Being $2011$ prime, calculate 2009! divided by 2011












1












$begingroup$


I'm working in the following excercise:




Being $2011$ prime, calculate $2009!$ divided by $2011$




By Wilson's theorem I have:



$$2010! equiv -1 mod 2011$$
$$2009! * 2010 equiv -1 * 2010 mod 2011$$
$$2009! equiv -2010 mod 2011 $$
$$2009! equiv 1 mod 2011$$



Checking the answer is correct but I'm not sure about step two, is that correct? any help will be really appreciated.






number-theory modular-arithmetic







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$endgroup$








  • 1




    $begingroup$
    It's perfectly correct.
    $endgroup$
    – Bernard
    Jan 30 at 22:21






  • 1




    $begingroup$
    No. It's not $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.
    $endgroup$
    – fleablood
    Jan 30 at 22:23






  • 1




    $begingroup$
    It should be noted that 2011 is prime
    $endgroup$
    – J. W. Tanner
    Jan 30 at 22:41






  • 1




    $begingroup$
    Perhaps a lemma to wilsons thereom should be: For $p > 2$ a prime. $(p-1)! equiv -1 pmod p$ so $(p-2)! = -(p-2)!(-1) equiv -(p-2)!(p-1)=-(p-1)! equiv -(-1) equiv 1 pmod p$.
    $endgroup$
    – fleablood
    Jan 30 at 22:49
















1












$begingroup$


I'm working in the following excercise:




Being $2011$ prime, calculate $2009!$ divided by $2011$




By Wilson's theorem I have:



$$2010! equiv -1 mod 2011$$
$$2009! * 2010 equiv -1 * 2010 mod 2011$$
$$2009! equiv -2010 mod 2011 $$
$$2009! equiv 1 mod 2011$$



Checking the answer is correct but I'm not sure about step two, is that correct? any help will be really appreciated.






number-theory modular-arithmetic







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It's perfectly correct.
    $endgroup$
    – Bernard
    Jan 30 at 22:21






  • 1




    $begingroup$
    No. It's not $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.
    $endgroup$
    – fleablood
    Jan 30 at 22:23






  • 1




    $begingroup$
    It should be noted that 2011 is prime
    $endgroup$
    – J. W. Tanner
    Jan 30 at 22:41






  • 1




    $begingroup$
    Perhaps a lemma to wilsons thereom should be: For $p > 2$ a prime. $(p-1)! equiv -1 pmod p$ so $(p-2)! = -(p-2)!(-1) equiv -(p-2)!(p-1)=-(p-1)! equiv -(-1) equiv 1 pmod p$.
    $endgroup$
    – fleablood
    Jan 30 at 22:49














1












1








1


2



$begingroup$


I'm working in the following excercise:




Being $2011$ prime, calculate $2009!$ divided by $2011$




By Wilson's theorem I have:



$$2010! equiv -1 mod 2011$$
$$2009! * 2010 equiv -1 * 2010 mod 2011$$
$$2009! equiv -2010 mod 2011 $$
$$2009! equiv 1 mod 2011$$



Checking the answer is correct but I'm not sure about step two, is that correct? any help will be really appreciated.






number-theory modular-arithmetic







share|cite|improve this question











$endgroup$




I'm working in the following excercise:




Being $2011$ prime, calculate $2009!$ divided by $2011$




By Wilson's theorem I have:



$$2010! equiv -1 mod 2011$$
$$2009! * 2010 equiv -1 * 2010 mod 2011$$
$$2009! equiv -2010 mod 2011 $$
$$2009! equiv 1 mod 2011$$



Checking the answer is correct but I'm not sure about step two, is that correct? any help will be really appreciated.






number-theory modular-arithmetic




number-theory modular-arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 30 at 22:42







mraz

















asked Jan 30 at 22:17









mrazmraz

450110




450110








  • 1




    $begingroup$
    It's perfectly correct.
    $endgroup$
    – Bernard
    Jan 30 at 22:21






  • 1




    $begingroup$
    No. It's not $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.
    $endgroup$
    – fleablood
    Jan 30 at 22:23






