Solving $472x ≡ 32 ;(text{mod } 92)$: How to solve when $a$ is greater than $m$
$begingroup$
I'm not sure where to begin here. I think that I can change the $472x$ to be $12x$ because they should be equal in mod $92$. Then, I believe I can simplify the equation to be: $3x ≡ 8 ; (text{mod } 23)$
modular-arithmetic
edited Dec 10 '18 at 20:56
MisterRiemann
5,8121625
asked Dec 10 '18 at 20:39
user52640user52640
453
$endgroup$
add a comment |
$begingroup$
I'm not sure where to begin here. I think that I can change the $472x$ to be $12x$ because they should be equal in mod $92$. Then, I believe I can simplify the equation to be: $3x ≡ 8 ; (text{mod } 23)$
modular-arithmetic
edited Dec 10 '18 at 20:56
MisterRiemann
5,8121625
asked Dec 10 '18 at 20:39
user52640user52640
453
$endgroup$
$begingroup$
You're almost done: $bmod 23!:, 3xequiv 8equiv -15,$ so $,xequiv -5 $
$endgroup$
– Bill Dubuque
Dec 26 '18 at 0:47
add a comment |
$begingroup$
I'm not sure where to begin here. I think that I can change the $472x$ to be $12x$ because they should be equal in mod $92$. Then, I believe I can simplify the equation to be: $3x ≡ 8 ; (text{mod } 23)$
modular-arithmetic
edited Dec 10 '18 at 20:56
MisterRiemann
5,8121625
asked Dec 10 '18 at 20:39
user52640user52640
453
$endgroup$
I'm not sure where to begin here. I think that I can change the $472x$ to be $12x$ because they should be equal in mod $92$. Then, I believe I can simplify the equation to be: $3x ≡ 8 ; (text{mod } 23)$
modular-arithmetic
modular-arithmetic
edited Dec 10 '18 at 20:56
MisterRiemann
5,8121625
asked Dec 10 '18 at 20:39
user52640user52640
453
edited Dec 10 '18 at 20:56
MisterRiemann
5,8121625
asked Dec 10 '18 at 20:39
user52640user52640
453
edited Dec 10 '18 at 20:56
MisterRiemann
5,8121625
edited Dec 10 '18 at 20:56
MisterRiemann
5,8121625
edited Dec 10 '18 at 20:56
MisterRiemann
5,8121625
5,8121625
asked Dec 10 '18 at 20:39
user52640user52640
453
asked Dec 10 '18 at 20:39
user52640user52640
453
asked Dec 10 '18 at 20:39
user52640user52640
453
453
$begingroup$
You're almost done: $bmod 23!:, 3xequiv 8equiv -15,$ so $,xequiv -5 $
$endgroup$
– Bill Dubuque
Dec 26 '18 at 0:47
add a comment |
$begingroup$
You're almost done: $bmod 23!:, 3xequiv 8equiv -15,$ so $,xequiv -5 $
$endgroup$
– Bill Dubuque
Dec 26 '18 at 0:47
$begingroup$
You're almost done: $bmod 23!:, 3xequiv 8equiv -15,$ so $,xequiv -5 $
$endgroup$
– Bill Dubuque
Dec 26 '18 at 0:47
$begingroup$
You're almost done: $bmod 23!:, 3xequiv 8equiv -15,$ so $,xequiv -5 $
$endgroup$
– Bill Dubuque
Dec 26 '18 at 0:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You're on the right track. Starting from
$$ 3x equiv 8 ; (text{mod } 23), $$
you should find the multiplicative inverse $3^{-1}$ of $3$ modulo $23$, i.e. the number satisfying
$$ 3^{-1} cdot 3 equiv 1 ; (text{mod } 23). $$
This can be done with the help of the Euclidean algorithm, or alternatively you could just notice that
$$ 8cdot 3 = 24 equiv 1 ; (text{mod } 23), $$
so $3^{-1} = 8$ in this case. Then you can multiply both sides of your equation by this multiplicative inverse to obtain
$$ x equiv 8cdot 3 x equiv 8 cdot 8 = 64 equiv 18 ; (text{mod } 23). $$
Hence $x = 18+23k$, for an integer $k$, so you get four incongruent solutions (modulo $92$) to your original equation, namely by inserting $k=0,1,2,3$.
