Detemine the unit digit of a number [duplicate]

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This question already has an answer here:



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

    9 answers



Find the unit digit of the number:
$$3^7005 times 6^8000$$
My turn:
$$3^7005times 6^8000=3^7005times 3^8000 times2^8000$$ $$3^13005 times 2^8000$$
But i could not go on ?






algebra-precalculus elementary-number-theory







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marked as duplicate by Bill Dubuque algebra-precalculus
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 Mar 8 at 15:14


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.

















  • $begingroup$
    Well, you can start by the fact that the last digit of $6^n$ is always $6$ for any positive whole number $n$. Then you only have the $3^7005$ to deal with.
    $endgroup$
    – Matti P.
    Mar 8 at 13:39















1












$begingroup$



This question already has an answer here:



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

    9 answers



Find the unit digit of the number:
$$3^7005 times 6^8000$$
My turn:
$$3^7005times 6^8000=3^7005times 3^8000 times2^8000$$ $$3^13005 times 2^8000$$
But i could not go on ?






algebra-precalculus elementary-number-theory







share|cite|improve this question











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marked as duplicate by Bill Dubuque algebra-precalculus
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 Mar 8 at 15:14


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.

















  • $begingroup$
    Well, you can start by the fact that the last digit of $6^n$ is always $6$ for any positive whole number $n$. Then you only have the $3^7005$ to deal with.
    $endgroup$
    – Matti P.
    Mar 8 at 13:39













1












1








1





$begingroup$



This question already has an answer here:



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

    9 answers



Find the unit digit of the number:
$$3^7005 times 6^8000$$
My turn:
$$3^7005times 6^8000=3^7005times 3^8000 times2^8000$$ $$3^13005 times 2^8000$$
But i could not go on ?






algebra-precalculus elementary-number-theory







share|cite|improve this question











$endgroup$





This question already has an answer here:



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

    9 answers



Find the unit digit of the number:
$$3^7005 times 6^8000$$
My turn:
$$3^7005times 6^8000=3^7005times 3^8000 times2^8000$$ $$3^13005 times 2^8000$$
But i could not go on ?





This question already has an answer here:



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

    9 answers






algebra-precalculus elementary-number-theory




algebra-precalculus elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 8 at 14:39









José Carlos Santos

172k23133241




172k23133241










asked Mar 8 at 13:36









Hussien MohamedHussien Mohamed

244




244




marked as duplicate by Bill Dubuque algebra-precalculus
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 Mar 8 at 15:14


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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 Mar 8 at 15:14


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.













  • $begingroup$
    Well, you can start by the fact that the last digit of $6^n$ is always $6$ for any positive whole number $n$. Then you only have the $3^7005$ to deal with.
    $endgroup$
    – Matti P.
    Mar 8 at 13:39
















  • $begingroup$
    Well, you can start by the fact that the last digit of $6^n$ is always $6$ for any positive whole number $n$. Then you only have the $3^7005$ to deal with.
    $endgroup$
    – Matti P.
    Mar 8 at 13:39















$begingroup$
Well, you can start by the fact that the last digit of $6^n$ is always $6$ for any positive whole number $n$. Then you only have the $3^7005$ to deal with.
$endgroup$
– Matti P.
Mar 8 at 13:39




$begingroup$
Well, you can start by the fact that the last digit of $6^n$ is always $6$ for any positive whole number $n$. Then you only have the $3^7005$ to deal with.
$endgroup$
– Matti P.
Mar 8 at 13:39










3 Answers
3






active

oldest

votes


















4












$begingroup$

The unit digit of the numbers of the powers of $3$ form a cyclic sequence: $1,3,9,7,1,3,9,7,ldots$ In particular, the unit digit of $3^n$ is $3$ if $n$ is of the form $4k+1$. And the unit digit of any power of $6$ is $6$. Since $7,005$ is of the form $4k+1$, the answer to your question is $8(=3times6pmod10)$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Agreed! I wonder why someone downvoted ...
    $endgroup$
    – Matti P.
    Mar 8 at 13:50










  • $begingroup$
    It happens all the time: :-(
    $endgroup$
    – José Carlos Santos
    Mar 8 at 13:51










  • $begingroup$
    Maybe the down-voter, like me, can't parse the sentence beginning with "In particular..."? Oh, I just got it. "of" is "if".
    $endgroup$
    – B. Goddard
    Mar 8 at 14:08










