nyuuzyou/wikis huggingface dataset:
Propositiones ad Acuendos Juvenes/Problems/27 - De Civitate Quadrangula

== by : Problem $27$ ==
;De Civitate Quadrangula: A Four-Sided Town
:A four-sided town measures $1100$ feet on one side
:and $1000$ feet on the other side;
:on one edge $600$
:and on the other edge $600$.
:I want to cover it with roofs of houses,
:each of which should be $40$ feet long
:and $30$ feet wide.
:How many dwellings can I make there?
== Solution ==
The two long sides add up to $2100$ feet.
Similarly the two short sides add up to $1200$ feet.
Then the half of $1200$ is $600$, and the half of $2100$ is $1050$.
Because each house is $40$ feet long and $30$ feet wide:
:take the $40$th part of $1050$ which is $26$
:and the $30$th part of $600$ which is $20$.
$20$ times $26$ is $520$.
That is how many houses can be built.
== Historical Note ==
== Sources ==
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Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/27 - De Civitate Quadrangula/Historical Note

== Historical Note on by : Problem $27$: De Civitate Quadrangula ==
The Egyptian Formula for Area of Quadrilateral is being used to estimate the total area of the town.
 makes no indication of actually trying to fit the houses into the town.
Because the shape of the town is not actually rectangular, the exercise is not trivial.
 reports that he has tried, but cannot actually get any more than $519$ in, and that is by turning the houses in either direction.
== Sources ==
* 
Category:Historical Notes
Propositiones ad Acuendos Juvenes/Problems/28 - De Civitate Triangula

== by : Problem $28$ ==
;De Civitate Triangula: A Triangular Town
:A triangular town measures $100$ feet along one side,
:$100$ feet along another side
:and $90$ feet along the front.
:I want to build houses there,
:each house being $20$ feet long
:and $10$ feet wide.
:How many houses can there be?
== Solution ==
Adding the two [[Definition:Side of Polygon|sides] makes $200$ feet.
The half of $200$ is $100$.
The front measures $90$ and the half of $90$ is $45$.
Because each house is $20$ feet long and $20$ feet wide:
:take one $20$th of $100$ which is $5$
:and take one $10$th of $40$ which is $4$.
Then $5$ times $4$ is $20$.
That is the number of houses the town should contain.
== Historical Note ==
== Sources ==
* 
* 
Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/28 - De Civitate Triangula/Historical Note

== Historical Note on by : Problem $28$: De Civitate Triangula ==
According to the Roman formula the area of the triangle is $4500$ square feet.
This is $22 \frac 1 2$ house areas.
The actual area is approximately $4019$ square feet, which is just over $20$ house areas.
The marking down of $45$ to $40$ in the calculation may be an attempt to compensate for the inaccuracy of the formula.
However, fitting them into a triangle is not easy.
 reports that he managed to fit $15$ in.
, on the other hand, managed $18$, but bent the walls of some of the houses slightly.
== Sources ==
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Category:Historical Notes
Propositiones ad Acuendos Juvenes/Problems/29 - De Civitate Rotunda

== by : Problem $29$ ==
;De Civitate Rotunda: A Round Town
:There is a round town, $8000$ feet in circumference.
:I want to build houses there,
:each house being $20$ feet long
:and $10$ feet wide.
:How many houses must it contain,
:each house being $30$ feet long
:and $20$ feet wide?
== Solution 1 ==
== Solution 2 ==
== Historical Note ==
== Sources ==
* 
* 
Category:Propositiones ad Acuendos Juvenes
Category:Propositiones ad Acuendos Juvenes/Problems/29 - De Civitate Rotunda
Propositiones ad Acuendos Juvenes/Problems/29 - De Civitate Rotunda/Historical Note

== Historical Note on by : Problem $29$: De Civitate Rotunda ==
The actual area of the town is approximately $8488$ house areas.
Using $\AA = \dfrac {C^2} {16}$ as in $25$: De Campo Rotundo, the area is approximately $6667$ house areas.
 assumes the circle contains a $1600 \times 2400$ rectangle, but such a circle could hold over $10 \, 000$ house areas.
 reports that he managed to fit $8307$ in, but suspects it may be possible to do better.
== Sources ==
* 
Category:Historical Notes
Propositiones ad Acuendos Juvenes/Problems/29 - De Civitate Rotunda/Solution 1

