Mixing
How to solve congruence modulo equations?
0
$begingroup$
While studying Affine Cipher in cryptography it tells that we need to solve a system of modulo congruence equations.
The equations are:
- $8alpha+betaequiv 15 pmod{26}$
- $5alpha+betaequiv 16 pmod{26}$
Could anyone tell how to solve these equations.
modular-arithmetic systems-of-equations congruences
asked May 11 '15 at 8:13
justinjustin
217113
$endgroup$
|
show 6 more comments
0
$begingroup$
While studying Affine Cipher in cryptography it tells that we need to solve a system of modulo congruence equations.
The equations are:
- $8alpha+betaequiv 15 pmod{26}$
- $5alpha+betaequiv 16 pmod{26}$
Could anyone tell how to solve these equations.
modular-arithmetic systems-of-equations congruences
asked May 11 '15 at 8:13
justinjustin
217113
$endgroup$
$begingroup$
$betaequiv 15-8alpha,Rightarrow, 5alpha + (15-8alpha)equiv 15-3alphaequiv 16iff 3alphaequiv -1equiv -27$ $stackrel{:3}iff alphaequiv -9equiv 17$ mod $26$. So $8(17)+betaequiv 15iff betaequiv 9$ mod $26$.
$endgroup$
– user26486
May 13 '15 at 14:44
$begingroup$
@user31415:That's a good answer.Could you tell me how $-1 equiv -27$ came?Also why always we place equivalence symbol in these equations when an equal symbol is enough to get $alpha$ and $beta$?
$endgroup$
– justin
May 14 '15 at 9:24
$begingroup$
$-1 = -27 + 26(1)$, so $-1$ and $-27$ give same remainders when divided by $26$. Definition: $aequiv bpmod{! n}!iff! nmid a-b$ or i.e. $a,b$ leave same remainders when divided by $n$. It is nitpicking about notation whether you can write equality symbols in place of equivalences there, but I would just write equivalences everywhere. I could later divide both sides by $3$ because $(26,3)=1$. It is because $26mid 3(alpha-(-9))!iff! 26mid alpha-(-9)$, which may make this division property more apparent. Remember Fundamental theorem of arithmetic.
$endgroup$
– user26486
May 14 '15 at 10:19
$begingroup$
@user31415:Okay that's right.Yeah I'll go with equivalence.Is it sure that for any numbers m and n if (m,n)=1 you could divide both sides of the equivalence by n?How did you get -27 in the R.H.S?Is it a guess or is there any strategy to follow for finding -27?Also couldn't get how you wrote $26mid 3(alpha-(-9))$ while I think we just need to consider $26mid 3alpha$?
$endgroup$
– justin
May 14 '15 at 12:21
$begingroup$
@user31415:Yeah got it.Could you tell how did you changed $26mid alpha-(-9)$ into $alphaequiv-9 equiv 17$ mod 26?Does this use the same logic you stated in previous comment?Also how to find the value of $alpha$ from $alphaequiv-9 equiv 17$ mod 26?Is it($alpha$) -9 or 17?
$endgroup$
– justin
May 14 '15 at 12:37
|
show 6 more comments
0
0
0
$begingroup$
While studying Affine Cipher in cryptography it tells that we need to solve a system of modulo congruence equations.
The equations are:
- $8alpha+betaequiv 15 pmod{26}$
- $5alpha+betaequiv 16 pmod{26}$
Could anyone tell how to solve these equations.
modular-arithmetic systems-of-equations congruences
asked May 11 '15 at 8:13
justinjustin
217113
$endgroup$
While studying Affine Cipher in cryptography it tells that we need to solve a system of modulo congruence equations.
