How many 5-digit numbers satisfy this criterion?












1














How many $5$-digit numbers have the property that the sum of its digits equals the product of its digits?






permutations







share|cite|improve this question
























  • Examples seem to be rare. I found $11125$ and $11133$ though.
    – SmileyCraft
    1 hour ago












  • 11222 is one of it.
    – Heroic24
    1 hour ago










  • I am convinced these are all up to permutations. So there are $20+10+10=40$ examples in total.
    – SmileyCraft
    1 hour ago
















1














How many $5$-digit numbers have the property that the sum of its digits equals the product of its digits?






permutations







share|cite|improve this question
























  • Examples seem to be rare. I found $11125$ and $11133$ though.
    – SmileyCraft
    1 hour ago












  • 11222 is one of it.
    – Heroic24
    1 hour ago










  • I am convinced these are all up to permutations. So there are $20+10+10=40$ examples in total.
    – SmileyCraft
    1 hour ago














1












1








1







How many $5$-digit numbers have the property that the sum of its digits equals the product of its digits?






permutations







share|cite|improve this question















How many $5$-digit numbers have the property that the sum of its digits equals the product of its digits?






permutations




permutations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 57 mins ago









David G. Stork

9,93521232




9,93521232










asked 1 hour ago









Heroic24

1627




1627












  • Examples seem to be rare. I found $11125$ and $11133$ though.
    – SmileyCraft
    1 hour ago












  • 11222 is one of it.
    – Heroic24
    1 hour ago










  • I am convinced these are all up to permutations. So there are $20+10+10=40$ examples in total.
    – SmileyCraft
    1 hour ago


















  • Examples seem to be rare. I found $11125$ and $11133$ though.
    – SmileyCraft
    1 hour ago












  • 11222 is one of it.
    – Heroic24
    1 hour ago










  • I am convinced these are all up to permutations. So there are $20+10+10=40$ examples in total.
    – SmileyCraft
    1 hour ago
















Examples seem to be rare. I found $11125$ and $11133$ though.
– SmileyCraft
1 hour ago






Examples seem to be rare. I found $11125$ and $11133$ though.
– SmileyCraft
1 hour ago














11222 is one of it.
– Heroic24
1 hour ago




11222 is one of it.
– Heroic24
1 hour ago












I am convinced these are all up to permutations. So there are $20+10+10=40$ examples in total.
– SmileyCraft
1 hour ago




I am convinced these are all up to permutations. So there are $20+10+10=40$ examples in total.
– SmileyCraft
1 hour ago










2 Answers
2






active

oldest

votes


















4














We want to find natural number solutions to $$x_1+x_2+cdots +x_5 = x_1cdot x_2cdots x_5.$$
Assume that $$x_1leq x_2leq x_3leq x_4 leq x_5.$$
Since $x_1+cdots +x_5 = x_1cdots x_5 < 5x_5$ (as $x_1=x_2=cdots =x_5$ is impossible) we have that $x_1x_2x_3x_4 leq 4.$ This implies that $(x_1, x_2, x_3, x_4)$ is one of the sequences
$$(1, 1, 1, 1), (1, 1, 1, 2), (1, 1, 1, 3), (1, 1, 1, 4),(1, 1, 2, 2).$$
The sequences $(1, 1, 1, 1)$ and $(1, 1, 1, 4)$ are not good. There is no number
$x_5$ for such sequences. From the remaining sequences we obtain all the solutions
$$(1, 1, 1, 2, 5), (1, 1, 1, 3, 3) text{ and } (1, 1, 2, 2, 2).$$



This diophantine equation has an unsolved problem associated with it. If $a(n)$ is the number of solutions then $a(n)=1$ when $n=114,174text{ or }444$ where $n>100.$






share|cite|improve this answer























  • Oh, I should have mentioned that $n>100.$
    – Hello_World
    1 hour ago



















0














A simple Mathematica program finds the $40$ cases:



${11125, 11133, 11152, 11215, 11222, 11251, 11313, 11331, 11512,
11521, 12115, 12122, 12151, 12212, 12221, 12511, 13113, 13131, 13311,
15112, 15121, 15211, 21115, 21122, 21151, 21212, 21221, 21511, 22112,
22121, 22211, 25111, 31113, 31131, 31311, 33111, 51112, 51121, 51211,
52111}$




myList = Table[i, {i, 10000, 99999}];

