How to calculate mod of number with big exponent [duplicate]












1












$begingroup$



This question already has an answer here:




  • How do I compute $a^b,bmod c$ by hand?

    8 answers



  • Mod of numbers with large exponents

    3 answers




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







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marked as duplicate by Bill Dubuque algebra-precalculus
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 3 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 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
    6 hours ago












  • $begingroup$
    What does the triple equal sign mean?
    $endgroup$
    – Sandi
    6 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
    6 hours ago












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
















1












$begingroup$



This question already has an answer here:




  • How do I compute $a^b,bmod c$ by hand?

    8 answers



  • Mod of numbers with large exponents

    3 answers




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











$endgroup$



marked as duplicate by Bill Dubuque algebra-precalculus
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 3 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 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
    6 hours ago












  • $begingroup$
    What does the triple equal sign mean?
    $endgroup$
    – Sandi
    6 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
    6 hours ago












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














1












1








1





$begingroup$



This question already has an answer here:




  • How do I compute $a^b,bmod c$ by hand?

    8 answers



  • Mod of numbers with large exponents

    3 answers




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











$endgroup$





This question already has an answer here:




  • How do I compute $a^b,bmod c$ by hand?

    8 answers



  • Mod of numbers with large exponents

    3 answers




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)?





This question already has an answer here:




  • How do I compute $a^b,bmod c$ by hand?

    8 answers



  • Mod of numbers with large exponents

    3 answers







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 6 hours ago







Sandi

















asked 6 hours ago









SandiSandi

262112




262112




marked as duplicate by Bill Dubuque algebra-precalculus
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 3 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Bill Dubuque algebra-precalculus
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 3 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 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
    6 hours ago












  • $begingroup$
    What does the triple equal sign mean?
    $endgroup$
    – Sandi
    6 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
    6 hours ago












  • $begingroup$
    @Sandi this is the common way to write congruences.
    $endgroup$
    – Mark
    6 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
    6 hours ago












  • $begingroup$
    What does the triple equal sign mean?
    $endgroup$
    – Sandi
    6 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
    6 hours ago












  • $begingroup$
    @Sandi this is the common way to write congruences.
    $endgroup$
    – Mark
    6 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
6 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
6 hours ago














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




$begingroup$
What does the triple equal sign mean?
$endgroup$
– Sandi
6 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
6 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
6 hours ago














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




$begingroup$
@Sandi this is the common way to write congruences.
$endgroup$
– Mark
6 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
    6 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
    6 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
    6 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
    6 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
    5 hours 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











$endgroup$




















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    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
      6 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
      6 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
      6 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
      6 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
      5 hours 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
      6 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
      6 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
      6 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
      6 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
      5 hours 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 6 hours ago









    J. W. Tanner

    2,0591117




    2,0591117










    answered 6 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
      6 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
      6 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
      6 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
      6 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
      5 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
      6 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
      6 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
      6 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
      6 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
      5 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
    6 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
    6 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
    6 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
    6 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
    6 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
    6 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
    6 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
    6 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
    5 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
    5 hours 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 5 hours ago

























        answered 5 hours ago









        Chris CusterChris Custer

        13.6k3827




        13.6k3827















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            Polycentropodidae Plectrocnemia conspersa Siyentipiko nga pagklasipika Ginhadi-an: Animalia Phylum: Arthropoda Ubosphylum: Hexapoda Klase: Insecta Orden: Trichoptera Labawbanay: Hydropsychoidea Banay: Polycentropodidae Binomial nga ngaran Polycentropodidae An Polycentropodidae [1] in uska familia han Insecta. An Polycentropodidae in nahilalakip ha familia nga Hydropsychoidea, ordo nga Trichoptera, classis nga Insecta, punoan nga Arthropoda, ngan regnum nga Animalia. [1] An familia nga Polycentropodidae in naglalakip hin 660 ka mga species, sumala ha Catalogue of Life [1] . An kladograma hini sumala ha Catalogue of Life [1] : Hydropsychoidea   Polycentropodidae  Adectophylax Antillopsyche Cernotina Cyrnellus Cyrnodes Cyrnopsis Cyrnus Eodipseudopsis Eoneureclipsis Holocentropus Kambaitipsyche Neucentropus
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            Magento 2 Error message: Invalid state change requested

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            2 1 I've created custom Rest API for adding and removing products from cart. I am able to add and delete products from my cart, but when I log in as a customer, I can't add or delete products from cart. API throws message: Invalid state change requested. How can I solve this? magento2 share | improve this question edited Mar 15 '18 at 13:02 Bare Feet 1,561 1 7 32 asked Mar 15 '18 at 12:50 Prabhu M. Prabhu M. 18 5
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            Paulmy

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            An Paulmy amo in usa ka komyun ha departamento han Indre-et-Loire ngan ha Rehiyon Sentro ha nasod han Fransya. k h l Mga komyun ha departamento han Indre-et-Loire Abilly  · Ambillou  · Amboise  · Anché  · Antogny-le-Tillac  · Artannes-sur-Indre  · Assay  · Athée-sur-Cher  · Autrèche  · Auzouer-en-Touraine  · Avoine  · Avon-les-Roches  · Avrillé-les-Ponceaux  · Azay-le-Rideau  · Azay-sur-Cher  · Azay-sur-Indre  · Ballan-Miré  · Barrou  · Beaulieu-lès-Loches  · Beaumont-en-Véron  · Beaumont-la-Ronce  · Beaumont-Village  · Benais  · Berthenay  · Betz-le-Château  · Bléré  · Bossay-sur-Claise  · Bossée  · Le Boulay  · Bourgueil  · Bournan  · Boussay  · Braslou  · Braye-sous-Faye  · Braye-sur-Maulne  · Brèches  · Bréhémont  · Bridoré  · Brizay  · Bueil-en-Touraine  · Candes-Saint-Martin  · Cangey  · La Celle-Guenand  · La Celle-Saint-Avant  · Céré-la-Ronde  · Cerelles  · Chambon  ·
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