How to calculate mod of number with big exponent












1












$begingroup$


I want to find



$$
5^{133} mod 8.
$$
I have noticed that $5^n mod 8 = 5$ when $n$ is uneven and 1 otherwise, which would lead me to say that $5^{133} mod 8 = 5$ But I don't know how to prove this. How can I prove that this is the case (or find another solution if it is not)?






algebra-precalculus arithmetic







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: $bmod 8!: color{#c00}{5^{large 2}equiv 1},Rightarrow, 5^{large 2n}equiv (color{#c00}{5^{large 2}})^{large n}equiv color{#c00}1^{large n}equiv 1 $
    $endgroup$
    – Bill Dubuque
    2 hours ago












  • $begingroup$
    What does the triple equal sign mean?
    $endgroup$
    – Sandi
    2 hours ago










  • $begingroup$
    25 leaves a remainder of 1 when divided by 8 so it means they are "equivalent" under mod 8
    $endgroup$
    – Eleven-Eleven
    2 hours ago












  • $begingroup$
    @Sandi this is the common way to write congruences.
    $endgroup$
    – Mark
    2 hours ago
















1












$begingroup$


I want to find



$$
5^{133} mod 8.
$$
I have noticed that $5^n mod 8 = 5$ when $n$ is uneven and 1 otherwise, which would lead me to say that $5^{133} mod 8 = 5$ But I don't know how to prove this. How can I prove that this is the case (or find another solution if it is not)?






algebra-precalculus arithmetic







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: $bmod 8!: color{#c00}{5^{large 2}equiv 1},Rightarrow, 5^{large 2n}equiv (color{#c00}{5^{large 2}})^{large n}equiv color{#c00}1^{large n}equiv 1 $
    $endgroup$
    – Bill Dubuque
    2 hours ago












  • $begingroup$
    What does the triple equal sign mean?
    $endgroup$
    – Sandi
    2 hours ago










  • $begingroup$
    25 leaves a remainder of 1 when divided by 8 so it means they are "equivalent" under mod 8
    $endgroup$
    – Eleven-Eleven
    2 hours ago












  • $begingroup$
    @Sandi this is the common way to write congruences.
    $endgroup$
    – Mark
    2 hours ago














1












1








1





$begingroup$


I want to find



$$
5^{133} mod 8.
$$
I have noticed that $5^n mod 8 = 5$ when $n$ is uneven and 1 otherwise, which would lead me to say that $5^{133} mod 8 = 5$ But I don't know how to prove this. How can I prove that this is the case (or find another solution if it is not)?






algebra-precalculus arithmetic







share|cite|improve this question











$endgroup$




I want to find



$$
5^{133} mod 8.
$$
I have noticed that $5^n mod 8 = 5$ when $n$ is uneven and 1 otherwise, which would lead me to say that $5^{133} mod 8 = 5$ But I don't know how to prove this. How can I prove that this is the case (or find another solution if it is not)?






algebra-precalculus arithmetic




algebra-precalculus arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago







Sandi

















asked 2 hours ago









SandiSandi

262112




262112








  • 1




    $begingroup$
    Hint: $bmod 8!: color{#c00}{5^{large 2}equiv 1},Rightarrow, 5^{large 2n}equiv (color{#c00}{5^{large 2}})^{large n}equiv color{#c00}1^{large n}equiv 1 $
    $endgroup$
    – Bill Dubuque
    2 hours ago












  • $begingroup$
    What does the triple equal sign mean?
    $endgroup$
    – Sandi
    2 hours ago










  • $begingroup$
    25 leaves a remainder of 1 when divided by 8 so it means they are "equivalent" under mod 8
    $endgroup$
    – Eleven-Eleven
    2 hours ago












  • $begingroup$
    @Sandi this is the common way to write congruences.
    $endgroup$
    – Mark
    2 hours ago














  • 1




    $begingroup$
    Hint: $bmod 8!: color{#c00}{5^{large 2}equiv 1},Rightarrow, 5^{large 2n}equiv (color{#c00}{5^{large 2}})^{large n}equiv color{#c00}1^{large n}equiv 1 $
    $endgroup$
    – Bill Dubuque
    2 hours ago












  • $begingroup$
    What does the triple equal sign mean?
    $endgroup$
    – Sandi
    2 hours ago










  • $begingroup$
    25 leaves a remainder of 1 when divided by 8 so it means they are "equivalent" under mod 8
    $endgroup$
    – Eleven-Eleven
    2 hours ago












