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smallest n digit number which is multiple of 7


Finding the smallest baseA big “smallest” numberWhat's the smallest number with first digit 1 that triples when this digit is moved to the end?Find the smallest natural number $n$Determine the smallest prime which does not divide any five digit number…Smallest multiple containing only 1's and 0'sSmallest number starting with N divisble by every non zero digit of NBy which convention 04 (for example) is no valid 2 digit number?What is the smallest number $n>2$ of this kind?What's the smallest three digit number that satisfies the following system of congruences?













2












$begingroup$


I want to find the smallest n digit number which is a multiple of $7$. For example, when $n=1$ then $7$ is the answer, if $n=2$ then $14$ is the answer, if $n=3$ then $105$ is the answer. What will be the answer for $n$th term?






sequences-and-series elementary-number-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Are you familiar with modular arithmetic?
    $endgroup$
    – Arthur
    7 hours ago










  • $begingroup$
    yes i am familiar.
    $endgroup$
    – sr123
    7 hours ago










  • $begingroup$
    $!bmod 7!:, 10^{large n-1}!+k,equiv, 0iff k,equiv,ldots $
    $endgroup$
    – Bill Dubuque
    7 hours ago


















2












$begingroup$


I want to find the smallest n digit number which is a multiple of $7$. For example, when $n=1$ then $7$ is the answer, if $n=2$ then $14$ is the answer, if $n=3$ then $105$ is the answer. What will be the answer for $n$th term?






sequences-and-series elementary-number-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Are you familiar with modular arithmetic?
    $endgroup$
    – Arthur
    7 hours ago










  • $begingroup$
    yes i am familiar.
    $endgroup$
    – sr123
    7 hours ago










  • $begingroup$
    $!bmod 7!:, 10^{large n-1}!+k,equiv, 0iff k,equiv,ldots $
    $endgroup$
    – Bill Dubuque
    7 hours ago
















2












2








2


2



$begingroup$


I want to find the smallest n digit number which is a multiple of $7$. For example, when $n=1$ then $7$ is the answer, if $n=2$ then $14$ is the answer, if $n=3$ then $105$ is the answer. What will be the answer for $n$th term?






sequences-and-series elementary-number-theory







share|cite|improve this question











$endgroup$




I want to find the smallest n digit number which is a multiple of $7$. For example, when $n=1$ then $7$ is the answer, if $n=2$ then $14$ is the answer, if $n=3$ then $105$ is the answer. What will be the answer for $n$th term?






sequences-and-series elementary-number-theory




sequences-and-series elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 6 hours ago









tarit goswami

2,4761522




2,4761522










asked 7 hours ago









sr123sr123

254




254








  • 1




    $begingroup$
    Are you familiar with modular arithmetic?
    $endgroup$
    – Arthur
    7 hours ago










  • $begingroup$
    yes i am familiar.
    $endgroup$
    – sr123
    7 hours ago










  • $begingroup$
    $!bmod 7!:, 10^{large n-1}!+k,equiv, 0iff k,equiv,ldots $
    $endgroup$
    – Bill Dubuque
    7 hours ago
















  • 1




    $begingroup$
    Are you familiar with modular arithmetic?
    $endgroup$
    – Arthur
    7 hours ago










  • $begingroup$
    yes i am familiar.
    $endgroup$
    – sr123
    7 hours ago










  • $begingroup$
    $!bmod 7!:, 10^{large n-1}!+k,equiv, 0iff k,equiv,ldots $
    $endgroup$
    – Bill Dubuque
    7 hours ago










1




1




$begingroup$
Are you familiar with modular arithmetic?
$endgroup$
– Arthur
7 hours ago




$begingroup$
Are you familiar with modular arithmetic?
$endgroup$
– Arthur
7 hours ago












$begingroup$
yes i am familiar.
$endgroup$
– sr123
7 hours ago




$begingroup$
yes i am familiar.
$endgroup$
– sr123
7 hours ago












$begingroup$
$!bmod 7!:, 10^{large n-1}!+k,equiv, 0iff k,equiv,ldots $
$endgroup$
– Bill Dubuque
7 hours ago






$begingroup$
$!bmod 7!:, 10^{large n-1}!+k,equiv, 0iff k,equiv,ldots $
$endgroup$
– Bill Dubuque
7 hours ago












3 Answers
3






active

oldest

votes


















3












$begingroup$

The $n$ th term of the sequence will be $$10^{n-1}+big(7-(10^{n-1}pmod{7})big)$$



Because to get a $n$ digit number you will need at least $10^{n-1}$, now suppose the number is of the form $10^{n-1}+k$ for some natural $k$, for divisibility we need $(10^{n-1}+k)equiv 0 pmod{7}$ and to satisfy the minimality condition we need $kin [0,6].$ And $7-10^{n-1}pmod{7}in [0,6]$, hence the result:
$$a_n=10^{n-1}+big(7-(10^{n-1}pmod{7})big)$$






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Hint The function $k mapsto m leftlceil frac{k}{m} rightrceil$ maps $k$ to the smallest integer multiple of $m$ at least as large as $k$.