  • 1




    $begingroup$
    It should be noted that 2011 is prime
    $endgroup$
    – J. W. Tanner
    Jan 30 at 22:41






  • 1




    $begingroup$
    Perhaps a lemma to wilsons thereom should be: For $p > 2$ a prime. $(p-1)! equiv -1 pmod p$ so $(p-2)! = -(p-2)!(-1) equiv -(p-2)!(p-1)=-(p-1)! equiv -(-1) equiv 1 pmod p$.
    $endgroup$
    – fleablood
    Jan 30 at 22:49














  • 1




    $begingroup$
    It's perfectly correct.
    $endgroup$
    – Bernard
    Jan 30 at 22:21






  • 1




    $begingroup$
    No. It's not $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.
    $endgroup$
    – fleablood
    Jan 30 at 22:23






  • 1




    $begingroup$
    It should be noted that 2011 is prime
    $endgroup$
    – J. W. Tanner
    Jan 30 at 22:41






  • 1




    $begingroup$
    Perhaps a lemma to wilsons thereom should be: For $p > 2$ a prime. $(p-1)! equiv -1 pmod p$ so $(p-2)! = -(p-2)!(-1) equiv -(p-2)!(p-1)=-(p-1)! equiv -(-1) equiv 1 pmod p$.
    $endgroup$
    – fleablood
    Jan 30 at 22:49








1




1




$begingroup$
It's perfectly correct.
$endgroup$
– Bernard
Jan 30 at 22:21




$begingroup$
It's perfectly correct.
$endgroup$
– Bernard
Jan 30 at 22:21




1




1




$begingroup$
No. It's not $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.
$endgroup$
– fleablood
Jan 30 at 22:23




$begingroup$
No. It's not $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.
$endgroup$
– fleablood
Jan 30 at 22:23




1




1




$begingroup$
It should be noted that 2011 is prime
$endgroup$
– J. W. Tanner
Jan 30 at 22:41




$begingroup$
It should be noted that 2011 is prime
$endgroup$
– J. W. Tanner
Jan 30 at 22:41




1




1




$begingroup$
Perhaps a lemma to wilsons thereom should be: For $p > 2$ a prime. $(p-1)! equiv -1 pmod p$ so $(p-2)! = -(p-2)!(-1) equiv -(p-2)!(p-1)=-(p-1)! equiv -(-1) equiv 1 pmod p$.
$endgroup$
– fleablood
Jan 30 at 22:49




$begingroup$
Perhaps a lemma to wilsons thereom should be: For $p > 2$ a prime. $(p-1)! equiv -1 pmod p$ so $(p-2)! = -(p-2)!(-1) equiv -(p-2)!(p-1)=-(p-1)! equiv -(-1) equiv 1 pmod p$.
$endgroup$
– fleablood
Jan 30 at 22:49










2 Answers
2






active

oldest

votes


















2












$begingroup$

No. It's not. $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.



And then when you had $2009!* 2010 equiv -2010$ you went to $2009! equiv -2010$ by simply dropping the $2010$ out of nowhere. You dropped it in for no reason. And then you dropped it out for no reason.



That won't work. Two wrongs frequently make a right and they did in this case. But they did so for the wrong reasons.



$$color{blue}{2010!} equiv color{blue}{-1}pmod{2011}$$



$$color{blue}{2009!*2010}equiv color{blue}{-1}*color{red}{2010} pmod{2011}$$ (This is wrong! The blue colors are all equivalent but the red $color{red}{2010}$ came from absolutely nowhere.)



$$color{orange}{2009!} equiv color{blue}{-}color{red}{2010}pmod{2011}$$



(This is wrong! The $color{blue}{2009!*2010}$ simply turned into $color{orange}{2009!}$ for no reason! What happened to the $color{blue}{2010}$? Where did it go?)



....



Instead do:



$2010! equiv -1 pmod {2011}$



$2009!*2010 = 2010! equiv -1 pmod {2011}$.



Now notice $2010 equiv -1 pmod {2011}$ so



$2009!*(-1) equiv - 1pmod {2011}$ and



$2009!*(-1)*(-1) equiv (-1)(-1) pmod {2011}$ and so



$2009! equiv 1 pmod {2011}$.






