Can divisibility rules for digits be generalized to sum of digits The 2019 Stack Overflow Developer Survey Results Are InDivisibility 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

How to create dashed lines/arrows in Illustrator

Inversion Puzzle

Idiomatic way to prevent slicing?

How are circuits which use complex ICs normally simulated?

Does it makes sense to buy a new cycle to learn riding?

Dual Citizen. Exited the US on Italian passport recently

Unbreakable Formation vs. Cry of the Carnarium

What is this 4-propeller plane?

Does duplicating a spell with Wish count as casting that spell?

Can't find the latex code for the ⍎ (down tack jot) symbol

What does Linus Torvalds mean when he says that Git "never ever" tracks a file?

Access elements in std::string where positon of string is greater than its size

Why is my p-value correlated to difference between means in two sample tests?

"Riffle" two strings

Why could you hear an Amstrad CPC working?

Why is the maximum length of OpenWrt’s root password 8 characters?

Patience, young "Padovan"

What is the use of option -o in the useradd command?

Why is it "Tumoren" and not "Tumore"?

Inline version of a function returns different value than non-inline version

Time travel alters history but people keep saying nothing's changed

Can distinct morphisms between curves induce the same morphism on singular cohomology?

Output the Arecibo Message

How was Skylab's orbit inclination chosen?



Can divisibility rules for digits be generalized to sum of digits



The 2019 Stack Overflow Developer Survey Results Are InDivisibility 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










share|cite|improve this question











$endgroup$











  • $begingroup$
    math.stackexchange.com/questions/328562/…
    $endgroup$
    – lab bhattacharjee
    Apr 6 at 16:20










  • $begingroup$
    Use pmod11 to produce $pmod11$. So aequiv bpmod11 produces $aequiv bpmod11$.
    $endgroup$
    – Arturo Magidin
    Apr 6 at 16:20















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










share|cite|improve this question











$endgroup$











  • $begingroup$
    math.stackexchange.com/questions/328562/…
    $endgroup$
    – lab bhattacharjee
    Apr 6 at 16:20










  • $begingroup$
    Use pmod11 to produce $pmod11$. So aequiv bpmod11 produces $aequiv bpmod11$.
    $endgroup$
    – Arturo Magidin
    Apr 6 at 16:20













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










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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 6 at 16:19







André Armatowski

















asked Apr 6 at 16:14









André ArmatowskiAndré Armatowski

263




263











  • $begingroup$
    math.stackexchange.com/questions/328562/…
    $endgroup$
    – lab bhattacharjee
    Apr 6 at 16:20










  • $begingroup$
    Use pmod11 to produce $pmod11$. So aequiv bpmod11 produces $aequiv bpmod11$.
    $endgroup$
    – Arturo Magidin
    Apr 6 at 16:20
















  • $begingroup$
    math.stackexchange.com/questions/328562/…
    $endgroup$
    – lab bhattacharjee
    Apr 6 at 16:20










  • $begingroup$
    Use pmod11 to produce $pmod11$. So aequiv bpmod11 produces $aequiv bpmod11$.
    $endgroup$
    – Arturo Magidin
    Apr 6 at 16:20















$begingroup$
math.stackexchange.com/questions/328562/…
$endgroup$
– lab bhattacharjee
Apr 6 at 16:20




$begingroup$
math.stackexchange.com/questions/328562/…
$endgroup$
– lab bhattacharjee
Apr 6 at 16:20












$begingroup$
Use pmod11 to produce $pmod11$. So aequiv bpmod11 produces $aequiv bpmod11$.
$endgroup$
– Arturo Magidin
Apr 6 at 16:20




$begingroup$
Use pmod11 to produce $pmod11$. So aequiv bpmod11 produces $aequiv bpmod11$.
$endgroup$
– Arturo Magidin
Apr 6 at 16:20










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 $$rm: mod b!+!1: color#c00bequiv -1 Rightarrow bf tilde rm P(b) = color#c00b^n P(1/color#c00b) equiv (color#c00-1)^n P(color#c00-1)equiv -P(-1),:$$ 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











$endgroup$




















    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
      Apr 6 at 16:20


















    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









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function ()
      return StackExchange.using("mathjaxEditing", function ()
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      );
      );
      , "mathjax-editing");