The equations are:
- $8alpha+betaequiv 15 pmod{26}$
- $5alpha+betaequiv 16 pmod{26}$
Could anyone tell how to solve these equations.
modular-arithmetic systems-of-equations congruences
modular-arithmetic systems-of-equations congruences
asked May 11 '15 at 8:13
justinjustin
217113
asked May 11 '15 at 8:13
justinjustin
217113
asked May 11 '15 at 8:13
justinjustin
217113
asked May 11 '15 at 8:13
justinjustin
217113
asked May 11 '15 at 8:13
justinjustin
217113
217113
$begingroup$
$betaequiv 15-8alpha,Rightarrow, 5alpha + (15-8alpha)equiv 15-3alphaequiv 16iff 3alphaequiv -1equiv -27$ $stackrel{:3}iff alphaequiv -9equiv 17$ mod $26$. So $8(17)+betaequiv 15iff betaequiv 9$ mod $26$.
$endgroup$
– user26486
May 13 '15 at 14:44
$begingroup$
@user31415:That's a good answer.Could you tell me how $-1 equiv -27$ came?Also why always we place equivalence symbol in these equations when an equal symbol is enough to get $alpha$ and $beta$?
$endgroup$
– justin
May 14 '15 at 9:24
$begingroup$
$-1 = -27 + 26(1)$, so $-1$ and $-27$ give same remainders when divided by $26$. Definition: $aequiv bpmod{! n}!iff! nmid a-b$ or i.e. $a,b$ leave same remainders when divided by $n$. It is nitpicking about notation whether you can write equality symbols in place of equivalences there, but I would just write equivalences everywhere. I could later divide both sides by $3$ because $(26,3)=1$. It is because $26mid 3(alpha-(-9))!iff! 26mid alpha-(-9)$, which may make this division property more apparent. Remember Fundamental theorem of arithmetic.
$endgroup$
– user26486
May 14 '15 at 10:19
$begingroup$
@user31415:Okay that's right.Yeah I'll go with equivalence.Is it sure that for any numbers m and n if (m,n)=1 you could divide both sides of the equivalence by n?How did you get -27 in the R.H.S?Is it a guess or is there any strategy to follow for finding -27?Also couldn't get how you wrote $26mid 3(alpha-(-9))$ while I think we just need to consider $26mid 3alpha$?
$endgroup$
– justin
May 14 '15 at 12:21
$begingroup$
@user31415:Yeah got it.Could you tell how did you changed $26mid alpha-(-9)$ into $alphaequiv-9 equiv 17$ mod 26?Does this use the same logic you stated in previous comment?Also how to find the value of $alpha$ from $alphaequiv-9 equiv 17$ mod 26?Is it($alpha$) -9 or 17?
$endgroup$
– justin
May 14 '15 at 12:37
|
show 6 more comments
$begingroup$
$betaequiv 15-8alpha,Rightarrow, 5alpha + (15-8alpha)equiv 15-3alphaequiv 16iff 3alphaequiv -1equiv -27$ $stackrel{:3}iff alphaequiv -9equiv 17$ mod $26$. So $8(17)+betaequiv 15iff betaequiv 9$ mod $26$.
$endgroup$
– user26486
May 13 '15 at 14:44
$begingroup$
@user31415:That's a good answer.Could you tell me how $-1 equiv -27$ came?Also why always we place equivalence symbol in these equations when an equal symbol is enough to get $alpha$ and $beta$?
$endgroup$
– justin
May 14 '15 at 9:24
$begingroup$
$-1 = -27 + 26(1)$, so $-1$ and $-27$ give same remainders when divided by $26$. Definition: $aequiv bpmod{! n}!iff! nmid a-b$ or i.e. $a,b$ leave same remainders when divided by $n$. It is nitpicking about notation whether you can write equality symbols in place of equivalences there, but I would just write equivalences everywhere. I could later divide both sides by $3$ because $(26,3)=1$. It is because $26mid 3(alpha-(-9))!iff! 26mid alpha-(-9)$, which may make this division property more apparent. Remember Fundamental theorem of arithmetic.
$endgroup$
– user26486
May 14 '15 at 10:19
$begingroup$
@user31415:Okay that's right.Yeah I'll go with equivalence.Is it sure that for any numbers m and n if (m,n)=1 you could divide both sides of the equivalence by n?How did you get -27 in the R.H.S?Is it a guess or is there any strategy to follow for finding -27?Also couldn't get how you wrote $26mid 3(alpha-(-9))$ while I think we just need to consider $26mid 3alpha$?