  • $begingroup$
    @Sandi this is the common way to write congruences.
    $endgroup$
    – Mark
    2 hours ago








1




1




$begingroup$
Hint: $bmod 8!: color{#c00}{5^{large 2}equiv 1},Rightarrow, 5^{large 2n}equiv (color{#c00}{5^{large 2}})^{large n}equiv color{#c00}1^{large n}equiv 1 $
$endgroup$
– Bill Dubuque
2 hours ago






$begingroup$
Hint: $bmod 8!: color{#c00}{5^{large 2}equiv 1},Rightarrow, 5^{large 2n}equiv (color{#c00}{5^{large 2}})^{large n}equiv color{#c00}1^{large n}equiv 1 $
$endgroup$
– Bill Dubuque
2 hours ago














$begingroup$
What does the triple equal sign mean?
$endgroup$
– Sandi
2 hours ago




$begingroup$
What does the triple equal sign mean?
$endgroup$
– Sandi
2 hours ago












$begingroup$
25 leaves a remainder of 1 when divided by 8 so it means they are "equivalent" under mod 8
$endgroup$
– Eleven-Eleven
2 hours ago






$begingroup$
25 leaves a remainder of 1 when divided by 8 so it means they are "equivalent" under mod 8
$endgroup$
– Eleven-Eleven
2 hours ago














$begingroup$
@Sandi this is the common way to write congruences.
$endgroup$
– Mark
2 hours ago




$begingroup$
@Sandi this is the common way to write congruences.
$endgroup$
– Mark
2 hours ago










2 Answers
2






active

oldest

votes


















6












$begingroup$

First of all it is easy to see that $5^2equiv 1$ (mod $8$). We also know that $133=66times 2+1$. Hence



$5^{133}equiv 5^{2times 66+1}equiv 5times (5^2)^{66}equiv 5times 1^{66}equiv 5$ (mod $8$).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Should the first triple-equal sign on the second row not be a normal equal sign? And the second one too?
    $endgroup$
    – Sandi
    2 hours ago








  • 1




    $begingroup$
    Yes, you can also write regular equality there. But if two numbers are equal in $mathbb{Z}$ then they are also congruent mod $n$ for any $ninmathbb{N}$, and because we only care about congruence here I decided to write congruence from the beginning.
    $endgroup$
    – Mark
    2 hours ago










  • $begingroup$
    I don't understand how you do the step $5 times (5^2)^{66} equiv 5 times 1^{66} (mathrm{mod} 8)$. Could you clarify please?
    $endgroup$
    – Sandi
    2 hours ago








  • 1




    $begingroup$
    $5^2equiv 1$(mod $8$), hence $(5^2)^{66}equiv 1^{66}equiv 1$(mod $8$). That's it. It all follows from a basic theorem on modular arithmetic: that if $aequiv b$(mod $n$) and $cequiv d$(mod $n$) then $abequiv cd$(mod $n$). In other words multiplication mod $n$ is well defined. (same thing can we said about sum mod $n$)
    $endgroup$
    – Mark
    2 hours ago












  • $begingroup$
    I found a similar thing in my book, but it said that $ac equiv bd (modn)$. Did you make a typo or is it a different theorem?
    $endgroup$
    – Sandi
    1 hour ago



















0












$begingroup$

"Highbrow method":



If you undertake to learn any number theory , there is Euler's theorem: $$a^{varphi (n)}cong1pmod n$$, for $a$ relatively prime to $n$.



($varphi$ is Euler's totient function, and just returns the number of positive integers less than $n$ that are relatively prime to $n$.)



Now $varphi (8)=4$. Hence $5^4cong1pmod8.$



Use this to reduce the exponent: $133=4×33+1implies5^{133}=(5^4)^{33}cdot 5cong5pmod8$.






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    First of all it is easy to see that $5^2equiv 1$ (mod $8$). We also know that $133=66times 2+1$. Hence



    $5^{133}equiv 5^{2times 66+1}equiv 5times (5^2)^{66}equiv 5times 1^{66}equiv 5$ (mod $8$).






