What's $1^3+2^3+3^3+cdots+99^3$ modulo $3$?












-1












$begingroup$


What's the remainder when the sum



$1^3+2^3+3^3+cdots+99^3$ is divided by $3$?



Background:



I saw this question on MSE but it was closed and I wanted to learn how to approach it. The help given to the asker had failed to bring the question up to an acceptable standard.



I can solve it but I doubt my methods are efficient.



One way is to start by removing every third term. Then the first two terms of the remaining sequence can be removed, and so on... a pattern may emerge.



Another way is to look at the expansion of $(x+1)^3$ and see what it does to the residues for each $x$, then sum over that by induction.



But I'm sure my inventions aren't very efficient.






algebra-precalculus elementary-number-theory modular-arithmetic







share|cite|improve this question









$endgroup$








  • 5




    $begingroup$
    Use $n^3equiv npmod3$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 27 '18 at 5:42










  • $begingroup$
    math.stackexchange.com/questions/1328798
    $endgroup$
    – Andrei
    Dec 27 '18 at 5:43






  • 1




    $begingroup$
    In general, note that you can handle the terms in groups of $3$ as $n + 3 equiv n pmod 3$, and also use that $n^3 equiv n pmod 3$ as Lord Shark the Unknown stated above.
    $endgroup$
    – John Omielan
    Dec 27 '18 at 5:46












  • $begingroup$
    @LordSharktheUnknown ah thanks. I guess the $(x+1)^3$ method would've revealed that.
    $endgroup$
    – user334732
    Dec 27 '18 at 5:46
















-1












$begingroup$


What's the remainder when the sum



$1^3+2^3+3^3+cdots+99^3$ is divided by $3$?



Background:



I saw this question on MSE but it was closed and I wanted to learn how to approach it. The help given to the asker had failed to bring the question up to an acceptable standard.



I can solve it but I doubt my methods are efficient.



One way is to start by removing every third term. Then the first two terms of the remaining sequence can be removed, and so on... a pattern may emerge.



Another way is to look at the expansion of $(x+1)^3$ and see what it does to the residues for each $x$, then sum over that by induction.



But I'm sure my inventions aren't very efficient.






algebra-precalculus elementary-number-theory modular-arithmetic







share|cite|improve this question









$endgroup$








  • 5




    $begingroup$
    Use $n^3equiv npmod3$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 27 '18 at 5:42










  • $begingroup$
    math.stackexchange.com/questions/1328798
    $endgroup$
    – Andrei
    Dec 27 '18 at 5:43






  • 1




    $begingroup$
    In general, note that you can handle the terms in groups of $3$ as $n + 3 equiv n pmod 3$, and also use that $n^3 equiv n pmod 3$ as Lord Shark the Unknown stated above.
    $endgroup$
    – John Omielan
    Dec 27 '18 at 5:46












  • $begingroup$
    @LordSharktheUnknown ah thanks. I guess the $(x+1)^3$ method would've revealed that.
    $endgroup$
    – user334732
    Dec 27 '18 at 5:46














-1












-1








-1





$begingroup$


What's the remainder when the sum



$1^3+2^3+3^3+cdots+99^3$ is divided by $3$?



Background:



I saw this question on MSE but it was closed and I wanted to learn how to approach it. The help given to the asker had failed to bring the question up to an acceptable standard.



I can solve it but I doubt my methods are efficient.



One way is to start by removing every third term. Then the first two terms of the remaining sequence can be removed, and so on... a pattern may emerge.



Another way is to look at the expansion of $(x+1)^3$ and see what it does to the residues for each $x$, then sum over that by induction.



But I'm sure my inventions aren't very efficient.






algebra-precalculus elementary-number-theory modular-arithmetic







share|cite|improve this question









$endgroup$




What's the remainder when the sum



$1^3+2^3+3^3+cdots+99^3$ is divided by $3$?



Background:



I saw this question on MSE but it was closed and I wanted to learn how to approach it. The help given to the asker had failed to bring the question up to an acceptable standard.



I can solve it but I doubt my methods are efficient.



