To determine the remainder of the division












-1












$begingroup$


To determine the remainder of the division of 3302 + 7200 with 5.



Is it correct if I find the remainder of the division separately for 3302 and 7200 and then add the two of the remainders?






discrete-mathematics modular-arithmetic







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What do you mean by the 'rest' of the division? Are you referring to the remainder?
    $endgroup$
    – D.B.
    Jan 14 at 0:22










  • $begingroup$
    Yes the remainder, sorry.
    $endgroup$
    – Viktor
    Jan 14 at 0:23
















-1












$begingroup$


To determine the remainder of the division of 3302 + 7200 with 5.



Is it correct if I find the remainder of the division separately for 3302 and 7200 and then add the two of the remainders?






discrete-mathematics modular-arithmetic







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What do you mean by the 'rest' of the division? Are you referring to the remainder?
    $endgroup$
    – D.B.
    Jan 14 at 0:22










  • $begingroup$
    Yes the remainder, sorry.
    $endgroup$
    – Viktor
    Jan 14 at 0:23














-1












-1








-1





$begingroup$


To determine the remainder of the division of 3302 + 7200 with 5.



Is it correct if I find the remainder of the division separately for 3302 and 7200 and then add the two of the remainders?






discrete-mathematics modular-arithmetic







share|cite|improve this question











$endgroup$




To determine the remainder of the division of 3302 + 7200 with 5.



Is it correct if I find the remainder of the division separately for 3302 and 7200 and then add the two of the remainders?






discrete-mathematics modular-arithmetic




discrete-mathematics modular-arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 3:08







Viktor

















asked Jan 14 at 0:20









ViktorViktor

477




477








  • 1




    $begingroup$
    What do you mean by the 'rest' of the division? Are you referring to the remainder?
    $endgroup$
    – D.B.
    Jan 14 at 0:22










  • $begingroup$
    Yes the remainder, sorry.
    $endgroup$
    – Viktor
    Jan 14 at 0:23














  • 1




    $begingroup$
    What do you mean by the 'rest' of the division? Are you referring to the remainder?
    $endgroup$
    – D.B.
    Jan 14 at 0:22










  • $begingroup$
    Yes the remainder, sorry.
    $endgroup$
    – Viktor
    Jan 14 at 0:23








1




1




$begingroup$
What do you mean by the 'rest' of the division? Are you referring to the remainder?
$endgroup$
– D.B.
Jan 14 at 0:22




$begingroup$
What do you mean by the 'rest' of the division? Are you referring to the remainder?
$endgroup$
– D.B.
Jan 14 at 0:22












$begingroup$
Yes the remainder, sorry.
$endgroup$
– Viktor
Jan 14 at 0:23




$begingroup$
Yes the remainder, sorry.
$endgroup$
– Viktor
Jan 14 at 0:23










3 Answers
3






active

oldest

votes


















0












$begingroup$

In general, you would add the remainders of $3^{302}/5$ and $7^{200}/5$ modulo $5$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The remainder of 3^302 mod 5 is 4 and the remainder of 7^200 mod 5 is 1.But 5 mod 5 is zero so the remainder is 0?
    $endgroup$
    – Viktor
    Jan 14 at 0:47












  • $begingroup$
    Yes. This is confirmed by @user289143 answer.
    $endgroup$
    – D.B.
    Jan 14 at 0:51



















2












$begingroup$

The last digit of the powers of $3$ are $3,9,7,1,3,9,..$ they are cyclic and repeating every four times, so the last digit of $3^{302}$ is the same as the last digit of $3^{302 (mod 4)}=3^2=9$.

Same argument for the last digit of the powers of $7$: $7,9,3,1,7,9...$. Therefore the last digit of $7^{200}$ is the same as the last digit of $7^0=1$.

