Divisibility rule












2












$begingroup$


Example: $2^1$=2 --> $2mid2$



If a number has their last digit divisible by 2, than the number is divisible by 2



$2^2$=4--> $4mid2$, $4mid4$



If a number has their last two digit divisible by 4, than the number is divisible by 4



$2^3$=8--> $8mid2$, $8mid4$ , $8mid8$



If a number has their last three digit divisible by 8, than the number is divisible by 8



and so on...




How would you word this into a conjecture?







divisibility exponentiation conjectures







share|cite|improve this question











$endgroup$












  • $begingroup$
    do you mean word the conjecture?, or the question title
    $endgroup$
    – Saketh Malyala
    Oct 27 '18 at 19:04












  • $begingroup$
    @SakethMalyala you are right. I worded that wrong.
    $endgroup$
    – DeNel
    Oct 27 '18 at 19:06






  • 2




    $begingroup$
    I mean, you can just say that for any natural $k$, if the last $k$ digits of a number are divisible by $2^k$, so is the whole number.
    $endgroup$
    – Don Thousand
    Oct 27 '18 at 19:07
















2












$begingroup$


Example: $2^1$=2 --> $2mid2$



If a number has their last digit divisible by 2, than the number is divisible by 2



$2^2$=4--> $4mid2$, $4mid4$



If a number has their last two digit divisible by 4, than the number is divisible by 4



$2^3$=8--> $8mid2$, $8mid4$ , $8mid8$



If a number has their last three digit divisible by 8, than the number is divisible by 8



and so on...




How would you word this into a conjecture?







divisibility exponentiation conjectures







share|cite|improve this question











$endgroup$












  • $begingroup$
    do you mean word the conjecture?, or the question title
    $endgroup$
    – Saketh Malyala
    Oct 27 '18 at 19:04












  • $begingroup$
    @SakethMalyala you are right. I worded that wrong.
    $endgroup$
    – DeNel
    Oct 27 '18 at 19:06






  • 2




    $begingroup$
    I mean, you can just say that for any natural $k$, if the last $k$ digits of a number are divisible by $2^k$, so is the whole number.
    $endgroup$
    – Don Thousand
    Oct 27 '18 at 19:07














2












2








2





$begingroup$


Example: $2^1$=2 --> $2mid2$



If a number has their last digit divisible by 2, than the number is divisible by 2



$2^2$=4--> $4mid2$, $4mid4$



If a number has their last two digit divisible by 4, than the number is divisible by 4



$2^3$=8--> $8mid2$, $8mid4$ , $8mid8$



If a number has their last three digit divisible by 8, than the number is divisible by 8



and so on...




How would you word this into a conjecture?







divisibility exponentiation conjectures







share|cite|improve this question











$endgroup$




Example: $2^1$=2 --> $2mid2$



If a number has their last digit divisible by 2, than the number is divisible by 2



$2^2$=4--> $4mid2$, $4mid4$



If a number has their last two digit divisible by 4, than the number is divisible by 4



$2^3$=8--> $8mid2$, $8mid4$ , $8mid8$



If a number has their last three digit divisible by 8, than the number is divisible by 8



and so on...




How would you word this into a conjecture?







divisibility exponentiation conjectures




divisibility exponentiation conjectures






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 28 '18 at 8:35







DeNel

















asked Oct 27 '18 at 19:03









DeNelDeNel

616




616












  • $begingroup$
    do you mean word the conjecture?, or the question title
    $endgroup$
    – Saketh Malyala
    Oct 27 '18 at 19:04












  • $begingroup$
    @SakethMalyala you are right. I worded that wrong.
    $endgroup$
    – DeNel
    Oct 27 '18 at 19:06






  • 2




    $begingroup$
    I mean, you can just say that for any natural $k$, if the last $k$ digits of a number are divisible by $2^k$, so is the whole number.
    $endgroup$
    – Don Thousand
    Oct 27 '18 at 19:07


















  • $begingroup$
    do you mean word the conjecture?, or the question title
    $endgroup$
    – Saketh Malyala
    Oct 27 '18 at 19:04












  • $begingroup$
    @SakethMalyala you are right. I worded that wrong.
    $endgroup$
    – DeNel
    Oct 27 '18 at 19:06






