!
              $endgroup$
              – André Armatowski
              2 days ago




              $begingroup$
              Very clean, totally escaped me!
              $endgroup$
              – André Armatowski
              2 days ago











              2












              $begingroup$

              You can easily push through what you were trying though. Say your numbers are $AB$ and $BA$. If $A+Bgt 10$, then write it as $A+B=10+c$; note that $0leq cleq 8$, because two digits cannot add to $19$. That means that when you do the carry, the second digit is $(c+1)$, and so $AB+BA$ will be a three digit number: $1$, then $c+1$, and then $c$. At this point, your test gives you $1-(c+1)+c = 0$, so you get a multiple of $11$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                You can easily push through what you were trying though. Say your numbers are $AB$ and $BA$. If $A+Bgt 10$, then write it as $A+B=10+c$; note that $0leq cleq 8$, because two digits cannot add to $19$. That means that when you do the carry, the second digit is $(c+1)$, and so $AB+BA$ will be a three digit number: $1$, then $c+1$, and then $c$. At this point, your test gives you $1-(c+1)+c = 0$, so you get a multiple of $11$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  You can easily push through what you were trying though. Say your numbers are $AB$ and $BA$. If $A+Bgt 10$, then write it as $A+B=10+c$; note that $0leq cleq 8$, because two digits cannot add to $19$. That means that when you do the carry, the second digit is $(c+1)$, and so $AB+BA$ will be a three digit number: $1$, then $c+1$, and then $c$. At this point, your test gives you $1-(c+1)+c = 0$, so you get a multiple of $11$.






                  share|cite|improve this answer









                  $endgroup$



                  You can easily push through what you were trying though. Say your numbers are $AB$ and $BA$. If $A+Bgt 10$, then write it as $A+B=10+c$; note that $0leq cleq 8$, because two digits cannot add to $19$. That means that when you do the carry, the second digit is $(c+1)$, and so $AB+BA$ will be a three digit number: $1$, then $c+1$, and then $c$. At this point, your test gives you $1-(c+1)+c = 0$, so you get a multiple of $11$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 days ago









                  Arturo MagidinArturo Magidin

                  266k34590921




                  266k34590921






























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