Thus the last digit of $3^{302}+7^{200}$ is $9+1=10=0 mod 5$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It's not 3^202 but 3^302
    $endgroup$
    – Viktor
    Jan 14 at 0:36












  • $begingroup$
    Oh sorry. Anyway $302 mod 4 = 202 mod 4 = 2 mod 4$ so the result is the same
    $endgroup$
    – user289143
    Jan 14 at 0:39










  • $begingroup$
    The remainder of 3^302 mod 5 is 4 and the remainder of 7^200 mod 5 is 1. So the result is 4 + 1 = 5. But 5 mod 5 is zero so the remainder is 0? Can I go this way?
    $endgroup$
    – Viktor
    Jan 14 at 0:46












  • $begingroup$
    Yes, it’s just another way of saying it
    $endgroup$
    – user289143
    Jan 14 at 0:51










  • $begingroup$
    It's an easier way to calculate for me. I only needed to be sure if I could find the remainders separately. Thank you.
    $endgroup$
    – Viktor
    Jan 14 at 0:55





















1












$begingroup$

$begin{align}bmod 5!: & 3^{302}+7^{200}\ equiv & 9^{151}+49^{100}\ equiv & (-1)^{151}!+(-1)^{100}\ equiv & {-1} + 1 end{align}$



using standard congruence rules, including the Sum Rule (which answers your query affirmatively).






share|cite|improve this answer











$endgroup$













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    3 Answers
    3






    active

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    3 Answers
    3






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    In general, you would add the remainders of $3^{302}/5$ and $7^{200}/5$ modulo $5$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The remainder of 3^302 mod 5 is 4 and the remainder of 7^200 mod 5 is 1.But 5 mod 5 is zero so the remainder is 0?
      $endgroup$
      – Viktor
      Jan 14 at 0:47












    • $begingroup$
      Yes. This is confirmed by @user289143 answer.
      $endgroup$
      – D.B.
      Jan 14 at 0:51
















    0












    $begingroup$

    In general, you would add the remainders of $3^{302}/5$ and $7^{200}/5$ modulo $5$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The remainder of 3^302 mod 5 is 4 and the remainder of 7^200 mod 5 is 1.But 5 mod 5 is zero so the remainder is 0?
      $endgroup$
      – Viktor
      Jan 14 at 0:47












    • $begingroup$
      Yes. This is confirmed by @user289143 answer.
      $endgroup$
      – D.B.
      Jan 14 at 0:51














    0












    0








    0





    $begingroup$

    In general, you would add the remainders of $3^{302}/5$ and $7^{200}/5$ modulo $5$.






    share|cite|improve this answer









    $endgroup$



    In general, you would add the remainders of $3^{302}/5$ and $7^{200}/5$ modulo $5$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 14 at 0:25









    D.B.D.B.

    1,27518




    1,27518












    • $begingroup$
      The remainder of 3^302 mod 5 is 4 and the remainder of 7^200 mod 5 is 1.But 5 mod 5 is zero so the remainder is 0?
      $endgroup$
      – Viktor
      Jan 14 at 0:47












    • $begingroup$
      Yes. This is confirmed by @user289143 answer.
      $endgroup$
      – D.B.
      Jan 14 at 0:51


















    • $begingroup$
      The remainder of 3^302 mod 5 is 4 and the remainder of 7^200 mod 5 is 1.But 5 mod 5 is zero so the remainder is 0?
      $endgroup$
      – Viktor
      Jan 14 at 0:47












    • $begingroup$
      Yes. This is confirmed by @user289143 answer.
      $endgroup$
      – D.B.
      Jan 14 at 0:51
















    $begingroup$
    The remainder of 3^302 mod 5 is 4 and the remainder of 7^200 mod 5 is 1.But 5 mod 5 is zero so the remainder is 0?
    $endgroup$
    – Viktor
    Jan 14 at 0:47






    $begingroup$
    The remainder of 3^302 mod 5 is 4 and the remainder of 7^200 mod 5 is 1.But 5 mod 5 is zero so the remainder is 0?
    $endgroup$
    – Viktor
    Jan 14 at 0:47














    $begingroup$
    Yes. This is confirmed by @user289143 answer.
    $endgroup$
    – D.B.
    Jan 14 at 0:51




    $begingroup$
    Yes. This is confirmed by @user289143 answer.
    $endgroup$
    – D.B.
    Jan 14 at 0:51











    2












    $begingroup$

    The last digit of the powers of $3$ are $3,9,7,1,3,9,..$ they are cyclic and repeating every four times, so the last digit of $3^{302}$ is the same as the last digit of $3^{302 (mod 4)}=3^2=9$.

    Same argument for the last digit of the powers of $7$: $7,9,3,1,7,9...$. Therefore the last digit of $7^{200}$ is the same as the last digit of $7^0=1$.