== by : Problem $29$ ==
== Solution ==
One circuit of the town is $8000$ feet.
Divided in proportion $3$ to $2$ gives $4800$ and $3200$.
These must contain the length and breadth of the houses.
So take from each one half and there remains:
:of the larger $2400$
:and of the smaller $1600$.
Divide $1600$ by $20$, and there are $80$ $20$s.
Again the greater, that is $2400$, divided into $30$s, gives $80$.
Take $80$ times $80$ and there are $6400$. 
This is the number of houses which can be built in the town according to the above instructions.
== Historical Note ==
== Sources ==
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Category:Propositiones ad Acuendos Juvenes/Problems/29 - De Civitate Rotunda
Propositiones ad Acuendos Juvenes/Problems/29 - De Civitate Rotunda/Solution 2

== by : Problem $29$ ==
== Solution ==
The boundary of the town is $8000$ feet.
Take a quarter of $8000$, getting $2000$.
Again take a third of $8000$, getting $2666$.
Take the half of $2000$, which is $1000$, and the half of $2666$, which is $1333$.
Not the thirtieth part of $1333$ is $44$.
Likewise the twentieth part of $1000$ is $50$.
$50$ times $44$ is $2200$.
Then form $2200$ times $4$ whch is $8800$.
This is the total number of houses.
== Historical Note ==
== Sources ==
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Category:Propositiones ad Acuendos Juvenes/Problems/29 - De Civitate Rotunda
Propositiones ad Acuendos Juvenes/Problems/2 - De Viro Ambulante in Via

== by : Problem $2$ ==
;De Viro Ambulante in Via: A Walker
:A man walking down the road saw others coming towards him, and he said to them,
:"I wish there were others with you, as many as you are,
:plus a quarter of the sum that would be,
:plus half of that last amount.
:Then with me there would be $100$ altogether."
:How many did he see on the road?
== Solution ==
:$36$.
== Proof ==
Let $x$ be the number seen on the road.
Then we have:
:$x + x + \dfrac {x + x} 4 + \dfrac 1 2 \paren {\dfrac {x + x} 4} + 1 = 100$
That is, after algebra:
:$\dfrac {11 x} 4 = 99$
Hence the result.
== Sources ==
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Category:Propositiones ad Acuendos Juvenes
Category:Heap Problems
Propositiones ad Acuendos Juvenes/Problems/30 - De Basilica

== by : Problem $30$ ==
;De Basilica: A Basilica
:A basilica is $240$ feet long and $120$ feet wide.
:It is paved with paving stones one foot $11$ inches long
:and $12$ inches, that is one foot, wide.
:How many stones are needed?
== Solution ==
It takes:
:$126$ paving stones to cover the length of $240$ feet
:and $120$ paving stones to cover the width of $120$ feet.
That is the number of paving stones required to pave the basilica.
== Sources ==
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Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/31 - De Canava

== by : Problem $31$ ==
;De Canava: A Wine Cellar
:There is a wine cellar $100$ feet long and $64$ feet wide.
:How many casks will it hold if each cask is $7$ feet long and $4$ feet wide across the middle,
:and there is one path $4$ feet wide?
== Solution ==
There are $14$ sevens in $100$, and $16$ fours in $64$.
But $4$ of those $64$ feet must be taken up by the path that runs the length of the cellar.
This leaves $60$ feet and there are $15$ fours in $60$.
Take $15$ times $14$ which is $210$.
That is the number of casks of the given length which can be contained in the wine cellar.
== Sources ==
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Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/32 - De Quodam Patrefamilias Distribuente Annonam