One way is to start by removing every third term. Then the first two terms of the remaining sequence can be removed, and so on... a pattern may emerge.



Another way is to look at the expansion of $(x+1)^3$ and see what it does to the residues for each $x$, then sum over that by induction.



But I'm sure my inventions aren't very efficient.






algebra-precalculus elementary-number-theory modular-arithmetic




algebra-precalculus elementary-number-theory modular-arithmetic






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 27 '18 at 5:40









user334732user334732

4,28411240




4,28411240








  • 5




    $begingroup$
    Use $n^3equiv npmod3$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 27 '18 at 5:42










  • $begingroup$
    math.stackexchange.com/questions/1328798
    $endgroup$
    – Andrei
    Dec 27 '18 at 5:43






  • 1




    $begingroup$
    In general, note that you can handle the terms in groups of $3$ as $n + 3 equiv n pmod 3$, and also use that $n^3 equiv n pmod 3$ as Lord Shark the Unknown stated above.
    $endgroup$
    – John Omielan
    Dec 27 '18 at 5:46












  • $begingroup$
    @LordSharktheUnknown ah thanks. I guess the $(x+1)^3$ method would've revealed that.
    $endgroup$
    – user334732
    Dec 27 '18 at 5:46














  • 5




    $begingroup$
    Use $n^3equiv npmod3$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 27 '18 at 5:42










  • $begingroup$
    math.stackexchange.com/questions/1328798
    $endgroup$
    – Andrei
    Dec 27 '18 at 5:43






  • 1




    $begingroup$
    In general, note that you can handle the terms in groups of $3$ as $n + 3 equiv n pmod 3$, and also use that $n^3 equiv n pmod 3$ as Lord Shark the Unknown stated above.
    $endgroup$
    – John Omielan
    Dec 27 '18 at 5:46












  • $begingroup$
    @LordSharktheUnknown ah thanks. I guess the $(x+1)^3$ method would've revealed that.
    $endgroup$
    – user334732
    Dec 27 '18 at 5:46








5




5




$begingroup$
Use $n^3equiv npmod3$.
$endgroup$
– Lord Shark the Unknown
Dec 27 '18 at 5:42




$begingroup$
Use $n^3equiv npmod3$.
$endgroup$
– Lord Shark the Unknown
Dec 27 '18 at 5:42












$begingroup$
math.stackexchange.com/questions/1328798
$endgroup$
– Andrei
Dec 27 '18 at 5:43




$begingroup$
math.stackexchange.com/questions/1328798
$endgroup$
– Andrei
Dec 27 '18 at 5:43




1




1




$begingroup$
In general, note that you can handle the terms in groups of $3$ as $n + 3 equiv n pmod 3$, and also use that $n^3 equiv n pmod 3$ as Lord Shark the Unknown stated above.
$endgroup$
– John Omielan
Dec 27 '18 at 5:46






$begingroup$
In general, note that you can handle the terms in groups of $3$ as $n + 3 equiv n pmod 3$, and also use that $n^3 equiv n pmod 3$ as Lord Shark the Unknown stated above.
$endgroup$
– John Omielan
Dec 27 '18 at 5:46














$begingroup$
@LordSharktheUnknown ah thanks. I guess the $(x+1)^3$ method would've revealed that.
$endgroup$
– user334732
Dec 27 '18 at 5:46




$begingroup$
@LordSharktheUnknown ah thanks. I guess the $(x+1)^3$ method would've revealed that.
$endgroup$
– user334732
Dec 27 '18 at 5:46










9 Answers
9






active

oldest

votes


















3












$begingroup$

By Fermat's little theorem $a^3equiv apmod3$. Then you have $sum_{k=1}^{99} k=frac{(99)(100)}2=50cdot 99equiv 0pmod3$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Hint:



    $n^3equiv npmod3$ as $n^3-n=n(n-1)(n+1)$ is the product of three consecutive integers



