Power tower last digits












1












$begingroup$


Can anyone solve the following problem of finding the 6th last digit from the right of the decimal representation of the following number:



$6^{6^{6^{6^{6^{6}}}}}$



Essentially it means reducing this modulo $10^6$ and supposedly Chinese Remainder Theorem should be used, but I have no idea how to solve this. Can anyone help?






modular-arithmetic problem-solving







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Can anyone solve the following problem of finding the 6th last digit from the right of the decimal representation of the following number:



    $6^{6^{6^{6^{6^{6}}}}}$



    Essentially it means reducing this modulo $10^6$ and supposedly Chinese Remainder Theorem should be used, but I have no idea how to solve this. Can anyone help?






    modular-arithmetic problem-solving







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Can anyone solve the following problem of finding the 6th last digit from the right of the decimal representation of the following number:



      $6^{6^{6^{6^{6^{6}}}}}$



      Essentially it means reducing this modulo $10^6$ and supposedly Chinese Remainder Theorem should be used, but I have no idea how to solve this. Can anyone help?






      modular-arithmetic problem-solving







      share|cite|improve this question









      $endgroup$




      Can anyone solve the following problem of finding the 6th last digit from the right of the decimal representation of the following number:



      $6^{6^{6^{6^{6^{6}}}}}$



      Essentially it means reducing this modulo $10^6$ and supposedly Chinese Remainder Theorem should be used, but I have no idea how to solve this. Can anyone help?






      modular-arithmetic problem-solving




      modular-arithmetic problem-solving






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 13 at 19:45









      Michael MuntaMichael Munta

      1048




      1048






















          1 Answer
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          $begingroup$

          We're interested in $6^x mod 10^6$; this is determined by $6^i mod 2^6$ and $6^i mod 5^6$. Mod $2^6$ it's easy: $6^x$ is divisible by $2^6$ if $x ge 6$. $6$ is coprime to $5^6$ with multiplicative order $3125 = 5^5$ mod $5^6$. In fact, the multiplicative order of $6$ mod $5^m$ seems to be $5^{m-1}$ (prove!). So
          $$eqalign{6 &equiv 1 mod 5cr 6^6 &equiv 6^1 = 6 mod 5^2cr 6^{6^6} &equiv 6^6 equiv 31 mod 5^3cr
          6^{6^{6^6}} &equiv 6^{31} equiv 531 mod 5^4cr
          6^{6^{6^{6^6}}} &equiv 6^{531}equiv 1156 mod 5^5cr
          6^{6^{6^{6^{6^6}}}} &equiv 6^{1156} equiv 4281 mod 5^6cr
          }$$
          and using Chinese Remainder, $$6^{6^{6^{6^{6^6}}}} equiv 238656 mod 10^6$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How did you do this mod $5^n$ without calculator? For example $6^{1156}$
            $endgroup$
            – Michael Munta
            Jan 13 at 21:14








          • 1




            $begingroup$
            I did it with a computer. It would be possible, though rather tedious, to do by hand using repeated squaring.
            $endgroup$
            – Robert Israel
            Jan 14 at 1:11










          • $begingroup$
            And this process of finding multiplicative order, does that have a name? What did you do there really?
            $endgroup$
            – Michael Munta
            Jan 14 at 6:35










          • $begingroup$
            Also I am interested in how did you reach the conclusion that $6^x$ is $0$ in mod $2^6$.
            $endgroup$
            – Michael Munta
            Jan 14 at 10:44










          • $begingroup$
            $6^x = 2^x cdot 3^x$ is divisible by $2^x$.
            $endgroup$
            – Robert Israel
            Jan 14 at 12:47













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          1 Answer
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          1 Answer
          1






          active

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          active

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          active

          oldest

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          2












          $begingroup$

          We're interested in $6^x mod 10^6$; this is determined by $6^i mod 2^6$ and $6^i mod 5^6$. Mod $2^6$ it's easy: $6^x$ is divisible by $2^6$ if $x ge 6$. $6$ is coprime to $5^6$ with multiplicative order $3125 = 5^5$ mod $5^6$. In fact, the multiplicative order of $6$ mod $5^m$ seems to be $5^{m-1}$ (prove!). So
          $$eqalign{6 &equiv 1 mod 5cr 6^6 &equiv 6^1 = 6 mod 5^2cr 6^{6^6} &equiv 6^6 equiv 31 mod 5^3cr
          6^{6^{6^6}} &equiv 6^{31} equiv 531 mod 5^4cr
          6^{6^{6^{6^6}}} &equiv 6^{531}equiv 1156 mod 5^5cr
          6^{6^{6^{6^{6^6}}}} &equiv 6^{1156} equiv 4281 mod 5^6cr
          }$$
          and using Chinese Remainder, $$6^{6^{6^{6^{6^6}}}} equiv 238656 mod 10^6$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How did you do this mod $5^n$ without calculator? For example $6^{1156}$
            $endgroup$
            – Michael Munta
            Jan 13 at 21:14








