Below, we have provided at least one solution to each of the 11 problems selected from the Pamiers text. Other methods of solution may be possible.

**Problem 1.** Two merchants want to barter together, planning to exchange their merchandise; and the first of them has wool and the other has cloth. The one who has cloth wants 24 *sous* for trading each bolt of cloth whose cash price is but 20 *sous*. I ask for how much the other must sell each *quintal* of wool that is worth but 12 *liras*, in order that he lose nothing.

The author of the Pamiers manuscript showed solutions similar to B and C below.

**Solution A:** a proportion using the augmented (“marked-up”) selling prices

Begin with the proportion,

20 *sous* : 24 *sous* :: 12 *liras* : ?

(This is like asking, “A marked-up price of 24 *sous* on a list price of 20 *sous* is equivalent to a marked-up price of how many *liras* on a list price of 12 *liras*?)

By the rule of three, the marked-up price is therefore (12 × 24) ÷ 20 = 288 ÷ 20 = 14 2/5 *liras*, or 14 *liras* 8 *sous*.

**Solution B:** a proportion using the markups

Begin with the proportion,

20 *sous* : 4 *sous* :: 12 *liras* : ?

(This is like asking, “A markup of 4 *sous* on a list price of 20 *sous* is equivalent to a markup of how many *liras* on a list price of 12 *liras*?)

By the rule of three, the markup is therefore (12 × 4) ÷ 20 = 48 ÷ 20 = 2 2/5 *liras*, or 2 *liras* 8 *sous*.

Augmenting the price of 12 *liras* by the markup of 2 *liras* 8 *sous* gives a marked-up price of 14 *liras* 8 *sous*.

**Solution C:** double false position using the markups

Begin with the proportion,

20 *sous* : 4 *sous* :: 12 *liras* : ?

Note 12 × 4 = 48.

First, suppose the markup is 2 *liras*. But 20 × 2 = 40, a deficit of 8.

Second, suppose the markup is 3 *liras*. But 20 × 3 = 60, an excess of 12.

By double false position, the correct markup is therefore

\[{\frac{(2\times12\,{\rm excess})+(3\times8\,{\rm deficit})}{12\,{\rm excess}\,+8\,{\rm deficit}}}={\frac{48}{20}}\]\[=2\frac{2}{5}\,{liras}=2\,{liras}\,\,8\,{sous}.\]

Augmenting the price of 12 *liras* by the markup of 2 *liras* 8 *sous* gives a marked-up price of 14 *liras* 8 *sous*.

**Problem 2.** If 3 *deniers* of Perpignan are worth 5 *deniers* of Montpellier, and 2 of Montpellier are worth 3 of those of Avignon, I ask how many of those of Perpignan are worth 12 of those of Avignon.

The author showed a solution similar to A below.

**Solution A:** composite rule of three \[\frac{5}{3}\times\frac{3}{2}=\frac{5}{2}\]

5 : 2 :: 12 : ?

(12 × 2) ÷ 5 = 24 ÷ 5 = 4 4/5* deniers*.

**Solution B:** chain rule

(3 × 2 × 12) ÷ (5 × 3) = 24 ÷ 5 = 4 4/5 *deniers*.

**Solution C:** algebra

Given \(3P=5M,\) and \(2M=3A,\) we have \[P={\frac{5}{3}}M,\quad M={\frac{3}{2}}A\] \[P={\frac{5}{3}}\left({\frac{3}{2}}{A}\right) = {\frac{5}{2}}A\] \[{\frac{2}{5}}P=A\] \[12\left({\frac{2}{5}}P\right)=12A\] \[{\frac{24}{5}}P=12A\]

Thus, 24/5 or 4 4/5* deniers* of Perpignan are worth 12 of those of Avignon.

**Problem 3.** A merchant paid 10 *liras*, which is 200 *sous*, for two manners of grain, namely wheat and oats, and he purchased each *setier* of wheat for 10 *sous* and each *setier* of oats for 5 *sous*. This merchant turned around and sold his grain, selling each *setier* of oats for 4 *sous* and each *setier* of wheat for 12 *sous*, and realized a profit of 10 *sous*. I ask how many *setiers* of wheat and how many of oats he purchased, and how much money he paid for each grain.

The author showed a solution similar to A below.