  • $begingroup$
    @B.Goddard Oops! I've edited my answer. Thank you.
    $endgroup$
    – José Carlos Santos
    Mar 8 at 14:15


















2












$begingroup$

We want the unit digit of the number $3^7005 cdot6^8000$. This is equivalent to asking for the residue of the number $3^7005 cdot6^8000$ modulo $10$. Hence, we want some number $n$ with $0 leq n leq 9$ such that
$$
n equiv 3^7005 cdot6^8000 pmod10
$$



Note that $3^4 equiv 1 pmod10$ and $6^n equiv 6 pmod10$ for all $ninmathbbN$. With this result, we have
beginalign
3^7005 cdot6^8000 &= 3^7004cdot 3^1 cdot 6^8000 \
&= left(3^4right)^1751cdot 3^1 cdot 6^8000\
&equiv1^1751cdot 3^1 cdot 6^1 pmod10 \
&= 3 cdot 6 \
&equiv 8 pmod10 \
endalign
Hence, the unit digit of $3^7005 cdot6^8000$ is $8$.






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    Right outcome, slight error in reasoning. $3^4equiv 1 mod 10; 3^7equiv 7 mod 10$. You figured it out right when you found $3^5equiv 3 mod 10$. Note that $4mid 7004$.
    $endgroup$
    – Keith Backman
    Mar 8 at 14:50










  • $begingroup$
    @KeithBackman I'm shocked I overlooked that. I've edited my response with your correction. Thanks for reviewing my answer so closely!
    $endgroup$
    – Brian
    Mar 8 at 14:54










  • $begingroup$
    Not to worry -- you should see some of the bloopers I've made!
    $endgroup$
    – Keith Backman
    Mar 8 at 15:56


















1












$begingroup$

If you think about the way you typically do multiplication, the unit digit of a product is the product of the unit digits.



Can you figure out the unit digit of the two terms in the product?



Hint: calculate $3^0, 3^1, 3^2, 3^3, 3^4, 3^5,ldots$ Do you see a pattern in the unit digits? Do you see why the pattern holds?






share|cite|improve this answer











$endgroup$



















    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    The unit digit of the numbers of the powers of $3$ form a cyclic sequence: $1,3,9,7,1,3,9,7,ldots$ In particular, the unit digit of $3^n$ is $3$ if $n$ is of the form $4k+1$. And the unit digit of any power of $6$ is $6$. Since $7,005$ is of the form $4k+1$, the answer to your question is $8(=3times6pmod10)$.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      Agreed! I wonder why someone downvoted ...
      $endgroup$
      – Matti P.
      Mar 8 at 13:50










    • $begingroup$
      It happens all the time: :-(
      $endgroup$
      – José Carlos Santos
      Mar 8 at 13:51










    • $begingroup$
      Maybe the down-voter, like me, can't parse the sentence beginning with "In particular..."? Oh, I just got it. "of" is "if".
      $endgroup$
      – B. Goddard
      Mar 8 at 14:08










    • $begingroup$
      @B.Goddard Oops! I've edited my answer. Thank you.
      $endgroup$
      – José Carlos Santos
      Mar 8 at 14:15















    4












    $begingroup$

    The unit digit of the numbers of the powers of $3$ form a cyclic sequence: $1,3,9,7,1,3,9,7,ldots$ In particular, the unit digit of $3^n$ is $3$ if $n$ is of the form $4k+1$. And the unit digit of any power of $6$ is $6$. Since $7,005$ is of the form $4k+1$, the answer to your question is $8(=3times6pmod10)$.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      Agreed! I wonder why someone downvoted ...
      $endgroup$
      – Matti P.
      Mar 8 at 13:50










    • $begingroup$
      It happens all the time: :-(
      $endgroup$
      – José Carlos Santos
      Mar 8 at 13:51










    • $begingroup$
      Maybe the down-voter, like me, can't parse the sentence beginning with "In particular..."? Oh, I just got it. "of" is "if".
      $endgroup$
      – B. Goddard
      Mar 8 at 14:08










    • $begingroup$
      @B.Goddard Oops! I've edited my answer. Thank you.
      $endgroup$
      – José Carlos Santos
      Mar 8 at 14:15













    4












    4








    4





    $begingroup$

    The unit digit of the numbers of the powers of $3$ form a cyclic sequence: $1,3,9,7,1,3,9,7,ldots$ In particular, the unit digit of $3^n$ is $3$ if $n$ is of the form $4k+1$. And the unit digit of any power of $6$ is $6$. Since $7,005$ is of the form $4k+1$, the answer to your question is $8(=3times6pmod10)$.






