An Accountant Seeks the Help of a MathematicianFour prisoners wearing black and white hatsDeduce the numbers (addition + subtraction)Determine $X$ and $Y$: sum, difference and ratioSum, Product and DifferenceFind the identity of the four troll brothers4 Gridded Prime NumbersHow does the hit-man assassinate his target using only two questions?My Friend, 4 kids and gold and silver coinsA Counterfeit $100 Bill among 93 BillsRanking teacher salaries in the Republic of al'Alghaz
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An Accountant Seeks the Help of a Mathematician
Four prisoners wearing black and white hatsDeduce the numbers (addition + subtraction)Determine $X$ and $Y$: sum, difference and ratioSum, Product and DifferenceFind the identity of the four troll brothers4 Gridded Prime NumbersHow does the hit-man assassinate his target using only two questions?My Friend, 4 kids and gold and silver coinsA Counterfeit $100 Bill among 93 BillsRanking teacher salaries in the Republic of al'Alghaz
$begingroup$
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?
logical-deduction number-theory
$endgroup$
add a comment |
$begingroup$
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?
logical-deduction number-theory
$endgroup$
add a comment |
$begingroup$
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?
logical-deduction number-theory
$endgroup$
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
logical-deduction number-theory
edited Mar 14 at 15:46
Hugh
2,26811127
2,26811127
asked Mar 14 at 15:32
user13242352user13242352
443
443
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
$endgroup$
3
$begingroup$
Your explanation reminds me of the famous Clue scene: youtube.com/watch?v=O5ROhf5Soqs
$endgroup$
– Kevin
Mar 14 at 16:57
add a comment |
$begingroup$
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.
$endgroup$
1
$begingroup$
In the second part, I think you are missing the possibility 2,3,3,4
$endgroup$
– Cain
Mar 14 at 21:56
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
3
$begingroup$
Your explanation reminds me of the famous Clue scene: youtube.com/watch?v=O5ROhf5Soqs
$endgroup$
– Kevin
Mar 14 at 16:57
add a comment |
$begingroup$
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.
$endgroup$
3
$begingroup$
Your explanation reminds me of the famous Clue scene: youtube.com/watch?v=O5ROhf5Soqs
$endgroup$
– Kevin
Mar 14 at 16:57
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
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
add a comment |
$begingroup$
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.
$endgroup$
1
$begingroup$
In the second part, I think you are missing the possibility 2,3,3,4
$endgroup$
– Cain
Mar 14 at 21:56
add a comment |
$begingroup$
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.
$endgroup$
1
$begingroup$
In the second part, I think you are missing the possibility 2,3,3,4
$endgroup$
– Cain
Mar 14 at 21:56
add a comment |
$begingroup$
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.
$endgroup$
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.
answered Mar 14 at 15:59
Rubio♦Rubio
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
add a comment |
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
add a comment |
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