    We have $sum_{r=1}^mr^3equivsum_{r=1}^mrpmod3$



    $equiv dfrac{m(m+1)}2$



    Alternatively $$sum_{r=1}^m r^3=dfrac{m^2(m+1)^2}4$$






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      begin{align}
      sum_{i=1}^{99}i^3 &= sum_{i=0}^{32} (3i+1)^3 + sum_{i=0}^{32} (3i+2)^3 + sum_{i=0}^{32} (3i+3)^3 \
      &equiv sum_{i=0}^{32} 1^3 + sum_{i=0}^{32} 2^3 + sum_{i=0}^{32} 0^3 pmod{3}\
      &equiv sum_{i=0}^{32} 1^3 + sum_{i=0}^{32} (-1)^3 + sum_{i=0}^{32} 0^3 pmod{3}\
      &equiv 0 pmod{3}
      end{align}






      share|cite|improve this answer









      $endgroup$





















        2












        $begingroup$

        Split the numbers from $1,ldots, 99$ in





        • $33$ with remainder $1$


        • $33$ with remainder $2$


        • $33$ with remainder $0$
          $$1^3 + 2^3 + cdots + 99^3 equiv 33cdot (1^3+2^3+0^3) equiv 0 mod 3$$






        share|cite|improve this answer









        $endgroup$





















          2












          $begingroup$

          Another method is to use the identity



          $$
          sum_{k=1}^nk^3=left(sum_{k=1}^nkright)^2
          $$
          So,
          $$
          sum_{k=1}^{99}k^3equivleft(sum_{k=1}^{99}kright)^2equivfrac{(99)^2(100)^2}{4}equiv0 (mod3)
          $$






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Yet another method is to use that $(x-1)^3=x^3-3x^2+3x-1$ and $(x+1)^3=x^3+3x^2+3x+1$, so $(x-1)^3+x^3+(x+1)^3 = 3x^3+6x$, which is divisible by three. This shows that any three consecutive numbers, when cubed, add up to zero modulus three.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              @Accumulation A small issue is that in the second equation, $left(x - 1right)^3$ should be $left(x + 1right)^3$ instead.
              $endgroup$
              – John Omielan
              Dec 27 '18 at 6:18





















            1












            $begingroup$

            From $a^{3} + b^{3} = (a+b)(a^{2}-ab+b^{2})$, you have $a+b|a^{3} +b^{3}$ and
            begin{align}
            &1^{3} + 2^{3} + 3^{3} + cdots + 97^{3} + 98^{3} + 99^{3} \
            &= (1^{3} + 2^{3}) + (4^{3} + 5^{3}) + cdots + (97^{3}+98^{3}) + (3^{3}+6^{3} + cdots + 99^{3})
            end{align}
            is a multiple of 3.






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              You can also calculate that sum using



              $sum_{i=0}^{i=n}i^3=dfrac{n^2(n+1)^2}{4}$



              Here: $sum_{i=0}^{i=99}i^3=dfrac{99^2(99+1)^2}{4}$ which is clearly divisible by 3.






              share|cite|improve this answer









              $endgroup$





















                1












                $begingroup$

                consider these observations:



                $$1 = 1 pmod 3$$
                $$2 = -1 pmod 3$$
                $$3 = 0 pmod 3$$



                This becomes:



                $$33(1^3+(-1)^3+0) equiv 33(1+(-1)+0) equiv 33(0) equiv 0 pmod 3$$






                share|cite|improve this answer









                $endgroup$













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                  9 Answers
                  9






                  active

                  oldest

                  votes








                  9 Answers
                  9






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes









                  3












                  $begingroup$

                  By Fermat's little theorem $a^3equiv apmod3$. Then you have $sum_{k=1}^{99} k=frac{(99)(100)}2=50cdot 99equiv 0pmod3$.






                  share|cite|improve this answer









                  $endgroup$


















                    3












                    $begingroup$

                    By Fermat's little theorem $a^3equiv apmod3$. Then you have $sum_{k=1}^{99} k=frac{(99)(100)}2=50cdot 99equiv 0pmod3$.