  • 2




    $begingroup$
    I mean, you can just say that for any natural $k$, if the last $k$ digits of a number are divisible by $2^k$, so is the whole number.
    $endgroup$
    – Don Thousand
    Oct 27 '18 at 19:07
















$begingroup$
do you mean word the conjecture?, or the question title
$endgroup$
– Saketh Malyala
Oct 27 '18 at 19:04






$begingroup$
do you mean word the conjecture?, or the question title
$endgroup$
– Saketh Malyala
Oct 27 '18 at 19:04














$begingroup$
@SakethMalyala you are right. I worded that wrong.
$endgroup$
– DeNel
Oct 27 '18 at 19:06




$begingroup$
@SakethMalyala you are right. I worded that wrong.
$endgroup$
– DeNel
Oct 27 '18 at 19:06




2




2




$begingroup$
I mean, you can just say that for any natural $k$, if the last $k$ digits of a number are divisible by $2^k$, so is the whole number.
$endgroup$
– Don Thousand
Oct 27 '18 at 19:07




$begingroup$
I mean, you can just say that for any natural $k$, if the last $k$ digits of a number are divisible by $2^k$, so is the whole number.
$endgroup$
– Don Thousand
Oct 27 '18 at 19:07










3 Answers
3






active

oldest

votes


















1












$begingroup$

Your conjuncture is essentially a true statement. The formulation of the statement is in @Ethan Bolker's answer. Here is also a prove for the statement:




Each number having $n+1$ digits can be written as follows:



$$10^na_n + 10^{n-1}a_{n-1} + cdots+10a_1 + a_0 tag 1$$




  • Notice that we can extract $10$ from the first $n$ terms and the
    number will be $10m+a_0$ where $m = (10^{n-1}a_n + 10^{n-2}a_{n-1} +cdots+a_1)$. Now, if $2$ divides this number, $2$ shall divide
    its form, $10m+a_0$. However $2|10$, hence, it will divide $10m$.
    Therefore $a_0$ must be divisible by $2$ so that the number is
    divisible by $2$.


  • For the general case where $2^k$ divides $n$ if and only if $2^k$
    divides the last $k$ digits, extract $10^k$ from the first $n-k+1$
    terms, and follow the previous argument given that $2^k$ divides $10^k$ as $10^k = 2^ktimes5^k$.








share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Your conjecture is (essentially) correct. You should be able to fill in the blank in




    If a number has their last ?? digit divisible by ??, than the number
    is divisible by $2^k$.




    But you are using the "$|$" symbol incorrectly. It means "divides", not "is divisible by", so
    $$
    4 mid8
    $$
    but
    $$
    8 nmid 4
    $$     .



    Finally, the statements you write after the arrows are true when you turn them around, but they don't prove the conjecture. Were you supposed to do that?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      we have been using "$|$" meaning "factor of". That's how I was taught. Maybe I was give the wrong information.
      $endgroup$
      – DeNel
      Oct 27 '18 at 19:54










    • $begingroup$
      @DeNel If that's what you were taught then the source of your information is out of sync with the standard usage.
      $endgroup$
      – Ethan Bolker
      Oct 27 '18 at 20:16










    • $begingroup$
      Well, $2$ is a factor of $10$, and $2|10$. I guess it is a matter of @DeNel’s application of the definition.
      $endgroup$
      – Lubin
      Oct 27 '18 at 21:44






    • 1




      $begingroup$
      @Lubin I hurried. THe reading "is a factor of" is correct, of course. But then $ 8 | 4$, as written, is obviously wrong.
      $endgroup$
      – Ethan Bolker
      Oct 28 '18 at 2:29



















    0












    $begingroup$

    Hint $ 2^kmid a!+!10^k biff 2^kmid a, $ by $, 2^kmid 10^k = 2^k 5^k$



    Better $, abmod 2^k = (underbrace{abmod 10^k}_{large {rm first} k {rm digits}})bmod{ 2^k}$



    Better $, aequiv bpmod{!10^k},Rightarrow, aequiv bpmod{!2^k}$



    Better $, aequiv bpmod{!mn},Rightarrow, aequiv bpmod{!n}, $ by $, nmid mnmid a-b$



    e.g. $bmod 1001!: color{#c00}{10^{large 3}}!equiv -1 ,Rightarrow, a=12,013,002equiv 12(color{#c00}{10^{large 3}})^{large 2}!+13(color{#c00}{10^{large 3}})+2equiv 12!-!13!+!2equiv 1$



    so $ 7!cdot!13=10^2!-!10!+!1mid 10^3!+!1,Rightarrow, a bmod 13 = (a bmod 1001)bmod 13 = 1bmod 13 = 1$.