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    The right side on the second line should still be $-1$ because you just split the product on the left. Now note that $2010 equiv -1 pmod {2011}$ so multiply both sides by $2010$
    $$2009! * 2010 equiv -1 pmod {2011}\2009!cdot 2010^2equiv -1cdot 2010 pmod {2011}\
    2009!equiv 1 pmod {2011}$$
    You also just removed a factor $2010$ from the left between the second and third lines, which canceled out the error from the first to the second.






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      No. It's not. $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.



      And then when you had $2009!* 2010 equiv -2010$ you went to $2009! equiv -2010$ by simply dropping the $2010$ out of nowhere. You dropped it in for no reason. And then you dropped it out for no reason.



      That won't work. Two wrongs frequently make a right and they did in this case. But they did so for the wrong reasons.



      $$color{blue}{2010!} equiv color{blue}{-1}pmod{2011}$$



      $$color{blue}{2009!*2010}equiv color{blue}{-1}*color{red}{2010} pmod{2011}$$ (This is wrong! The blue colors are all equivalent but the red $color{red}{2010}$ came from absolutely nowhere.)



      $$color{orange}{2009!} equiv color{blue}{-}color{red}{2010}pmod{2011}$$



      (This is wrong! The $color{blue}{2009!*2010}$ simply turned into $color{orange}{2009!}$ for no reason! What happened to the $color{blue}{2010}$? Where did it go?)



      ....



      Instead do:



      $2010! equiv -1 pmod {2011}$



      $2009!*2010 = 2010! equiv -1 pmod {2011}$.



      Now notice $2010 equiv -1 pmod {2011}$ so



      $2009!*(-1) equiv - 1pmod {2011}$ and



      $2009!*(-1)*(-1) equiv (-1)(-1) pmod {2011}$ and so



      $2009! equiv 1 pmod {2011}$.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        No. It's not. $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.



        And then when you had $2009!* 2010 equiv -2010$ you went to $2009! equiv -2010$ by simply dropping the $2010$ out of nowhere. You dropped it in for no reason. And then you dropped it out for no reason.



        That won't work. Two wrongs frequently make a right and they did in this case. But they did so for the wrong reasons.



        $$color{blue}{2010!} equiv color{blue}{-1}pmod{2011}$$



        $$color{blue}{2009!*2010}equiv color{blue}{-1}*color{red}{2010} pmod{2011}$$ (This is wrong! The blue colors are all equivalent but the red $color{red}{2010}$ came from absolutely nowhere.)



        $$color{orange}{2009!} equiv color{blue}{-}color{red}{2010}pmod{2011}$$



        (This is wrong! The $color{blue}{2009!*2010}$ simply turned into $color{orange}{2009!}$ for no reason! What happened to the $color{blue}{2010}$? Where did it go?)



        ....



        Instead do:



        $2010! equiv -1 pmod {2011}$



        $2009!*2010 = 2010! equiv -1 pmod {2011}$.



        Now notice $2010 equiv -1 pmod {2011}$ so



        $2009!*(-1) equiv - 1pmod {2011}$ and



        $2009!*(-1)*(-1) equiv (-1)(-1) pmod {2011}$ and so



        $2009! equiv 1 pmod {2011}$.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          No. It's not. $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.



          And then when you had $2009!* 2010 equiv -2010$ you went to $2009! equiv -2010$ by simply dropping the $2010$ out of nowhere. You dropped it in for no reason. And then you dropped it out for no reason.



          That won't work. Two wrongs frequently make a right and they did in this case. But they did so for the wrong reasons.



          $$color{blue}{2010!} equiv color{blue}{-1}pmod{2011}$$



          $$color{blue}{2009!*2010}equiv color{blue}{-1}*color{red}{2010} pmod{2011}$$ (This is wrong! The blue colors are all equivalent but the red $color{red}{2010}$ came from absolutely nowhere.)



          $$color{orange}{2009!} equiv color{blue}{-}color{red}{2010}pmod{2011}$$



          (This is wrong! The $color{blue}{2009!*2010}$ simply turned into $color{orange}{2009!}$ for no reason! What happened to the $color{blue}{2010}$? Where did it go?)



          ....