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Should the first triple-equal sign on the second row not be a normal equal sign? And the second one too?
      $endgroup$
      – Sandi
      2 hours ago








    • 1




      $begingroup$
      Yes, you can also write regular equality there. But if two numbers are equal in $mathbb{Z}$ then they are also congruent mod $n$ for any $ninmathbb{N}$, and because we only care about congruence here I decided to write congruence from the beginning.
      $endgroup$
      – Mark
      2 hours ago










    • $begingroup$
      I don't understand how you do the step $5 times (5^2)^{66} equiv 5 times 1^{66} (mathrm{mod} 8)$. Could you clarify please?
      $endgroup$
      – Sandi
      2 hours ago








    • 1




      $begingroup$
      $5^2equiv 1$(mod $8$), hence $(5^2)^{66}equiv 1^{66}equiv 1$(mod $8$). That's it. It all follows from a basic theorem on modular arithmetic: that if $aequiv b$(mod $n$) and $cequiv d$(mod $n$) then $abequiv cd$(mod $n$). In other words multiplication mod $n$ is well defined. (same thing can we said about sum mod $n$)
      $endgroup$
      – Mark
      2 hours ago












    • $begingroup$
      I found a similar thing in my book, but it said that $ac equiv bd (modn)$. Did you make a typo or is it a different theorem?
      $endgroup$
      – Sandi
      1 hour ago
















    6












    $begingroup$

    First of all it is easy to see that $5^2equiv 1$ (mod $8$). We also know that $133=66times 2+1$. Hence



    $5^{133}equiv 5^{2times 66+1}equiv 5times (5^2)^{66}equiv 5times 1^{66}equiv 5$ (mod $8$).






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Should the first triple-equal sign on the second row not be a normal equal sign? And the second one too?
      $endgroup$
      – Sandi
      2 hours ago








    • 1




      $begingroup$
      Yes, you can also write regular equality there. But if two numbers are equal in $mathbb{Z}$ then they are also congruent mod $n$ for any $ninmathbb{N}$, and because we only care about congruence here I decided to write congruence from the beginning.
      $endgroup$
      – Mark
      2 hours ago










    • $begingroup$
      I don't understand how you do the step $5 times (5^2)^{66} equiv 5 times 1^{66} (mathrm{mod} 8)$. Could you clarify please?
      $endgroup$
      – Sandi
      2 hours ago








    • 1




      $begingroup$
      $5^2equiv 1$(mod $8$), hence $(5^2)^{66}equiv 1^{66}equiv 1$(mod $8$). That's it. It all follows from a basic theorem on modular arithmetic: that if $aequiv b$(mod $n$) and $cequiv d$(mod $n$) then $abequiv cd$(mod $n$). In other words multiplication mod $n$ is well defined. (same thing can we said about sum mod $n$)
      $endgroup$
      – Mark
      2 hours ago












    • $begingroup$
      I found a similar thing in my book, but it said that $ac equiv bd (modn)$. Did you make a typo or is it a different theorem?
      $endgroup$
      – Sandi
      1 hour ago














    6












    6








    6





    $begingroup$

    First of all it is easy to see that $5^2equiv 1$ (mod $8$). We also know that $133=66times 2+1$. Hence



    $5^{133}equiv 5^{2times 66+1}equiv 5times (5^2)^{66}equiv 5times 1^{66}equiv 5$ (mod $8$).






    share|cite|improve this answer











    $endgroup$



    First of all it is easy to see that $5^2equiv 1$ (mod $8$). We also know that $133=66times 2+1$. Hence



    $5^{133}equiv 5^{2times 66+1}equiv 5times (5^2)^{66}equiv 5times 1^{66}equiv 5$ (mod $8$).







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 2 hours ago









    J. W. Tanner

    2,0491116




    2,0491116










    answered 2 hours ago









    MarkMark

    8,113420




    8,113420












    • $begingroup$
      Should the first triple-equal sign on the second row not be a normal equal sign? And the second one too?
      $endgroup$
      – Sandi
      2 hours ago








    • 1




      $begingroup$
      Yes, you can also write regular equality there. But if two numbers are equal in $mathbb{Z}$ then they are also congruent mod $n$ for any $ninmathbb{N}$, and because we only care about congruence here I decided to write congruence from the beginning.
      $endgroup$
      – Mark
      2 hours ago










    • $begingroup$
      I don't understand how you do the step $5 times (5^2)^{66} equiv 5 times 1^{66} (mathrm{mod} 8)$. Could you clarify please?
      $endgroup$
      – Sandi
      2 hours ago








    • 1




      $begingroup$
      $5^2equiv 1$(mod $8$), hence $(5^2)^{66}equiv 1^{66}equiv 1$(mod $8$). That's it. It all follows from a basic theorem on modular arithmetic: that if $aequiv b$(mod $n$) and $cequiv d$(mod $n$) then $abequiv cd$(mod $n$). In other words multiplication mod $n$ is well defined. (same thing can we said about sum mod $n$)
      $endgroup$
      – Mark
      2 hours ago