== by : Problem $32$ ==
;De Quodam Patrefamilias Distribuente Annonam: A Lord of the Manor Distributing Grain
:A gentleman has a household of $20$ persons and orders that they be given $20$ measures of grain.
:He directs that:
::each man should receive $3$ measures,
::each woman $2$ measures,
::and each child $\frac 1 2$ a measure.
:How many men, women and children must there be?
== Solution ==
:$1$ man, $5$ women and $14$ children.
== Proof ==
Let $m$, $w$ and $c$ denote the number of men, women and children respectively.
We have:
We note that $5 m$ is a multiple of $5$.
Hence $3 w$ also has to be a multiple of $5$.
Thus $w$ has to be a multiple of $5$.
Hence the following possible solutions for $m$ and $w$:
It is implicit that there are at least some women in the household, so the solution:
:$m = 4, w = 0, c = 16$
is usually ruled out.
Hence we have:
:$m = 1, w = 5, c = 14$
== Sources ==
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Category:Propositiones ad Acuendos Juvenes
Category:One Hundred Fowls Problems
Propositiones ad Acuendos Juvenes/Problems/33 - De Alio Patrefamilias Erogante Suae Familiae Annonam

== by : Problem $33$ ==
;De Alio Patrefamilias Erogante Suae Familiae Annonam: Another Landlord Apportioning Grain
:A gentleman has a household of $30$ persons and directs that they be given $30$ measures of grain.
:He directs that:
::each man should receive $3$ measures,
::each woman $2$ measures,
::and each child $\frac 1 2$ a measure.
:How many men, women and children must there be?
=== $\text {33 a}$: Variant ===
== Solution ==
:$3$ men, $5$ women and $22$ children.
== Proof ==
Let $m$, $w$ and $c$ denote the number of men, women and children respectively.
We have:
We note that $5 m$ is a multiple of $5$.
Hence $3 w$ also has to be a multiple of $5$.
Thus $w$ has to be a multiple of $5$.
Hence the following possible solutions for $m$ and $w$:
It is implicit that there are at least some men and some women in the household, so the solutions:
:$m = 6, w = 0, c = 24$
:$m = 0, w = 10, c = 20$
are usually ruled out.
Hence we have:
:$m = 3, w = 5, c = 22$
== Sources ==
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Category:Propositiones ad Acuendos Juvenes
Category:One Hundred Fowls Problems
Propositiones ad Acuendos Juvenes/Problems/33a - De Alio Patrefamilias Erogante Suae Familiae Annonam: Variant

== by : Problem $\text {33 a}$ ==
;De Alio Patrefamilias Erogante Suae Familiae Annonam: Another Landlord Apportioning Grain
:A gentleman has a household of $90$ persons and ordered that they be given $90$ measures of grain.
:He directs that:
::each man should receive $3$ measures,
::each woman $2$ measures,
::and each child $\frac 1 2$ a measure.
:How many men, women and children must there be?
== Solution ==
There are $5$ solutions:
:$15$ men, $5$ women and $70$ children
:$12$ men, $10$ women and $68$ children
:$9$ men, $15$ women and $66$ children
:$6$ men, $20$ women and $64$ children
:$3$ men, $25$ women and $62$ children.
The solution given by is:
:$6$ men, $20$ women and $64$ children.
== Proof ==
Let $m$, $w$ and $c$ denote the number of men, women and children respectively.
We have:
We note that $5 m$ is a multiple of $5$.
Hence $3 w$ also has to be a multiple of $5$.
Thus $w$ has to be a multiple of $5$.
Hence the following possible solutions for $m$ and $w$:
Hence the following possible solutions for $m$ and $w$:
It is implicit that there are at least some men and some women in the household, so the solutions:
:$m = 18, w = 0, c = 72$
:$m = 0, w = 30, c = 60$
are usually ruled out.
Hence we have the following solutions:
:$m = 15, w = 5, c = 70$
:$m = 12, w = 10, c = 68$
:$m = 10, w = 15, c = 66$
:$m = 6, w = 20, c = 64$
:$m = 3, w = 25, c = 62$
This can be expressed as:
where $n = 1, 2, \ldots, 5$.
== Sources ==
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Category:Propositiones ad Acuendos Juvenes
Category:One Hundred Fowls Problems
Propositiones ad Acuendos Juvenes/Problems/34 - Altera de Patrefamilias Partiente Familiae Suae Annonam