$endgroup$
– justin
May 14 '15 at 12:21
$begingroup$
@user31415:Yeah got it.Could you tell how did you changed $26mid alpha-(-9)$ into $alphaequiv-9 equiv 17$ mod 26?Does this use the same logic you stated in previous comment?Also how to find the value of $alpha$ from $alphaequiv-9 equiv 17$ mod 26?Is it($alpha$) -9 or 17?
$endgroup$
– justin
May 14 '15 at 12:37
$begingroup$
$betaequiv 15-8alpha,Rightarrow, 5alpha + (15-8alpha)equiv 15-3alphaequiv 16iff 3alphaequiv -1equiv -27$ $stackrel{:3}iff alphaequiv -9equiv 17$ mod $26$. So $8(17)+betaequiv 15iff betaequiv 9$ mod $26$.
$endgroup$
– user26486
May 13 '15 at 14:44
$begingroup$
$betaequiv 15-8alpha,Rightarrow, 5alpha + (15-8alpha)equiv 15-3alphaequiv 16iff 3alphaequiv -1equiv -27$ $stackrel{:3}iff alphaequiv -9equiv 17$ mod $26$. So $8(17)+betaequiv 15iff betaequiv 9$ mod $26$.
$endgroup$
– user26486
May 13 '15 at 14:44
$begingroup$
@user31415:That's a good answer.Could you tell me how $-1 equiv -27$ came?Also why always we place equivalence symbol in these equations when an equal symbol is enough to get $alpha$ and $beta$?
$endgroup$
– justin
May 14 '15 at 9:24
$begingroup$
@user31415:That's a good answer.Could you tell me how $-1 equiv -27$ came?Also why always we place equivalence symbol in these equations when an equal symbol is enough to get $alpha$ and $beta$?
$endgroup$
– justin
May 14 '15 at 9:24
$begingroup$
$-1 = -27 + 26(1)$, so $-1$ and $-27$ give same remainders when divided by $26$. Definition: $aequiv bpmod{! n}!iff! nmid a-b$ or i.e. $a,b$ leave same remainders when divided by $n$. It is nitpicking about notation whether you can write equality symbols in place of equivalences there, but I would just write equivalences everywhere. I could later divide both sides by $3$ because $(26,3)=1$. It is because $26mid 3(alpha-(-9))!iff! 26mid alpha-(-9)$, which may make this division property more apparent. Remember Fundamental theorem of arithmetic.
$endgroup$
– user26486
May 14 '15 at 10:19
$begingroup$
$-1 = -27 + 26(1)$, so $-1$ and $-27$ give same remainders when divided by $26$. Definition: $aequiv bpmod{! n}!iff! nmid a-b$ or i.e. $a,b$ leave same remainders when divided by $n$. It is nitpicking about notation whether you can write equality symbols in place of equivalences there, but I would just write equivalences everywhere. I could later divide both sides by $3$ because $(26,3)=1$. It is because $26mid 3(alpha-(-9))!iff! 26mid alpha-(-9)$, which may make this division property more apparent. Remember Fundamental theorem of arithmetic.
$endgroup$
– user26486
May 14 '15 at 10:19
$begingroup$
@user31415:Okay that's right.Yeah I'll go with equivalence.Is it sure that for any numbers m and n if (m,n)=1 you could divide both sides of the equivalence by n?How did you get -27 in the R.H.S?Is it a guess or is there any strategy to follow for finding -27?Also couldn't get how you wrote $26mid 3(alpha-(-9))$ while I think we just need to consider $26mid 3alpha$?
$endgroup$
– justin
May 14 '15 at 12:21
$begingroup$
@user31415:Okay that's right.Yeah I'll go with equivalence.Is it sure that for any numbers m and n if (m,n)=1 you could divide both sides of the equivalence by n?How did you get -27 in the R.H.S?Is it a guess or is there any strategy to follow for finding -27?Also couldn't get how you wrote $26mid 3(alpha-(-9))$ while I think we just need to consider $26mid 3alpha$?