    For an integer $n > 0$ the smallest $n$-digit number is $10^{n - 1}$, so the the smallest $n$-digit number that is a multiple of $7$ is $$7leftlceilfrac{10^{n - 1}}{7}rightrceil.$$ The sequence begins $$7, 14, 105, 1001, 10003, ldots .$$ Since $10^{n - 1} equiv 3^{n - 1} pmod 7$ and $3$ has order $6$ in $Bbb Z_7^times$, (except for the first term, $7$) the smallest $n$-digit number that is a multiple of $7$ is $10^{n - 1} + r(n)$, where $r(n)$ depends only on the value of $n pmod 6$, i.e., the trailing digits $4, 5, 1, 3, ldots$ repeat every six terms.







    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      Solve



      $$(10^n+c)bmod 7=0$$ for the smallest positive $c$.



      You have



      $$(10^n+c)bmod7=(10^nbmod 7+cbmod 7)bmod 7=0$$



      so that $$c=7-10^nbmod 7.$$






      share|cite|improve this answer











      $endgroup$














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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        The $n$ th term of the sequence will be $$10^{n-1}+big(7-(10^{n-1}pmod{7})big)$$



        Because to get a $n$ digit number you will need at least $10^{n-1}$, now suppose the number is of the form $10^{n-1}+k$ for some natural $k$, for divisibility we need $(10^{n-1}+k)equiv 0 pmod{7}$ and to satisfy the minimality condition we need $kin [0,6].$ And $7-10^{n-1}pmod{7}in [0,6]$, hence the result:
        $$a_n=10^{n-1}+big(7-(10^{n-1}pmod{7})big)$$






        share|cite|improve this answer











        $endgroup$


















          3












          $begingroup$

          The $n$ th term of the sequence will be $$10^{n-1}+big(7-(10^{n-1}pmod{7})big)$$



          Because to get a $n$ digit number you will need at least $10^{n-1}$, now suppose the number is of the form $10^{n-1}+k$ for some natural $k$, for divisibility we need $(10^{n-1}+k)equiv 0 pmod{7}$ and to satisfy the minimality condition we need $kin [0,6].$ And $7-10^{n-1}pmod{7}in [0,6]$, hence the result:
          $$a_n=10^{n-1}+big(7-(10^{n-1}pmod{7})big)$$






          share|cite|improve this answer











          $endgroup$
















            3












            3








            3





            $begingroup$

            The $n$ th term of the sequence will be $$10^{n-1}+big(7-(10^{n-1}pmod{7})big)$$



            Because to get a $n$ digit number you will need at least $10^{n-1}$, now suppose the number is of the form $10^{n-1}+k$ for some natural $k$, for divisibility we need $(10^{n-1}+k)equiv 0 pmod{7}$ and to satisfy the minimality condition we need $kin [0,6].$ And $7-10^{n-1}pmod{7}in [0,6]$, hence the result:
            $$a_n=10^{n-1}+big(7-(10^{n-1}pmod{7})big)$$






            share|cite|improve this answer











            $endgroup$



            The $n$ th term of the sequence will be $$10^{n-1}+big(7-(10^{n-1}pmod{7})big)$$



            Because to get a $n$ digit number you will need at least $10^{n-1}$, now suppose the number is of the form $10^{n-1}+k$ for some natural $k$, for divisibility we need $(10^{n-1}+k)equiv 0 pmod{7}$ and to satisfy the minimality condition we need $kin [0,6].$ And $7-10^{n-1}pmod{7}in [0,6]$, hence the result:
            $$a_n=10^{n-1}+big(7-(10^{n-1}pmod{7})big)$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 6 hours ago

























            answered 7 hours ago









            tarit goswamitarit goswami

            2,4761522




            2,4761522























                2












                $begingroup$

                Hint The function $k mapsto m leftlceil frac{k}{m} rightrceil$ maps $k$ to the smallest integer multiple of $m$ at least as large as $k$.




