Can divisibility rules for digits be generalized to sum of digits












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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
    4 hours ago










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










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














2












2








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











$endgroup$




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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edited 4 hours ago







André Armatowski

















asked 4 hours ago









André ArmatowskiAndré Armatowski

233




233












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










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


















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










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
















$begingroup$
math.stackexchange.com/questions/328562/…
$endgroup$
– lab bhattacharjee
4 hours ago




$begingroup$
math.stackexchange.com/questions/328562/…
$endgroup$
– lab bhattacharjee
4 hours ago












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




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










3 Answers
3






active

oldest

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1












$begingroup$

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 $rm {bar P}(x) = x^n P(1/x).,$ If $rm:n:$ is odd the Polynomial Congruence Rule yields $$rm: mod b!+!1: color{#c00}{bequiv -1} Rightarrow {bar P}(b) = color{#c00}b^n P(1/color{#c00}b) equiv (color{#c00}{-1})^n P(color{#c00}{-1})equiv {-}P(-1),:$$ therefore we conclude that $rm P(b) + bar P(b)equiv P(-1)-P(-1)equiv 0.,$ 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











$endgroup$





















    3












    $begingroup$

    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
      4 hours ago



















    1












    $begingroup$

    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

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      3 Answers
      3






      active

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      active

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      1












      $begingroup$

      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 $rm {bar P}(x) = x^n P(1/x).,$ If $rm:n:$ is odd the Polynomial Congruence Rule yields $$rm: mod b!+!1: color{#c00}{bequiv -1} Rightarrow {bar P}(b) = color{#c00}b^n P(1/color{#c00}b) equiv (color{#c00}{-1})^n P(color{#c00}{-1})equiv {-}P(-1),:$$ therefore we conclude that $rm P(b) + bar P(b)equiv P(-1)-P(-1)equiv 0.,$ 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











      $endgroup$


















        1












        $begingroup$

        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 $rm {bar P}(x) = x^n P(1/x).,$ If $rm:n:$ is odd the Polynomial Congruence Rule yields $$rm: mod b!+!1: color{#c00}{bequiv -1} Rightarrow {bar P}(b) = color{#c00}b^n P(1/color{#c00}b) equiv (color{#c00}{-1})^n P(color{#c00}{-1})equiv {-}P(-1),:$$ therefore we conclude that $rm P(b) + bar P(b)equiv P(-1)-P(-1)equiv 0.,$ 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











        $endgroup$
















          1












          1








          1





          $begingroup$

          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 $rm {bar P}(x) = x^n P(1/x).,$ If $rm:n:$ is odd the Polynomial Congruence Rule yields $$rm: mod b!+!1: color{#c00}{bequiv -1} Rightarrow {bar P}(b) = color{#c00}b^n P(1/color{#c00}b) equiv (color{#c00}{-1})^n P(color{#c00}{-1})equiv {-}P(-1),:$$ therefore we conclude that $rm P(b) + bar P(b)equiv P(-1)-P(-1)equiv 0.,$ 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 $rm {bar P}(x) = x^n P(1/x).,$ If $rm:n:$ is odd the Polynomial Congruence Rule yields $$rm: mod b!+!1: color{#c00}{bequiv -1} Rightarrow {bar P}(b) = color{#c00}b^n P(1/color{#c00}b) equiv (color{#c00}{-1})^n P(color{#c00}{-1})equiv {-}P(-1),:$$ therefore we conclude that $rm P(b) + bar P(b)equiv P(-1)-P(-1)equiv 0.,$ 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 3 mins ago

























          answered 3 hours ago









          Bill DubuqueBill Dubuque

          213k29196654




          213k29196654























              3












              $begingroup$

              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
                4 hours ago
















              3












              $begingroup$

              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
                4 hours ago














              3












              3








              3





              $begingroup$

              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 4 hours ago

























              answered 4 hours ago









              lulululu

              43.4k25080




              43.4k25080












              • $begingroup$
                Very clean, totally escaped me!
                $endgroup$
                – André Armatowski
                4 hours ago


















              • $begingroup$
                Very clean, totally escaped me!
                $endgroup$
                – André Armatowski
                4 hours ago
















              $begingroup$
              Very clean, totally escaped me!
              $endgroup$
              – André Armatowski
              4 hours ago




              $begingroup$
              Very clean, totally escaped me!
              $endgroup$
              – André Armatowski
              4 hours ago











              1












              $begingroup$

              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









              $endgroup$


















                1












                $begingroup$

                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









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  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 3 hours ago









                  Arturo MagidinArturo Magidin

                  266k34590920




                  266k34590920






























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                      No ethernet ip address in my vocore2

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                      0 I just flashed my vocore2u with openwrt 18.06.1, and plug it with an ethernet cable to access network while setting up wireless access-point for my phone. The AP is working and connected to the network, but the ip in my phone is in the subnet of another router. And there's no ip assigned to my vocore2 so I cannot ssh into it. After connecting to vocore2 with serial terminal, ping 8.8.8.8 is also failed. How can I make my vocore2 gain ip from the router and setting up the network well? root@OpenWrt:/etc/config# ifconfig apcli0 Link encap:Ethernet HWaddr BA:D8:12:07:01:71 inet6 addr: fe80::b8d8:12ff:fe07:171/64 Scope:Link UP BROADCAST RUNNING MULTICAST MTU:1500 Metric:1 RX packets:0 errors:0 dropped:0 overruns:0 frame:0 TX packets:0 erro
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