**Solution A:** double false position

First, suppose he sold 10 *setiers* of wheat at 10 *sous*, thus 100 *sous* total.

Then he paid 200 – 100 = 100 *sous* for the oats at 5 *sous,*

and thus purchased 100 ÷ 5 = 20 *setiers* of oats.

So his sales proceeds are

(10 *setiers* of wheat at 12 *sous*) + (20 *setiers* of oats at 4 *sous*) = 120 + 80 = 200 *sous*

and his profits are therefore

200 – 200 = no *sous*, instead of the stated 10 *sous* (thus, deficit 10).

Likewise, suppose he sold 15 *setiers* of wheat. Then we get

200 – 150 = 50

50 ÷ 5 = 10

(15 × 12) + (10 × 4) = 220

and his profits are therefore

220 – 200 = 20 *sous*, instead of the stated 10 *sous* (thus, excess 10).

By double false position, the correct amount is thus \[{\frac{(10\,{setiers}\,\times10\,{sous}\,{\rm excess})+{(15\,{setiers}\,\times10\,{sous}\,{\rm deficit})}}{10\,{sous}\,{\rm excess}+10\,{sous}\,{\rm deficit}}}=\frac{250}{20}=12\frac{1}{2}\] *setiers* of wheat. Therefore, he paid 12 1/2 × 10 = 125 *sous* for the wheat and 200 – 125 = 75 *sous* for the oats, and so the quantity of oats purchased was 75 ÷ 5 = 15 *setiers*.

Check. The proceeds were thus (12 1/2 × 12) + (15 × 4) = 210 *sous*, a profit of 210 – 200 = 10 *sous*.

**Solution B:** algebra (simple elimination)

Let \(x=\) the quantity of wheat, and \(y=\) the quantity of oats. Then \[10x+5y=200\] \[12x+4y=210.\]

Dividing the equations by \(5\) and \(4,\) respectively, we get \[2x+y=40\] \[3x+y=52.5.\]

Subtracting the first equation from the second eliminates the second variable, leaving \[x=12.5,\]

so \(12.5\) *setiers* of wheat were purchased for \(12.5\times 10=125\) *sous.*

Back-substituting this result yields \[2(12.5)+y=40\] \[25+y=40\] \[y=15,\]

so \(15\) *setiers* of oats were purchased for \(15\times 5=75\) *sous.*

**Solution C:** matrix algebra

Given \[\left[ {\begin{array}{cc}

10&5 \\

12&4 \\ \end{array}}\right]\left[ {\begin{array}{c}

x\\

y\\ \end{array}}\right]=\left[ {\begin{array}{c}

200\\

210\\ \end{array}}\right],\]

the quantities of wheat and oats purchased were \[\left[ {\begin{array}{c}

x\\

y\\ \end{array}}\right]={{\left[ {\begin{array}{cc}

10&5 \\

12&4 \\ \end{array}}\right]}^{-1}}\left[ {\begin{array}{c}

200\\

210\\ \end{array}}\right]=\left[ {\begin{array}{c}

12.5\\

15\\ \end{array}}\right];\]

that is, 12.5 *setiers* of wheat and 15* setiers* of grain, and so the amounts paid were 10 × 12.5 = 125 *sous* and 5 × 15 = 75 *sous*, respectively.

**Problem 4.** A merchant purchased three pieces of cloth that cost him [a total of] 30 *motos*, and doesn’t know with certainty what each of the pieces cost, but does know that the second cost double the first and 4 more; the third cost three times as much as the second, less 7. I ask what each one cost.

The author showed a solution similar to A below.

**Solution A:** double false position

Suppose the first piece cost 3 *motos*.

Then the second piece cost (2 × 3) + 4 = 10 *motos*,

and the third piece cost (3 × 10) – 7 = 23 *motos*.

But 3 + 10 + 23 = 36 *motos*, an excess of 6.

Suppose the first piece cost 4 *motos*.

Then the second piece cost (2 × 4) + 4 = 12 *motos*,

and the third piece cost (3 × 12) – 7 = 29 *motos*.

But 4 + 12 + 29 = 45 *motos*, an excess of 15.

