Find all three digit numbers which are divisible by groups of its digits












1












$begingroup$


How can I find all three-digit numbers which:




  • Do not contain a $0$ digit

  • Have different digits

  • Are divisible by below described groups of its own digits


The number passing first two conditions should be divisible by two-digit group of its own digits, which are made by omitting one of the number's digits.



For example:



number = $132$



It has only non-zero digits

It has different digits

And it should be divisible by $13$, $12$, and $32$. (omitting one digit)



Thanks a lot in advance for helping me finding these!






combinatorics divisibility decimal-expansion







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  • 1




    $begingroup$
    What have you tried?
    $endgroup$
    – Robert Z
    1 hour ago
















1












$begingroup$


How can I find all three-digit numbers which:




  • Do not contain a $0$ digit

  • Have different digits

  • Are divisible by below described groups of its own digits


The number passing first two conditions should be divisible by two-digit group of its own digits, which are made by omitting one of the number's digits.



For example:



number = $132$



It has only non-zero digits

It has different digits

And it should be divisible by $13$, $12$, and $32$. (omitting one digit)



Thanks a lot in advance for helping me finding these!






combinatorics divisibility decimal-expansion







share|cite|improve this question







New contributor




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







$endgroup$








  • 1




    $begingroup$
    What have you tried?
    $endgroup$
    – Robert Z
    1 hour ago














1












1








1





$begingroup$


How can I find all three-digit numbers which:




  • Do not contain a $0$ digit

  • Have different digits

  • Are divisible by below described groups of its own digits


The number passing first two conditions should be divisible by two-digit group of its own digits, which are made by omitting one of the number's digits.



For example:



number = $132$



It has only non-zero digits

It has different digits

And it should be divisible by $13$, $12$, and $32$. (omitting one digit)



Thanks a lot in advance for helping me finding these!






combinatorics divisibility decimal-expansion







share|cite|improve this question







New contributor




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







$endgroup$




How can I find all three-digit numbers which:




  • Do not contain a $0$ digit

  • Have different digits

  • Are divisible by below described groups of its own digits


The number passing first two conditions should be divisible by two-digit group of its own digits, which are made by omitting one of the number's digits.



For example:



number = $132$



It has only non-zero digits

It has different digits

And it should be divisible by $13$, $12$, and $32$. (omitting one digit)



Thanks a lot in advance for helping me finding these!






combinatorics divisibility decimal-expansion




combinatorics divisibility decimal-expansion






share|cite|improve this question







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Xxx Ddd is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 1 hour ago









Xxx DddXxx Ddd

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173




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New contributor





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  • 1




    $begingroup$
    What have you tried?
    $endgroup$
    – Robert Z
    1 hour ago














  • 1




    $begingroup$
    What have you tried?
    $endgroup$
    – Robert Z
    1 hour ago








1




1




$begingroup$
What have you tried?
$endgroup$
– Robert Z
1 hour ago




$begingroup$
What have you tried?
$endgroup$
– Robert Z
1 hour ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

It's actually never possible to find such numbers since for a three digit number $[abc]$
$$10a+b mid 100a+10b+ciff frac{100a+10b+c}{10a+b}in mathbb Z$$
However
$$frac{100a+10b+c}{10a+b}=frac{10·(10a+b)+c}{10a+b}=10+frac{c}{10a+b}notin mathbb Z$$
Which is the desired contradiction since




$$a,b,cin {ninmathbb N: 1≤n≤9}$$ Therefore $$c<10a+b$$and hence $$10a+bnmid c$$







share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    A number $$abc$$ formed by the non-zero digits $a,b,c$ can never be divisible by $$ab$$ formed by $a$ and $b$ because if we divide by this number, the residue is $c$ which is non-zero and smaller than the number $ab$.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      2












      $begingroup$

      It's actually never possible to find such numbers since for a three digit number $[abc]$
      $$10a+b mid 100a+10b+ciff frac{100a+10b+c}{10a+b}in mathbb Z$$
      However
      $$frac{100a+10b+c}{10a+b}=frac{10·(10a+b)+c}{10a+b}=10+frac{c}{10a+b}notin mathbb Z$$
      Which is the desired contradiction since




      $$a,b,cin {ninmathbb N: 1≤n≤9}$$ Therefore $$c<10a+b$$and hence $$10a+bnmid c$$







      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        It's actually never possible to find such numbers since for a three digit number $[abc]$
        $$10a+b mid 100a+10b+ciff frac{100a+10b+c}{10a+b}in mathbb Z$$
        However
        $$frac{100a+10b+c}{10a+b}=frac{10·(10a+b)+c}{10a+b}=10+frac{c}{10a+b}notin mathbb Z$$
        Which is the desired contradiction since




        $$a,b,cin {ninmathbb N: 1≤n≤9}$$ Therefore $$c<10a+b$$and hence $$10a+bnmid c$$







        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          It's actually never possible to find such numbers since for a three digit number $[abc]$
          $$10a+b mid 100a+10b+ciff frac{100a+10b+c}{10a+b}in mathbb Z$$
          However
          $$frac{100a+10b+c}{10a+b}=frac{10·(10a+b)+c}{10a+b}=10+frac{c}{10a+b}notin mathbb Z$$
          Which is the desired contradiction since




          $$a,b,cin {ninmathbb N: 1≤n≤9}$$ Therefore $$c<10a+b$$and hence $$10a+bnmid c$$







          share|cite|improve this answer









          $endgroup$



          It's actually never possible to find such numbers since for a three digit number $[abc]$
          $$10a+b mid 100a+10b+ciff frac{100a+10b+c}{10a+b}in mathbb Z$$
          However
          $$frac{100a+10b+c}{10a+b}=frac{10·(10a+b)+c}{10a+b}=10+frac{c}{10a+b}notin mathbb Z$$
          Which is the desired contradiction since




          $$a,b,cin {ninmathbb N: 1≤n≤9}$$ Therefore $$c<10a+b$$and hence $$10a+bnmid c$$








          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          Dr. MathvaDr. Mathva

          997316




          997316























              2












              $begingroup$

              A number $$abc$$ formed by the non-zero digits $a,b,c$ can never be divisible by $$ab$$ formed by $a$ and $b$ because if we divide by this number, the residue is $c$ which is non-zero and smaller than the number $ab$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                A number $$abc$$ formed by the non-zero digits $a,b,c$ can never be divisible by $$ab$$ formed by $a$ and $b$ because if we divide by this number, the residue is $c$ which is non-zero and smaller than the number $ab$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  A number $$abc$$ formed by the non-zero digits $a,b,c$ can never be divisible by $$ab$$ formed by $a$ and $b$ because if we divide by this number, the residue is $c$ which is non-zero and smaller than the number $ab$.






                  share|cite|improve this answer









                  $endgroup$



                  A number $$abc$$ formed by the non-zero digits $a,b,c$ can never be divisible by $$ab$$ formed by $a$ and $b$ because if we divide by this number, the residue is $c$ which is non-zero and smaller than the number $ab$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  PeterPeter

                  47k1039125




                  47k1039125






















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