$endgroup$
– justin
May 14 '15 at 12:21
$begingroup$
@user31415:Yeah got it.Could you tell how did you changed $26mid alpha-(-9)$ into $alphaequiv-9 equiv 17$ mod 26?Does this use the same logic you stated in previous comment?Also how to find the value of $alpha$ from $alphaequiv-9 equiv 17$ mod 26?Is it($alpha$) -9 or 17?
$endgroup$
– justin
May 14 '15 at 12:37
$begingroup$
@user31415:Yeah got it.Could you tell how did you changed $26mid alpha-(-9)$ into $alphaequiv-9 equiv 17$ mod 26?Does this use the same logic you stated in previous comment?Also how to find the value of $alpha$ from $alphaequiv-9 equiv 17$ mod 26?Is it($alpha$) -9 or 17?
$endgroup$
– justin
May 14 '15 at 12:37
|
show 6 more comments
1 Answer
1
active
oldest
votes
0
$begingroup$
On subtraction, $$3alphaequiv-1pmod{26}equiv-1+2cdot26$$
As $(3,26)=1,$ $$alphaequiv17pmod{26}$$
and $$betaequiv15-8alphapmod{26}equiv?$$
answered May 11 '15 at 8:15
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
$begingroup$
:Could you tell how did you get (-1+2.26) on the R.H.S in the first line of the answer?
$endgroup$
– justin
May 11 '15 at 9:48
$begingroup$
@justin, As $52equiv0pmod{26},51equiv-1$
$endgroup$
– lab bhattacharjee
May 11 '15 at 10:14
$begingroup$
:Could you tell how did you get 52?Do you mean to say that when we subtract these equations we would get $52 equiv 0 pmod{26}$?
$endgroup$
– justin
May 11 '15 at 10:23
$begingroup$
@justin, We need to find a value of $y$ such that $3|(26y-1)$ . By observation, $y=2$ is one of such values
$endgroup$
– lab bhattacharjee
May 11 '15 at 10:24
$begingroup$
:Could you tell why do we need to get a value of y for $3|(26y-1)$ when the equation is just $3alphaequiv-1pmod{26}$ where there is no y?
$endgroup$
– justin
May 11 '15 at 10:44
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
});
}
});
draft saved
draft discarded
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%2f1276790%2fhow-to-solve-congruence-modulo-equations%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
$begingroup$
On subtraction, $$3alphaequiv-1pmod{26}equiv-1+2cdot26$$
As $(3,26)=1,$ $$alphaequiv17pmod{26}$$
and $$betaequiv15-8alphapmod{26}equiv?$$
answered May 11 '15 at 8:15
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
$begingroup$
:Could you tell how did you get (-1+2.26) on the R.H.S in the first line of the answer?
$endgroup$
– justin
May 11 '15 at 9:48
$begingroup$
@justin, As $52equiv0pmod{26},51equiv-1$
$endgroup$
– lab bhattacharjee
May 11 '15 at 10:14
$begingroup$
:Could you tell how did you get 52?Do you mean to say that when we subtract these equations we would get $52 equiv 0 pmod{26}$?
$endgroup$
– justin
May 11 '15 at 10:23
$begingroup$
@justin, We need to find a value of $y$ such that $3|(26y-1)$ . By observation, $y=2$ is one of such values
$endgroup$
– lab bhattacharjee
May 11 '15 at 10:24
$begingroup$
:Could you tell why do we need to get a value of y for $3|(26y-1)$ when the equation is just $3alphaequiv-1pmod{26}$ where there is no y?
$endgroup$
– justin
May 11 '15 at 10:44
add a comment |
0
$begingroup$
On subtraction, $$3alphaequiv-1pmod{26}equiv-1+2cdot26$$
As $(3,26)=1,$ $$alphaequiv17pmod{26}$$
and $$betaequiv15-8alphapmod{26}equiv?$$
answered May 11 '15 at 8:15
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
$begingroup$
:Could you tell how did you get (-1+2.26) on the R.H.S in the first line of the answer?