Select[myList, 

 Times @@ RealDigits[#][[1]] == Total@RealDigits[#][[1]] &]





share|cite|improve this answer























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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    4














    We want to find natural number solutions to $$x_1+x_2+cdots +x_5 = x_1cdot x_2cdots x_5.$$
    Assume that $$x_1leq x_2leq x_3leq x_4 leq x_5.$$
    Since $x_1+cdots +x_5 = x_1cdots x_5 < 5x_5$ (as $x_1=x_2=cdots =x_5$ is impossible) we have that $x_1x_2x_3x_4 leq 4.$ This implies that $(x_1, x_2, x_3, x_4)$ is one of the sequences
    $$(1, 1, 1, 1), (1, 1, 1, 2), (1, 1, 1, 3), (1, 1, 1, 4),(1, 1, 2, 2).$$
    The sequences $(1, 1, 1, 1)$ and $(1, 1, 1, 4)$ are not good. There is no number
    $x_5$ for such sequences. From the remaining sequences we obtain all the solutions
    $$(1, 1, 1, 2, 5), (1, 1, 1, 3, 3) text{ and } (1, 1, 2, 2, 2).$$



    This diophantine equation has an unsolved problem associated with it. If $a(n)$ is the number of solutions then $a(n)=1$ when $n=114,174text{ or }444$ where $n>100.$






    share|cite|improve this answer























    • Oh, I should have mentioned that $n>100.$
      – Hello_World
      1 hour ago
















    4














    We want to find natural number solutions to $$x_1+x_2+cdots +x_5 = x_1cdot x_2cdots x_5.$$
    Assume that $$x_1leq x_2leq x_3leq x_4 leq x_5.$$
    Since $x_1+cdots +x_5 = x_1cdots x_5 < 5x_5$ (as $x_1=x_2=cdots =x_5$ is impossible) we have that $x_1x_2x_3x_4 leq 4.$ This implies that $(x_1, x_2, x_3, x_4)$ is one of the sequences
    $$(1, 1, 1, 1), (1, 1, 1, 2), (1, 1, 1, 3), (1, 1, 1, 4),(1, 1, 2, 2).$$
    The sequences $(1, 1, 1, 1)$ and $(1, 1, 1, 4)$ are not good. There is no number
    $x_5$ for such sequences. From the remaining sequences we obtain all the solutions
    $$(1, 1, 1, 2, 5), (1, 1, 1, 3, 3) text{ and } (1, 1, 2, 2, 2).$$



    This diophantine equation has an unsolved problem associated with it. If $a(n)$ is the number of solutions then $a(n)=1$ when $n=114,174text{ or }444$ where $n>100.$






    share|cite|improve this answer























    • Oh, I should have mentioned that $n>100.$
      – Hello_World
      1 hour ago














    4












    4








    4






    We want to find natural number solutions to $$x_1+x_2+cdots +x_5 = x_1cdot x_2cdots x_5.$$
    Assume that $$x_1leq x_2leq x_3leq x_4 leq x_5.$$
    Since $x_1+cdots +x_5 = x_1cdots x_5 < 5x_5$ (as $x_1=x_2=cdots =x_5$ is impossible) we have that $x_1x_2x_3x_4 leq 4.$ This implies that $(x_1, x_2, x_3, x_4)$ is one of the sequences
    $$(1, 1, 1, 1), (1, 1, 1, 2), (1, 1, 1, 3), (1, 1, 1, 4),(1, 1, 2, 2).$$
    The sequences $(1, 1, 1, 1)$ and $(1, 1, 1, 4)$ are not good. There is no number
    $x_5$ for such sequences. From the remaining sequences we obtain all the solutions
    $$(1, 1, 1, 2, 5), (1, 1, 1, 3, 3) text{ and } (1, 1, 2, 2, 2).$$