    share|cite|improve this answer











    $endgroup$



    The unit digit of the numbers of the powers of $3$ form a cyclic sequence: $1,3,9,7,1,3,9,7,ldots$ In particular, the unit digit of $3^n$ is $3$ if $n$ is of the form $4k+1$. And the unit digit of any power of $6$ is $6$. Since $7,005$ is of the form $4k+1$, the answer to your question is $8(=3times6pmod10)$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Mar 8 at 14:14

























    answered Mar 8 at 13:39









    José Carlos SantosJosé Carlos Santos

    172k23133241




    172k23133241











    • $begingroup$
      Agreed! I wonder why someone downvoted ...
      $endgroup$
      – Matti P.
      Mar 8 at 13:50










    • $begingroup$
      It happens all the time: :-(
      $endgroup$
      – José Carlos Santos
      Mar 8 at 13:51










    • $begingroup$
      Maybe the down-voter, like me, can't parse the sentence beginning with "In particular..."? Oh, I just got it. "of" is "if".
      $endgroup$
      – B. Goddard
      Mar 8 at 14:08










    • $begingroup$
      @B.Goddard Oops! I've edited my answer. Thank you.
      $endgroup$
      – José Carlos Santos
      Mar 8 at 14:15
















    • $begingroup$
      Agreed! I wonder why someone downvoted ...
      $endgroup$
      – Matti P.
      Mar 8 at 13:50










    • $begingroup$
      It happens all the time: :-(
      $endgroup$
      – José Carlos Santos
      Mar 8 at 13:51










    • $begingroup$
      Maybe the down-voter, like me, can't parse the sentence beginning with "In particular..."? Oh, I just got it. "of" is "if".
      $endgroup$
      – B. Goddard
      Mar 8 at 14:08










    • $begingroup$
      @B.Goddard Oops! I've edited my answer. Thank you.
      $endgroup$
      – José Carlos Santos
      Mar 8 at 14:15















    $begingroup$
    Agreed! I wonder why someone downvoted ...
    $endgroup$
    – Matti P.
    Mar 8 at 13:50




    $begingroup$
    Agreed! I wonder why someone downvoted ...
    $endgroup$
    – Matti P.
    Mar 8 at 13:50












    $begingroup$
    It happens all the time: :-(
    $endgroup$
    – José Carlos Santos
    Mar 8 at 13:51




    $begingroup$
    It happens all the time: :-(
    $endgroup$
    – José Carlos Santos
    Mar 8 at 13:51












    $begingroup$
    Maybe the down-voter, like me, can't parse the sentence beginning with "In particular..."? Oh, I just got it. "of" is "if".
    $endgroup$
    – B. Goddard
    Mar 8 at 14:08




    $begingroup$
    Maybe the down-voter, like me, can't parse the sentence beginning with "In particular..."? Oh, I just got it. "of" is "if".
    $endgroup$
    – B. Goddard
    Mar 8 at 14:08












    $begingroup$
    @B.Goddard Oops! I've edited my answer. Thank you.
    $endgroup$
    – José Carlos Santos
    Mar 8 at 14:15




    $begingroup$
    @B.Goddard Oops! I've edited my answer. Thank you.
    $endgroup$
    – José Carlos Santos
    Mar 8 at 14:15











    2












    $begingroup$

    We want the unit digit of the number $3^7005 cdot6^8000$. This is equivalent to asking for the residue of the number $3^7005 cdot6^8000$ modulo $10$. Hence, we want some number $n$ with $0 leq n leq 9$ such that
    $$
    n equiv 3^7005 cdot6^8000 pmod10
    $$



    Note that $3^4 equiv 1 pmod10$ and $6^n equiv 6 pmod10$ for all $ninmathbbN$. With this result, we have
    beginalign
    3^7005 cdot6^8000 &= 3^7004cdot 3^1 cdot 6^8000 \
    &= left(3^4right)^1751cdot 3^1 cdot 6^8000\
    &equiv1^1751cdot 3^1 cdot 6^1 pmod10 \
    &= 3 cdot 6 \
    &equiv 8 pmod10 \
    endalign
    Hence, the unit digit of $3^7005 cdot6^8000$ is $8$.