                    share|cite|improve this answer









                    $endgroup$
















                      3












                      3








                      3





                      $begingroup$

                      By Fermat's little theorem $a^3equiv apmod3$. Then you have $sum_{k=1}^{99} k=frac{(99)(100)}2=50cdot 99equiv 0pmod3$.






                      share|cite|improve this answer









                      $endgroup$



                      By Fermat's little theorem $a^3equiv apmod3$. Then you have $sum_{k=1}^{99} k=frac{(99)(100)}2=50cdot 99equiv 0pmod3$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 27 '18 at 5:46









                      Chris CusterChris Custer

                      13.8k3827




                      13.8k3827























                          2












                          $begingroup$

                          Hint:



                          $n^3equiv npmod3$ as $n^3-n=n(n-1)(n+1)$ is the product of three consecutive integers



                          We have $sum_{r=1}^mr^3equivsum_{r=1}^mrpmod3$



                          $equiv dfrac{m(m+1)}2$



                          Alternatively $$sum_{r=1}^m r^3=dfrac{m^2(m+1)^2}4$$






                          share|cite|improve this answer









                          $endgroup$


















                            2












                            $begingroup$

                            Hint:



                            $n^3equiv npmod3$ as $n^3-n=n(n-1)(n+1)$ is the product of three consecutive integers



                            We have $sum_{r=1}^mr^3equivsum_{r=1}^mrpmod3$



                            $equiv dfrac{m(m+1)}2$



                            Alternatively $$sum_{r=1}^m r^3=dfrac{m^2(m+1)^2}4$$






                            share|cite|improve this answer









                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              Hint:



                              $n^3equiv npmod3$ as $n^3-n=n(n-1)(n+1)$ is the product of three consecutive integers



                              We have $sum_{r=1}^mr^3equivsum_{r=1}^mrpmod3$



                              $equiv dfrac{m(m+1)}2$



                              Alternatively $$sum_{r=1}^m r^3=dfrac{m^2(m+1)^2}4$$






                              share|cite|improve this answer









                              $endgroup$



                              Hint:



                              $n^3equiv npmod3$ as $n^3-n=n(n-1)(n+1)$ is the product of three consecutive integers



                              We have $sum_{r=1}^mr^3equivsum_{r=1}^mrpmod3$



                              $equiv dfrac{m(m+1)}2$



                              Alternatively $$sum_{r=1}^m r^3=dfrac{m^2(m+1)^2}4$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 27 '18 at 5:44









                              lab bhattacharjeelab bhattacharjee

                              226k15157275




                              226k15157275























                                  2












                                  $begingroup$

                                  begin{align}
                                  sum_{i=1}^{99}i^3 &= sum_{i=0}^{32} (3i+1)^3 + sum_{i=0}^{32} (3i+2)^3 + sum_{i=0}^{32} (3i+3)^3 \
                                  &equiv sum_{i=0}^{32} 1^3 + sum_{i=0}^{32} 2^3 + sum_{i=0}^{32} 0^3 pmod{3}\
                                  &equiv sum_{i=0}^{32} 1^3 + sum_{i=0}^{32} (-1)^3 + sum_{i=0}^{32} 0^3 pmod{3}\
                                  &equiv 0 pmod{3}
                                  end{align}






                                  share|cite|improve this answer









                                  $endgroup$


















                                    2












                                    $begingroup$

                                    begin{align}
                                    sum_{i=1}^{99}i^3 &= sum_{i=0}^{32} (3i+1)^3 + sum_{i=0}^{32} (3i+2)^3 + sum_{i=0}^{32} (3i+3)^3 \
                                    &equiv sum_{i=0}^{32} 1^3 + sum_{i=0}^{32} 2^3 + sum_{i=0}^{32} 0^3 pmod{3}\
                                    &equiv sum_{i=0}^{32} 1^3 + sum_{i=0}^{32} (-1)^3 + sum_{i=0}^{32} 0^3 pmod{3}\
                                    &equiv 0 pmod{3}
                                    end{align}