By double false position, the true position is thus \[{\frac{(3\times15\,{\rm excess})-{(4\times\,6\,{\rm excess})}}{15\,{\rm excess}-6\,{\rm excess}}}=\frac{21}{9}.\]

Therefore, the first piece cost \(\frac{21}{9}\) or \(2\frac{1}{3}\) *motos,*

the second piece cost \(\left(2\times2\frac{1}{3}\right)+4=8\frac{2}{3}\) *motos,*

and the third piece cost \(\left(3\times8\frac{2}{3}\right)-7=19\) *motos*.

Check. The total cost was thus \(2\frac{1}{3}+8\frac{2}{3}+19=30\) *motos*.

**Solution B:** algebra

Let *x* = the cost of the first piece.

Then the cost of the second piece is 2*x* + 4,

and the cost of the third piece is 3(2*x* + 4) – 7.

Since the total cost of the three pieces is 30 *motos*, we have:

*x* + (2*x* + 4) + (3(2*x* + 4) – 7) = 30.

*x* + (2*x* + 4) + (6*x* + 12 – 7) = 30.

9*x* + 9 = 30

9*x* = 21

*x* = \(\frac{21}{9}\) or \(2\frac{1}{3}\) *motos,* etc. as in Solution A above.

**Problem 5.** Three merchants formed a company together. The first advanced 200 *motos* and stayed for 15 months, the second advanced 94 *motos* and stayed for 17 months, the third advanced 38 *motos* and stayed for 10 months; and at the end they realized a profit of 400 [*motos*]. I ask how it should be divided [among them].

**Solution by proportion:**

200 *motos* × 15 months = 3000

94 *motos* × 17 months = 1598

38 *motos* × 10 months = 380

Total = 4978.

The lattice or *gelosia* technique proceeds as follows for the second of the three multiplications noted above.

Using the rule of three, the proportional shares of the profit are, respectively,

4978 : 3000 :: 400 : ? implies 400 × 3000 ÷ 4978 = \(241\frac{151}{2489}\)* motos*

4978 : 1598 :: 400 : ? implies 400 × 1598 ÷ 4978 = \(128\frac{1008}{2489}\)* motos*

4978 : 380 :: 400 : ? implies 400 × 380 ÷ 4978 = \(30\frac{70}{131}\)* motos.*

Check. The total profit is thus \(241\frac{151}{2489}+128\frac{1008}{2489}+30\frac{70}{131}=400\)* motos.*

**Problem 6.** A merchant gave 600 *liras* to a factor who had 200 *liras* of his own, by such arrangement that he work with that 800 *liras* for 5 years, and at the end of that time [said the merchant] we will divide in half the principal and the profit. It so happened that the factor spent none of the 200 *liras*, but had made use of the 600 *liras* of the merchant; and at the end of 5 years had realized 2400 *liras*, counting principal and profit. I ask how the division should be carried out— considering that the factor spent none of what he should have spent— in order that the merchant not be deceived.

**Solution by proportion:**

By agreeing in advance to divide their proceeds in half, the two men were valuing equally their prospective contributions: 600 *liras* from the merchant, and 200* liras* plus 5 years of labor from the factor. Therefore, the value of the factor’s five years of labor was 600 – 200 = 400 *liras*.

The factor’s 200* liras* of money that ended up not being needed for production cannot be thought of as part of the investment, since throughout the five years the factor was free to use that money in any way that pleased him. Therefore, the total value invested by the two men together was 600 + 400* *= 1000* liras*, and the total proceeds on the investment itself were 2400 – 200* *= 2200* liras*. These 2200* liras* must be divided between the two men in proportion to their respective shares in the investment, namely 600 and 400* liras*. Using the rule of three,

merchant’s share, 1000 : 600 :: 2200 : ? implies 2200 × 600 ÷ 1000 = 1320* liras*

factor’s share, 1000 : 400 :: 2200 : ? implies 2200 × 400 ÷ 1000 = 880* liras*.

(The factor also retains his unused 200* liras* for a total of 1080* liras*.)

Check. The total proceeds were thus 1320 + 880 = 2200* liras*.

**Problem 7.** Two merchants went to the fair; the first had 20 sacks of wool, for which he paid the duty collector 1 sack of the wool, and the collector returned to him 2 *liras*; the second had 60 sacks of wool, for which he paid the collector 2 sacks of wool plus 6 *liras*. I ask the value of each sack of the wool, and how much was paid the collector per sack.

The author showed a solution similar to A below.

**Solution A:** double false position

First, suppose the value of each sack is 4 *liras*.

