Prove that $1982cdot 1983|underbrace{22 ldots2}_{1980}$












1














Prove that $1982cdot 1983|underbrace{22 ldots2}_{1980}$



Since $underbrace{22 ldots2}_{1980}=frac{2(10^{1980}-1)}{10-1}$ then I need to show that $9cdot 1982cdot 1983| 2(10^{1980}-1)$, or $9cdot 991cdot 1983|(10^{1980}-1)$ since $(9,991,1983)=1$ if I show that $9|(10^{1980}-1)$, $991|(10^{1980}-1)$, $1983|(10^{1980}-1)$ then I finish the task.



Since $10equiv 1 (mod{9})$ then $10^{1980}equiv 1(mod{9})$ so $10^{1980}-1equiv 0 (mod{9})$ so we show that $9|(10^{1980}-1)$.



In $991|(10^{1980}-1)$ we can use Euler's function since $(991,10)=1$ so $10^{varphi (991)}equiv 1 (mod 991)$ so $10^{990}equiv 1 (mod{9})$ then if I square $10^{2cdot 990}equiv 1 (mod{991})$ so $10^{1980}-1equiv 0 (mod{991})$.



Then $1983|(10^{1980}-1)$ $1983=3cdot 661$ $3|10^{1980}-1$ it is trivial, and we can you Euler's theorem then $10^{660}equiv 1(mod{661})$ and then $10^{3cdot 660}equiv 1 (mod{661})$ so $10^{1980}-1equiv 0 (mod{661})$.



But I just not so sure when you show something like this and number $x,y,z in mathbb N$ (x,y,z)=1 and if say (y,x)=3 then that not change anything in task, if we want to show that x|n y|n and z|n? If you understand what I ask, here $(9,1983)=3$, but $(9,1982,1983)=1$






discrete-mathematics modular-arithmetic divisibility







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  • 1




    You need a backslash before "underbrace".
    – Shaun
    Nov 23 at 6:58
















1














Prove that $1982cdot 1983|underbrace{22 ldots2}_{1980}$



Since $underbrace{22 ldots2}_{1980}=frac{2(10^{1980}-1)}{10-1}$ then I need to show that $9cdot 1982cdot 1983| 2(10^{1980}-1)$, or $9cdot 991cdot 1983|(10^{1980}-1)$ since $(9,991,1983)=1$ if I show that $9|(10^{1980}-1)$, $991|(10^{1980}-1)$, $1983|(10^{1980}-1)$ then I finish the task.



Since $10equiv 1 (mod{9})$ then $10^{1980}equiv 1(mod{9})$ so $10^{1980}-1equiv 0 (mod{9})$ so we show that $9|(10^{1980}-1)$.



In $991|(10^{1980}-1)$ we can use Euler's function since $(991,10)=1$ so $10^{varphi (991)}equiv 1 (mod 991)$ so $10^{990}equiv 1 (mod{9})$ then if I square $10^{2cdot 990}equiv 1 (mod{991})$ so $10^{1980}-1equiv 0 (mod{991})$.



Then $1983|(10^{1980}-1)$ $1983=3cdot 661$ $3|10^{1980}-1$ it is trivial, and we can you Euler's theorem then $10^{660}equiv 1(mod{661})$ and then $10^{3cdot 660}equiv 1 (mod{661})$ so $10^{1980}-1equiv 0 (mod{661})$.



But I just not so sure when you show something like this and number $x,y,z in mathbb N$ (x,y,z)=1 and if say (y,x)=3 then that not change anything in task, if we want to show that x|n y|n and z|n? If you understand what I ask, here $(9,1983)=3$, but $(9,1982,1983)=1$






discrete-mathematics modular-arithmetic divisibility







share|cite|improve this question




















  • 1




    You need a backslash before "underbrace".
    – Shaun
    Nov 23 at 6:58














1












1








1







Prove that $1982cdot 1983|underbrace{22 ldots2}_{1980}$



Since $underbrace{22 ldots2}_{1980}=frac{2(10^{1980}-1)}{10-1}$ then I need to show that $9cdot 1982cdot 1983| 2(10^{1980}-1)$, or $9cdot 991cdot 1983|(10^{1980}-1)$ since $(9,991,1983)=1$ if I show that $9|(10^{1980}-1)$, $991|(10^{1980}-1)$, $1983|(10^{1980}-1)$ then I finish the task.



Since $10equiv 1 (mod{9})$ then $10^{1980}equiv 1(mod{9})$ so $10^{1980}-1equiv 0 (mod{9})$ so we show that $9|(10^{1980}-1)$.



In $991|(10^{1980}-1)$ we can use Euler's function since $(991,10)=1$ so $10^{varphi (991)}equiv 1 (mod 991)$ so $10^{990}equiv 1 (mod{9})$ then if I square $10^{2cdot 990}equiv 1 (mod{991})$ so $10^{1980}-1equiv 0 (mod{991})$.



Then $1983|(10^{1980}-1)$ $1983=3cdot 661$ $3|10^{1980}-1$ it is trivial, and we can you Euler's theorem then $10^{660}equiv 1(mod{661})$ and then $10^{3cdot 660}equiv 1 (mod{661})$ so $10^{1980}-1equiv 0 (mod{661})$.