                                    share|cite|improve this answer









                                    $endgroup$
















                                      2












                                      2








                                      2





                                      $begingroup$

                                      begin{align}
                                      sum_{i=1}^{99}i^3 &= sum_{i=0}^{32} (3i+1)^3 + sum_{i=0}^{32} (3i+2)^3 + sum_{i=0}^{32} (3i+3)^3 \
                                      &equiv sum_{i=0}^{32} 1^3 + sum_{i=0}^{32} 2^3 + sum_{i=0}^{32} 0^3 pmod{3}\
                                      &equiv sum_{i=0}^{32} 1^3 + sum_{i=0}^{32} (-1)^3 + sum_{i=0}^{32} 0^3 pmod{3}\
                                      &equiv 0 pmod{3}
                                      end{align}






                                      share|cite|improve this answer









                                      $endgroup$



                                      begin{align}
                                      sum_{i=1}^{99}i^3 &= sum_{i=0}^{32} (3i+1)^3 + sum_{i=0}^{32} (3i+2)^3 + sum_{i=0}^{32} (3i+3)^3 \
                                      &equiv sum_{i=0}^{32} 1^3 + sum_{i=0}^{32} 2^3 + sum_{i=0}^{32} 0^3 pmod{3}\
                                      &equiv sum_{i=0}^{32} 1^3 + sum_{i=0}^{32} (-1)^3 + sum_{i=0}^{32} 0^3 pmod{3}\
                                      &equiv 0 pmod{3}
                                      end{align}







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 27 '18 at 5:48









                                      Siong Thye GohSiong Thye Goh

                                      102k1466118




                                      102k1466118























                                          2












                                          $begingroup$

                                          Split the numbers from $1,ldots, 99$ in





                                          • $33$ with remainder $1$


                                          • $33$ with remainder $2$


                                          • $33$ with remainder $0$
                                            $$1^3 + 2^3 + cdots + 99^3 equiv 33cdot (1^3+2^3+0^3) equiv 0 mod 3$$






                                          share|cite|improve this answer









                                          $endgroup$


















                                            2












                                            $begingroup$

                                            Split the numbers from $1,ldots, 99$ in





                                            • $33$ with remainder $1$


                                            • $33$ with remainder $2$


                                            • $33$ with remainder $0$
                                              $$1^3 + 2^3 + cdots + 99^3 equiv 33cdot (1^3+2^3+0^3) equiv 0 mod 3$$






                                            share|cite|improve this answer









                                            $endgroup$
















                                              2












                                              2








                                              2





                                              $begingroup$

                                              Split the numbers from $1,ldots, 99$ in





                                              • $33$ with remainder $1$


                                              • $33$ with remainder $2$


                                              • $33$ with remainder $0$
                                                $$1^3 + 2^3 + cdots + 99^3 equiv 33cdot (1^3+2^3+0^3) equiv 0 mod 3$$






                                              share|cite|improve this answer









                                              $endgroup$



                                              Split the numbers from $1,ldots, 99$ in





                                              • $33$ with remainder $1$


                                              • $33$ with remainder $2$


                                              • $33$ with remainder $0$
                                                $$1^3 + 2^3 + cdots + 99^3 equiv 33cdot (1^3+2^3+0^3) equiv 0 mod 3$$







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered Dec 27 '18 at 5:51









                                              trancelocationtrancelocation

                                              12.3k1826




                                              12.3k1826























                                                  2












                                                  $begingroup$

                                                  Another method is to use the identity



                                                  $$
                                                  sum_{k=1}^nk^3=left(sum_{k=1}^nkright)^2
                                                  $$
                                                  So,
                                                  $$
                                                  sum_{k=1}^{99}k^3equivleft(sum_{k=1}^{99}kright)^2equivfrac{(99)^2(100)^2}{4}equiv0 (mod3)
                                                  $$






                                                  share|cite|improve this answer









                                                  $endgroup$


















                                                    2












                                                    $begingroup$

                                                    Another method is to use the identity



                                                    $$
                                                    sum_{k=1}^nk^3=left(sum_{k=1}^nkright)^2
                                                    $$
                                                    So,
                                                    $$
                                                    sum_{k=1}^{99}k^3equivleft(sum_{k=1}^{99}kright)^2equivfrac{(99)^2(100)^2}{4}equiv0 (mod3)
                                                    $$