Then the tax on 20 sacks, being 1 sack minus 2 *liras*, is 4 – 2 = 2 *liras*.

Thus, the tax on 60 sacks is 3 × 2 = 6 *liras*,

instead of the stated 2 sacks plus 6 *liras*, or (2 × 4) + 6 = 14 *liras*.

Therefore, the deficit is 14 – 6 = 8 *liras*.

Second, suppose the value of each sack is 8 *liras*.

Then the tax on 20 sacks, being 1 sack minus 2 *liras*, is 8 – 2 = 6 *liras*.

Thus, the tax on 60 sacks is 3 × 6 = 18 *liras*,

instead of the stated 2 sacks plus 6 *liras*, or (2 × 8) + 6 = 22 *liras*.

Therefore, the deficit is 22 – 18 = 4 *liras*.

By double false position, the correct value of each sack is therefore \[{\frac{(8\times8\,{\rm deficit})-{(4\times4\,{\rm deficit})}}{8\,{\rm deficit}-4\,{\rm deficit}}}=\frac{48}{4}=12\,\,{liras}.\]

To find the tax rate, recall that the tax on 20 sacks was 1 sack minus 2 *liras*, or 12 – 2 = 10 *liras*. Thus, the tax rate was 10 *liras* ÷ 20 sacks = ½ *lira *per sack.

Check. The tax on 60 sacks was 2 sacks plus 6 *liras* = (2 × 12) + 6 = 30 *liras*, for a rate of 30 *liras* ÷ 60 sacks = ½ *lira *per sack.

**Solution B:** algebraic style, by student Alex Wolstencroft (Schoolcraft College)

We are given:

tax on 20 sacks = 1 sack – 2* liras*

tax on 60 sacks = 2 sacks + 6* liras*.

Multiplying the first equation by 3, we get

tax on 60 sacks = 3 sacks – 6* liras*.

Comparing with the second equation,

2 sacks + 6* liras* = 3 sacks – 6* liras*

2 sacks = 3 sacks – 12* liras*

1 sack = 12* liras*.

Putting this in the first equation,

tax on 20 sacks = 12* liras* – 2* liras*

tax on 20 sacks = 10* liras*

tax on 1 sack = 10/20* lira* = ½ *lira*.

**Solution C:** symbolic algebra

Let *x* = the value of each sack, in *liras*, and *y* = the tax on each sack, in *liras*.

Then we are given:

20*y* = *x* – 2

60*y* = 2*x* + 6

Multiplying the first equation by 3, we get

60*y* = 3*x* – 6.

Comparing with the second equation,

3*x* – 6 = 2*x* + 6

*x* = 12 *liras *per sack

*y* = (*x* – 2)/20 = 10/20 = ½ *lira *per sack.

**Problem 8.** A man entered a fair, and the first day doubled all his money and spent 1 *gros*; likewise on the second tripled all his money that he had left and spent 2 *gros*; likewise on the third quadrupled his money that he had left and spent 2 *gros*, and found that he had nothing but 3 *gros*. I ask how much money he carried [to the fair].

The author showed a solution similar to A below.

**Solution A:** double false position

First, suppose he entered with 1 *gros*.

After Day 1: (2 × 1) – 1 = 1 *gros*

After Day 2: (3 × 1) – 2 = 1 *gros*

After Day 3: (4 × 1) – 2 = 2 *gros* instead of the stated 3 *gros* (deficit 1).

Second, suppose he entered with 2 *gros*.

After Day 1: (2 × 2) – 1 = 3 *gros*

After Day 2: (3 × 3) – 2 = 7 *gros*

After Day 3: (4 × 7) – 2 = 26 *gros* instead of the stated 3 *gros* (excess 23).

Therefore, by double false position, the correct amount with which he entered the fair is\[{\frac{(1\times23\,{\rm excess})+{(2\times1\,{\rm deficit})}}{23\,{\rm excess}+1\,{\rm deficit}}}=\frac{25}{24}=1\frac{1}{24}\,\,{gros}.\]

Check.

After Day 1: (2 × 1 1/24) – 1 = 1 1/12 *gros*

After Day 2: (3 × 1 1/12) – 2 = 1 1/4 *gros*

After Day 3: (4 × 1 1/4) – 2 = 3 *gros*.

