An Accountant Seeks the Help of a Mathematician












7












$begingroup$


The accountant complaints to the mathematician:




“I lent money to five other faculty members and still haven’t been paid back. You are one of them; the other four owe me 12 dollars altogether, but I don’t remember how much each person owes me separately.”



“Are the debts in whole dollars?”



“Yes. I do remember that the four other debts multiplied together equals your debt. Do you remember how much you owe me?”



“Yes, but I still haven’t figured out how much each of the other four owe you.”



“Wait! The statistician is the one who owes the least.”



“That does it. Now I know the amount of each debt.”




What are the debts and how did the mathematician determine them?










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$endgroup$

















    7












    $begingroup$


    The accountant complaints to the mathematician:




    “I lent money to five other faculty members and still haven’t been paid back. You are one of them; the other four owe me 12 dollars altogether, but I don’t remember how much each person owes me separately.”



    “Are the debts in whole dollars?”



    “Yes. I do remember that the four other debts multiplied together equals your debt. Do you remember how much you owe me?”



    “Yes, but I still haven’t figured out how much each of the other four owe you.”



    “Wait! The statistician is the one who owes the least.”



    “That does it. Now I know the amount of each debt.”




    What are the debts and how did the mathematician determine them?










    share|improve this question









    New contributor




    user13242352 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      7












      7








      7





      $begingroup$


      The accountant complaints to the mathematician:




      “I lent money to five other faculty members and still haven’t been paid back. You are one of them; the other four owe me 12 dollars altogether, but I don’t remember how much each person owes me separately.”



      “Are the debts in whole dollars?”



      “Yes. I do remember that the four other debts multiplied together equals your debt. Do you remember how much you owe me?”



      “Yes, but I still haven’t figured out how much each of the other four owe you.”



      “Wait! The statistician is the one who owes the least.”



      “That does it. Now I know the amount of each debt.”




      What are the debts and how did the mathematician determine them?










      share|improve this question









      New contributor




      user13242352 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      The accountant complaints to the mathematician:




      “I lent money to five other faculty members and still haven’t been paid back. You are one of them; the other four owe me 12 dollars altogether, but I don’t remember how much each person owes me separately.”



      “Are the debts in whole dollars?”



      “Yes. I do remember that the four other debts multiplied together equals your debt. Do you remember how much you owe me?”



      “Yes, but I still haven’t figured out how much each of the other four owe you.”



      “Wait! The statistician is the one who owes the least.”



      “That does it. Now I know the amount of each debt.”




      What are the debts and how did the mathematician determine them?







      logical-deduction number-theory






      share|improve this question









      New contributor




      user13242352 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









      New contributor




      user13242352 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|improve this question




      share|improve this question








      edited yesterday









      Hugh

      2,26811127




      2,26811127






      New contributor




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      asked yesterday









      user13242352user13242352

      443




      443




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      New contributor





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          2 Answers
          2






          active

          oldest

          votes


















          15












          $begingroup$

          The mathematician owes




          48 dollars




          and the other faculty members owe




          4, 4, 3 and 1 dollar.




          Explanation:




          We know the total debt of the other four faculty members is 12, so we're looking for partitions of 12 in four parts. Let's list them and compute the product (which would be the mathematician's debt):




          Table:




          9 + 1 + 1 + 1 = 12 --> 9
          8 + 2 + 1 + 1 = 12 --> 16
          7 + 3 + 1 + 1 = 12 --> 21
          7 + 2 + 2 + 1 = 12 --> 28
          6 + 4 + 1 + 1 = 12 --> 24
          6 + 3 + 2 + 1 = 12 --> 36
          6 + 2 + 2 + 2 = 12 --> 48
          5 + 5 + 1 + 1 = 12 --> 25
          5 + 4 + 2 + 1 = 12 --> 40
          5 + 3 + 3 + 1 = 12 --> 45
          5 + 3 + 2 + 2 = 12 --> 60
          4 + 4 + 3 + 1 = 12 --> 48
          4 + 4 + 2 + 2 = 12 --> 64
          4 + 3 + 3 + 2 = 12 --> 72
          3 + 3 + 3 + 3 = 12 --> 81

          Now, the mathematician knows his own debt, so if he's unsure of the distribution of the others, there must be multiple partitions of 12 in four parts with a product equal to his debt. That's only the case for 48, which can be split as $6 times 2 times 2 times 2$ and $4 times 4 times 3 times 1$. Since there is a single least number among those (corresponding to the statistician's debt), it must be the second option.







          share|improve this answer











          $endgroup$









          • 3




            $begingroup$
            Your explanation reminds me of the famous Clue scene: youtube.com/watch?v=O5ROhf5Soqs
            $endgroup$
            – Kevin
            yesterday



















          5












          $begingroup$

          The mathematician owes




          $48




          Because




          nothing is said that the individual debts are unique, only that the lowest one must be.