    Thus the last digit of $3^{302}+7^{200}$ is $9+1=10=0 mod 5$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      It's not 3^202 but 3^302
      $endgroup$
      – Viktor
      Jan 14 at 0:36












    • $begingroup$
      Oh sorry. Anyway $302 mod 4 = 202 mod 4 = 2 mod 4$ so the result is the same
      $endgroup$
      – user289143
      Jan 14 at 0:39










    • $begingroup$
      The remainder of 3^302 mod 5 is 4 and the remainder of 7^200 mod 5 is 1. So the result is 4 + 1 = 5. But 5 mod 5 is zero so the remainder is 0? Can I go this way?
      $endgroup$
      – Viktor
      Jan 14 at 0:46












    • $begingroup$
      Yes, it’s just another way of saying it
      $endgroup$
      – user289143
      Jan 14 at 0:51










    • $begingroup$
      It's an easier way to calculate for me. I only needed to be sure if I could find the remainders separately. Thank you.
      $endgroup$
      – Viktor
      Jan 14 at 0:55


















    2












    $begingroup$

    The last digit of the powers of $3$ are $3,9,7,1,3,9,..$ they are cyclic and repeating every four times, so the last digit of $3^{302}$ is the same as the last digit of $3^{302 (mod 4)}=3^2=9$.

    Same argument for the last digit of the powers of $7$: $7,9,3,1,7,9...$. Therefore the last digit of $7^{200}$ is the same as the last digit of $7^0=1$.

    Thus the last digit of $3^{302}+7^{200}$ is $9+1=10=0 mod 5$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      It's not 3^202 but 3^302
      $endgroup$
      – Viktor
      Jan 14 at 0:36












    • $begingroup$
      Oh sorry. Anyway $302 mod 4 = 202 mod 4 = 2 mod 4$ so the result is the same
      $endgroup$
      – user289143
      Jan 14 at 0:39










    • $begingroup$
      The remainder of 3^302 mod 5 is 4 and the remainder of 7^200 mod 5 is 1. So the result is 4 + 1 = 5. But 5 mod 5 is zero so the remainder is 0? Can I go this way?
      $endgroup$
      – Viktor
      Jan 14 at 0:46












    • $begingroup$
      Yes, it’s just another way of saying it
      $endgroup$
      – user289143
      Jan 14 at 0:51










    • $begingroup$
      It's an easier way to calculate for me. I only needed to be sure if I could find the remainders separately. Thank you.
      $endgroup$
      – Viktor
      Jan 14 at 0:55
















    2












    2








    2





    $begingroup$

    The last digit of the powers of $3$ are $3,9,7,1,3,9,..$ they are cyclic and repeating every four times, so the last digit of $3^{302}$ is the same as the last digit of $3^{302 (mod 4)}=3^2=9$.

    Same argument for the last digit of the powers of $7$: $7,9,3,1,7,9...$. Therefore the last digit of $7^{200}$ is the same as the last digit of $7^0=1$.

    Thus the last digit of $3^{302}+7^{200}$ is $9+1=10=0 mod 5$






    share|cite|improve this answer











    $endgroup$



    The last digit of the powers of $3$ are $3,9,7,1,3,9,..$ they are cyclic and repeating every four times, so the last digit of $3^{302}$ is the same as the last digit of $3^{302 (mod 4)}=3^2=9$.

    Same argument for the last digit of the powers of $7$: $7,9,3,1,7,9...$. Therefore the last digit of $7^{200}$ is the same as the last digit of $7^0=1$.

    Thus the last digit of $3^{302}+7^{200}$ is $9+1=10=0 mod 5$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 14 at 0:39

























    answered Jan 14 at 0:29









    user289143user289143

    1,000313




    1,000313












    • $begingroup$
      It's not 3^202 but 3^302
      $endgroup$
      – Viktor
      Jan 14 at 0:36












    • $begingroup$
      Oh sorry. Anyway $302 mod 4 = 202 mod 4 = 2 mod 4$ so the result is the same
      $endgroup$
      – user289143
      Jan 14 at 0:39










    • $begingroup$
      The remainder of 3^302 mod 5 is 4 and the remainder of 7^200 mod 5 is 1. So the result is 4 + 1 = 5. But 5 mod 5 is zero so the remainder is 0? Can I go this way?
      $endgroup$
      – Viktor
      Jan 14 at 0:46












    • $begingroup$
      Yes, it’s just another way of saying it
      $endgroup$
      – user289143
      Jan 14 at 0:51










    • $begingroup$
      It's an easier way to calculate for me. I only needed to be sure if I could find the remainders separately. Thank you.
      $endgroup$
      – Viktor
      Jan 14 at 0:55




