== by : Problem $34$ ==
;Altera de Patrefamilias Partiente Familiae Suae Annonam: Another Landowner Dividing Grain among his Household
:A gentleman has a household of $200$ persons and orders that they be given $100$ measures of grain.
:He directs that:
::each man should receive $3$ measures,
::each woman $2$ measures,
::and each child $\frac 1 2$ a measure.
:How many men, women and children must there be?
== Solution ==
There are $6$ solutions:
:$17$ men, $5$ women and $78$ children
:$14$ men, $10$ women and $76$ children
:$11$ men, $15$ women and $74$ children
:$8$ men, $20$ women and $72$ children
:$5$ men, $25$ women and $70$ children
:$2$ men, $30$ women and $68$ children.
The solution given by is:
:$11$ men, $15$ women and $74$ children.
== Proof ==
Let $m$, $w$ and $c$ denote the number of men, women and children respectively.
We have:
We note that $5 m$ is a multiple of $5$.
Hence $3 w$ also has to be a multiple of $5$.
Thus $w$ has to be a multiple of $5$.
Hence the following possible solutions for $m$ and $w$:
It is implicit that there are at least some women in the household, so the solution:
:$m = 20, w = 0, c = 80$
is usually ruled out.
Hence we have the following solutions:
:$m = 17, w = 5, c = 78$
:$m = 14, w = 10, c = 76$
:$m = 11, w = 15, c = 74$
:$m = 8, w = 20, c = 72$
:$m = 5, w = 25, c = 70$
:$m = 2, w = 30, c = 68$
This can be expressed as:
where $n = 1, 2, \ldots, 6$.
== Sources ==
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Category:Propositiones ad Acuendos Juvenes
Category:One Hundred Fowls Problems
Propositiones ad Acuendos Juvenes/Problems/35 - De Obitu Cuiusdam Patrisfamilias

== by : Problem $35$ ==
;De Obitu Cuiusdam Patrisfamilias: A Dying Man's Will
:A dying man left $960$ shillings and a pregnant wife.
:He directed that:
::if a boy is born he should receive $\frac 9 {12}$ of the estate and the mother $\frac 3 {12}$.
::However if a daughter is born she would receive $\frac 7 {12}$ of the estate and the mother $\frac 5 {12}$.
:It happened however that twins were born -- a boy and a girl.
:How much should the mother receive, how much the son, how much the daughter?
== Solution ==
$9$ and $3$ make $12$, and there are $12$ troy ounces to the troy pound.
Similarly, $5$ and $7$ make $12$.
Twice $12$ is $24$.
$24$ (troy) ounces makes $2$ (troy) pounds, that is, $40$ shillings.
Divide $960$ into $24$ parts.
Each part will be $40$ shillings.
Then take $9$ parts of $40$ shillings.
The son receives these $9$ $40$s, which is $18$ pounds, making $360$ shillings.
The mother takes $3$ parts compared to the son and $5$ parts compared to the daughter, and $3$ and $5$ make $8$.
So, take $8$ parts of $40$.
The mother receives $8$ of those $40$s, making $16$ pounds, or $320$ shillings.
Then take what remains, which is $7$ parts, that is $7$ parts of $40$, which is $14$ pounds or $280$ shillings.
This is what the daughter receives.
Adding $360$ and $320$ and $280$ gives $960$ shillings, or $48$ pounds.
== Historical Note ==
== Sources ==
* 
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* : 'Propositions to Sharpen Up the Young': $84$
:::although cites this as being Problem $\text {XXV}$ (which is $25$)
Category:Propositiones ad Acuendos Juvenes
Category:Testament Problems
Propositiones ad Acuendos Juvenes/Problems/35 - De Obitu Cuiusdam Patrisfamilias/Historical Note