          So any of the following distributions of debts by the four remaining are possible, and thus the following possibilities for what the mathematician owes:


          1,2,2,7 → $28

          1,2,3,6 → $36

          1,2,4,5 → $40

          1,3,3,5 → $45

          1,3,4,4 → $48




          Now the mathematician says at first that




          the other totals are not known to them until the accountant mentions that the statistician owes the least, thus requiring there to be a distinct "least". Prior to that discovery, ...


          The mathematician can't owe $45 because there is only one way to sum to 12 and multiply to 45.

          The mathematician can't owe $36 because there is only one way to sum to 12 and multiply to 36.

          The mathematician can't owe $28 because there is only one way to sum to 12 and multiply to 28.

          The mathematician could owe $48 because 2,2,2,6 is another valid solution that sums to 12 and multiplies to 48, but is ruled out when the discovery is made.







          share|improve this answer









          $endgroup$









          • 1




            $begingroup$
            In the second part, I think you are missing the possibility 2,3,3,4
            $endgroup$
            – Cain
            22 hours ago











          Your 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

          votes









          15












          $begingroup$

          The mathematician owes




          48 dollars




          and the other faculty members owe




          4, 4, 3 and 1 dollar.




          Explanation:




          We know the total debt of the other four faculty members is 12, so we're looking for partitions of 12 in four parts. Let's list them and compute the product (which would be the mathematician's debt):




          Table:




          9 + 1 + 1 + 1 = 12 --> 9
          8 + 2 + 1 + 1 = 12 --> 16
          7 + 3 + 1 + 1 = 12 --> 21
          7 + 2 + 2 + 1 = 12 --> 28
          6 + 4 + 1 + 1 = 12 --> 24
          6 + 3 + 2 + 1 = 12 --> 36
          6 + 2 + 2 + 2 = 12 --> 48
          5 + 5 + 1 + 1 = 12 --> 25
          5 + 4 + 2 + 1 = 12 --> 40
          5 + 3 + 3 + 1 = 12 --> 45
          5 + 3 + 2 + 2 = 12 --> 60
          4 + 4 + 3 + 1 = 12 --> 48
          4 + 4 + 2 + 2 = 12 --> 64
          4 + 3 + 3 + 2 = 12 --> 72
          3 + 3 + 3 + 3 = 12 --> 81

          Now, the mathematician knows his own debt, so if he's unsure of the distribution of the others, there must be multiple partitions of 12 in four parts with a product equal to his debt. That's only the case for 48, which can be split as $6 times 2 times 2 times 2$ and $4 times 4 times 3 times 1$. Since there is a single least number among those (corresponding to the statistician's debt), it must be the second option.







          share|improve this answer











          $endgroup$









          • 3




            $begingroup$
            Your explanation reminds me of the famous Clue scene: youtube.com/watch?v=O5ROhf5Soqs
            $endgroup$
            – Kevin
            yesterday
















          15












          $begingroup$

          The mathematician owes




          48 dollars




          and the other faculty members owe




          4, 4, 3 and 1 dollar.




          Explanation:




          We know the total debt of the other four faculty members is 12, so we're looking for partitions of 12 in four parts. Let's list them and compute the product (which would be the mathematician's debt):




          Table:




          9 + 1 + 1 + 1 = 12 --> 9
          8 + 2 + 1 + 1 = 12 --> 16
          7 + 3 + 1 + 1 = 12 --> 21
          7 + 2 + 2 + 1 = 12 --> 28
          6 + 4 + 1 + 1 = 12 --> 24
          6 + 3 + 2 + 1 = 12 --> 36
          6 + 2 + 2 + 2 = 12 --> 48
          5 + 5 + 1 + 1 = 12 --> 25
          5 + 4 + 2 + 1 = 12 --> 40
          5 + 3 + 3 + 1 = 12 --> 45
          5 + 3 + 2 + 2 = 12 --> 60
          4 + 4 + 3 + 1 = 12 --> 48
          4 + 4 + 2 + 2 = 12 --> 64
          4 + 3 + 3 + 2 = 12 --> 72
          3 + 3 + 3 + 3 = 12 --> 81

          Now, the mathematician knows his own debt, so if he's unsure of the distribution of the others, there must be multiple partitions of 12 in four parts with a product equal to his debt. That's only the case for 48, which can be split as $6 times 2 times 2 times 2$ and $4 times 4 times 3 times 1$. Since there is a single least number among those (corresponding to the statistician's debt), it must be the second option.







          share|improve this answer











          $endgroup$









          • 3




            $begingroup$
            Your explanation reminds me of the famous Clue scene: youtube.com/watch?v=O5ROhf5Soqs
            $endgroup$
            – Kevin
            yesterday














          15












          15








          15





          $begingroup$

          The mathematician owes




          48 dollars




          and the other faculty members owe




          4, 4, 3 and 1 dollar.