          • 1




            $begingroup$
            I did it with a computer. It would be possible, though rather tedious, to do by hand using repeated squaring.
            $endgroup$
            – Robert Israel
            Jan 14 at 1:11










          • $begingroup$
            And this process of finding multiplicative order, does that have a name? What did you do there really?
            $endgroup$
            – Michael Munta
            Jan 14 at 6:35










          • $begingroup$
            Also I am interested in how did you reach the conclusion that $6^x$ is $0$ in mod $2^6$.
            $endgroup$
            – Michael Munta
            Jan 14 at 10:44










          • $begingroup$
            $6^x = 2^x cdot 3^x$ is divisible by $2^x$.
            $endgroup$
            – Robert Israel
            Jan 14 at 12:47


















          2












          $begingroup$

          We're interested in $6^x mod 10^6$; this is determined by $6^i mod 2^6$ and $6^i mod 5^6$. Mod $2^6$ it's easy: $6^x$ is divisible by $2^6$ if $x ge 6$. $6$ is coprime to $5^6$ with multiplicative order $3125 = 5^5$ mod $5^6$. In fact, the multiplicative order of $6$ mod $5^m$ seems to be $5^{m-1}$ (prove!). So
          $$eqalign{6 &equiv 1 mod 5cr 6^6 &equiv 6^1 = 6 mod 5^2cr 6^{6^6} &equiv 6^6 equiv 31 mod 5^3cr
          6^{6^{6^6}} &equiv 6^{31} equiv 531 mod 5^4cr
          6^{6^{6^{6^6}}} &equiv 6^{531}equiv 1156 mod 5^5cr
          6^{6^{6^{6^{6^6}}}} &equiv 6^{1156} equiv 4281 mod 5^6cr
          }$$
          and using Chinese Remainder, $$6^{6^{6^{6^{6^6}}}} equiv 238656 mod 10^6$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How did you do this mod $5^n$ without calculator? For example $6^{1156}$
            $endgroup$
            – Michael Munta
            Jan 13 at 21:14








          • 1




            $begingroup$
            I did it with a computer. It would be possible, though rather tedious, to do by hand using repeated squaring.
            $endgroup$
            – Robert Israel
            Jan 14 at 1:11










          • $begingroup$
            And this process of finding multiplicative order, does that have a name? What did you do there really?
            $endgroup$
            – Michael Munta
            Jan 14 at 6:35










          • $begingroup$
            Also I am interested in how did you reach the conclusion that $6^x$ is $0$ in mod $2^6$.
            $endgroup$
            – Michael Munta
            Jan 14 at 10:44










          • $begingroup$
            $6^x = 2^x cdot 3^x$ is divisible by $2^x$.
            $endgroup$
            – Robert Israel
            Jan 14 at 12:47
















          2












          2








          2





          $begingroup$

          We're interested in $6^x mod 10^6$; this is determined by $6^i mod 2^6$ and $6^i mod 5^6$. Mod $2^6$ it's easy: $6^x$ is divisible by $2^6$ if $x ge 6$. $6$ is coprime to $5^6$ with multiplicative order $3125 = 5^5$ mod $5^6$. In fact, the multiplicative order of $6$ mod $5^m$ seems to be $5^{m-1}$ (prove!). So
          $$eqalign{6 &equiv 1 mod 5cr 6^6 &equiv 6^1 = 6 mod 5^2cr 6^{6^6} &equiv 6^6 equiv 31 mod 5^3cr
          6^{6^{6^6}} &equiv 6^{31} equiv 531 mod 5^4cr
          6^{6^{6^{6^6}}} &equiv 6^{531}equiv 1156 mod 5^5cr
          6^{6^{6^{6^{6^6}}}} &equiv 6^{1156} equiv 4281 mod 5^6cr
          }$$
          and using Chinese Remainder, $$6^{6^{6^{6^{6^6}}}} equiv 238656 mod 10^6$$