    In congruence language: $ aequiv 1pmod{!13j!=!!10^3!+!1},Rightarrow, aequiv 1pmod{!13}$



    That's the idea behind one divisibility test for $13$.






    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









      1












      $begingroup$

      Your conjuncture is essentially a true statement. The formulation of the statement is in @Ethan Bolker's answer. Here is also a prove for the statement:




      Each number having $n+1$ digits can be written as follows:



      $$10^na_n + 10^{n-1}a_{n-1} + cdots+10a_1 + a_0 tag 1$$




      • Notice that we can extract $10$ from the first $n$ terms and the
        number will be $10m+a_0$ where $m = (10^{n-1}a_n + 10^{n-2}a_{n-1} +cdots+a_1)$. Now, if $2$ divides this number, $2$ shall divide
        its form, $10m+a_0$. However $2|10$, hence, it will divide $10m$.
        Therefore $a_0$ must be divisible by $2$ so that the number is
        divisible by $2$.


      • For the general case where $2^k$ divides $n$ if and only if $2^k$
        divides the last $k$ digits, extract $10^k$ from the first $n-k+1$
        terms, and follow the previous argument given that $2^k$ divides $10^k$ as $10^k = 2^ktimes5^k$.








      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        Your conjuncture is essentially a true statement. The formulation of the statement is in @Ethan Bolker's answer. Here is also a prove for the statement:




        Each number having $n+1$ digits can be written as follows:



        $$10^na_n + 10^{n-1}a_{n-1} + cdots+10a_1 + a_0 tag 1$$




        • Notice that we can extract $10$ from the first $n$ terms and the
          number will be $10m+a_0$ where $m = (10^{n-1}a_n + 10^{n-2}a_{n-1} +cdots+a_1)$. Now, if $2$ divides this number, $2$ shall divide
          its form, $10m+a_0$. However $2|10$, hence, it will divide $10m$.
          Therefore $a_0$ must be divisible by $2$ so that the number is
          divisible by $2$.


        • For the general case where $2^k$ divides $n$ if and only if $2^k$
          divides the last $k$ digits, extract $10^k$ from the first $n-k+1$
          terms, and follow the previous argument given that $2^k$ divides $10^k$ as $10^k = 2^ktimes5^k$.








        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          Your conjuncture is essentially a true statement. The formulation of the statement is in @Ethan Bolker's answer. Here is also a prove for the statement:




          Each number having $n+1$ digits can be written as follows:



          $$10^na_n + 10^{n-1}a_{n-1} + cdots+10a_1 + a_0 tag 1$$




          • Notice that we can extract $10$ from the first $n$ terms and the
            number will be $10m+a_0$ where $m = (10^{n-1}a_n + 10^{n-2}a_{n-1} +cdots+a_1)$. Now, if $2$ divides this number, $2$ shall divide
            its form, $10m+a_0$. However $2|10$, hence, it will divide $10m$.
            Therefore $a_0$ must be divisible by $2$ so that the number is
            divisible by $2$.


          • For the general case where $2^k$ divides $n$ if and only if $2^k$
            divides the last $k$ digits, extract $10^k$ from the first $n-k+1$
            terms, and follow the previous argument given that $2^k$ divides $10^k$ as $10^k = 2^ktimes5^k$.








          share|cite|improve this answer











          $endgroup$



          Your conjuncture is essentially a true statement. The formulation of the statement is in @Ethan Bolker's answer. Here is also a prove for the statement:




          Each number having $n+1$ digits can be written as follows:



          $$10^na_n + 10^{n-1}a_{n-1} + cdots+10a_1 + a_0 tag 1$$




          • Notice that we can extract $10$ from the first $n$ terms and the
            number will be $10m+a_0$ where $m = (10^{n-1}a_n + 10^{n-2}a_{n-1} +cdots+a_1)$. Now, if $2$ divides this number, $2$ shall divide
            its form, $10m+a_0$. However $2|10$, hence, it will divide $10m$.
            Therefore $a_0$ must be divisible by $2$ so that the number is
            divisible by $2$.