          Instead do:



          $2010! equiv -1 pmod {2011}$



          $2009!*2010 = 2010! equiv -1 pmod {2011}$.



          Now notice $2010 equiv -1 pmod {2011}$ so



          $2009!*(-1) equiv - 1pmod {2011}$ and



          $2009!*(-1)*(-1) equiv (-1)(-1) pmod {2011}$ and so



          $2009! equiv 1 pmod {2011}$.






          share|cite|improve this answer











          $endgroup$



          No. It's not. $2009*2010 = 2010! equiv -1 pmod {2011}$. Why did you multiply it by $2010$ when you don't know $2009! not equiv -1 pmod {2011}$.



          And then when you had $2009!* 2010 equiv -2010$ you went to $2009! equiv -2010$ by simply dropping the $2010$ out of nowhere. You dropped it in for no reason. And then you dropped it out for no reason.



          That won't work. Two wrongs frequently make a right and they did in this case. But they did so for the wrong reasons.



          $$color{blue}{2010!} equiv color{blue}{-1}pmod{2011}$$



          $$color{blue}{2009!*2010}equiv color{blue}{-1}*color{red}{2010} pmod{2011}$$ (This is wrong! The blue colors are all equivalent but the red $color{red}{2010}$ came from absolutely nowhere.)



          $$color{orange}{2009!} equiv color{blue}{-}color{red}{2010}pmod{2011}$$



          (This is wrong! The $color{blue}{2009!*2010}$ simply turned into $color{orange}{2009!}$ for no reason! What happened to the $color{blue}{2010}$? Where did it go?)



          ....



          Instead do:



          $2010! equiv -1 pmod {2011}$



          $2009!*2010 = 2010! equiv -1 pmod {2011}$.



          Now notice $2010 equiv -1 pmod {2011}$ so



          $2009!*(-1) equiv - 1pmod {2011}$ and



          $2009!*(-1)*(-1) equiv (-1)(-1) pmod {2011}$ and so



          $2009! equiv 1 pmod {2011}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 30 at 22:45

























          answered Jan 30 at 22:29









          fleabloodfleablood

          73.8k22891




          73.8k22891























              3












              $begingroup$

              The right side on the second line should still be $-1$ because you just split the product on the left. Now note that $2010 equiv -1 pmod {2011}$ so multiply both sides by $2010$
              $$2009! * 2010 equiv -1 pmod {2011}\2009!cdot 2010^2equiv -1cdot 2010 pmod {2011}\
              2009!equiv 1 pmod {2011}$$
              You also just removed a factor $2010$ from the left between the second and third lines, which canceled out the error from the first to the second.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                The right side on the second line should still be $-1$ because you just split the product on the left. Now note that $2010 equiv -1 pmod {2011}$ so multiply both sides by $2010$
                $$2009! * 2010 equiv -1 pmod {2011}\2009!cdot 2010^2equiv -1cdot 2010 pmod {2011}\
                2009!equiv 1 pmod {2011}$$
                You also just removed a factor $2010$ from the left between the second and third lines, which canceled out the error from the first to the second.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  The right side on the second line should still be $-1$ because you just split the product on the left. Now note that $2010 equiv -1 pmod {2011}$ so multiply both sides by $2010$
                  $$2009! * 2010 equiv -1 pmod {2011}\2009!cdot 2010^2equiv -1cdot 2010 pmod {2011}\
                  2009!equiv 1 pmod {2011}$$
                  You also just removed a factor $2010$ from the left between the second and third lines, which canceled out the error from the first to the second.






                  share|cite|improve this answer









                  $endgroup$



                  The right side on the second line should still be $-1$ because you just split the product on the left. Now note that $2010 equiv -1 pmod {2011}$ so multiply both sides by $2010$
                  $$2009! * 2010 equiv -1 pmod {2011}\2009!cdot 2010^2equiv -1cdot 2010 pmod {2011}\
                  2009!equiv 1 pmod {2011}$$
                  You also just removed a factor $2010$ from the left between the second and third lines, which canceled out the error from the first to the second.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 30 at 22:22









                  Ross MillikanRoss Millikan

                  301k24200375




                  301k24200375






























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