                                                    share|cite|improve this answer









                                                    $endgroup$
















                                                      2












                                                      2








                                                      2





                                                      $begingroup$

                                                      Another method is to use the identity



                                                      $$
                                                      sum_{k=1}^nk^3=left(sum_{k=1}^nkright)^2
                                                      $$
                                                      So,
                                                      $$
                                                      sum_{k=1}^{99}k^3equivleft(sum_{k=1}^{99}kright)^2equivfrac{(99)^2(100)^2}{4}equiv0 (mod3)
                                                      $$






                                                      share|cite|improve this answer









                                                      $endgroup$



                                                      Another method is to use the identity



                                                      $$
                                                      sum_{k=1}^nk^3=left(sum_{k=1}^nkright)^2
                                                      $$
                                                      So,
                                                      $$
                                                      sum_{k=1}^{99}k^3equivleft(sum_{k=1}^{99}kright)^2equivfrac{(99)^2(100)^2}{4}equiv0 (mod3)
                                                      $$







                                                      share|cite|improve this answer












                                                      share|cite|improve this answer



                                                      share|cite|improve this answer










                                                      answered Dec 27 '18 at 6:18









                                                      GödelGödel

                                                      1,440319




                                                      1,440319























                                                          2












                                                          $begingroup$

                                                          Yet another method is to use that $(x-1)^3=x^3-3x^2+3x-1$ and $(x+1)^3=x^3+3x^2+3x+1$, so $(x-1)^3+x^3+(x+1)^3 = 3x^3+6x$, which is divisible by three. This shows that any three consecutive numbers, when cubed, add up to zero modulus three.






                                                          share|cite|improve this answer











                                                          $endgroup$









                                                          • 1




                                                            $begingroup$
                                                            @Accumulation A small issue is that in the second equation, $left(x - 1right)^3$ should be $left(x + 1right)^3$ instead.
                                                            $endgroup$
                                                            – John Omielan
                                                            Dec 27 '18 at 6:18


















                                                          2












                                                          $begingroup$

                                                          Yet another method is to use that $(x-1)^3=x^3-3x^2+3x-1$ and $(x+1)^3=x^3+3x^2+3x+1$, so $(x-1)^3+x^3+(x+1)^3 = 3x^3+6x$, which is divisible by three. This shows that any three consecutive numbers, when cubed, add up to zero modulus three.






                                                          share|cite|improve this answer











                                                          $endgroup$









                                                          • 1




                                                            $begingroup$
                                                            @Accumulation A small issue is that in the second equation, $left(x - 1right)^3$ should be $left(x + 1right)^3$ instead.
                                                            $endgroup$
                                                            – John Omielan
                                                            Dec 27 '18 at 6:18
















                                                          2












                                                          2








                                                          2





                                                          $begingroup$

                                                          Yet another method is to use that $(x-1)^3=x^3-3x^2+3x-1$ and $(x+1)^3=x^3+3x^2+3x+1$, so $(x-1)^3+x^3+(x+1)^3 = 3x^3+6x$, which is divisible by three. This shows that any three consecutive numbers, when cubed, add up to zero modulus three.






                                                          share|cite|improve this answer











                                                          $endgroup$



                                                          Yet another method is to use that $(x-1)^3=x^3-3x^2+3x-1$ and $(x+1)^3=x^3+3x^2+3x+1$, so $(x-1)^3+x^3+(x+1)^3 = 3x^3+6x$, which is divisible by three. This shows that any three consecutive numbers, when cubed, add up to zero modulus three.