    share|cite|improve this answer











    $endgroup$








    • 1




      $begingroup$
      Right outcome, slight error in reasoning. $3^4equiv 1 mod 10; 3^7equiv 7 mod 10$. You figured it out right when you found $3^5equiv 3 mod 10$. Note that $4mid 7004$.
      $endgroup$
      – Keith Backman
      Mar 8 at 14:50










    • $begingroup$
      @KeithBackman I'm shocked I overlooked that. I've edited my response with your correction. Thanks for reviewing my answer so closely!
      $endgroup$
      – Brian
      Mar 8 at 14:54










    • $begingroup$
      Not to worry -- you should see some of the bloopers I've made!
      $endgroup$
      – Keith Backman
      Mar 8 at 15:56















    2












    $begingroup$

    We want the unit digit of the number $3^7005 cdot6^8000$. This is equivalent to asking for the residue of the number $3^7005 cdot6^8000$ modulo $10$. Hence, we want some number $n$ with $0 leq n leq 9$ such that
    $$
    n equiv 3^7005 cdot6^8000 pmod10
    $$



    Note that $3^4 equiv 1 pmod10$ and $6^n equiv 6 pmod10$ for all $ninmathbbN$. With this result, we have
    beginalign
    3^7005 cdot6^8000 &= 3^7004cdot 3^1 cdot 6^8000 \
    &= left(3^4right)^1751cdot 3^1 cdot 6^8000\
    &equiv1^1751cdot 3^1 cdot 6^1 pmod10 \
    &= 3 cdot 6 \
    &equiv 8 pmod10 \
    endalign
    Hence, the unit digit of $3^7005 cdot6^8000$ is $8$.






    share|cite|improve this answer











    $endgroup$








    • 1




      $begingroup$
      Right outcome, slight error in reasoning. $3^4equiv 1 mod 10; 3^7equiv 7 mod 10$. You figured it out right when you found $3^5equiv 3 mod 10$. Note that $4mid 7004$.
      $endgroup$
      – Keith Backman
      Mar 8 at 14:50










    • $begingroup$
      @KeithBackman I'm shocked I overlooked that. I've edited my response with your correction. Thanks for reviewing my answer so closely!
      $endgroup$
      – Brian
      Mar 8 at 14:54










    • $begingroup$
      Not to worry -- you should see some of the bloopers I've made!
      $endgroup$
      – Keith Backman
      Mar 8 at 15:56













    2












    2








    2





    $begingroup$

    We want the unit digit of the number $3^7005 cdot6^8000$. This is equivalent to asking for the residue of the number $3^7005 cdot6^8000$ modulo $10$. Hence, we want some number $n$ with $0 leq n leq 9$ such that
    $$
    n equiv 3^7005 cdot6^8000 pmod10
    $$



    Note that $3^4 equiv 1 pmod10$ and $6^n equiv 6 pmod10$ for all $ninmathbbN$. With this result, we have
    beginalign
    3^7005 cdot6^8000 &= 3^7004cdot 3^1 cdot 6^8000 \
    &= left(3^4right)^1751cdot 3^1 cdot 6^8000\
    &equiv1^1751cdot 3^1 cdot 6^1 pmod10 \
    &= 3 cdot 6 \
    &equiv 8 pmod10 \
    endalign
    Hence, the unit digit of $3^7005 cdot6^8000$ is $8$.






    share|cite|improve this answer











    $endgroup$



    We want the unit digit of the number $3^7005 cdot6^8000$. This is equivalent to asking for the residue of the number $3^7005 cdot6^8000$ modulo $10$. Hence, we want some number $n$ with $0 leq n leq 9$ such that
    $$
    n equiv 3^7005 cdot6^8000 pmod10
    $$



    Note that $3^4 equiv 1 pmod10$ and $6^n equiv 6 pmod10$ for all $ninmathbbN$. With this result, we have
    beginalign
    3^7005 cdot6^8000 &= 3^7004cdot 3^1 cdot 6^8000 \
    &= left(3^4right)^1751cdot 3^1 cdot 6^8000\
    &equiv1^1751cdot 3^1 cdot 6^1 pmod10 \
    &= 3 cdot 6 \
    &equiv 8 pmod10 \
    endalign
    Hence, the unit digit of $3^7005 cdot6^8000$ is $8$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Mar 8 at 14:53

























    answered Mar 8 at 14:12









    BrianBrian

    1,268216




    1,268216







    • 1




      $begingroup$
      Right outcome, slight error in reasoning. $3^4equiv 1 mod 10; 3^7equiv 7 mod 10$. You figured it out right when you found $3^5equiv 3 mod 10$. Note that $4mid 7004$.
      $endgroup$
      – Keith Backman
      Mar 8 at 14:50










