What is the remainder of $1^1$+$2^2$+$3^3$+…+$2020^{2020}$ if divided by 10? (Without Calculator) [closed]












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Is there any way to calculate the remainder of $1^1$ + $2^2$ + $3^3$ + ... + $2020^{2020}$ when divided by 10 without calculator?






elementary-number-theory modular-arithmetic







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closed as off-topic by Eevee Trainer, heropup, Jyrki Lahtonen, naveen dankal, verret Jan 9 at 20:53


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, heropup, Jyrki Lahtonen, naveen dankal, verret

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Probably the first thing that comes to mind to me is to realize that, for any of the numbers of the sum, only its one's digit to the power is the relevant bit with respect to the sum modulo 10. There's probably something more elaborate but that's a start at least.
    $endgroup$
    – Eevee Trainer
    Jan 9 at 2:51










  • $begingroup$
    See this question and the answer there math.stackexchange.com/questions/2578034
    $endgroup$
    – JimmyK4542
    Jan 9 at 2:57










  • $begingroup$
    Also compare with math.stackexchange.com/questions/2578034/…
    $endgroup$
    – David K
    Jan 9 at 5:04










  • $begingroup$
    And also a duplicate of this. Quick to find with Approach0.
    $endgroup$
    – Jyrki Lahtonen
    Jan 9 at 5:39
















0












$begingroup$


Is there any way to calculate the remainder of $1^1$ + $2^2$ + $3^3$ + ... + $2020^{2020}$ when divided by 10 without calculator?






elementary-number-theory modular-arithmetic







share|cite|improve this question









$endgroup$



closed as off-topic by Eevee Trainer, heropup, Jyrki Lahtonen, naveen dankal, verret Jan 9 at 20:53


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, heropup, Jyrki Lahtonen, naveen dankal, verret

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Probably the first thing that comes to mind to me is to realize that, for any of the numbers of the sum, only its one's digit to the power is the relevant bit with respect to the sum modulo 10. There's probably something more elaborate but that's a start at least.
    $endgroup$
    – Eevee Trainer
    Jan 9 at 2:51










  • $begingroup$
    See this question and the answer there math.stackexchange.com/questions/2578034
    $endgroup$
    – JimmyK4542
    Jan 9 at 2:57










  • $begingroup$
    Also compare with math.stackexchange.com/questions/2578034/…
    $endgroup$
    – David K
    Jan 9 at 5:04










  • $begingroup$
    And also a duplicate of this. Quick to find with Approach0.
    $endgroup$
    – Jyrki Lahtonen
    Jan 9 at 5:39














0












0








0





$begingroup$


Is there any way to calculate the remainder of $1^1$ + $2^2$ + $3^3$ + ... + $2020^{2020}$ when divided by 10 without calculator?






elementary-number-theory modular-arithmetic







share|cite|improve this question









$endgroup$




Is there any way to calculate the remainder of $1^1$ + $2^2$ + $3^3$ + ... + $2020^{2020}$ when divided by 10 without calculator?






elementary-number-theory modular-arithmetic




elementary-number-theory modular-arithmetic






share|cite|improve this question













share|cite|improve this question











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asked Jan 9 at 2:48









ShromiShromi

3069




3069




closed as off-topic by Eevee Trainer, heropup, Jyrki Lahtonen, naveen dankal, verret Jan 9 at 20:53


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, heropup, Jyrki Lahtonen, naveen dankal, verret

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Eevee Trainer, heropup, Jyrki Lahtonen, naveen dankal, verret Jan 9 at 20:53


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, heropup, Jyrki Lahtonen, naveen dankal, verret

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Probably the first thing that comes to mind to me is to realize that, for any of the numbers of the sum, only its one's digit to the power is the relevant bit with respect to the sum modulo 10. There's probably something more elaborate but that's a start at least.
    $endgroup$
    – Eevee Trainer
    Jan 9 at 2:51










  • $begingroup$
    See this question and the answer there math.stackexchange.com/questions/2578034
    $endgroup$
    – JimmyK4542
    Jan 9 at 2:57