$endgroup$
– justin
May 11 '15 at 9:48
$begingroup$
@justin, As $52equiv0pmod{26},51equiv-1$
$endgroup$
– lab bhattacharjee
May 11 '15 at 10:14
$begingroup$
:Could you tell how did you get 52?Do you mean to say that when we subtract these equations we would get $52 equiv 0 pmod{26}$?
$endgroup$
– justin
May 11 '15 at 10:23
$begingroup$
@justin, We need to find a value of $y$ such that $3|(26y-1)$ . By observation, $y=2$ is one of such values
$endgroup$
– lab bhattacharjee
May 11 '15 at 10:24
$begingroup$
:Could you tell why do we need to get a value of y for $3|(26y-1)$ when the equation is just $3alphaequiv-1pmod{26}$ where there is no y?
$endgroup$
– justin
May 11 '15 at 10:44
add a comment |
0
0
0
$begingroup$
On subtraction, $$3alphaequiv-1pmod{26}equiv-1+2cdot26$$
As $(3,26)=1,$ $$alphaequiv17pmod{26}$$
and $$betaequiv15-8alphapmod{26}equiv?$$
answered May 11 '15 at 8:15
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
On subtraction, $$3alphaequiv-1pmod{26}equiv-1+2cdot26$$
As $(3,26)=1,$ $$alphaequiv17pmod{26}$$
and $$betaequiv15-8alphapmod{26}equiv?$$
answered May 11 '15 at 8:15
lab bhattacharjeelab bhattacharjee
227k15158275
answered May 11 '15 at 8:15
lab bhattacharjeelab bhattacharjee
227k15158275
answered May 11 '15 at 8:15
lab bhattacharjeelab bhattacharjee
227k15158275
answered May 11 '15 at 8:15
lab bhattacharjeelab bhattacharjee
227k15158275
227k15158275
$begingroup$
:Could you tell how did you get (-1+2.26) on the R.H.S in the first line of the answer?
$endgroup$
– justin
May 11 '15 at 9:48
$begingroup$
@justin, As $52equiv0pmod{26},51equiv-1$
$endgroup$
– lab bhattacharjee
May 11 '15 at 10:14
$begingroup$
:Could you tell how did you get 52?Do you mean to say that when we subtract these equations we would get $52 equiv 0 pmod{26}$?
$endgroup$
– justin
May 11 '15 at 10:23
$begingroup$
@justin, We need to find a value of $y$ such that $3|(26y-1)$ . By observation, $y=2$ is one of such values
$endgroup$
– lab bhattacharjee
May 11 '15 at 10:24
$begingroup$
:Could you tell why do we need to get a value of y for $3|(26y-1)$ when the equation is just $3alphaequiv-1pmod{26}$ where there is no y?
$endgroup$
– justin
May 11 '15 at 10:44
add a comment |
$begingroup$
:Could you tell how did you get (-1+2.26) on the R.H.S in the first line of the answer?
$endgroup$
– justin
May 11 '15 at 9:48
$begingroup$
@justin, As $52equiv0pmod{26},51equiv-1$
$endgroup$
– lab bhattacharjee
May 11 '15 at 10:14
$begingroup$
:Could you tell how did you get 52?Do you mean to say that when we subtract these equations we would get $52 equiv 0 pmod{26}$?
$endgroup$
– justin
May 11 '15 at 10:23
$begingroup$
@justin, We need to find a value of $y$ such that $3|(26y-1)$ . By observation, $y=2$ is one of such values
$endgroup$
– lab bhattacharjee
May 11 '15 at 10:24
$begingroup$
:Could you tell why do we need to get a value of y for $3|(26y-1)$ when the equation is just $3alphaequiv-1pmod{26}$ where there is no y?
$endgroup$
– justin
May 11 '15 at 10:44
$begingroup$
:Could you tell how did you get (-1+2.26) on the R.H.S in the first line of the answer?
$endgroup$
– justin
May 11 '15 at 9:48
$begingroup$
:Could you tell how did you get (-1+2.26) on the R.H.S in the first line of the answer?
