Find some digits of factorial 17












7












$begingroup$


$17!$ is equal to $$35568x428096y00$$
Both x and y, are digits. Find x,y.



So, $$17!=2^{15}*3^6*5^3*7^2*11*13*17=(2^3*5^3)*2^{12}*3^6*7^2*11*13*17$$
If there`s a product of $(2*5)^3$



Then this number has $3$ zeros at the end, so $y=0$



How do I find the $x$ now?










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    7












    $begingroup$


    $17!$ is equal to $$35568x428096y00$$
    Both x and y, are digits. Find x,y.



    So, $$17!=2^{15}*3^6*5^3*7^2*11*13*17=(2^3*5^3)*2^{12}*3^6*7^2*11*13*17$$
    If there`s a product of $(2*5)^3$



    Then this number has $3$ zeros at the end, so $y=0$



    How do I find the $x$ now?










    share|cite|improve this question









    New contributor




    a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      7












      7








      7





      $begingroup$


      $17!$ is equal to $$35568x428096y00$$
      Both x and y, are digits. Find x,y.



      So, $$17!=2^{15}*3^6*5^3*7^2*11*13*17=(2^3*5^3)*2^{12}*3^6*7^2*11*13*17$$
      If there`s a product of $(2*5)^3$



      Then this number has $3$ zeros at the end, so $y=0$



      How do I find the $x$ now?










      share|cite|improve this question









      New contributor




      a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      $17!$ is equal to $$35568x428096y00$$
      Both x and y, are digits. Find x,y.



      So, $$17!=2^{15}*3^6*5^3*7^2*11*13*17=(2^3*5^3)*2^{12}*3^6*7^2*11*13*17$$
      If there`s a product of $(2*5)^3$



      Then this number has $3$ zeros at the end, so $y=0$



      How do I find the $x$ now?







      elementary-number-theory factorial






      share|cite|improve this question









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      share|cite|improve this question









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      share|cite|improve this question




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      edited 3 hours ago









      J. W. Tanner

      2,6271217




      2,6271217






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      asked 3 hours ago









      a_man_with_no_namea_man_with_no_name

      823




      823




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          2 Answers
          2






          active

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          13












          $begingroup$

          HINT $17!$ is divisible by $9$. What is an easy test for divisibility by 9, involving the digits of a number?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            The sum od its digits Has to be divisible by 9
            $endgroup$
            – a_man_with_no_name
            3 hours ago










          • $begingroup$
            How could i miss it!
            $endgroup$
            – a_man_with_no_name
            3 hours ago



















          5












          $begingroup$

          The alternating sum of digits must be divisible by $11$, i.e., $11mid 18-x$. It follows that $x=7$.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

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            2 Answers
            2






            active

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            active

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            active

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            13












            $begingroup$

            HINT $17!$ is divisible by $9$. What is an easy test for divisibility by 9, involving the digits of a number?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              The sum od its digits Has to be divisible by 9
              $endgroup$
              – a_man_with_no_name
              3 hours ago










            • $begingroup$
              How could i miss it!
              $endgroup$
              – a_man_with_no_name
              3 hours ago
















            13












            $begingroup$

            HINT $17!$ is divisible by $9$. What is an easy test for divisibility by 9, involving the digits of a number?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              The sum od its digits Has to be divisible by 9
              $endgroup$
              – a_man_with_no_name
              3 hours ago










            • $begingroup$
              How could i miss it!
              $endgroup$
              – a_man_with_no_name
              3 hours ago














            13












            13








            13





            $begingroup$

            HINT $17!$ is divisible by $9$. What is an easy test for divisibility by 9, involving the digits of a number?






            share|cite|improve this answer









            $endgroup$



            HINT $17!$ is divisible by $9$. What is an easy test for divisibility by 9, involving the digits of a number?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 3 hours ago









            Mark FischlerMark Fischler

            32.7k12252




            32.7k12252












            • $begingroup$
              The sum od its digits Has to be divisible by 9
              $endgroup$
              – a_man_with_no_name
              3 hours ago










            • $begingroup$
              How could i miss it!
              $endgroup$
              – a_man_with_no_name
              3 hours ago


















            • $begingroup$
              The sum od its digits Has to be divisible by 9
              $endgroup$
              – a_man_with_no_name
              3 hours ago










            • $begingroup$
              How could i miss it!
              $endgroup$
              – a_man_with_no_name
              3 hours ago
















            $begingroup$
            The sum od its digits Has to be divisible by 9
            $endgroup$
            – a_man_with_no_name
            3 hours ago




            $begingroup$
            The sum od its digits Has to be divisible by 9
            $endgroup$
            – a_man_with_no_name
            3 hours ago












            $begingroup$
            How could i miss it!
            $endgroup$
            – a_man_with_no_name
            3 hours ago




            $begingroup$
            How could i miss it!
            $endgroup$
            – a_man_with_no_name
            3 hours ago











            5












            $begingroup$

            The alternating sum of digits must be divisible by $11$, i.e., $11mid 18-x$. It follows that $x=7$.






            share|cite|improve this answer









            $endgroup$


















              5












              $begingroup$

              The alternating sum of digits must be divisible by $11$, i.e., $11mid 18-x$. It follows that $x=7$.






              share|cite|improve this answer









              $endgroup$
















                5












                5








                5





                $begingroup$

                The alternating sum of digits must be divisible by $11$, i.e., $11mid 18-x$. It follows that $x=7$.






                share|cite|improve this answer









                $endgroup$



                The alternating sum of digits must be divisible by $11$, i.e., $11mid 18-x$. It follows that $x=7$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 3 hours ago









                Dietrich BurdeDietrich Burde

                80k647103




                80k647103






















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