But I just not so sure when you show something like this and number $x,y,z in mathbb N$ (x,y,z)=1 and if say (y,x)=3 then that not change anything in task, if we want to show that x|n y|n and z|n? If you understand what I ask, here $(9,1983)=3$, but $(9,1982,1983)=1$






discrete-mathematics modular-arithmetic divisibility







share|cite|improve this question















Prove that $1982cdot 1983|underbrace{22 ldots2}_{1980}$



Since $underbrace{22 ldots2}_{1980}=frac{2(10^{1980}-1)}{10-1}$ then I need to show that $9cdot 1982cdot 1983| 2(10^{1980}-1)$, or $9cdot 991cdot 1983|(10^{1980}-1)$ since $(9,991,1983)=1$ if I show that $9|(10^{1980}-1)$, $991|(10^{1980}-1)$, $1983|(10^{1980}-1)$ then I finish the task.



Since $10equiv 1 (mod{9})$ then $10^{1980}equiv 1(mod{9})$ so $10^{1980}-1equiv 0 (mod{9})$ so we show that $9|(10^{1980}-1)$.



In $991|(10^{1980}-1)$ we can use Euler's function since $(991,10)=1$ so $10^{varphi (991)}equiv 1 (mod 991)$ so $10^{990}equiv 1 (mod{9})$ then if I square $10^{2cdot 990}equiv 1 (mod{991})$ so $10^{1980}-1equiv 0 (mod{991})$.



Then $1983|(10^{1980}-1)$ $1983=3cdot 661$ $3|10^{1980}-1$ it is trivial, and we can you Euler's theorem then $10^{660}equiv 1(mod{661})$ and then $10^{3cdot 660}equiv 1 (mod{661})$ so $10^{1980}-1equiv 0 (mod{661})$.



But I just not so sure when you show something like this and number $x,y,z in mathbb N$ (x,y,z)=1 and if say (y,x)=3 then that not change anything in task, if we want to show that x|n y|n and z|n? If you understand what I ask, here $(9,1983)=3$, but $(9,1982,1983)=1$






discrete-mathematics modular-arithmetic divisibility




discrete-mathematics modular-arithmetic divisibility






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edited Nov 23 at 7:04

























asked Nov 23 at 6:57









Marko Škorić

70310




70310








  • 1




    You need a backslash before "underbrace".
    – Shaun
    Nov 23 at 6:58














  • 1




    You need a backslash before "underbrace".
    – Shaun
    Nov 23 at 6:58








1




1




You need a backslash before "underbrace".
– Shaun
Nov 23 at 6:58




You need a backslash before "underbrace".
– Shaun
Nov 23 at 6:58










2 Answers
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While $gcd(9,1982,1983)=1$, you have $gcd(9,1983)ne 1$. You should rather factor
$ 9cdot 1982cdot 1983$ as $27cdot 661cdot 991$.






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    0














    $$1982cdot 1983|underbrace{22 ldots2}_{1980}$$
    $$2cdot991cdot 1983|underbrace{left(11 ldots1right)}_{1980}cdot2$$


    This implies that $gcd geq 2$ and therefore it is $1982cdot 1983|underbrace{22 ldots2}_{1980}$






    share|cite|improve this answer























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      2 Answers
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      2 Answers
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      While $gcd(9,1982,1983)=1$, you have $gcd(9,1983)ne 1$. You should rather factor
      $ 9cdot 1982cdot 1983$ as $27cdot 661cdot 991$.






      share|cite|improve this answer


























        2














        While $gcd(9,1982,1983)=1$, you have $gcd(9,1983)ne 1$. You should rather factor
        $ 9cdot 1982cdot 1983$ as $27cdot 661cdot 991$.






        share|cite|improve this answer
























          2












          2








          2






          While $gcd(9,1982,1983)=1$, you have $gcd(9,1983)ne 1$. You should rather factor
          $ 9cdot 1982cdot 1983$ as $27cdot 661cdot 991$.






          share|cite|improve this answer












          While $gcd(9,1982,1983)=1$, you have $gcd(9,1983)ne 1$. You should rather factor
          $ 9cdot 1982cdot 1983$ as $27cdot 661cdot 991$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 23 at 7:16









          Hagen von Eitzen

          275k21268495




          275k21268495























              0














              $$1982cdot 1983|underbrace{22 ldots2}_{1980}$$
              $$2cdot991cdot 1983|underbrace{left(11 ldots1right)}_{1980}cdot2$$


              This implies that $gcd geq 2$ and therefore it is $1982cdot 1983|underbrace{22 ldots2}_{1980}$






              share|cite|improve this answer




























                0














                $$1982cdot 1983|underbrace{22 ldots2}_{1980}$$
                $$2cdot991cdot 1983|underbrace{left(11 ldots1right)}_{1980}cdot2$$


                This implies that $gcd geq 2$ and therefore it is $1982cdot 1983|underbrace{22 ldots2}_{1980}$






                share|cite|improve this answer


























                  0












                  0








                  0






                  $$1982cdot 1983|underbrace{22 ldots2}_{1980}$$
                  $$2cdot991cdot 1983|underbrace{left(11 ldots1right)}_{1980}cdot2$$


                  This implies that $gcd geq 2$ and therefore it is $1982cdot 1983|underbrace{22 ldots2}_{1980}$






                  share|cite|improve this answer














                  $$1982cdot 1983|underbrace{22 ldots2}_{1980}$$
                  $$2cdot991cdot 1983|underbrace{left(11 ldots1right)}_{1980}cdot2$$


                  This implies that $gcd geq 2$ and therefore it is $1982cdot 1983|underbrace{22 ldots2}_{1980}$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 23 at 8:48

























                  answered Nov 23 at 7:37









                  Darío A. Gutiérrez

                  2,59141530




                  2,59141530






























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