$endgroup$
– justin
May 11 '15 at 9:48
$begingroup$
@justin, As $52equiv0pmod{26},51equiv-1$
$endgroup$
– lab bhattacharjee
May 11 '15 at 10:14
$begingroup$
@justin, As $52equiv0pmod{26},51equiv-1$
$endgroup$
– lab bhattacharjee
May 11 '15 at 10:14
$begingroup$
:Could you tell how did you get 52?Do you mean to say that when we subtract these equations we would get $52 equiv 0 pmod{26}$?
$endgroup$
– justin
May 11 '15 at 10:23
$begingroup$
:Could you tell how did you get 52?Do you mean to say that when we subtract these equations we would get $52 equiv 0 pmod{26}$?
$endgroup$
– justin
May 11 '15 at 10:23
$begingroup$
@justin, We need to find a value of $y$ such that $3|(26y-1)$ . By observation, $y=2$ is one of such values
$endgroup$
– lab bhattacharjee
May 11 '15 at 10:24
$begingroup$
@justin, We need to find a value of $y$ such that $3|(26y-1)$ . By observation, $y=2$ is one of such values
$endgroup$
– lab bhattacharjee
May 11 '15 at 10:24
$begingroup$
:Could you tell why do we need to get a value of y for $3|(26y-1)$ when the equation is just $3alphaequiv-1pmod{26}$ where there is no y?
$endgroup$
– justin
May 11 '15 at 10:44
$begingroup$
:Could you tell why do we need to get a value of y for $3|(26y-1)$ when the equation is just $3alphaequiv-1pmod{26}$ where there is no y?
$endgroup$
– justin
May 11 '15 at 10:44
add a comment |
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
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%2f1276790%2fhow-to-solve-congruence-modulo-equations%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
$begingroup$
$betaequiv 15-8alpha,Rightarrow, 5alpha + (15-8alpha)equiv 15-3alphaequiv 16iff 3alphaequiv -1equiv -27$ $stackrel{:3}iff alphaequiv -9equiv 17$ mod $26$. So $8(17)+betaequiv 15iff betaequiv 9$ mod $26$.
$endgroup$
– user26486
May 13 '15 at 14:44
$begingroup$
@user31415:That's a good answer.Could you tell me how $-1 equiv -27$ came?Also why always we place equivalence symbol in these equations when an equal symbol is enough to get $alpha$ and $beta$?
$endgroup$
– justin
May 14 '15 at 9:24
$begingroup$
$-1 = -27 + 26(1)$, so $-1$ and $-27$ give same remainders when divided by $26$. Definition: $aequiv bpmod{! n}!iff! nmid a-b$ or i.e. $a,b$ leave same remainders when divided by $n$. It is nitpicking about notation whether you can write equality symbols in place of equivalences there, but I would just write equivalences everywhere. I could later divide both sides by $3$ because $(26,3)=1$. It is because $26mid 3(alpha-(-9))!iff! 26mid alpha-(-9)$, which may make this division property more apparent. Remember Fundamental theorem of arithmetic.
$endgroup$
– user26486
May 14 '15 at 10:19
$begingroup$
@user31415:Okay that's right.Yeah I'll go with equivalence.Is it sure that for any numbers m and n if (m,n)=1 you could divide both sides of the equivalence by n?How did you get -27 in the R.H.S?Is it a guess or is there any strategy to follow for finding -27?Also couldn't get how you wrote $26mid 3(alpha-(-9))$ while I think we just need to consider $26mid 3alpha$?
$endgroup$
– justin
May 14 '15 at 12:21
$begingroup$
@user31415:Yeah got it.Could you tell how did you changed $26mid alpha-(-9)$ into $alphaequiv-9 equiv 17$ mod 26?Does this use the same logic you stated in previous comment?Also how to find the value of $alpha$ from $alphaequiv-9 equiv 17$ mod 26?Is it($alpha$) -9 or 17?
$endgroup$
– justin
May 14 '15 at 12:37