**Solution B:** inversion (“working backward”)

Before Day 3: (3 + 2) ÷ 4 = 5/4 *gros*

Before Day 2: (5/4 + 2) ÷ 3 = 13/12 *gros*

Before Day 1: (13/12 + 1) ÷ 2 = 25/24 or 1 1/24 *gros**.*

**Solution C:** algebra

Let \(x =\) the amount with which he entered the fair, in *gros*.

Then we are given \(4\left[3\left(2x-1\right)-2\right]-2=3,\)

so \[12(2x-1)-8-2=3\] \[24x-12-8-2=3\] \[24x=25\] \[x=\frac{25}{24}=1\frac{1}{24}\,\,{gros}.\]

**Problem 9.** A vat [of wine] has three spigots, the first bigger than the others, and in such a manner that with the biggest open, all the wine empties in 3 hours; and with the medium one open, it empties in 4 hours; and with the smallest open, it empties in 6 hours. I ask when all the spigots are open together, in how much time will the wine empty.

The author showed a solution similar to A below.

**Solution A:** single false position

Suppose the time needed to empty 1 vat is 12 hours (chosen purely for convenience, since 12 is divisible by 3, 4, and 6).

In that time, the three spigots together would drain \[\left({\frac{1\,{\rm vat}}{3\,{\rm hrs}}}\times12\,{\rm hrs}\right)+\left({\frac{1\,{\rm vat}}{4\,{\rm hrs}}}\times12\,{\rm hrs}\right)+\left({\frac{1\,{\rm vat}}{6\,{\rm hrs}}}\times12\,{\rm hrs}\right)=9\,\,{\rm vats},\]

instead of the correct 1 vat. Thus, the supposition was too large by a factor of 9, indicating that the correct answer is 12 ÷ 9 = 1 1/3 hrs, or 80 mins.

**Solution B:** additive rates

Since the rates (vats/hour) are additive, the net rate of drainage is \[\frac{1}{3}+\frac{1}{4}+\frac{1}{6}=\frac{9}{12}=\frac{3}{4}\frac{\rm vats}{\rm hour}.\]

Thus, the net time is \(\frac{4}{3}\frac{\rm hours}{\rm vat}\) or \(1\frac{1}{3}\frac{\rm hours}{\rm vat}.\)

**Problem 10.** Two men leave on the same day and the same hour from two cities, say for example the first leaves Béziers for Barcelona and the other goes from Barcelona to Béziers. And the one that leaves Béziers does the route in 7 days, and the other takes 9 days. I ask how much time is required before they see each other, and [if the cities are separated by 63 leagues] how far each has traveled.

The author showed a solution similar to A below.

**Solution A:** single false position

Suppose 4 days are required before the men see each other (chosen for convenience, since 4 is half of the 8-day average of the 7 and 9 days required by each man, respectively, to traverse the entire route alone).

In that time, the two men together would cover \[\left({\frac{1\,{\rm route}}{7\,{\rm days}}}\times4\,{\rm days}\right)+\left({\frac{1\,{\rm route}}{9\,{\rm days}}}\times4\,{\rm days}\right)=\frac{4}{7}+\frac{4}{9}=\frac{64}{63}\,\, {\rm route},\]

instead of the correct 1 route. Thus, the supposition was too large by a factor of 64/63, indicating that the correct answer is 4 days ÷ (64/63) = 63/16 or 3 15/16 days.

The distance covered by each man can then be found by multiplying rate by time:

The first man covers \[{\frac{1\,{\rm route}}{7\,{\rm days}}}\times{{\frac{63}{16}}\,{\rm days}}={\frac{9}{16}\,{\rm route}}={{\frac{9}{16}}\left(63\,{\rm leagues}\right)}={35\frac{7}{16}\,{\rm leagues}}.\]

The second man covers \[{\frac{1\,{\rm route}}{9\,{\rm days}}}\times{{\frac{63}{16}}\,{\rm days}}={\frac{7}{16}\,{\rm route}}={{\frac{7}{16}}\left(63\,{\rm leagues}\right)}={27\frac{9}{16}\,{\rm leagues}}.\]

Check. The total distance covered by the two men is thus 35 7/16 + 27 9/16 = 63 leagues.