                                                          share|cite|improve this answer














                                                          share|cite|improve this answer



                                                          share|cite|improve this answer








                                                          edited Dec 27 '18 at 6:19









                                                          abiessu

                                                          6,68721541




                                                          6,68721541










                                                          answered Dec 27 '18 at 6:08









                                                          AcccumulationAcccumulation

                                                          7,0192619




                                                          7,0192619








                                                          • 1




                                                            $begingroup$
                                                            @Accumulation A small issue is that in the second equation, $left(x - 1right)^3$ should be $left(x + 1right)^3$ instead.
                                                            $endgroup$
                                                            – John Omielan
                                                            Dec 27 '18 at 6:18
















                                                          • 1




                                                            $begingroup$
                                                            @Accumulation A small issue is that in the second equation, $left(x - 1right)^3$ should be $left(x + 1right)^3$ instead.
                                                            $endgroup$
                                                            – John Omielan
                                                            Dec 27 '18 at 6:18










                                                          1




                                                          1




                                                          $begingroup$
                                                          @Accumulation A small issue is that in the second equation, $left(x - 1right)^3$ should be $left(x + 1right)^3$ instead.
                                                          $endgroup$
                                                          – John Omielan
                                                          Dec 27 '18 at 6:18






                                                          $begingroup$
                                                          @Accumulation A small issue is that in the second equation, $left(x - 1right)^3$ should be $left(x + 1right)^3$ instead.
                                                          $endgroup$
                                                          – John Omielan
                                                          Dec 27 '18 at 6:18













                                                          1












                                                          $begingroup$

                                                          From $a^{3} + b^{3} = (a+b)(a^{2}-ab+b^{2})$, you have $a+b|a^{3} +b^{3}$ and
                                                          begin{align}
                                                          &1^{3} + 2^{3} + 3^{3} + cdots + 97^{3} + 98^{3} + 99^{3} \
                                                          &= (1^{3} + 2^{3}) + (4^{3} + 5^{3}) + cdots + (97^{3}+98^{3}) + (3^{3}+6^{3} + cdots + 99^{3})
                                                          end{align}
                                                          is a multiple of 3.






                                                          share|cite|improve this answer









                                                          $endgroup$


















                                                            1












                                                            $begingroup$

                                                            From $a^{3} + b^{3} = (a+b)(a^{2}-ab+b^{2})$, you have $a+b|a^{3} +b^{3}$ and
                                                            begin{align}
                                                            &1^{3} + 2^{3} + 3^{3} + cdots + 97^{3} + 98^{3} + 99^{3} \
                                                            &= (1^{3} + 2^{3}) + (4^{3} + 5^{3}) + cdots + (97^{3}+98^{3}) + (3^{3}+6^{3} + cdots + 99^{3})
                                                            end{align}
                                                            is a multiple of 3.






                                                            share|cite|improve this answer









                                                            $endgroup$
















                                                              1












                                                              1








                                                              1





                                                              $begingroup$

                                                              From $a^{3} + b^{3} = (a+b)(a^{2}-ab+b^{2})$, you have $a+b|a^{3} +b^{3}$ and
                                                              begin{align}
                                                              &1^{3} + 2^{3} + 3^{3} + cdots + 97^{3} + 98^{3} + 99^{3} \
                                                              &= (1^{3} + 2^{3}) + (4^{3} + 5^{3}) + cdots + (97^{3}+98^{3}) + (3^{3}+6^{3} + cdots + 99^{3})
                                                              end{align}
                                                              is a multiple of 3.






                                                              share|cite|improve this answer









                                                              $endgroup$



                                                              From $a^{3} + b^{3} = (a+b)(a^{2}-ab+b^{2})$, you have $a+b|a^{3} +b^{3}$ and
                                                              begin{align}
                                                              &1^{3} + 2^{3} + 3^{3} + cdots + 97^{3} + 98^{3} + 99^{3} \
                                                              &= (1^{3} + 2^{3}) + (4^{3} + 5^{3}) + cdots + (97^{3}+98^{3}) + (3^{3}+6^{3} + cdots + 99^{3})
                                                              end{align}
                                                              is a multiple of 3.