          • For the general case where $2^k$ divides $n$ if and only if $2^k$
            divides the last $k$ digits, extract $10^k$ from the first $n-k+1$
            terms, and follow the previous argument given that $2^k$ divides $10^k$ as $10^k = 2^ktimes5^k$.









          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Oct 27 '18 at 21:22

























          answered Oct 27 '18 at 19:36









          Maged SaeedMaged Saeed

          8471417




          8471417























              0












              $begingroup$

              Your conjecture is (essentially) correct. You should be able to fill in the blank in




              If a number has their last ?? digit divisible by ??, than the number
              is divisible by $2^k$.




              But you are using the "$|$" symbol incorrectly. It means "divides", not "is divisible by", so
              $$
              4 mid8
              $$
              but
              $$
              8 nmid 4
              $$               .



              Finally, the statements you write after the arrows are true when you turn them around, but they don't prove the conjecture. Were you supposed to do that?






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                we have been using "$|$" meaning "factor of". That's how I was taught. Maybe I was give the wrong information.
                $endgroup$
                – DeNel
                Oct 27 '18 at 19:54










              • $begingroup$
                @DeNel If that's what you were taught then the source of your information is out of sync with the standard usage.
                $endgroup$
                – Ethan Bolker
                Oct 27 '18 at 20:16










              • $begingroup$
                Well, $2$ is a factor of $10$, and $2|10$. I guess it is a matter of @DeNel’s application of the definition.
                $endgroup$
                – Lubin
                Oct 27 '18 at 21:44






              • 1




                $begingroup$
                @Lubin I hurried. THe reading "is a factor of" is correct, of course. But then $ 8 | 4$, as written, is obviously wrong.
                $endgroup$
                – Ethan Bolker
                Oct 28 '18 at 2:29
















              0












              $begingroup$

              Your conjecture is (essentially) correct. You should be able to fill in the blank in




              If a number has their last ?? digit divisible by ??, than the number
              is divisible by $2^k$.




              But you are using the "$|$" symbol incorrectly. It means "divides", not "is divisible by", so
              $$
              4 mid8
              $$
              but
              $$
              8 nmid 4
              $$               .



              Finally, the statements you write after the arrows are true when you turn them around, but they don't prove the conjecture. Were you supposed to do that?






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                we have been using "$|$" meaning "factor of". That's how I was taught. Maybe I was give the wrong information.
                $endgroup$
                – DeNel
                Oct 27 '18 at 19:54










              • $begingroup$
                @DeNel If that's what you were taught then the source of your information is out of sync with the standard usage.
                $endgroup$
                – Ethan Bolker
                Oct 27 '18 at 20:16










              • $begingroup$
                Well, $2$ is a factor of $10$, and $2|10$. I guess it is a matter of @DeNel’s application of the definition.
                $endgroup$
                – Lubin
                Oct 27 '18 at 21:44






              • 1




                $begingroup$
                @Lubin I hurried. THe reading "is a factor of" is correct, of course. But then $ 8 | 4$, as written, is obviously wrong.
                $endgroup$
                – Ethan Bolker
                Oct 28 '18 at 2:29














              0












              0








              0





              $begingroup$

              Your conjecture is (essentially) correct. You should be able to fill in the blank in




              If a number has their last ?? digit divisible by ??, than the number
              is divisible by $2^k$.




              But you are using the "$|$" symbol incorrectly. It means "divides", not "is divisible by", so
              $$
              4 mid8
              $$
              but
              $$
              8 nmid 4
              $$               .



              Finally, the statements you write after the arrows are true when you turn them around, but they don't prove the conjecture. Were you supposed to do that?






              share|cite|improve this answer











              $endgroup$



              Your conjecture is (essentially) correct. You should be able to fill in the blank in




              If a number has their last ?? digit divisible by ??, than the number
              is divisible by $2^k$.




              But you are using the "$|$" symbol incorrectly. It means "divides", not "is divisible by", so
              $$
              4 mid8
              $$
              but
              $$
              8 nmid 4
              $$               .



              Finally, the statements you write after the arrows are true when you turn them around, but they don't prove the conjecture. Were you supposed to do that?







