How many 5-digit numbers satisfy this criterion? [on hold]












0














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










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put on hold as off-topic by Saad, Hanul Jeon, José Carlos Santos, Abcd, amWhy 17 hours ago


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  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Hanul Jeon, José Carlos Santos, Abcd, amWhy

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












  • 11222 is one of it.
    – Heroic24
    yesterday










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
















0














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










share|cite|improve this question















put on hold as off-topic by Saad, Hanul Jeon, José Carlos Santos, Abcd, amWhy 17 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Hanul Jeon, José Carlos Santos, Abcd, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.













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












  • 11222 is one of it.
    – Heroic24
    yesterday










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














0












0








0


1





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










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







elementary-number-theory






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edited 17 hours ago









amWhy

192k28225439




192k28225439










asked yesterday









Heroic24

1637




1637




put on hold as off-topic by Saad, Hanul Jeon, José Carlos Santos, Abcd, amWhy 17 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Hanul Jeon, José Carlos Santos, Abcd, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by Saad, Hanul Jeon, José Carlos Santos, Abcd, amWhy 17 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Hanul Jeon, José Carlos Santos, Abcd, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.












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












  • 11222 is one of it.
    – Heroic24
    yesterday










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


















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












  • 11222 is one of it.
    – Heroic24
    yesterday










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
















Examples seem to be rare. I found $11125$ and $11133$ though.
– SmileyCraft
yesterday






Examples seem to be rare. I found $11125$ and $11133$ though.
– SmileyCraft
yesterday














11222 is one of it.
– Heroic24
yesterday




11222 is one of it.
– Heroic24
yesterday












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




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










2 Answers
2






active

oldest

votes


















5














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
    yesterday



















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






























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5














    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
      yesterday
















    5














    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
      yesterday














    5












    5








    5






    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 yesterday

























    answered yesterday









    Hello_World

    4,11621630




    4,11621630












    • Oh, I should have mentioned that $n>100.$
      – Hello_World
      yesterday


















    • Oh, I should have mentioned that $n>100.$
      – Hello_World
      yesterday
















    Oh, I should have mentioned that $n>100.$
    – Hello_World
    yesterday




    Oh, I should have mentioned that $n>100.$
    – Hello_World
    yesterday











    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 yesterday

























        answered yesterday









        David G. Stork

        9,96021232




        9,96021232















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