          Explanation:




          We know the total debt of the other four faculty members is 12, so we're looking for partitions of 12 in four parts. Let's list them and compute the product (which would be the mathematician's debt):




          Table:




          9 + 1 + 1 + 1 = 12 --> 9
          8 + 2 + 1 + 1 = 12 --> 16
          7 + 3 + 1 + 1 = 12 --> 21
          7 + 2 + 2 + 1 = 12 --> 28
          6 + 4 + 1 + 1 = 12 --> 24
          6 + 3 + 2 + 1 = 12 --> 36
          6 + 2 + 2 + 2 = 12 --> 48
          5 + 5 + 1 + 1 = 12 --> 25
          5 + 4 + 2 + 1 = 12 --> 40
          5 + 3 + 3 + 1 = 12 --> 45
          5 + 3 + 2 + 2 = 12 --> 60
          4 + 4 + 3 + 1 = 12 --> 48
          4 + 4 + 2 + 2 = 12 --> 64
          4 + 3 + 3 + 2 = 12 --> 72
          3 + 3 + 3 + 3 = 12 --> 81

          Now, the mathematician knows his own debt, so if he's unsure of the distribution of the others, there must be multiple partitions of 12 in four parts with a product equal to his debt. That's only the case for 48, which can be split as $6 times 2 times 2 times 2$ and $4 times 4 times 3 times 1$. Since there is a single least number among those (corresponding to the statistician's debt), it must be the second option.







          share|improve this answer











          $endgroup$



          The mathematician owes




          48 dollars




          and the other faculty members owe




          4, 4, 3 and 1 dollar.




          Explanation:




          We know the total debt of the other four faculty members is 12, so we're looking for partitions of 12 in four parts. Let's list them and compute the product (which would be the mathematician's debt):




          Table:




          9 + 1 + 1 + 1 = 12 --> 9
          8 + 2 + 1 + 1 = 12 --> 16
          7 + 3 + 1 + 1 = 12 --> 21
          7 + 2 + 2 + 1 = 12 --> 28
          6 + 4 + 1 + 1 = 12 --> 24
          6 + 3 + 2 + 1 = 12 --> 36
          6 + 2 + 2 + 2 = 12 --> 48
          5 + 5 + 1 + 1 = 12 --> 25
          5 + 4 + 2 + 1 = 12 --> 40
          5 + 3 + 3 + 1 = 12 --> 45
          5 + 3 + 2 + 2 = 12 --> 60
          4 + 4 + 3 + 1 = 12 --> 48
          4 + 4 + 2 + 2 = 12 --> 64
          4 + 3 + 3 + 2 = 12 --> 72
          3 + 3 + 3 + 3 = 12 --> 81

          Now, the mathematician knows his own debt, so if he's unsure of the distribution of the others, there must be multiple partitions of 12 in four parts with a product equal to his debt. That's only the case for 48, which can be split as $6 times 2 times 2 times 2$ and $4 times 4 times 3 times 1$. Since there is a single least number among those (corresponding to the statistician's debt), it must be the second option.








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited yesterday









          Yakk

          1706




          1706










          answered yesterday









          GlorfindelGlorfindel

          14k45185




          14k45185








          • 3




            $begingroup$
            Your explanation reminds me of the famous Clue scene: youtube.com/watch?v=O5ROhf5Soqs
            $endgroup$
            – Kevin
            yesterday














          • 3




            $begingroup$
            Your explanation reminds me of the famous Clue scene: youtube.com/watch?v=O5ROhf5Soqs
            $endgroup$
            – Kevin
            yesterday








          3




          3




          $begingroup$
          Your explanation reminds me of the famous Clue scene: youtube.com/watch?v=O5ROhf5Soqs
          $endgroup$
          – Kevin
          yesterday




          $begingroup$
          Your explanation reminds me of the famous Clue scene: youtube.com/watch?v=O5ROhf5Soqs
          $endgroup$
          – Kevin
          yesterday











          5












          $begingroup$

          The mathematician owes




          $48




          Because




          nothing is said that the individual debts are unique, only that the lowest one must be.

          So any of the following distributions of debts by the four remaining are possible, and thus the following possibilities for what the mathematician owes:


          1,2,2,7 → $28

          1,2,3,6 → $36

          1,2,4,5 → $40

          1,3,3,5 → $45

          1,3,4,4 → $48




          Now the mathematician says at first that




          the other totals are not known to them until the accountant mentions that the statistician owes the least, thus requiring there to be a distinct "least". Prior to that discovery, ...