    • $begingroup$
      I found a similar thing in my book, but it said that $ac equiv bd (modn)$. Did you make a typo or is it a different theorem?
      $endgroup$
      – Sandi
      1 hour ago


















    • $begingroup$
      Should the first triple-equal sign on the second row not be a normal equal sign? And the second one too?
      $endgroup$
      – Sandi
      2 hours ago








    • 1




      $begingroup$
      Yes, you can also write regular equality there. But if two numbers are equal in $mathbb{Z}$ then they are also congruent mod $n$ for any $ninmathbb{N}$, and because we only care about congruence here I decided to write congruence from the beginning.
      $endgroup$
      – Mark
      2 hours ago










    • $begingroup$
      I don't understand how you do the step $5 times (5^2)^{66} equiv 5 times 1^{66} (mathrm{mod} 8)$. Could you clarify please?
      $endgroup$
      – Sandi
      2 hours ago








    • 1




      $begingroup$
      $5^2equiv 1$(mod $8$), hence $(5^2)^{66}equiv 1^{66}equiv 1$(mod $8$). That's it. It all follows from a basic theorem on modular arithmetic: that if $aequiv b$(mod $n$) and $cequiv d$(mod $n$) then $abequiv cd$(mod $n$). In other words multiplication mod $n$ is well defined. (same thing can we said about sum mod $n$)
      $endgroup$
      – Mark
      2 hours ago












    • $begingroup$
      I found a similar thing in my book, but it said that $ac equiv bd (modn)$. Did you make a typo or is it a different theorem?
      $endgroup$
      – Sandi
      1 hour ago
















    $begingroup$
    Should the first triple-equal sign on the second row not be a normal equal sign? And the second one too?
    $endgroup$
    – Sandi
    2 hours ago






    $begingroup$
    Should the first triple-equal sign on the second row not be a normal equal sign? And the second one too?
    $endgroup$
    – Sandi
    2 hours ago






    1




    1




    $begingroup$
    Yes, you can also write regular equality there. But if two numbers are equal in $mathbb{Z}$ then they are also congruent mod $n$ for any $ninmathbb{N}$, and because we only care about congruence here I decided to write congruence from the beginning.
    $endgroup$
    – Mark
    2 hours ago




    $begingroup$
    Yes, you can also write regular equality there. But if two numbers are equal in $mathbb{Z}$ then they are also congruent mod $n$ for any $ninmathbb{N}$, and because we only care about congruence here I decided to write congruence from the beginning.
    $endgroup$
    – Mark
    2 hours ago












    $begingroup$
    I don't understand how you do the step $5 times (5^2)^{66} equiv 5 times 1^{66} (mathrm{mod} 8)$. Could you clarify please?
    $endgroup$
    – Sandi
    2 hours ago






    $begingroup$
    I don't understand how you do the step $5 times (5^2)^{66} equiv 5 times 1^{66} (mathrm{mod} 8)$. Could you clarify please?
    $endgroup$
    – Sandi
    2 hours ago






    1




    1




    $begingroup$
    $5^2equiv 1$(mod $8$), hence $(5^2)^{66}equiv 1^{66}equiv 1$(mod $8$). That's it. It all follows from a basic theorem on modular arithmetic: that if $aequiv b$(mod $n$) and $cequiv d$(mod $n$) then $abequiv cd$(mod $n$). In other words multiplication mod $n$ is well defined. (same thing can we said about sum mod $n$)
    $endgroup$
    – Mark
    2 hours ago






    $begingroup$
    $5^2equiv 1$(mod $8$), hence $(5^2)^{66}equiv 1^{66}equiv 1$(mod $8$). That's it. It all follows from a basic theorem on modular arithmetic: that if $aequiv b$(mod $n$) and $cequiv d$(mod $n$) then $abequiv cd$(mod $n$). In other words multiplication mod $n$ is well defined. (same thing can we said about sum mod $n$)
    $endgroup$
    – Mark
    2 hours ago














    $begingroup$
    I found a similar thing in my book, but it said that $ac equiv bd (modn)$. Did you make a typo or is it a different theorem?
    $endgroup$
    – Sandi
    1 hour ago




    $begingroup$
    I found a similar thing in my book, but it said that $ac equiv bd (modn)$. Did you make a typo or is it a different theorem?
    $endgroup$
    – Sandi
    1 hour ago











    0












    $begingroup$

    "Highbrow method":



    If you undertake to learn any number theory , there is Euler's theorem: $$a^{varphi (n)}cong1pmod n$$, for $a$ relatively prime to $n$.