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Oct 27 '18 at 19:09









              Arthur

              112k7109191




              112k7109191










              answered Oct 27 '18 at 19:08









              Ethan BolkerEthan Bolker

              42.2k548111




              42.2k548111












              • $begingroup$
                we have been using "$|$" meaning "factor of". That's how I was taught. Maybe I was give the wrong information.
                $endgroup$
                – DeNel
                Oct 27 '18 at 19:54










              • $begingroup$
                @DeNel If that's what you were taught then the source of your information is out of sync with the standard usage.
                $endgroup$
                – Ethan Bolker
                Oct 27 '18 at 20:16










              • $begingroup$
                Well, $2$ is a factor of $10$, and $2|10$. I guess it is a matter of @DeNel’s application of the definition.
                $endgroup$
                – Lubin
                Oct 27 '18 at 21:44






              • 1




                $begingroup$
                @Lubin I hurried. THe reading "is a factor of" is correct, of course. But then $ 8 | 4$, as written, is obviously wrong.
                $endgroup$
                – Ethan Bolker
                Oct 28 '18 at 2:29


















              • $begingroup$
                we have been using "$|$" meaning "factor of". That's how I was taught. Maybe I was give the wrong information.
                $endgroup$
                – DeNel
                Oct 27 '18 at 19:54










              • $begingroup$
                @DeNel If that's what you were taught then the source of your information is out of sync with the standard usage.
                $endgroup$
                – Ethan Bolker
                Oct 27 '18 at 20:16










              • $begingroup$
                Well, $2$ is a factor of $10$, and $2|10$. I guess it is a matter of @DeNel’s application of the definition.
                $endgroup$
                – Lubin
                Oct 27 '18 at 21:44






              • 1




                $begingroup$
                @Lubin I hurried. THe reading "is a factor of" is correct, of course. But then $ 8 | 4$, as written, is obviously wrong.
                $endgroup$
                – Ethan Bolker
                Oct 28 '18 at 2:29
















              $begingroup$
              we have been using "$|$" meaning "factor of". That's how I was taught. Maybe I was give the wrong information.
              $endgroup$
              – DeNel
              Oct 27 '18 at 19:54




              $begingroup$
              we have been using "$|$" meaning "factor of". That's how I was taught. Maybe I was give the wrong information.
              $endgroup$
              – DeNel
              Oct 27 '18 at 19:54












              $begingroup$
              @DeNel If that's what you were taught then the source of your information is out of sync with the standard usage.
              $endgroup$
              – Ethan Bolker
              Oct 27 '18 at 20:16




              $begingroup$
              @DeNel If that's what you were taught then the source of your information is out of sync with the standard usage.
              $endgroup$
              – Ethan Bolker
              Oct 27 '18 at 20:16












              $begingroup$
              Well, $2$ is a factor of $10$, and $2|10$. I guess it is a matter of @DeNel’s application of the definition.
              $endgroup$
              – Lubin
              Oct 27 '18 at 21:44




              $begingroup$
              Well, $2$ is a factor of $10$, and $2|10$. I guess it is a matter of @DeNel’s application of the definition.
              $endgroup$
              – Lubin
              Oct 27 '18 at 21:44




              1




              1




              $begingroup$
              @Lubin I hurried. THe reading "is a factor of" is correct, of course. But then $ 8 | 4$, as written, is obviously wrong.
              $endgroup$
              – Ethan Bolker
              Oct 28 '18 at 2:29




              $begingroup$
              @Lubin I hurried. THe reading "is a factor of" is correct, of course. But then $ 8 | 4$, as written, is obviously wrong.
              $endgroup$
              – Ethan Bolker
              Oct 28 '18 at 2:29











              0












              $begingroup$

              Hint $ 2^kmid a!+!10^k biff 2^kmid a, $ by $, 2^kmid 10^k = 2^k 5^k$



              Better $, abmod 2^k = (underbrace{abmod 10^k}_{large {rm first} k {rm digits}})bmod{ 2^k}$



              Better $, aequiv bpmod{!10^k},Rightarrow, aequiv bpmod{!2^k}$



              Better $, aequiv bpmod{!mn},Rightarrow, aequiv bpmod{!n}, $ by $, nmid mnmid a-b$



              e.g. $bmod 1001!: color{#c00}{10^{large 3}}!equiv -1 ,Rightarrow, a=12,013,002equiv 12(color{#c00}{10^{large 3}})^{large 2}!+13(color{#c00}{10^{large 3}})+2equiv 12!-!13!+!2equiv 1$



              so $ 7!cdot!13=10^2!-!10!+!1mid 10^3!+!1,Rightarrow, a bmod 13 = (a bmod 1001)bmod 13 = 1bmod 13 = 1$.