== Historical Note on by : Problem $35$: De Obitu Cuiusdam Patrisfamilias ==
 discusses the general testament problem in his of $1925$.
He explains this type of problem and its legal origins in the law of the Roman Empire.
Commentators have argued that 's solution shows that he does not understand the law.
's take on this refers back to the "original translator", who suggests adding the original fractions they expected.
Thus we have:
:$\dfrac 3 4 + \dfrac 7 {12} + \dfrac 1 3$ (which was the mother's average expectation) for a total of $\dfrac 5 3$.
Multiplying their original expectations by $\dfrac 3 5$, this leaves:
:the mother with $432$ shillings
:the boy with $336$ shillings
:the girl with $192$ shillings.
== Sources ==
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* : 'Propositions to Sharpen Up the Young': $84$
Category:Historical Notes
Propositiones ad Acuendos Juvenes/Problems/36 - De Salutatione Cuiusdam Senis ad Puerum

== by : Problem $36$ ==
;De Salutatione Cuiusdam Senis ad Puerum: An Old Man Greeting a Boy
:An old man greeted a boy as follows:
:"May you live long --
::as long as you have lived so far,
::and as long again as your age would be then,
::and then to three times that age,
::and let God add one year more
::and you will be $100$.
:How old was the boy at the time?
== Solution ==
:$8$ years and $3$ months.
== Proof ==
Let $x$ years be the age of the boy at the time.
Then we have:
That is, $8$ years and $3$ months.
== Sources ==
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Category:Propositiones ad Acuendos Juvenes
Category:Heap Problems
Propositiones ad Acuendos Juvenes/Problems/37 - De Quodam Homine Volenti Aedificare Domum

== by : Problem $37$ ==
;De Quodam Homine Volenti Aedificare Domum: A Man Wanting to Build a House
:A man wanting to build a house contracted with $6$ builders,
:$5$ of whom were master builders.
:and the $6$th an apprentice,
:to build it for him.
:He agreed to pay them a total of $25$ pence a day,
:the apprentice to get half the rate of a master builder.
:How much did each receive a day?
== Solution ==
Each of the master builders earns $4 \frac 6 {11}$ pence per day.
The apprentice earns $2 \frac 3 {11}$ pence per day.
== Proof ==
Let $x$ pence denote the amount one master builder earns in one day.
Then we have:
It follows that the apprentice earns half that, that is $2 \frac 3 {11}$ pence, per day.
== Sources ==
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Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/38 - De Quodam Emptore in Animalibus Centum

== by : Problem $38$ ==
;De Quodam Emptore in Animalibus Centum: A Man Buying a Hundred Animals
:A man wanted to buy $100$ assorted animals for $100$ shillings.
:He was willing to pay $3$ shillings for a horse,
:$1$ shilling for an ox,
:and $1$ shilling for $24$ sheep.
:How many horses, oxen and sheep did he buy?
== Solution ==
:$23$ horses, $29$ oxen and $48$ sheep.
== Proof ==
Let $h$, $o$ and $s$ denote the number of horses, oxen and sheep respectively.
We have:
Note that both $h$ and $o$ need to be (strictly) positive.
We need to find possible values of $h$ such that $2300 - 71 h$ is divisible by $23$.
This can happen only when $h$ itself is divisible by $23$.
It is implicit that there are at least some horses are bought, so the solution:
:$h = 0, o = 100, s = 0$
is usually ruled out.
Hence we have:
:$h = 23, o = 29, s = 48$
== Sources ==
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Category:Propositiones ad Acuendos Juvenes
Category:One Hundred Fowls Problems
Propositiones ad Acuendos Juvenes/Problems/39 - De Quodam Emptore in Oriente