    ($varphi$ is Euler's totient function, and just returns the number of positive integers less than $n$ that are relatively prime to $n$.)



    Now $varphi (8)=4$. Hence $5^4cong1pmod8.$



    Use this to reduce the exponent: $133=4×33+1implies5^{133}=(5^4)^{33}cdot 5cong5pmod8$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      "Highbrow method":



      If you undertake to learn any number theory , there is Euler's theorem: $$a^{varphi (n)}cong1pmod n$$, for $a$ relatively prime to $n$.



      ($varphi$ is Euler's totient function, and just returns the number of positive integers less than $n$ that are relatively prime to $n$.)



      Now $varphi (8)=4$. Hence $5^4cong1pmod8.$



      Use this to reduce the exponent: $133=4×33+1implies5^{133}=(5^4)^{33}cdot 5cong5pmod8$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        "Highbrow method":



        If you undertake to learn any number theory , there is Euler's theorem: $$a^{varphi (n)}cong1pmod n$$, for $a$ relatively prime to $n$.



        ($varphi$ is Euler's totient function, and just returns the number of positive integers less than $n$ that are relatively prime to $n$.)



        Now $varphi (8)=4$. Hence $5^4cong1pmod8.$



        Use this to reduce the exponent: $133=4×33+1implies5^{133}=(5^4)^{33}cdot 5cong5pmod8$.






        share|cite|improve this answer











        $endgroup$



        "Highbrow method":



        If you undertake to learn any number theory , there is Euler's theorem: $$a^{varphi (n)}cong1pmod n$$, for $a$ relatively prime to $n$.



        ($varphi$ is Euler's totient function, and just returns the number of positive integers less than $n$ that are relatively prime to $n$.)



        Now $varphi (8)=4$. Hence $5^4cong1pmod8.$



        Use this to reduce the exponent: $133=4×33+1implies5^{133}=(5^4)^{33}cdot 5cong5pmod8$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 1 hour ago

























        answered 1 hour ago









        Chris CusterChris Custer

        13.6k3827




        13.6k3827






























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            Gregoriusmis

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            Een afbeelding van de Gregoriusmis uit een getijdenboek gemaakt in de zuidelijke Nederlanden tussen 1500 en 1525 Gregoriusmis, Meester van het Akense Altaar, ca. 1500-1525. Collectie Museum Catharijneconvent De Gregoriusmis of de Mis van Sint Gregorius is een motief in katholieke kunst, dat het eerst verscheen in de late Middeleeuwen en dat nog steeds werd gebruikt tijdens de contrareformatie. Paus Gregorius I (ca. 540-604) is afgebeeld, de H. Mis opdragend, terwijl Christus als de Man van Smarten verschijnt op het altaar dat voor Sint Gregorius staat. Volgens de legende had St. Gregorius gebeden om een teken voor een betwijfelaar van de transsubstantiatieleer, waarop de lijdende Christus tijdens het opdragen van de Mis verscheen om hem te overtuigen en waarbij zijn bloed in de altaarkelk liep. Vaak zijn ook de passiewerktuigen afgebeeld. Zie de categorie Mass of Saint Gregory van Wikimedia Commons voor mediabestanden over dit onderwerp. This page is onl...
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            Lijst van personen overleden op 24 maart

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            Dit is een lijst van personen die overleden zijn op 24 maart . De inhoud van deze pagina is gebaseerd op gegevens in Wikidata. Als er onvolkomenheden in deze lijst zitten, gelieve dan aldaar de gegevens aan te passen. Achter iedere persoon is hiervoor een link aanwezig met de aanduiding bewerken. Wijzigingen in Wikidata worden met enige (on)regelmatigheid geautomatiseerd overgenomen naar deze pagina. Anna van Egmond (1533-1558) overleden op 24 maart 1558 Frederik August van Nassau-Usingen overleden op 24 maart 1816 Jules Verne overleden op 24 maart 1905 Richard Widmark overleden op 24 maart 2008 Johan Cruijff overleden op 24 maart 2016 809 - Haroen ar-Rashid, politicus uit het kalifaat van de Abbasiden (geboren in 766) bewerken 832 - Wulfred, Aartsbisschop van Canterbury, priester bewerken 1275 - Beatrix van Engeland, hertog, aristocraat uit het Koninkrijk Engeland (geboren in 1242) bewerken 1284 - Hugo III van Cyprus, lijst van koningen van ...
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