**Solution B:** additive rates

Since the rates (routes/day) are additive, the net rate of speed is \[\frac{1}{7}+\frac{1}{9}=\frac{16}{63}\frac{\rm routes}{\rm day}.\]

Thus, the net time is 63/16 or 3 15/16 days/route. The distance covered by each man can then be found by multiplying rate by time, as in Solution A above.

**Solution C:** algebra

Using subscripts \(1\) and \(2,\) respectively, to distinguish quantities for the two men, their speeds (rates) are:

\[r_1 =\frac{1\,\rm route}{7\,\rm days}=\frac{63\,\rm leagues}{7\,\rm days}=9\frac{\rm leagues}{\rm day}\]

\[r_2 =\frac{1\,\rm route}{9\,\rm days}=\frac{63\,\rm leagues}{9\,\rm days}=7\frac{\rm leagues}{\rm day}.\]

Since the distances covered by the two men must total 63 leagues, we have \(d_1+d_2=63\) leagues, and so the time \(t,\) in days, required before the men see each other satisfies: \[r_1t+r_2t=63\,\,{\rm leagues}\] \[9t+7t=63\] \[16t=63,\] so that \(t=\frac{63}{16}\) or \(3\frac{15}{16}\) days. The distance covered by each man can then be found by multiplying rate by time, as in Solution A above.

**Problem 11.** There are 5 men who [each] want to purchase a piece of cloth in such manner [i.e., at such a cost] that the first asks of all the others 1/2 of all the gold and silver they carry, the second asks 1/3, the third asks 1/4, the fourth [asks] 1/5 and the fifth asks 1/6. I ask what the piece cost and what each of them carried.

All solutions differ from one another by an arbitrary multiplicative constant (scaling factor). The author showed, without justification, a rote solution similar to A below. I follow that with my own justification of the rote procedure, and then a solution using matrix algebra.

**Solution A:** rote procedure

Start with the multiplier 60 (chosen purely for convenience, since 60 is divisible by 2, 3, 4, and 5).

Divide this multiplier by the complements of the stated fractions 1/2, 1/3, …, 1/6:

60 ÷ (1 – 1/2) = 120

60 ÷ (1 – 1/3) = 90

60 ÷ (1 – 1/4) = 80

60 ÷ (1 – 1/5) = 75

60 ÷ (1 – 1/6) = 72.

Add these quotients and divide by one less than the number of men involved:

(120 + 90 + 80 + 75 + 72) ÷ (5 – 1) = 437 ÷ 4 = 109¼,

which gives the total sum carried by the men. The money that each man carried individually is found by subtracting each of the above quotients from this number, one by one:

109¼ – 120 = -10¾

109¼ – 90 = 19¼

109¼ – 80 = 29¼

109¼ – 75 = 34¼

109¼ – 72 = 37¼.

The price of the piece of cloth can be found by subtracting the chosen multiplier from this same number:

109¼ – 60 = 49¼.

Check. We check that each man can meet the price according to his claim:

-10¾ + 1/2(120) = 49¼

19¼ + 1/3(90) = 49¼

29¼ + 1/4(80) = 49¼

34¼ + 1/5(75) = 49¼

37¼ + 1/6(72) = 49¼.

**Justification for the rote procedure:**

Let \(x_i =\) the money that man \(i\) carries

\({\overline{x_i}}=\) the money that the other 4 men carry

\(T=x_i +{\overline{x_i}}=\) the total money carried

\(c=\) the cost of the piece of cloth

\(x = T – c =\) the excess money.

Taking man 5 as an example, we start with the given condition: \[x_5 +\frac{1}{6}{\overline{x_5}}=c\] \[x_5 +{\overline{x_5}}=c+\frac{5}{6}{\overline{x_5}}\] \[T=c+\frac{5}{6}{\overline{x_5}}\] \[x=\frac{5}{6}{\overline{x_5}}\] \[{\frac{x}{\frac{5}{6}}}={\overline{x_5}}.\]

Similarly, \[{\frac{x}{\frac{4}{5}}}={\overline{x_4}},\,\,{\rm etc.}\] Summing all 5 such equations, we get \[{\frac{x}{\frac{1}{2}}}+\cdots+{\frac{x}{\frac{5}{6}}}={\overline{x_1}}+\cdots+{\overline{x_5}}.\]