answered Dec 10 '18 at 20:46
MisterRiemannMisterRiemann
5,8121625
$endgroup$
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%2f3034431%2fsolving-472x-%25e2%2589%25a1-32-textmod-92-how-to-solve-when-a-is-greater-than-m%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
$begingroup$
You're on the right track. Starting from
$$ 3x equiv 8 ; (text{mod } 23), $$
you should find the multiplicative inverse $3^{-1}$ of $3$ modulo $23$, i.e. the number satisfying
$$ 3^{-1} cdot 3 equiv 1 ; (text{mod } 23). $$
This can be done with the help of the Euclidean algorithm, or alternatively you could just notice that
$$ 8cdot 3 = 24 equiv 1 ; (text{mod } 23), $$
so $3^{-1} = 8$ in this case. Then you can multiply both sides of your equation by this multiplicative inverse to obtain
$$ x equiv 8cdot 3 x equiv 8 cdot 8 = 64 equiv 18 ; (text{mod } 23). $$
Hence $x = 18+23k$, for an integer $k$, so you get four incongruent solutions (modulo $92$) to your original equation, namely by inserting $k=0,1,2,3$.
answered Dec 10 '18 at 20:46
MisterRiemannMisterRiemann
5,8121625
$endgroup$
add a comment |
$begingroup$
You're on the right track. Starting from
$$ 3x equiv 8 ; (text{mod } 23), $$
you should find the multiplicative inverse $3^{-1}$ of $3$ modulo $23$, i.e. the number satisfying
$$ 3^{-1} cdot 3 equiv 1 ; (text{mod } 23). $$
This can be done with the help of the Euclidean algorithm, or alternatively you could just notice that
$$ 8cdot 3 = 24 equiv 1 ; (text{mod } 23), $$
so $3^{-1} = 8$ in this case. Then you can multiply both sides of your equation by this multiplicative inverse to obtain
$$ x equiv 8cdot 3 x equiv 8 cdot 8 = 64 equiv 18 ; (text{mod } 23). $$
Hence $x = 18+23k$, for an integer $k$, so you get four incongruent solutions (modulo $92$) to your original equation, namely by inserting $k=0,1,2,3$.
answered Dec 10 '18 at 20:46
MisterRiemannMisterRiemann
5,8121625
$endgroup$
add a comment |
$begingroup$
You're on the right track. Starting from
$$ 3x equiv 8 ; (text{mod } 23), $$
you should find the multiplicative inverse $3^{-1}$ of $3$ modulo $23$, i.e. the number satisfying
$$ 3^{-1} cdot 3 equiv 1 ; (text{mod } 23). $$
This can be done with the help of the Euclidean algorithm, or alternatively you could just notice that
$$ 8cdot 3 = 24 equiv 1 ; (text{mod } 23), $$
so $3^{-1} = 8$ in this case. Then you can multiply both sides of your equation by this multiplicative inverse to obtain
$$ x equiv 8cdot 3 x equiv 8 cdot 8 = 64 equiv 18 ; (text{mod } 23). $$
Hence $x = 18+23k$, for an integer $k$, so you get four incongruent solutions (modulo $92$) to your original equation, namely by inserting $k=0,1,2,3$.
answered Dec 10 '18 at 20:46
MisterRiemannMisterRiemann
5,8121625
$endgroup$
You're on the right track. Starting from
$$ 3x equiv 8 ; (text{mod } 23), $$
you should find the multiplicative inverse $3^{-1}$ of $3$ modulo $23$, i.e. the number satisfying
$$ 3^{-1} cdot 3 equiv 1 ; (text{mod } 23). $$
This can be done with the help of the Euclidean algorithm, or alternatively you could just notice that
$$ 8cdot 3 = 24 equiv 1 ; (text{mod } 23), $$
so $3^{-1} = 8$ in this case. Then you can multiply both sides of your equation by this multiplicative inverse to obtain
$$ x equiv 8cdot 3 x equiv 8 cdot 8 = 64 equiv 18 ; (text{mod } 23). $$
Hence $x = 18+23k$, for an integer $k$, so you get four incongruent solutions (modulo $92$) to your original equation, namely by inserting $k=0,1,2,3$.
answered Dec 10 '18 at 20:46
MisterRiemannMisterRiemann
5,8121625
answered Dec 10 '18 at 20:46
MisterRiemannMisterRiemann
5,8121625
answered Dec 10 '18 at 20:46
MisterRiemannMisterRiemann
5,8121625
answered Dec 10 '18 at 20:46
MisterRiemannMisterRiemann
5,8121625
5,8121625
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%2f3034431%2fsolving-472x-%25e2%2589%25a1-32-textmod-92-how-to-solve-when-a-is-greater-than-m%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
$begingroup$
You're almost done: $bmod 23!:, 3xequiv 8equiv -15,$ so $,xequiv -5 $
$endgroup$
– Bill Dubuque
Dec 26 '18 at 0:47