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Can divisibility rules for digits be generalized to sum of digits

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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

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

edited 14 hours ago

André Armatowski

asked 14 hours ago

André ArmatowskiAndré Armatowski

263

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

edited 14 hours ago

André Armatowski

asked 14 hours ago

André ArmatowskiAndré Armatowski

263

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use pmod{11} to produce $pmod{11}$. So aequiv bpmod{11} produces $aequiv bpmod{11}$.
  $endgroup$
  – Arturo Magidin
  14 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 14 hours ago

André Armatowski

asked 14 hours ago

André ArmatowskiAndré Armatowski

263

$endgroup$

share|cite|improve this question

edited 14 hours ago

André Armatowski

asked 14 hours ago

André ArmatowskiAndré Armatowski

263

share|cite|improve this question

edited 14 hours ago

André Armatowski

asked 14 hours ago

André ArmatowskiAndré Armatowski

263

asked 14 hours ago

André ArmatowskiAndré Armatowski

263

asked 14 hours ago

André ArmatowskiAndré Armatowski

263

3 Answers
3

active

oldest

votes

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

edited 10 hours ago

answered 14 hours ago

Bill DubuqueBill Dubuque

213k29196654

$endgroup$

add a comment | 

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

edited 14 hours ago

answered 14 hours ago

lulululu

43.5k25081

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

answered 14 hours ago

Arturo MagidinArturo Magidin

266k34590920

$endgroup$

add a comment | 

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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

edited 10 hours ago

answered 14 hours ago

Bill DubuqueBill Dubuque

213k29196654

$endgroup$

add a comment | 

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

edited 10 hours ago

answered 14 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 10 hours ago

answered 14 hours ago

Bill DubuqueBill Dubuque

213k29196654

$endgroup$

share|cite|improve this answer

edited 10 hours ago

answered 14 hours ago

Bill DubuqueBill Dubuque

213k29196654

answered 14 hours ago

Bill DubuqueBill Dubuque

213k29196654

answered 14 hours ago

Bill DubuqueBill Dubuque

213k29196654

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

edited 14 hours ago

answered 14 hours ago

lulululu

43.5k25081

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

edited 14 hours ago

answered 14 hours ago

lulululu

43.5k25081

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very clean, totally escaped me!
  $endgroup$
  – André Armatowski
  14 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 14 hours ago

answered 14 hours ago

lulululu

43.5k25081

$endgroup$

share|cite|improve this answer

edited 14 hours ago

answered 14 hours ago

lulululu

43.5k25081

answered 14 hours ago

lulululu

43.5k25081

answered 14 hours ago

lulululu

43.5k25081

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

answered 14 hours ago

Arturo MagidinArturo Magidin

266k34590920

$endgroup$

add a comment | 

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

answered 14 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 14 hours ago

Arturo MagidinArturo Magidin

266k34590920

$endgroup$

share|cite|improve this answer

answered 14 hours ago

Arturo MagidinArturo Magidin

266k34590920

answered 14 hours ago

Arturo MagidinArturo Magidin

266k34590920

answered 14 hours ago

Arturo MagidinArturo Magidin

266k34590920