Find X in the Equation












0












$begingroup$


I'm not a mathematician and I have forgotten about some basics in mathematics.
I have this equation:



$$x^y pmod z = w$$



Given $y, z,$ and $w,$ how will I find $x$? How will I get the equation for $x$?






modular-arithmetic linear-transformations education prime-factorization







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    (old joke): I found $x$, it is inside the parentheses: $$(boxed{x}^y) mod z = w$$
    $endgroup$
    – Michael
    Sep 17 '18 at 4:22








  • 1




    $begingroup$
    @Michael Lol.Anyways how will I get the equation for x?
    $endgroup$
    – Eliyah
    Sep 17 '18 at 4:28






  • 1




    $begingroup$
    Please refer to how to ask a good question to help you improve formatting and your tags
    $endgroup$
    – Chase Ryan Taylor
    Sep 17 '18 at 5:00
















0












$begingroup$


I'm not a mathematician and I have forgotten about some basics in mathematics.
I have this equation:



$$x^y pmod z = w$$



Given $y, z,$ and $w,$ how will I find $x$? How will I get the equation for $x$?






modular-arithmetic linear-transformations education prime-factorization







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    (old joke): I found $x$, it is inside the parentheses: $$(boxed{x}^y) mod z = w$$
    $endgroup$
    – Michael
    Sep 17 '18 at 4:22








  • 1




    $begingroup$
    @Michael Lol.Anyways how will I get the equation for x?
    $endgroup$
    – Eliyah
    Sep 17 '18 at 4:28






  • 1




    $begingroup$
    Please refer to how to ask a good question to help you improve formatting and your tags
    $endgroup$
    – Chase Ryan Taylor
    Sep 17 '18 at 5:00














0












0








0





$begingroup$


I'm not a mathematician and I have forgotten about some basics in mathematics.
I have this equation:



$$x^y pmod z = w$$



Given $y, z,$ and $w,$ how will I find $x$? How will I get the equation for $x$?






modular-arithmetic linear-transformations education prime-factorization







share|cite|improve this question











$endgroup$




I'm not a mathematician and I have forgotten about some basics in mathematics.
I have this equation:



$$x^y pmod z = w$$



Given $y, z,$ and $w,$ how will I find $x$? How will I get the equation for $x$?






modular-arithmetic linear-transformations education prime-factorization




modular-arithmetic linear-transformations education prime-factorization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 13:40









amWhy

192k28225439




192k28225439










asked Sep 17 '18 at 4:13









EliyahEliyah

1236




1236








  • 1




    $begingroup$
    (old joke): I found $x$, it is inside the parentheses: $$(boxed{x}^y) mod z = w$$
    $endgroup$
    – Michael
    Sep 17 '18 at 4:22








  • 1




    $begingroup$
    @Michael Lol.Anyways how will I get the equation for x?
    $endgroup$
    – Eliyah
    Sep 17 '18 at 4:28






  • 1




    $begingroup$
    Please refer to how to ask a good question to help you improve formatting and your tags
    $endgroup$
    – Chase Ryan Taylor
    Sep 17 '18 at 5:00














  • 1




    $begingroup$
    (old joke): I found $x$, it is inside the parentheses: $$(boxed{x}^y) mod z = w$$
    $endgroup$
    – Michael
    Sep 17 '18 at 4:22








  • 1




    $begingroup$
    @Michael Lol.Anyways how will I get the equation for x?
    $endgroup$
    – Eliyah
    Sep 17 '18 at 4:28






  • 1




    $begingroup$
    Please refer to how to ask a good question to help you improve formatting and your tags
    $endgroup$
    – Chase Ryan Taylor
    Sep 17 '18 at 5:00








1




1




$begingroup$
(old joke): I found $x$, it is inside the parentheses: $$(boxed{x}^y) mod z = w$$
$endgroup$
– Michael
Sep 17 '18 at 4:22






$begingroup$
(old joke): I found $x$, it is inside the parentheses: $$(boxed{x}^y) mod z = w$$
$endgroup$
– Michael
Sep 17 '18 at 4:22






1




1




$begingroup$
@Michael Lol.Anyways how will I get the equation for x?
$endgroup$
– Eliyah
Sep 17 '18 at 4:28




$begingroup$
@Michael Lol.Anyways how will I get the equation for x?
$endgroup$
– Eliyah
Sep 17 '18 at 4:28




1




1




$begingroup$
Please refer to how to ask a good question to help you improve formatting and your tags
$endgroup$
– Chase Ryan Taylor
Sep 17 '18 at 5:00




$begingroup$
Please refer to how to ask a good question to help you improve formatting and your tags
$endgroup$
– Chase Ryan Taylor
Sep 17 '18 at 5:00










1 Answer
1






active

oldest

votes


















2












$begingroup$

I assume here that $x,y,z,w$ are integers (and $y$ is a positive integer). Then your equation means:
$$ x^y = rz+w$$
for some integer $r$. Now note that if $x$ has this property, then so does $x+kz$ for any integer $k$. So it suffices to search for $x in {0, 1, 2, ..., z-1}$.



