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
    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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share|cite|improve this question













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share|cite|improve this question








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

votes


















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









    $endgroup$














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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

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      active

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      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











      $endgroup$


















        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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                      濃尾地震

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                      濃尾地震 『岐阜市街大地震之図』 歌川国利画 本震 発生日 1891年（明治24年）10月28日 発生時刻 6時38分50秒（JST） 震央 日本 岐阜県本巣郡西根尾村（現・本巣市） 北緯35度35分 東経136度20分（ 地図 ） 規模    マグニチュード（M）8.0 最大震度    震度7： 注1 福井県今立郡鯖江町、愛知県葉栗郡大田島村、東春日井郡勝川村 地震の種類 直下型地震 被害 死傷者数 死者7,273人、負傷者17,175人 注1：当時の震度階級では「激烈」 プロジェクト:地球科学 プロジェクト:災害 テンプレートを表示 濃尾地震 （のうびじしん）は、1891年（明治24年）10月28日に濃尾地方で発生した、日本史上最大の内陸地殻内地震。「 美濃・尾張地震 （みの・おわりじしん）」とも呼ばれている。辛卯の年に発生したことから 辛卯震災 と呼んでいる報告書もある。 目次 1 概要 1.1 震源断層 2 被害 3 各地の震度 4 前兆現象 5 報道 6 学術的な意義 7 地震防災 8 脚注 9 関連項目 10 外部リンク 概要 濃尾地震発生当時の根尾谷断層 濃尾地震を引き起こした根尾谷断層 写真中央を斜めに走る段差が根尾谷断層 濃尾地震は、1891年10月28日6時38分50秒に発生した。震源は、岐阜県本巣郡西根尾村（現・本巣市）、北緯35度35分、東経136度20分付近。河角廣 (1951) は岐阜市付近（北緯35.6度、東経136.6度）に震央を仮定し規模 M K = 7.0 を与え [1] 、マグニチュードは M = 8.4 に換算されているが、明治・大正期の地震については0.5程度大きく見積もられているとされる [2] 。また、震央距離と震度との関係など当時のデータから後にM8.0 [3] とも推定される。アメリカ地質調査所 (USGS) でも8.0としている [4] 。「根尾谷断層帯」が活動した典型的な内陸地殻内地震（いわゆる直下型地震）であり、これは記録が残っている日本の内陸...
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