    • $begingroup$
      It's not 3^202 but 3^302
      $endgroup$
      – Viktor
      Jan 14 at 0:36












    • $begingroup$
      Oh sorry. Anyway $302 mod 4 = 202 mod 4 = 2 mod 4$ so the result is the same
      $endgroup$
      – user289143
      Jan 14 at 0:39










    • $begingroup$
      The remainder of 3^302 mod 5 is 4 and the remainder of 7^200 mod 5 is 1. So the result is 4 + 1 = 5. But 5 mod 5 is zero so the remainder is 0? Can I go this way?
      $endgroup$
      – Viktor
      Jan 14 at 0:46












    • $begingroup$
      Yes, it’s just another way of saying it
      $endgroup$
      – user289143
      Jan 14 at 0:51










    • $begingroup$
      It's an easier way to calculate for me. I only needed to be sure if I could find the remainders separately. Thank you.
      $endgroup$
      – Viktor
      Jan 14 at 0:55


















    $begingroup$
    It's not 3^202 but 3^302
    $endgroup$
    – Viktor
    Jan 14 at 0:36






    $begingroup$
    It's not 3^202 but 3^302
    $endgroup$
    – Viktor
    Jan 14 at 0:36














    $begingroup$
    Oh sorry. Anyway $302 mod 4 = 202 mod 4 = 2 mod 4$ so the result is the same
    $endgroup$
    – user289143
    Jan 14 at 0:39




    $begingroup$
    Oh sorry. Anyway $302 mod 4 = 202 mod 4 = 2 mod 4$ so the result is the same
    $endgroup$
    – user289143
    Jan 14 at 0:39












    $begingroup$
    The remainder of 3^302 mod 5 is 4 and the remainder of 7^200 mod 5 is 1. So the result is 4 + 1 = 5. But 5 mod 5 is zero so the remainder is 0? Can I go this way?
    $endgroup$
    – Viktor
    Jan 14 at 0:46






    $begingroup$
    The remainder of 3^302 mod 5 is 4 and the remainder of 7^200 mod 5 is 1. So the result is 4 + 1 = 5. But 5 mod 5 is zero so the remainder is 0? Can I go this way?
    $endgroup$
    – Viktor
    Jan 14 at 0:46














    $begingroup$
    Yes, it’s just another way of saying it
    $endgroup$
    – user289143
    Jan 14 at 0:51




    $begingroup$
    Yes, it’s just another way of saying it
    $endgroup$
    – user289143
    Jan 14 at 0:51












    $begingroup$
    It's an easier way to calculate for me. I only needed to be sure if I could find the remainders separately. Thank you.
    $endgroup$
    – Viktor
    Jan 14 at 0:55






    $begingroup$
    It's an easier way to calculate for me. I only needed to be sure if I could find the remainders separately. Thank you.
    $endgroup$
    – Viktor
    Jan 14 at 0:55













    1












    $begingroup$

    $begin{align}bmod 5!: & 3^{302}+7^{200}\ equiv & 9^{151}+49^{100}\ equiv & (-1)^{151}!+(-1)^{100}\ equiv & {-1} + 1 end{align}$



    using standard congruence rules, including the Sum Rule (which answers your query affirmatively).






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      $begin{align}bmod 5!: & 3^{302}+7^{200}\ equiv & 9^{151}+49^{100}\ equiv & (-1)^{151}!+(-1)^{100}\ equiv & {-1} + 1 end{align}$



      using standard congruence rules, including the Sum Rule (which answers your query affirmatively).






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        $begin{align}bmod 5!: & 3^{302}+7^{200}\ equiv & 9^{151}+49^{100}\ equiv & (-1)^{151}!+(-1)^{100}\ equiv & {-1} + 1 end{align}$



        using standard congruence rules, including the Sum Rule (which answers your query affirmatively).






        share|cite|improve this answer











        $endgroup$



        $begin{align}bmod 5!: & 3^{302}+7^{200}\ equiv & 9^{151}+49^{100}\ equiv & (-1)^{151}!+(-1)^{100}\ equiv & {-1} + 1 end{align}$



        using standard congruence rules, including the Sum Rule (which answers your query affirmatively).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 14 at 2:18

























        answered Jan 14 at 1:54









        Bill DubuqueBill Dubuque

        212k29195654




        212k29195654






























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