  • $begingroup$
    Also compare with math.stackexchange.com/questions/2578034/…
    $endgroup$
    – David K
    Jan 9 at 5:04










  • $begingroup$
    And also a duplicate of this. Quick to find with Approach0.
    $endgroup$
    – Jyrki Lahtonen
    Jan 9 at 5:39


















  • $begingroup$
    Probably the first thing that comes to mind to me is to realize that, for any of the numbers of the sum, only its one's digit to the power is the relevant bit with respect to the sum modulo 10. There's probably something more elaborate but that's a start at least.
    $endgroup$
    – Eevee Trainer
    Jan 9 at 2:51










  • $begingroup$
    See this question and the answer there math.stackexchange.com/questions/2578034
    $endgroup$
    – JimmyK4542
    Jan 9 at 2:57










  • $begingroup$
    Also compare with math.stackexchange.com/questions/2578034/…
    $endgroup$
    – David K
    Jan 9 at 5:04










  • $begingroup$
    And also a duplicate of this. Quick to find with Approach0.
    $endgroup$
    – Jyrki Lahtonen
    Jan 9 at 5:39
















$begingroup$
Probably the first thing that comes to mind to me is to realize that, for any of the numbers of the sum, only its one's digit to the power is the relevant bit with respect to the sum modulo 10. There's probably something more elaborate but that's a start at least.
$endgroup$
– Eevee Trainer
Jan 9 at 2:51




$begingroup$
Probably the first thing that comes to mind to me is to realize that, for any of the numbers of the sum, only its one's digit to the power is the relevant bit with respect to the sum modulo 10. There's probably something more elaborate but that's a start at least.
$endgroup$
– Eevee Trainer
Jan 9 at 2:51












$begingroup$
See this question and the answer there math.stackexchange.com/questions/2578034
$endgroup$
– JimmyK4542
Jan 9 at 2:57




$begingroup$
See this question and the answer there math.stackexchange.com/questions/2578034
$endgroup$
– JimmyK4542
Jan 9 at 2:57












$begingroup$
Also compare with math.stackexchange.com/questions/2578034/…
$endgroup$
– David K
Jan 9 at 5:04




$begingroup$
Also compare with math.stackexchange.com/questions/2578034/…
$endgroup$
– David K
Jan 9 at 5:04












$begingroup$
And also a duplicate of this. Quick to find with Approach0.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 5:39




$begingroup$
And also a duplicate of this. Quick to find with Approach0.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 5:39










2 Answers
2






active

oldest

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3












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Hint $bmod 10!: n^{largecolor{#c00}{k+4j}}equiv n^{large k},$ for $,k>0,,$ by $, n^{large k+4j}!-n^{large k} equiv n^{large k}(n^{large 4j}!-1) equiv 0, $
by here.



So $,(n!+!20)^{large n+20}!equiv n^{largecolor{#c00}{ n+4cdot 5}}!equiv n^{large n},$ for $,n>0,$ so $!bmod 10$ the summands repeat in cycles of length $20,,$ so we need only compute a single cycle sum, which is easy using above, e.g. $,17^{large 17}!equiv 7^{largecolor{#c00}{ 1+4cdot 4}}!equiv 7^{large 1}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Note: $ 20 = {rm lcm}(10,color{#c00}4),$ explains the cycle length. $ $
    $endgroup$
    – Bill Dubuque
    Jan 9 at 5:16



















0












$begingroup$

You may split it into $mod 2$ and $mod 5$:



$mod 2$:
begin{eqnarray*} sum_{n=1}^{2020}n^n
& equiv_2 & 1010cdot 1 \
& equiv_2 & 0 \
end{eqnarray*}
$mod 5$:
begin{eqnarray*} sum_{n=1}^{2020}n^n
& equiv_5 & sum_{k=0}^{403}left((5k+1)^{5k+1} + (5k+2)^{5k+2}+(5k+3)^{5k+3}+(5k+4)^{5k+4} right) \
& stackrel{mbox{Fermat}}{equiv_5} & sum_{k=0}^{403}left(1 + 2^{k+2}+3^{k+3}+4^{k+4} right)\
& equiv_5 & sum_{k=0}^{403}left(1 + 2^{k+2}+(-2)^{k+3}+(-1)^{k+4} right)\
& equiv_5 & -1 + sum_{k=0}^{403}left(2^{k+2}+(-2)^{k+3}right) + 0\
& equiv_5 & -1 - sum_{k=0}^{403}2^{k+2} \
& equiv_5 & -1 + sum_{k=0}^{403}2^{k} \
& equiv_5 & -1 + 2^{404} - 1 \
& stackrel{mbox{Fermat}}{equiv_5} & -1 \
end{eqnarray*}
So, $sum_{n=1}^{2020}n^n equiv_2 0$ and $sum_{n=1}^{2020}n^nequiv_5 -1 equiv_5 4 Rightarrow boxed{sum_{n=1}^{2020}n^n equiv_{10} 4}$






share|cite|improve this answer









$endgroup$




















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Hint $bmod 10!: n^{largecolor{#c00}{k+4j}}equiv n^{large k},$ for $,k>0,,$ by $, n^{large k+4j}!-n^{large k} equiv n^{large k}(n^{large 4j}!-1) equiv 0, $
    by here.



    So $,(n!+!20)^{large n+20}!equiv n^{largecolor{#c00}{ n+4cdot 5}}!equiv n^{large n},$ for $,n>0,$ so $!bmod 10$ the summands repeat in cycles of length $20,,$ so we need only compute a single cycle sum, which is easy using above, e.g. $,17^{large 17}!equiv 7^{largecolor{#c00}{ 1+4cdot 4}}!equiv 7^{large 1}$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Note: $ 20 = {rm lcm}(10,color{#c00}4),$ explains the cycle length. $ $
      $endgroup$
      – Bill Dubuque
      Jan 9 at 5:16
















    3












    $begingroup$

    Hint $bmod 10!: n^{largecolor{#c00}{k+4j}}equiv n^{large k},$ for $,k>0,,$ by $, n^{large k+4j}!-n^{large k} equiv n^{large k}(n^{large 4j}!-1) equiv 0, $
    by here.



    So $,(n!+!20)^{large n+20}!equiv n^{largecolor{#c00}{ n+4cdot 5}}!equiv n^{large n},$ for $,n>0,$ so $!bmod 10$ the summands repeat in cycles of length $20,,$ so we need only compute a single cycle sum, which is easy using above, e.g. $,17^{large 17}!equiv 7^{largecolor{#c00}{ 1+4cdot 4}}!equiv 7^{large 1}$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Note: $ 20 = {rm lcm}(10,color{#c00}4),$ explains the cycle length. $ $
      $endgroup$
      – Bill Dubuque
      Jan 9 at 5:16














    3












    3








    3





    $begingroup$

    Hint $bmod 10!: n^{largecolor{#c00}{k+4j}}equiv n^{large k},$ for $,k>0,,$ by $, n^{large k+4j}!-n^{large k} equiv n^{large k}(n^{large 4j}!-1) equiv 0, $
    by here.



    So $,(n!+!20)^{large n+20}!equiv n^{largecolor{#c00}{ n+4cdot 5}}!equiv n^{large n},$ for $,n>0,$ so $!bmod 10$ the summands repeat in cycles of length $20,,$ so we need only compute a single cycle sum, which is easy using above, e.g. $,17^{large 17}!equiv 7^{largecolor{#c00}{ 1+4cdot 4}}!equiv 7^{large 1}$.






    share|cite|improve this answer











    $endgroup$



    Hint $bmod 10!: n^{largecolor{#c00}{k+4j}}equiv n^{large k},$ for $,k>0,,$ by $, n^{large k+4j}!-n^{large k} equiv n^{large k}(n^{large 4j}!-1) equiv 0, $
    by here.