              In congruence language: $ aequiv 1pmod{!13j!=!!10^3!+!1},Rightarrow, aequiv 1pmod{!13}$



              That's the idea behind one divisibility test for $13$.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Hint $ 2^kmid a!+!10^k biff 2^kmid a, $ by $, 2^kmid 10^k = 2^k 5^k$



                Better $, abmod 2^k = (underbrace{abmod 10^k}_{large {rm first} k {rm digits}})bmod{ 2^k}$



                Better $, aequiv bpmod{!10^k},Rightarrow, aequiv bpmod{!2^k}$



                Better $, aequiv bpmod{!mn},Rightarrow, aequiv bpmod{!n}, $ by $, nmid mnmid a-b$



                e.g. $bmod 1001!: color{#c00}{10^{large 3}}!equiv -1 ,Rightarrow, a=12,013,002equiv 12(color{#c00}{10^{large 3}})^{large 2}!+13(color{#c00}{10^{large 3}})+2equiv 12!-!13!+!2equiv 1$



                so $ 7!cdot!13=10^2!-!10!+!1mid 10^3!+!1,Rightarrow, a bmod 13 = (a bmod 1001)bmod 13 = 1bmod 13 = 1$.



                In congruence language: $ aequiv 1pmod{!13j!=!!10^3!+!1},Rightarrow, aequiv 1pmod{!13}$



                That's the idea behind one divisibility test for $13$.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint $ 2^kmid a!+!10^k biff 2^kmid a, $ by $, 2^kmid 10^k = 2^k 5^k$



                  Better $, abmod 2^k = (underbrace{abmod 10^k}_{large {rm first} k {rm digits}})bmod{ 2^k}$



                  Better $, aequiv bpmod{!10^k},Rightarrow, aequiv bpmod{!2^k}$



                  Better $, aequiv bpmod{!mn},Rightarrow, aequiv bpmod{!n}, $ by $, nmid mnmid a-b$



                  e.g. $bmod 1001!: color{#c00}{10^{large 3}}!equiv -1 ,Rightarrow, a=12,013,002equiv 12(color{#c00}{10^{large 3}})^{large 2}!+13(color{#c00}{10^{large 3}})+2equiv 12!-!13!+!2equiv 1$



                  so $ 7!cdot!13=10^2!-!10!+!1mid 10^3!+!1,Rightarrow, a bmod 13 = (a bmod 1001)bmod 13 = 1bmod 13 = 1$.



                  In congruence language: $ aequiv 1pmod{!13j!=!!10^3!+!1},Rightarrow, aequiv 1pmod{!13}$



                  That's the idea behind one divisibility test for $13$.






                  share|cite|improve this answer











                  $endgroup$



                  Hint $ 2^kmid a!+!10^k biff 2^kmid a, $ by $, 2^kmid 10^k = 2^k 5^k$



                  Better $, abmod 2^k = (underbrace{abmod 10^k}_{large {rm first} k {rm digits}})bmod{ 2^k}$



                  Better $, aequiv bpmod{!10^k},Rightarrow, aequiv bpmod{!2^k}$



                  Better $, aequiv bpmod{!mn},Rightarrow, aequiv bpmod{!n}, $ by $, nmid mnmid a-b$



                  e.g. $bmod 1001!: color{#c00}{10^{large 3}}!equiv -1 ,Rightarrow, a=12,013,002equiv 12(color{#c00}{10^{large 3}})^{large 2}!+13(color{#c00}{10^{large 3}})+2equiv 12!-!13!+!2equiv 1$



                  so $ 7!cdot!13=10^2!-!10!+!1mid 10^3!+!1,Rightarrow, a bmod 13 = (a bmod 1001)bmod 13 = 1bmod 13 = 1$.



                  In congruence language: $ aequiv 1pmod{!13j!=!!10^3!+!1},Rightarrow, aequiv 1pmod{!13}$



                  That's the idea behind one divisibility test for $13$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 25 '18 at 17:05

























                  answered Oct 27 '18 at 19:36









                  Bill DubuqueBill Dubuque

                  209k29191634




                  209k29191634






























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