          share|cite|improve this answer









          $endgroup$



          We're interested in $6^x mod 10^6$; this is determined by $6^i mod 2^6$ and $6^i mod 5^6$. Mod $2^6$ it's easy: $6^x$ is divisible by $2^6$ if $x ge 6$. $6$ is coprime to $5^6$ with multiplicative order $3125 = 5^5$ mod $5^6$. In fact, the multiplicative order of $6$ mod $5^m$ seems to be $5^{m-1}$ (prove!). So
          $$eqalign{6 &equiv 1 mod 5cr 6^6 &equiv 6^1 = 6 mod 5^2cr 6^{6^6} &equiv 6^6 equiv 31 mod 5^3cr
          6^{6^{6^6}} &equiv 6^{31} equiv 531 mod 5^4cr
          6^{6^{6^{6^6}}} &equiv 6^{531}equiv 1156 mod 5^5cr
          6^{6^{6^{6^{6^6}}}} &equiv 6^{1156} equiv 4281 mod 5^6cr
          }$$
          and using Chinese Remainder, $$6^{6^{6^{6^{6^6}}}} equiv 238656 mod 10^6$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 13 at 21:01









          Robert IsraelRobert Israel

          323k23212466




          323k23212466












          • $begingroup$
            How did you do this mod $5^n$ without calculator? For example $6^{1156}$
            $endgroup$
            – Michael Munta
            Jan 13 at 21:14








          • 1




            $begingroup$
            I did it with a computer. It would be possible, though rather tedious, to do by hand using repeated squaring.
            $endgroup$
            – Robert Israel
            Jan 14 at 1:11










          • $begingroup$
            And this process of finding multiplicative order, does that have a name? What did you do there really?
            $endgroup$
            – Michael Munta
            Jan 14 at 6:35










          • $begingroup$
            Also I am interested in how did you reach the conclusion that $6^x$ is $0$ in mod $2^6$.
            $endgroup$
            – Michael Munta
            Jan 14 at 10:44










          • $begingroup$
            $6^x = 2^x cdot 3^x$ is divisible by $2^x$.
            $endgroup$
            – Robert Israel
            Jan 14 at 12:47




















          • $begingroup$
            How did you do this mod $5^n$ without calculator? For example $6^{1156}$
            $endgroup$
            – Michael Munta
            Jan 13 at 21:14








          • 1




            $begingroup$
            I did it with a computer. It would be possible, though rather tedious, to do by hand using repeated squaring.
            $endgroup$
            – Robert Israel
            Jan 14 at 1:11










          • $begingroup$
            And this process of finding multiplicative order, does that have a name? What did you do there really?
            $endgroup$
            – Michael Munta
            Jan 14 at 6:35










          • $begingroup$
            Also I am interested in how did you reach the conclusion that $6^x$ is $0$ in mod $2^6$.
            $endgroup$
            – Michael Munta
            Jan 14 at 10:44










          • $begingroup$
            $6^x = 2^x cdot 3^x$ is divisible by $2^x$.
            $endgroup$
            – Robert Israel
            Jan 14 at 12:47


















          $begingroup$
          How did you do this mod $5^n$ without calculator? For example $6^{1156}$
          $endgroup$
          – Michael Munta
          Jan 13 at 21:14






          $begingroup$
          How did you do this mod $5^n$ without calculator? For example $6^{1156}$
          $endgroup$
          – Michael Munta
          Jan 13 at 21:14






          1




          1




          $begingroup$
          I did it with a computer. It would be possible, though rather tedious, to do by hand using repeated squaring.
          $endgroup$
          – Robert Israel
          Jan 14 at 1:11




          $begingroup$
          I did it with a computer. It would be possible, though rather tedious, to do by hand using repeated squaring.
          $endgroup$
          – Robert Israel
          Jan 14 at 1:11












          $begingroup$
          And this process of finding multiplicative order, does that have a name? What did you do there really?
          $endgroup$
          – Michael Munta
          Jan 14 at 6:35




          $begingroup$
          And this process of finding multiplicative order, does that have a name? What did you do there really?
          $endgroup$
          – Michael Munta
          Jan 14 at 6:35












          $begingroup$
          Also I am interested in how did you reach the conclusion that $6^x$ is $0$ in mod $2^6$.
          $endgroup$
          – Michael Munta
          Jan 14 at 10:44




          $begingroup$
          Also I am interested in how did you reach the conclusion that $6^x$ is $0$ in mod $2^6$.
          $endgroup$
          – Michael Munta
          Jan 14 at 10:44












          $begingroup$
          $6^x = 2^x cdot 3^x$ is divisible by $2^x$.
          $endgroup$
          – Robert Israel
          Jan 14 at 12:47






          $begingroup$
          $6^x = 2^x cdot 3^x$ is divisible by $2^x$.
          $endgroup$
          – Robert Israel
          Jan 14 at 12:47




















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