Since the money that any one man carries is included in 4 of the 5 terms on the right-hand side, the total on the right is \(4T:\) \[{\frac{x}{\frac{1}{2}}}+\cdots+{\frac{x}{\frac{5}{6}}}=4T\] \[\frac{1}{4}\left[{\frac{x}{\frac{1}{2}}}+\cdots+{\frac{x}{\frac{5}{6}}}\right]=T\]

The parameter \(x\) is an arbitrary multiplicative constant, whose choice determines each \(x_i\) and thus their total, \(T.\) To determine the money that any man carries, say man 5, \[x_5=T-{\overline{x_5}}\] \[x_5={\frac{1}{4}}{\left[{\frac{x}{\frac{1}{2}}}+\cdots+{\frac{x}{\frac{5}{6}}}\right]}-{\frac{x}{\frac{5}{6}}},\]

and to determine the cost, \[c=T-x\] \[c=\frac{1}{4}\left[{\frac{x}{\frac{1}{2}}}+\cdots+{\frac{x}{\frac{5}{6}}}\right]-x.\]

**Solution B:** matrix algebra

Let \(x_i=\) the money that man \(i\) carries and \(c =\) the cost of the piece of cloth.

The given conditions can be thought of as a system of 5 linear equations and 5 unknowns, with a common parameter on the right-hand side: \[ {\begin{array}{r} x_1+\frac{1}{2}\left(x_2+x_3+x_4+x_5\right)=c\\ x_2+\frac{1}{3}\left(x_1+x_3+x_4+x_5\right)=c\\ x_3+\frac{1}{4}\left(x_1+x_2+x_4+x_5\right)=c\\ x_4+\frac{1}{5}\left(x_1+x_2+x_3+x_5\right)=c\\ x_5+\frac{1}{6}\left(x_1+x_2+x_3+x_4\right)=c\\ \end{array}}\]

In matrix form, this becomes: \[\left[ {\begin{array}{ccccc} 1&{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}} \\{\frac{1}{3}}&1&{\frac{1}{3}}&{\frac{1}{3}}&{\frac{1}{3}} \\{\frac{1}{4}}&{\frac{1}{4}}&1&{\frac{1}{4}}&{\frac{1}{4}} \\{\frac{1}{5}}&{\frac{1}{5}}&{\frac{1}{5}}&1&{\frac{1}{5}} \\{\frac{1}{6}}&{\frac{1}{6}}&{\frac{1}{6}}&{\frac{1}{6}}&1 \\ \end{array}}\right] \left[ {\begin{array}{l} x_1\\ x_2\\ x_3\\ x_4\\ x_5\\ \end{array}}\right] =c \left[ {\begin{array}{c} 1\\ 1\\ 1\\ 1\\ 1\\ \end{array}}\right]\]

\[\left[ {\begin{array}{l} x_1\\ x_2\\ x_3\\ x_4\\ x_5\\ \end{array}}\right] = c{{\left[ {\begin{array}{ccccc} 1&1/2&1/2&1/2&1/2 \\ 1/3&1&1/3&1/3&1/3 \\ 1/4&1/4&1&1/4&1/4 \\ 1/5&1/5&1/5&1&1/5 \\ 1/6&1/6&1/6&1/6&1 \\ \end{array}}\right]}^{-1}}\left[ {\begin{array}{c} 1\\ 1\\ 1\\ 1\\ 1\\ \end{array}}\right]\]

\[\left[ {\begin{array}{l} x_1\\ x_2\\ x_3\\ x_4\\ x_5\\ \end{array}}\right]={\frac{c}{394}}\left[ {\begin{array}{rrrrr} 548&-180&-160&-150&-144 \\ -120&501&-80&-75&-72 \\ -80&-60&472&-50&-48 \\ -60&-45&-40&455&-36 \\ -48&-36&-32&-30&444 \\\end{array}}\right]\left[ {\begin{array}{c} 1\\ 1\\ 1\\ 1\\ 1\\ \end{array}}\right]\] \[\left[{\begin{array}{l} {x_1} \\ {x_2} \\ {x_3} \\ {x_4} \\ {x_5} \\ \end{array}}\right]={\frac{c}{197}}\left[ {\begin{array}{r} -43 \\ 77 \\ 117 \\ 137 \\ 149 \\ \end{array}}\right].\]

If the cost \(c\) is chosen to be \(\frac{197}{4} = 49\frac{1}{4}\), this yields the sample solution found by rote in the Pamiers manuscript, as summarized above.