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?










share|improve this question











$endgroup$

















    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











    $endgroup$















      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











      $endgroup$




      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















      share|improve this question













      share|improve this question




      share|improve this question








      edited Mar 14 at 15:46









      Hugh

      2,26811127




      2,26811127










      asked Mar 14 at 15:32









      user13242352user13242352

      443




      443






















          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
            Mar 14 at 16:57



















          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
            Mar 14 at 21:56











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          2 Answers
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          active

          oldest

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          2 Answers
          2






          active

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          active

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          active

          oldest

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          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
            Mar 14 at 16:57
















          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
            Mar 14 at 16:57














          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 Mar 14 at 19:31









          Yakk

          1706




          1706










          answered Mar 14 at 15:45









          GlorfindelGlorfindel

          14.1k45185




          14.1k45185








          • 3




            $begingroup$
            Your explanation reminds me of the famous Clue scene: youtube.com/watch?v=O5ROhf5Soqs
            $endgroup$
            – Kevin
            Mar 14 at 16:57














          • 3




            $begingroup$
            Your explanation reminds me of the famous Clue scene: youtube.com/watch?v=O5ROhf5Soqs
            $endgroup$
            – Kevin
            Mar 14 at 16:57








          3




          3




          $begingroup$
          Your explanation reminds me of the famous Clue scene: youtube.com/watch?v=O5ROhf5Soqs
          $endgroup$
          – Kevin
          Mar 14 at 16:57




          $begingroup$
          Your explanation reminds me of the famous Clue scene: youtube.com/watch?v=O5ROhf5Soqs
          $endgroup$
          – Kevin
          Mar 14 at 16:57











          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
            Mar 14 at 21:56
















          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
            Mar 14 at 21:56














          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 Mar 14 at 15:59









          RubioRubio

          30.4k567188




          30.4k567188








          • 1




            $begingroup$
            In the second part, I think you are missing the possibility 2,3,3,4
            $endgroup$
            – Cain
            Mar 14 at 21:56














          • 1




            $begingroup$
            In the second part, I think you are missing the possibility 2,3,3,4
            $endgroup$
            – Cain
            Mar 14 at 21:56








          1




          1




          $begingroup$
          In the second part, I think you are missing the possibility 2,3,3,4
          $endgroup$
          – Cain
          Mar 14 at 21:56




          $begingroup$
          In the second part, I think you are missing the possibility 2,3,3,4
          $endgroup$
          – Cain
          Mar 14 at 21:56


















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