Example:
$$x^{11} mod 5 = 2 $$



Try $x in {0, 1, 2, 3, 4}$:
begin{align}
0^{11} = 0 : (mod 5)\
1^{11} = 1 : (mod 5) \
2^{11} = 3 : (mod 5) \
boxed{3^{11} = 2} : (mod 5) \
4^{11} = 4 : (mod 5)
end{align}
So then solutions are $x=3+5k$ for all integers $k$.




The $mod z$ operation is equal to the remainder when we divide by $z$. So for example




  • $Rem[26/5] = 1$ if and only if $26-1$ is divisible by $5$.

  • $Rem[83/9] = 2$ if and only if $83-2$ is divisible by $9$.

  • $Rem[26/7] = 5$ if and only if $26 - 5$ is divisible by $7$.

  • $Rem[n/z] = w$ if and only if $n-w$ is divisible by $z$.


In particular $Rem[n/z]=w$ if and only if $n-w= rz$ for some integer $r$.



If we take $n=x^y$ then we see that $Rem[x^y/z]=w$ if and only if $x^y - w = rz$ for some integer $r$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much but how did u get the rz term. Or what is the opposite operation of a modulo? Can you explain it step by step. Thank you again.
    $endgroup$
    – Eliyah
    Sep 17 '18 at 4:49








  • 1




    $begingroup$
    See my added content on this.
    $endgroup$
    – Michael
    Sep 17 '18 at 4:56











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%2f2919791%2ffind-x-in-the-equation%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









2












$begingroup$

I assume here that $x,y,z,w$ are integers (and $y$ is a positive integer). Then your equation means:
$$ x^y = rz+w$$
for some integer $r$. Now note that if $x$ has this property, then so does $x+kz$ for any integer $k$. So it suffices to search for $x in {0, 1, 2, ..., z-1}$.



Example:
$$x^{11} mod 5 = 2 $$



Try $x in {0, 1, 2, 3, 4}$:
begin{align}
0^{11} = 0 : (mod 5)\
1^{11} = 1 : (mod 5) \
2^{11} = 3 : (mod 5) \
boxed{3^{11} = 2} : (mod 5) \
4^{11} = 4 : (mod 5)
end{align}
So then solutions are $x=3+5k$ for all integers $k$.




The $mod z$ operation is equal to the remainder when we divide by $z$. So for example




  • $Rem[26/5] = 1$ if and only if $26-1$ is divisible by $5$.

  • $Rem[83/9] = 2$ if and only if $83-2$ is divisible by $9$.

  • $Rem[26/7] = 5$ if and only if $26 - 5$ is divisible by $7$.

  • $Rem[n/z] = w$ if and only if $n-w$ is divisible by $z$.


In particular $Rem[n/z]=w$ if and only if $n-w= rz$ for some integer $r$.



If we take $n=x^y$ then we see that $Rem[x^y/z]=w$ if and only if $x^y - w = rz$ for some integer $r$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much but how did u get the rz term. Or what is the opposite operation of a modulo? Can you explain it step by step. Thank you again.
    $endgroup$
    – Eliyah
    Sep 17 '18 at 4:49








  • 1




    $begingroup$
    See my added content on this.
    $endgroup$
    – Michael
    Sep 17 '18 at 4:56
















2












$begingroup$

I assume here that $x,y,z,w$ are integers (and $y$ is a positive integer). Then your equation means:
$$ x^y = rz+w$$
for some integer $r$. Now note that if $x$ has this property, then so does $x+kz$ for any integer $k$. So it suffices to search for $x in {0, 1, 2, ..., z-1}$.



Example:
$$x^{11} mod 5 = 2 $$



Try $x in {0, 1, 2, 3, 4}$:
begin{align}
0^{11} = 0 : (mod 5)\
1^{11} = 1 : (mod 5) \
2^{11} = 3 : (mod 5) \
boxed{3^{11} = 2} : (mod 5) \
4^{11} = 4 : (mod 5)
end{align}
So then solutions are $x=3+5k$ for all integers $k$.




The $mod z$ operation is equal to the remainder when we divide by $z$. So for example




  • $Rem[26/5] = 1$ if and only if $26-1$ is divisible by $5$.

  • $Rem[83/9] = 2$ if and only if $83-2$ is divisible by $9$.

  • $Rem[26/7] = 5$ if and only if $26 - 5$ is divisible by $7$.

  • $Rem[n/z] = w$ if and only if $n-w$ is divisible by $z$.


In particular $Rem[n/z]=w$ if and only if $n-w= rz$ for some integer $r$.



