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Can divisibility rules for digits be generalized to sum of digits

Divisibility by 7 rule, and Congruence Arithmetic LawsWhy is $9$ special in testing divisiblity by $9$ by summing decimal digits? (casting out nines)Divisibility criteria for $7,11,13,17,19$Divisibility Rules for Bases other than $10$divisibility for numbers like 13,17 and 19 - Compartmentalization methodTrying to prove a congruence for Stirling numbers of the second kindThe following is a necessary condition for a number to be prime, from its digit expansion. Has it been referred somewhere?Let N be a four digit number, and N' be N with its digits reversed. Prove that N-N' is divisble by 9. Prove that N+N' is divisble by 11.Digit-sum division check in base-$n$Rules of thumb for divisibilityDivisibility by 7 involving grouping and alternating sumDivisibility of a 7-digit number by 21Divisibility Rule Proof about Special Numbers

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

edited 4 hours ago

André Armatowski

asked 4 hours ago

André ArmatowskiAndré Armatowski

233

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

edited 4 hours ago

André Armatowski

asked 4 hours ago

André ArmatowskiAndré Armatowski

233

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 4 hours ago

André Armatowski

asked 4 hours ago

André ArmatowskiAndré Armatowski

233

$endgroup$

share|cite|improve this question

edited 4 hours ago

André Armatowski

asked 4 hours ago

André ArmatowskiAndré Armatowski

233

share|cite|improve this question

edited 4 hours ago

André Armatowski

asked 4 hours ago

André ArmatowskiAndré Armatowski

233

asked 4 hours ago

André ArmatowskiAndré Armatowski

233

asked 4 hours ago

André ArmatowskiAndré Armatowski

233

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 $rm {bf tilde {rm P}}(x) = x^n P(1/x).,$ 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

edited 8 mins ago

answered 3 hours ago

Bill DubuqueBill Dubuque

213k29196654

$endgroup$

add a comment | 

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

edited 4 hours ago

answered 4 hours ago

lulululu

43.4k25080

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

answered 4 hours ago

Arturo MagidinArturo Magidin

266k34590920

$endgroup$

add a comment | 

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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 $rm {bf tilde {rm P}}(x) = x^n P(1/x).,$ 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

edited 8 mins ago

answered 3 hours ago

Bill DubuqueBill Dubuque

213k29196654

$endgroup$

add a comment | 

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 $rm {bf tilde {rm P}}(x) = x^n P(1/x).,$ 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

edited 8 mins ago

answered 3 hours ago

Bill DubuqueBill Dubuque

213k29196654

$endgroup$

add a comment | 

$begingroup$

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

edited 8 mins ago

answered 3 hours ago

Bill DubuqueBill Dubuque

213k29196654

$endgroup$

share|cite|improve this answer

edited 8 mins ago

answered 3 hours ago

Bill DubuqueBill Dubuque

213k29196654

answered 3 hours ago

Bill DubuqueBill Dubuque

213k29196654

answered 3 hours ago

Bill DubuqueBill Dubuque

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

edited 4 hours ago

answered 4 hours ago

lulululu

43.4k25080

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

edited 4 hours ago

answered 4 hours ago

lulululu

43.4k25080

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

edited 4 hours ago

answered 4 hours ago

lulululu

43.4k25080

$endgroup$

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

lulululu

43.4k25080

answered 4 hours ago

lulululu

43.4k25080

answered 4 hours ago

lulululu

43.4k25080

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

answered 4 hours ago

Arturo MagidinArturo Magidin

266k34590920

$endgroup$

add a comment | 

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

answered 4 hours ago

Arturo MagidinArturo Magidin

266k34590920

$endgroup$

add a comment | 

$begingroup$

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

answered 4 hours ago

Arturo MagidinArturo Magidin

266k34590920

$endgroup$

share|cite|improve this answer

answered 4 hours ago

Arturo MagidinArturo Magidin

266k34590920

answered 4 hours ago

Arturo MagidinArturo Magidin

266k34590920

answered 4 hours ago

Arturo MagidinArturo Magidin

266k34590920