    This diophantine equation has an unsolved problem associated with it. If $a(n)$ is the number of solutions then $a(n)=1$ when $n=114,174text{ or }444$ where $n>100.$






    share|cite|improve this answer














    We want to find natural number solutions to $$x_1+x_2+cdots +x_5 = x_1cdot x_2cdots x_5.$$
    Assume that $$x_1leq x_2leq x_3leq x_4 leq x_5.$$
    Since $x_1+cdots +x_5 = x_1cdots x_5 < 5x_5$ (as $x_1=x_2=cdots =x_5$ is impossible) we have that $x_1x_2x_3x_4 leq 4.$ This implies that $(x_1, x_2, x_3, x_4)$ is one of the sequences
    $$(1, 1, 1, 1), (1, 1, 1, 2), (1, 1, 1, 3), (1, 1, 1, 4),(1, 1, 2, 2).$$
    The sequences $(1, 1, 1, 1)$ and $(1, 1, 1, 4)$ are not good. There is no number
    $x_5$ for such sequences. From the remaining sequences we obtain all the solutions
    $$(1, 1, 1, 2, 5), (1, 1, 1, 3, 3) text{ and } (1, 1, 2, 2, 2).$$



    This diophantine equation has an unsolved problem associated with it. If $a(n)$ is the number of solutions then $a(n)=1$ when $n=114,174text{ or }444$ where $n>100.$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 1 hour ago

























    answered 1 hour ago









    Hello_World

    4,02821630




    4,02821630












    • Oh, I should have mentioned that $n>100.$
      – Hello_World
      1 hour ago


















    • Oh, I should have mentioned that $n>100.$
      – Hello_World
      1 hour ago
















    Oh, I should have mentioned that $n>100.$
    – Hello_World
    1 hour ago




    Oh, I should have mentioned that $n>100.$
    – Hello_World
    1 hour ago











    0














    A simple Mathematica program finds the $40$ cases:



    ${11125, 11133, 11152, 11215, 11222, 11251, 11313, 11331, 11512,
    11521, 12115, 12122, 12151, 12212, 12221, 12511, 13113, 13131, 13311,
    15112, 15121, 15211, 21115, 21122, 21151, 21212, 21221, 21511, 22112,
    22121, 22211, 25111, 31113, 31131, 31311, 33111, 51112, 51121, 51211,
    52111}$




    myList = Table[i, {i, 10000, 99999}];
    
Select[myList, 
    
 Times @@ RealDigits[#][[1]] == Total@RealDigits[#][[1]] &]





    share|cite|improve this answer




























      0














      A simple Mathematica program finds the $40$ cases:



      ${11125, 11133, 11152, 11215, 11222, 11251, 11313, 11331, 11512,
      11521, 12115, 12122, 12151, 12212, 12221, 12511, 13113, 13131, 13311,
      15112, 15121, 15211, 21115, 21122, 21151, 21212, 21221, 21511, 22112,
      22121, 22211, 25111, 31113, 31131, 31311, 33111, 51112, 51121, 51211,
      52111}$




      myList = Table[i, {i, 10000, 99999}];
      
Select[myList, 
      
 Times @@ RealDigits[#][[1]] == Total@RealDigits[#][[1]] &]





      share|cite|improve this answer


























        0












        0








        0






        A simple Mathematica program finds the $40$ cases:



        ${11125, 11133, 11152, 11215, 11222, 11251, 11313, 11331, 11512,
        11521, 12115, 12122, 12151, 12212, 12221, 12511, 13113, 13131, 13311,
        15112, 15121, 15211, 21115, 21122, 21151, 21212, 21221, 21511, 22112,
        22121, 22211, 25111, 31113, 31131, 31311, 33111, 51112, 51121, 51211,
        52111}$




        myList = Table[i, {i, 10000, 99999}];
        
Select[myList, 
        
 Times @@ RealDigits[#][[1]] == Total@RealDigits[#][[1]] &]





        share|cite|improve this answer














        A simple Mathematica program finds the $40$ cases:



        ${11125, 11133, 11152, 11215, 11222, 11251, 11313, 11331, 11512,
        11521, 12115, 12122, 12151, 12212, 12221, 12511, 13113, 13131, 13311,
        15112, 15121, 15211, 21115, 21122, 21151, 21212, 21221, 21511, 22112,
        22121, 22211, 25111, 31113, 31131, 31311, 33111, 51112, 51121, 51211,
        52111}$




        myList = Table[i, {i, 10000, 99999}];
        
Select[myList, 
        
 Times @@ RealDigits[#][[1]] == Total@RealDigits[#][[1]] &]






        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 58 mins ago

























        answered 1 hour ago









        David G. Stork

        9,93521232




        9,93521232






























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