If we take $n=x^y$ then we see that $Rem[x^y/z]=w$ if and only if $x^y - w = rz$ for some integer $r$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much but how did u get the rz term. Or what is the opposite operation of a modulo? Can you explain it step by step. Thank you again.
    $endgroup$
    – Eliyah
    Sep 17 '18 at 4:49








  • 1




    $begingroup$
    See my added content on this.
    $endgroup$
    – Michael
    Sep 17 '18 at 4:56














2












2








2





$begingroup$

I assume here that $x,y,z,w$ are integers (and $y$ is a positive integer). Then your equation means:
$$ x^y = rz+w$$
for some integer $r$. Now note that if $x$ has this property, then so does $x+kz$ for any integer $k$. So it suffices to search for $x in {0, 1, 2, ..., z-1}$.



Example:
$$x^{11} mod 5 = 2 $$



Try $x in {0, 1, 2, 3, 4}$:
begin{align}
0^{11} = 0 : (mod 5)\
1^{11} = 1 : (mod 5) \
2^{11} = 3 : (mod 5) \
boxed{3^{11} = 2} : (mod 5) \
4^{11} = 4 : (mod 5)
end{align}
So then solutions are $x=3+5k$ for all integers $k$.




The $mod z$ operation is equal to the remainder when we divide by $z$. So for example




  • $Rem[26/5] = 1$ if and only if $26-1$ is divisible by $5$.

  • $Rem[83/9] = 2$ if and only if $83-2$ is divisible by $9$.

  • $Rem[26/7] = 5$ if and only if $26 - 5$ is divisible by $7$.

  • $Rem[n/z] = w$ if and only if $n-w$ is divisible by $z$.


In particular $Rem[n/z]=w$ if and only if $n-w= rz$ for some integer $r$.



If we take $n=x^y$ then we see that $Rem[x^y/z]=w$ if and only if $x^y - w = rz$ for some integer $r$.






share|cite|improve this answer











$endgroup$



I assume here that $x,y,z,w$ are integers (and $y$ is a positive integer). Then your equation means:
$$ x^y = rz+w$$
for some integer $r$. Now note that if $x$ has this property, then so does $x+kz$ for any integer $k$. So it suffices to search for $x in {0, 1, 2, ..., z-1}$.



Example:
$$x^{11} mod 5 = 2 $$



Try $x in {0, 1, 2, 3, 4}$:
begin{align}
0^{11} = 0 : (mod 5)\
1^{11} = 1 : (mod 5) \
2^{11} = 3 : (mod 5) \
boxed{3^{11} = 2} : (mod 5) \
4^{11} = 4 : (mod 5)
end{align}
So then solutions are $x=3+5k$ for all integers $k$.




The $mod z$ operation is equal to the remainder when we divide by $z$. So for example




  • $Rem[26/5] = 1$ if and only if $26-1$ is divisible by $5$.

  • $Rem[83/9] = 2$ if and only if $83-2$ is divisible by $9$.

  • $Rem[26/7] = 5$ if and only if $26 - 5$ is divisible by $7$.

  • $Rem[n/z] = w$ if and only if $n-w$ is divisible by $z$.


In particular $Rem[n/z]=w$ if and only if $n-w= rz$ for some integer $r$.



If we take $n=x^y$ then we see that $Rem[x^y/z]=w$ if and only if $x^y - w = rz$ for some integer $r$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Sep 17 '18 at 5:01

























answered Sep 17 '18 at 4:42









MichaelMichael

12.8k11429




12.8k11429












  • $begingroup$
    Thank you so much but how did u get the rz term. Or what is the opposite operation of a modulo? Can you explain it step by step. Thank you again.
    $endgroup$
    – Eliyah
    Sep 17 '18 at 4:49








  • 1




    $begingroup$
    See my added content on this.
    $endgroup$
    – Michael
    Sep 17 '18 at 4:56


















  • $begingroup$
    Thank you so much but how did u get the rz term. Or what is the opposite operation of a modulo? Can you explain it step by step. Thank you again.
    $endgroup$
    – Eliyah
    Sep 17 '18 at 4:49








  • 1




    $begingroup$
    See my added content on this.
    $endgroup$
    – Michael
    Sep 17 '18 at 4:56
















$begingroup$
Thank you so much but how did u get the rz term. Or what is the opposite operation of a modulo? Can you explain it step by step. Thank you again.
$endgroup$
– Eliyah
Sep 17 '18 at 4:49






$begingroup$
Thank you so much but how did u get the rz term. Or what is the opposite operation of a modulo? Can you explain it step by step. Thank you again.
$endgroup$
– Eliyah
Sep 17 '18 at 4:49






1




1




$begingroup$
See my added content on this.
$endgroup$
– Michael
Sep 17 '18 at 4:56




$begingroup$
See my added content on this.
$endgroup$
– Michael
Sep 17 '18 at 4:56


















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%2f2919791%2ffind-x-in-the-equation%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







-

Popular posts from this blog

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
Read more