    • $begingroup$
      @KeithBackman I'm shocked I overlooked that. I've edited my response with your correction. Thanks for reviewing my answer so closely!
      $endgroup$
      – Brian
      Mar 8 at 14:54










    • $begingroup$
      Not to worry -- you should see some of the bloopers I've made!
      $endgroup$
      – Keith Backman
      Mar 8 at 15:56












    • 1




      $begingroup$
      Right outcome, slight error in reasoning. $3^4equiv 1 mod 10; 3^7equiv 7 mod 10$. You figured it out right when you found $3^5equiv 3 mod 10$. Note that $4mid 7004$.
      $endgroup$
      – Keith Backman
      Mar 8 at 14:50










    • $begingroup$
      @KeithBackman I'm shocked I overlooked that. I've edited my response with your correction. Thanks for reviewing my answer so closely!
      $endgroup$
      – Brian
      Mar 8 at 14:54










    • $begingroup$
      Not to worry -- you should see some of the bloopers I've made!
      $endgroup$
      – Keith Backman
      Mar 8 at 15:56







    1




    1




    $begingroup$
    Right outcome, slight error in reasoning. $3^4equiv 1 mod 10; 3^7equiv 7 mod 10$. You figured it out right when you found $3^5equiv 3 mod 10$. Note that $4mid 7004$.
    $endgroup$
    – Keith Backman
    Mar 8 at 14:50




    $begingroup$
    Right outcome, slight error in reasoning. $3^4equiv 1 mod 10; 3^7equiv 7 mod 10$. You figured it out right when you found $3^5equiv 3 mod 10$. Note that $4mid 7004$.
    $endgroup$
    – Keith Backman
    Mar 8 at 14:50












    $begingroup$
    @KeithBackman I'm shocked I overlooked that. I've edited my response with your correction. Thanks for reviewing my answer so closely!
    $endgroup$
    – Brian
    Mar 8 at 14:54




    $begingroup$
    @KeithBackman I'm shocked I overlooked that. I've edited my response with your correction. Thanks for reviewing my answer so closely!
    $endgroup$
    – Brian
    Mar 8 at 14:54












    $begingroup$
    Not to worry -- you should see some of the bloopers I've made!
    $endgroup$
    – Keith Backman
    Mar 8 at 15:56




    $begingroup$
    Not to worry -- you should see some of the bloopers I've made!
    $endgroup$
    – Keith Backman
    Mar 8 at 15:56











    1












    $begingroup$

    If you think about the way you typically do multiplication, the unit digit of a product is the product of the unit digits.



    Can you figure out the unit digit of the two terms in the product?



    Hint: calculate $3^0, 3^1, 3^2, 3^3, 3^4, 3^5,ldots$ Do you see a pattern in the unit digits? Do you see why the pattern holds?






    share|cite|improve this answer











    $endgroup$

















      1












      $begingroup$

      If you think about the way you typically do multiplication, the unit digit of a product is the product of the unit digits.



      Can you figure out the unit digit of the two terms in the product?



      Hint: calculate $3^0, 3^1, 3^2, 3^3, 3^4, 3^5,ldots$ Do you see a pattern in the unit digits? Do you see why the pattern holds?






      share|cite|improve this answer











      $endgroup$















        1












        1








        1





        $begingroup$

        If you think about the way you typically do multiplication, the unit digit of a product is the product of the unit digits.



        Can you figure out the unit digit of the two terms in the product?



        Hint: calculate $3^0, 3^1, 3^2, 3^3, 3^4, 3^5,ldots$ Do you see a pattern in the unit digits? Do you see why the pattern holds?






        share|cite|improve this answer











        $endgroup$



        If you think about the way you typically do multiplication, the unit digit of a product is the product of the unit digits.



        Can you figure out the unit digit of the two terms in the product?



        Hint: calculate $3^0, 3^1, 3^2, 3^3, 3^4, 3^5,ldots$ Do you see a pattern in the unit digits? Do you see why the pattern holds?







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Mar 8 at 13:52









        NickD

        1,2021512




        1,2021512










        answered Mar 8 at 13:41









        Daniel McLauryDaniel McLaury

        16.1k33081




        16.1k33081












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