      StackExchange.ready(function()
      var channelOptions =
      tags: "".split(" "),
      id: "69"
      ;
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function()
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled)
      StackExchange.using("snippets", function()
      createEditor();
      );

      else
      createEditor();

      );

      function createEditor()
      StackExchange.prepareEditor(
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader:
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      ,
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      );



      );













      draft saved

      draft discarded


















      StackExchange.ready(
      function ()
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3177167%2fcan-divisibility-rules-for-digits-be-generalized-to-sum-of-digits%23new-answer', 'question_page');

      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      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 $$rm: mod b!+!1: color#c00bequiv -1 Rightarrow bf tilde rm P(b) = color#c00b^n P(1/color#c00b) equiv (color#c00-1)^n P(color#c00-1)equiv -P(-1),:$$ 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 $$rm: mod b!+!1: color#c00bequiv -1 Rightarrow bf tilde rm P(b) = color#c00b^n P(1/color#c00b) equiv (color#c00-1)^n P(color#c00-1)equiv -P(-1),:$$ 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 $$rm: mod b!+!1: color#c00bequiv -1 Rightarrow bf tilde rm P(b) = color#c00b^n P(1/color#c00b) equiv (color#c00-1)^n P(color#c00-1)equiv -P(-1),:$$ 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 $$rm: mod b!+!1: color#c00bequiv -1 Rightarrow bf tilde rm P(b) = color#c00b^n P(1/color#c00b) equiv (color#c00-1)^n P(color#c00-1)equiv -P(-1),:$$ 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 Apr 6 at 20:23

























          answered Apr 6 at 16:34









          Bill DubuqueBill Dubuque

          214k29196656




          214k29196656





















              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
                Apr 6 at 16:20















              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
                Apr 6 at 16:20













              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 Apr 6 at 16:20

























              answered Apr 6 at 16:19









              lulululu

              43.6k25081




              43.6k25081











              • $begingroup$
                Very clean, totally escaped me!
                $endgroup$
                – André Armatowski
                Apr 6 at 16:20
















              • $begingroup$
                Very clean, totally escaped me!
                $endgroup$
                – André Armatowski
                Apr 6 at 16:20















              $begingroup$
              Very clean, totally escaped me!
              $endgroup$
              – André Armatowski
              Apr 6 at 16:20




              $begingroup$
              Very clean, totally escaped me!
              $endgroup$
              – André Armatowski
              Apr 6 at 16:20











              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 Apr 6 at 16:25









                  Arturo MagidinArturo Magidin

                  266k34590921




                  266k34590921



























                      draft saved

                      draft discarded
















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid


                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.

                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3177167%2fcan-divisibility-rules-for-digits-be-generalized-to-sum-of-digits%23new-answer', 'question_page');

                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -divisibility

                      Popular posts from this blog

                      Frič See also Navigation menuinternal link

                      Identify plant with long narrow paired leaves and reddish stems Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?What is this plant with long sharp leaves? Is it a weed?What is this 3ft high, stalky plant, with mid sized narrow leaves?What is this young shrub with opposite ovate, crenate leaves and reddish stems?What is this plant with large broad serrated leaves?Identify this upright branching weed with long leaves and reddish stemsPlease help me identify this bulbous plant with long, broad leaves and white flowersWhat is this small annual with narrow gray/green leaves and rust colored daisy-type flowers?What is this chilli plant?Does anyone know what type of chilli plant this is?Help identify this plant

                      fontconfig warning: “/etc/fonts/fonts.conf”, line 100: unknown “element blank” The 2019 Stack Overflow Developer Survey Results Are In“tar: unrecognized option --warning” during 'apt-get install'How to fix Fontconfig errorHow do I figure out which font file is chosen for a system generic font alias?Why are some apt-get-installed fonts being ignored by fc-list, xfontsel, etc?Reload settings in /etc/fonts/conf.dTaking 30 seconds longer to boot after upgrade from jessie to stretchHow to match multiple font names with a single <match> element?Adding a custom font to fontconfigRemoving fonts from fontconfig <match> resultsBroken fonts after upgrading Firefox ESR to latest Firefox