          The mathematician can't owe $45 because there is only one way to sum to 12 and multiply to 45.

          The mathematician can't owe $36 because there is only one way to sum to 12 and multiply to 36.

          The mathematician can't owe $28 because there is only one way to sum to 12 and multiply to 28.

          The mathematician could owe $48 because 2,2,2,6 is another valid solution that sums to 12 and multiplies to 48, but is ruled out when the discovery is made.







          share|improve this answer









          $endgroup$









          • 1




            $begingroup$
            In the second part, I think you are missing the possibility 2,3,3,4
            $endgroup$
            – Cain
            22 hours ago
















          5












          $begingroup$

          The mathematician owes




          $48




          Because




          nothing is said that the individual debts are unique, only that the lowest one must be.

          So any of the following distributions of debts by the four remaining are possible, and thus the following possibilities for what the mathematician owes:


          1,2,2,7 → $28

          1,2,3,6 → $36

          1,2,4,5 → $40

          1,3,3,5 → $45

          1,3,4,4 → $48




          Now the mathematician says at first that




          the other totals are not known to them until the accountant mentions that the statistician owes the least, thus requiring there to be a distinct "least". Prior to that discovery, ...


          The mathematician can't owe $45 because there is only one way to sum to 12 and multiply to 45.

          The mathematician can't owe $36 because there is only one way to sum to 12 and multiply to 36.

          The mathematician can't owe $28 because there is only one way to sum to 12 and multiply to 28.

          The mathematician could owe $48 because 2,2,2,6 is another valid solution that sums to 12 and multiplies to 48, but is ruled out when the discovery is made.







          share|improve this answer









          $endgroup$









          • 1




            $begingroup$
            In the second part, I think you are missing the possibility 2,3,3,4
            $endgroup$
            – Cain
            22 hours ago














          5












          5








          5





          $begingroup$

          The mathematician owes




          $48




          Because




          nothing is said that the individual debts are unique, only that the lowest one must be.

          So any of the following distributions of debts by the four remaining are possible, and thus the following possibilities for what the mathematician owes:


          1,2,2,7 → $28

          1,2,3,6 → $36

          1,2,4,5 → $40

          1,3,3,5 → $45

          1,3,4,4 → $48




          Now the mathematician says at first that




          the other totals are not known to them until the accountant mentions that the statistician owes the least, thus requiring there to be a distinct "least". Prior to that discovery, ...


          The mathematician can't owe $45 because there is only one way to sum to 12 and multiply to 45.

          The mathematician can't owe $36 because there is only one way to sum to 12 and multiply to 36.

          The mathematician can't owe $28 because there is only one way to sum to 12 and multiply to 28.

          The mathematician could owe $48 because 2,2,2,6 is another valid solution that sums to 12 and multiplies to 48, but is ruled out when the discovery is made.







          share|improve this answer









          $endgroup$



          The mathematician owes




          $48




          Because




          nothing is said that the individual debts are unique, only that the lowest one must be.

          So any of the following distributions of debts by the four remaining are possible, and thus the following possibilities for what the mathematician owes:


          1,2,2,7 → $28

          1,2,3,6 → $36

          1,2,4,5 → $40

          1,3,3,5 → $45

          1,3,4,4 → $48




          Now the mathematician says at first that




          the other totals are not known to them until the accountant mentions that the statistician owes the least, thus requiring there to be a distinct "least". Prior to that discovery, ...


          The mathematician can't owe $45 because there is only one way to sum to 12 and multiply to 45.

          The mathematician can't owe $36 because there is only one way to sum to 12 and multiply to 36.

          The mathematician can't owe $28 because there is only one way to sum to 12 and multiply to 28.

          The mathematician could owe $48 because 2,2,2,6 is another valid solution that sums to 12 and multiplies to 48, but is ruled out when the discovery is made.








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered yesterday









          RubioRubio

          29.8k566184




          29.8k566184








          • 1




            $begingroup$
            In the second part, I think you are missing the possibility 2,3,3,4
            $endgroup$
            – Cain
            22 hours ago














          • 1




            $begingroup$
            In the second part, I think you are missing the possibility 2,3,3,4
            $endgroup$
            – Cain
            22 hours ago








          1




          1




          $begingroup$
          In the second part, I think you are missing the possibility 2,3,3,4
          $endgroup$
          – Cain
          22 hours ago




          $begingroup$
          In the second part, I think you are missing the possibility 2,3,3,4
          $endgroup$
          – Cain
          22 hours ago










          user13242352 is a new contributor. Be nice, and check out our Code of Conduct.










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