                For an integer $n > 0$ the smallest $n$-digit number is $10^{n - 1}$, so the the smallest $n$-digit number that is a multiple of $7$ is $$7leftlceilfrac{10^{n - 1}}{7}rightrceil.$$ The sequence begins $$7, 14, 105, 1001, 10003, ldots .$$ Since $10^{n - 1} equiv 3^{n - 1} pmod 7$ and $3$ has order $6$ in $Bbb Z_7^times$, (except for the first term, $7$) the smallest $n$-digit number that is a multiple of $7$ is $10^{n - 1} + r(n)$, where $r(n)$ depends only on the value of $n pmod 6$, i.e., the trailing digits $4, 5, 1, 3, ldots$ repeat every six terms.







                share|cite|improve this answer











                $endgroup$


















                  2












                  $begingroup$

                  Hint The function $k mapsto m leftlceil frac{k}{m} rightrceil$ maps $k$ to the smallest integer multiple of $m$ at least as large as $k$.




                  For an integer $n > 0$ the smallest $n$-digit number is $10^{n - 1}$, so the the smallest $n$-digit number that is a multiple of $7$ is $$7leftlceilfrac{10^{n - 1}}{7}rightrceil.$$ The sequence begins $$7, 14, 105, 1001, 10003, ldots .$$ Since $10^{n - 1} equiv 3^{n - 1} pmod 7$ and $3$ has order $6$ in $Bbb Z_7^times$, (except for the first term, $7$) the smallest $n$-digit number that is a multiple of $7$ is $10^{n - 1} + r(n)$, where $r(n)$ depends only on the value of $n pmod 6$, i.e., the trailing digits $4, 5, 1, 3, ldots$ repeat every six terms.







                  share|cite|improve this answer











                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Hint The function $k mapsto m leftlceil frac{k}{m} rightrceil$ maps $k$ to the smallest integer multiple of $m$ at least as large as $k$.




                    For an integer $n > 0$ the smallest $n$-digit number is $10^{n - 1}$, so the the smallest $n$-digit number that is a multiple of $7$ is $$7leftlceilfrac{10^{n - 1}}{7}rightrceil.$$ The sequence begins $$7, 14, 105, 1001, 10003, ldots .$$ Since $10^{n - 1} equiv 3^{n - 1} pmod 7$ and $3$ has order $6$ in $Bbb Z_7^times$, (except for the first term, $7$) the smallest $n$-digit number that is a multiple of $7$ is $10^{n - 1} + r(n)$, where $r(n)$ depends only on the value of $n pmod 6$, i.e., the trailing digits $4, 5, 1, 3, ldots$ repeat every six terms.







                    share|cite|improve this answer











                    $endgroup$



                    Hint The function $k mapsto m leftlceil frac{k}{m} rightrceil$ maps $k$ to the smallest integer multiple of $m$ at least as large as $k$.




                    For an integer $n > 0$ the smallest $n$-digit number is $10^{n - 1}$, so the the smallest $n$-digit number that is a multiple of $7$ is $$7leftlceilfrac{10^{n - 1}}{7}rightrceil.$$ The sequence begins $$7, 14, 105, 1001, 10003, ldots .$$ Since $10^{n - 1} equiv 3^{n - 1} pmod 7$ and $3$ has order $6$ in $Bbb Z_7^times$, (except for the first term, $7$) the smallest $n$-digit number that is a multiple of $7$ is $10^{n - 1} + r(n)$, where $r(n)$ depends only on the value of $n pmod 6$, i.e., the trailing digits $4, 5, 1, 3, ldots$ repeat every six terms.








                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 5 hours ago

























                    answered 7 hours ago









                    TravisTravis

                    64.9k769152




                    64.9k769152























                        1












                        $begingroup$

                        Solve



                        $$(10^n+c)bmod 7=0$$ for the smallest positive $c$.



                        You have



                        $$(10^n+c)bmod7=(10^nbmod 7+cbmod 7)bmod 7=0$$



                        so that $$c=7-10^nbmod 7.$$






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          Solve



                          $$(10^n+c)bmod 7=0$$ for the smallest positive $c$.



                          You have



                          $$(10^n+c)bmod7=(10^nbmod 7+cbmod 7)bmod 7=0$$



                          so that $$c=7-10^nbmod 7.$$






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Solve



                            $$(10^n+c)bmod 7=0$$ for the smallest positive $c$.



                            You have



                            $$(10^n+c)bmod7=(10^nbmod 7+cbmod 7)bmod 7=0$$



                            so that $$c=7-10^nbmod 7.$$






                            share|cite|improve this answer











                            $endgroup$



                            Solve



                            $$(10^n+c)bmod 7=0$$ for the smallest positive $c$.



                            You have



                            $$(10^n+c)bmod7=(10^nbmod 7+cbmod 7)bmod 7=0$$



                            so that $$c=7-10^nbmod 7.$$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 6 hours ago

























                            answered 7 hours ago









                            Yves DaoustYves Daoust

                            134k676233




                            134k676233






























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