Can divisibility rules for digits be generalized to sum of digits
$begingroup$
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
asked 4 hours ago
André ArmatowskiAndré Armatowski
233
$endgroup$
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
asked 4 hours ago
André ArmatowskiAndré Armatowski
233
$endgroup$
$begingroup$
math.stackexchange.com/questions/328562/…
$endgroup$
– lab bhattacharjee
4 hours ago
$begingroup$
Usepmod{11}to produce $pmod{11}$. Soaequiv 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
asked 4 hours ago
André ArmatowskiAndré Armatowski
233
$endgroup$
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
asked 4 hours ago
André ArmatowskiAndré Armatowski
233
asked 4 hours ago
André ArmatowskiAndré Armatowski
233
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
233
$begingroup$
math.stackexchange.com/questions/328562/…
$endgroup$
– lab bhattacharjee
4 hours ago
$begingroup$
Usepmod{11}to produce $pmod{11}$. Soaequiv bpmod{11}produces $aequiv bpmod{11}$.
$endgroup$
– Arturo Magidin
4 hours ago
add a comment |
$begingroup$
math.stackexchange.com/questions/328562/…
$endgroup$
– lab bhattacharjee
4 hours ago
$begingroup$
Usepmod{11}to produce $pmod{11}$. Soaequiv 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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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).
answered 3 hours ago
Bill DubuqueBill Dubuque
213k29196654
$endgroup$
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.
answered 4 hours ago
lulululu
43.4k25080
$endgroup$
$begingroup$
Very clean, totally escaped me!
$endgroup$
– André Armatowski
4 hours ago
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 $11$.
answered 3 hours ago
Arturo MagidinArturo Magidin
266k34590920
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
$begingroup$
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).
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 $rm {bar P}(x) = x^n P(1/x).,$ 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).
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 $rm {bar P}(x) = x^n P(1/x).,$ 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).
answered 3 hours ago
Bill DubuqueBill Dubuque
213k29196654
$endgroup$
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).
answered 3 hours ago
Bill DubuqueBill Dubuque
213k29196654
edited 3 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
213k29196654
add a comment |
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.
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.
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.
answered 4 hours ago
lulululu
43.4k25080
$endgroup$
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.
answered 4 hours ago
lulululu
43.4k25080
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
43.4k25080
$begingroup$
Very clean, totally escaped me!
$endgroup$
– André Armatowski
4 hours ago
add a comment |
$begingroup$
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
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 $11$.
answered 3 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 $11$.
answered 3 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 $11$.
answered 3 hours ago
Arturo MagidinArturo Magidin
266k34590920
$endgroup$
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 $11$.
answered 3 hours ago
Arturo MagidinArturo Magidin
266k34590920
answered 3 hours ago
Arturo MagidinArturo Magidin
266k34590920
answered 3 hours ago
Arturo MagidinArturo Magidin
266k34590920
answered 3 hours ago
Arturo MagidinArturo Magidin
266k34590920
266k34590920
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
$begingroup$
math.stackexchange.com/questions/328562/…
$endgroup$
– lab bhattacharjee
4 hours ago
$begingroup$
Use
pmod{11}to produce $pmod{11}$. Soaequiv bpmod{11}produces $aequiv bpmod{11}$.$endgroup$
– Arturo Magidin
4 hours ago