    So $,(n!+!20)^{large n+20}!equiv n^{largecolor{#c00}{ n+4cdot 5}}!equiv n^{large n},$ for $,n>0,$ so $!bmod 10$ the summands repeat in cycles of length $20,,$ so we need only compute a single cycle sum, which is easy using above, e.g. $,17^{large 17}!equiv 7^{largecolor{#c00}{ 1+4cdot 4}}!equiv 7^{large 1}$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 9 at 5:56

























    answered Jan 9 at 3:59









    Bill DubuqueBill Dubuque

    211k29193647




    211k29193647












    • $begingroup$
      Note: $ 20 = {rm lcm}(10,color{#c00}4),$ explains the cycle length. $ $
      $endgroup$
      – Bill Dubuque
      Jan 9 at 5:16


















    • $begingroup$
      Note: $ 20 = {rm lcm}(10,color{#c00}4),$ explains the cycle length. $ $
      $endgroup$
      – Bill Dubuque
      Jan 9 at 5:16
















    $begingroup$
    Note: $ 20 = {rm lcm}(10,color{#c00}4),$ explains the cycle length. $ $
    $endgroup$
    – Bill Dubuque
    Jan 9 at 5:16




    $begingroup$
    Note: $ 20 = {rm lcm}(10,color{#c00}4),$ explains the cycle length. $ $
    $endgroup$
    – Bill Dubuque
    Jan 9 at 5:16











    0












    $begingroup$

    You may split it into $mod 2$ and $mod 5$:



    $mod 2$:
    begin{eqnarray*} sum_{n=1}^{2020}n^n
    & equiv_2 & 1010cdot 1 \
    & equiv_2 & 0 \
    end{eqnarray*}
    $mod 5$:
    begin{eqnarray*} sum_{n=1}^{2020}n^n
    & equiv_5 & sum_{k=0}^{403}left((5k+1)^{5k+1} + (5k+2)^{5k+2}+(5k+3)^{5k+3}+(5k+4)^{5k+4} right) \
    & stackrel{mbox{Fermat}}{equiv_5} & sum_{k=0}^{403}left(1 + 2^{k+2}+3^{k+3}+4^{k+4} right)\
    & equiv_5 & sum_{k=0}^{403}left(1 + 2^{k+2}+(-2)^{k+3}+(-1)^{k+4} right)\
    & equiv_5 & -1 + sum_{k=0}^{403}left(2^{k+2}+(-2)^{k+3}right) + 0\
    & equiv_5 & -1 - sum_{k=0}^{403}2^{k+2} \
    & equiv_5 & -1 + sum_{k=0}^{403}2^{k} \
    & equiv_5 & -1 + 2^{404} - 1 \
    & stackrel{mbox{Fermat}}{equiv_5} & -1 \
    end{eqnarray*}
    So, $sum_{n=1}^{2020}n^n equiv_2 0$ and $sum_{n=1}^{2020}n^nequiv_5 -1 equiv_5 4 Rightarrow boxed{sum_{n=1}^{2020}n^n equiv_{10} 4}$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      You may split it into $mod 2$ and $mod 5$:



      $mod 2$:
      begin{eqnarray*} sum_{n=1}^{2020}n^n
      & equiv_2 & 1010cdot 1 \
      & equiv_2 & 0 \
      end{eqnarray*}
      $mod 5$:
      begin{eqnarray*} sum_{n=1}^{2020}n^n
      & equiv_5 & sum_{k=0}^{403}left((5k+1)^{5k+1} + (5k+2)^{5k+2}+(5k+3)^{5k+3}+(5k+4)^{5k+4} right) \
      & stackrel{mbox{Fermat}}{equiv_5} & sum_{k=0}^{403}left(1 + 2^{k+2}+3^{k+3}+4^{k+4} right)\
      & equiv_5 & sum_{k=0}^{403}left(1 + 2^{k+2}+(-2)^{k+3}+(-1)^{k+4} right)\
      & equiv_5 & -1 + sum_{k=0}^{403}left(2^{k+2}+(-2)^{k+3}right) + 0\
      & equiv_5 & -1 - sum_{k=0}^{403}2^{k+2} \
      & equiv_5 & -1 + sum_{k=0}^{403}2^{k} \
      & equiv_5 & -1 + 2^{404} - 1 \
      & stackrel{mbox{Fermat}}{equiv_5} & -1 \
      end{eqnarray*}
      So, $sum_{n=1}^{2020}n^n equiv_2 0$ and $sum_{n=1}^{2020}n^nequiv_5 -1 equiv_5 4 Rightarrow boxed{sum_{n=1}^{2020}n^n equiv_{10} 4}$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        You may split it into $mod 2$ and $mod 5$:



        $mod 2$:
        begin{eqnarray*} sum_{n=1}^{2020}n^n
        & equiv_2 & 1010cdot 1 \
        & equiv_2 & 0 \
        end{eqnarray*}
        $mod 5$:
        begin{eqnarray*} sum_{n=1}^{2020}n^n
        & equiv_5 & sum_{k=0}^{403}left((5k+1)^{5k+1} + (5k+2)^{5k+2}+(5k+3)^{5k+3}+(5k+4)^{5k+4} right) \
        & stackrel{mbox{Fermat}}{equiv_5} & sum_{k=0}^{403}left(1 + 2^{k+2}+3^{k+3}+4^{k+4} right)\
        & equiv_5 & sum_{k=0}^{403}left(1 + 2^{k+2}+(-2)^{k+3}+(-1)^{k+4} right)\
        & equiv_5 & -1 + sum_{k=0}^{403}left(2^{k+2}+(-2)^{k+3}right) + 0\
        & equiv_5 & -1 - sum_{k=0}^{403}2^{k+2} \
        & equiv_5 & -1 + sum_{k=0}^{403}2^{k} \
        & equiv_5 & -1 + 2^{404} - 1 \
        & stackrel{mbox{Fermat}}{equiv_5} & -1 \
        end{eqnarray*}
        So, $sum_{n=1}^{2020}n^n equiv_2 0$ and $sum_{n=1}^{2020}n^nequiv_5 -1 equiv_5 4 Rightarrow boxed{sum_{n=1}^{2020}n^n equiv_{10} 4}$






        share|cite|improve this answer









        $endgroup$



        You may split it into $mod 2$ and $mod 5$:



        $mod 2$:
        begin{eqnarray*} sum_{n=1}^{2020}n^n
        & equiv_2 & 1010cdot 1 \
        & equiv_2 & 0 \
        end{eqnarray*}
        $mod 5$:
        begin{eqnarray*} sum_{n=1}^{2020}n^n
        & equiv_5 & sum_{k=0}^{403}left((5k+1)^{5k+1} + (5k+2)^{5k+2}+(5k+3)^{5k+3}+(5k+4)^{5k+4} right) \
        & stackrel{mbox{Fermat}}{equiv_5} & sum_{k=0}^{403}left(1 + 2^{k+2}+3^{k+3}+4^{k+4} right)\
        & equiv_5 & sum_{k=0}^{403}left(1 + 2^{k+2}+(-2)^{k+3}+(-1)^{k+4} right)\
        & equiv_5 & -1 + sum_{k=0}^{403}left(2^{k+2}+(-2)^{k+3}right) + 0\
        & equiv_5 & -1 - sum_{k=0}^{403}2^{k+2} \
        & equiv_5 & -1 + sum_{k=0}^{403}2^{k} \
        & equiv_5 & -1 + 2^{404} - 1 \
        & stackrel{mbox{Fermat}}{equiv_5} & -1 \
        end{eqnarray*}
        So, $sum_{n=1}^{2020}n^n equiv_2 0$ and $sum_{n=1}^{2020}n^nequiv_5 -1 equiv_5 4 Rightarrow boxed{sum_{n=1}^{2020}n^n equiv_{10} 4}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 9 at 5:25









        trancelocationtrancelocation

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        12.3k1826















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