                                                              share|cite|improve this answer












                                                              share|cite|improve this answer



                                                              share|cite|improve this answer










                                                              answered Dec 27 '18 at 5:49









                                                              Seewoo LeeSeewoo Lee

                                                              6,954927




                                                              6,954927























                                                                  1












                                                                  $begingroup$

                                                                  You can also calculate that sum using



                                                                  $sum_{i=0}^{i=n}i^3=dfrac{n^2(n+1)^2}{4}$



                                                                  Here: $sum_{i=0}^{i=99}i^3=dfrac{99^2(99+1)^2}{4}$ which is clearly divisible by 3.






                                                                  share|cite|improve this answer









                                                                  $endgroup$


















                                                                    1












                                                                    $begingroup$

                                                                    You can also calculate that sum using



                                                                    $sum_{i=0}^{i=n}i^3=dfrac{n^2(n+1)^2}{4}$



                                                                    Here: $sum_{i=0}^{i=99}i^3=dfrac{99^2(99+1)^2}{4}$ which is clearly divisible by 3.






                                                                    share|cite|improve this answer









                                                                    $endgroup$
















                                                                      1












                                                                      1








                                                                      1





                                                                      $begingroup$

                                                                      You can also calculate that sum using



                                                                      $sum_{i=0}^{i=n}i^3=dfrac{n^2(n+1)^2}{4}$



                                                                      Here: $sum_{i=0}^{i=99}i^3=dfrac{99^2(99+1)^2}{4}$ which is clearly divisible by 3.






                                                                      share|cite|improve this answer









                                                                      $endgroup$



                                                                      You can also calculate that sum using



                                                                      $sum_{i=0}^{i=n}i^3=dfrac{n^2(n+1)^2}{4}$



                                                                      Here: $sum_{i=0}^{i=99}i^3=dfrac{99^2(99+1)^2}{4}$ which is clearly divisible by 3.







                                                                      share|cite|improve this answer












                                                                      share|cite|improve this answer



                                                                      share|cite|improve this answer










                                                                      answered Dec 27 '18 at 7:46









                                                                      harshit54harshit54

                                                                      348113




                                                                      348113























                                                                          1












                                                                          $begingroup$

                                                                          consider these observations:



                                                                          $$1 = 1 pmod 3$$
                                                                          $$2 = -1 pmod 3$$
                                                                          $$3 = 0 pmod 3$$



                                                                          This becomes:



                                                                          $$33(1^3+(-1)^3+0) equiv 33(1+(-1)+0) equiv 33(0) equiv 0 pmod 3$$






                                                                          share|cite|improve this answer









                                                                          $endgroup$


















                                                                            1












                                                                            $begingroup$

                                                                            consider these observations:



                                                                            $$1 = 1 pmod 3$$
                                                                            $$2 = -1 pmod 3$$
                                                                            $$3 = 0 pmod 3$$



                                                                            This becomes:



                                                                            $$33(1^3+(-1)^3+0) equiv 33(1+(-1)+0) equiv 33(0) equiv 0 pmod 3$$






                                                                            share|cite|improve this answer









                                                                            $endgroup$
















                                                                              1












                                                                              1








                                                                              1





                                                                              $begingroup$

                                                                              consider these observations:



                                                                              $$1 = 1 pmod 3$$
                                                                              $$2 = -1 pmod 3$$
                                                                              $$3 = 0 pmod 3$$



                                                                              This becomes:



                                                                              $$33(1^3+(-1)^3+0) equiv 33(1+(-1)+0) equiv 33(0) equiv 0 pmod 3$$






                                                                              share|cite|improve this answer









                                                                              $endgroup$



                                                                              consider these observations:



                                                                              $$1 = 1 pmod 3$$
                                                                              $$2 = -1 pmod 3$$
                                                                              $$3 = 0 pmod 3$$



                                                                              This becomes:



                                                                              $$33(1^3+(-1)^3+0) equiv 33(1+(-1)+0) equiv 33(0) equiv 0 pmod 3$$







                                                                              share|cite|improve this answer












                                                                              share|cite|improve this answer



                                                                              share|cite|improve this answer










                                                                              answered Dec 27 '18 at 14:57









                                                                              Maged SaeedMaged Saeed

                                                                              8671417




                                                                              8671417






























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