== by : Problem $39$ ==
;De Quodam Emptore in Oriente: An Oriental Merchant
:A man in the east wanted to buy $100$ assorted animals for $100$ shillings.
:He ordered his servant to pay $5$ shillings for a camel,
:$1$ shilling for a donkey,
:and $1$ shilling for $20$ sheep.
:How many camels, donkeys and sheep did he buy?
== Solution ==
:$19$ camels, $1$ donkey and $80$ sheep.
== Proof ==
Let $c$, $d$ and $s$ denote the number of camels, donkeys and sheep respectively.
We have:
Note that both $c$ and $d$ need to be (strictly) positive.
We need to find possible values of $c$ such that $2300 - 71 c$ is divisible by $19$.
This can happen only when $c$ itself is divisible by $19$.
It is implicit that there are at least some camels are bought, so the solution:
:$c = 0, d = 100, s = 0$
is usually ruled out.
Hence we have:
:$c = 19, d = 1, s = 80$
== Historical Note ==
== Sources ==
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Category:Propositiones ad Acuendos Juvenes
Category:One Hundred Fowls Problems
Propositiones ad Acuendos Juvenes/Problems/39 - De Quodam Emptore in Oriente/Historical Note

== Historical Note on by : Problem $39$: De Quodam Emptore in Oriente ==
This is numerically the same as a problem given by $\text {c. 900}$, except that the latter uses ducks, hens and sparrows.
 discusses the One Hundred Fowls problems at some length, even investigating $4$ and $5$ different types of bird.
Thus it may be that this problem was circulating in both Europe and Arabia for some time before presents it here.
 points out that may very well never have seen a camel, unless when he visited Rome.
However, he would of course have read about them in the Bible.
== Sources ==
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Category:Historical Notes
Propositiones ad Acuendos Juvenes/Problems/3 - De Duobus Proficistentibus Visis Ciconiis

== by : Problem $3$ ==
;De Duobus Proficistentibus Visis Ciconiis: Two Travellers and a Flock of Storks
:Two walkers saw some storks and wondered how many there were.
:Conferring, they decided:
::If there were the same number again,
::and again,
::and then a half and a third of the sum that would make,
::plus two more,
::that would make $100$.
:How many storks were seen?
== Solution ==
:$28$.
== Proof ==
Let $x$ be the of storks seen.
Then we have:
:$x + x + x + \dfrac 1 2 \times \dfrac 1 3 \times \paren {x + x + x} + 2 = 100$
That is, after algebra:
:$\dfrac {21 x} 6 = 98$
Hence the result.
== Sources ==
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Category:Propositiones ad Acuendos Juvenes
Category:Heap Problems
Propositiones ad Acuendos Juvenes/Problems/40 - De Homine et Ovibus in Monte Pascentibus

== by : Problem $40$ ==
;De Homine et Ovibus in Monte Pascentibus: A Man and some Sheep Grazing on the Hillside
:A man looked at some sheep grazing on the hillside, and said,
:"I wish I had these,
:and as many again,
:plus half of half that total,
:plus half of that last amount;
:then, counting myself, I would take $100$ to my home.
:How many sheep did he see?
== Solution ==
:$36$.
== Proof ==
Let $x$ be the number of sheep seen.
Then we have:
:$x + x + \dfrac {x + x} 4 + \dfrac 1 2 \paren {\dfrac {x + x} 4} + 1 = 100$
That is, after algebra:
:$\dfrac {11 x} 4 = 99$
Hence the result.
== Historical Note ==
== Sources ==
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Category:Propositiones ad Acuendos Juvenes
Category:Heap Problems
Propositiones ad Acuendos Juvenes/Problems/41 - De Sode et Scrofa

== by : Problem $41$ ==
;De Sode et Scrofa: A Breeding Sow and a Pigsty
:A farmer created a new yard in which he put a breeding sow,
:which produced a litter of $7$ piglets in the centre,
:which with their mother makes $8$.
:They all bear litters of $7$ piglets each in the first corner of the yard,
:then all of them bear litters of $7$ piglets each in the next corner
:and so on for all corners.
:Finally they all bear litters of $7$ in the centre pigsty.
:How many pigs are there now altogether, including the mothers?
== Solution ==
:$262 \, 144$.
== Proof ==
At the first birth, in the centre pigsty, there are $7$ piglets with the mother, making $8$.
In the first corner there are $8 \times 8 = 64$.
In the second corner there are $64 \times 8 = 512$.
In the third corner there are $512 \times 8 = 4096$.
In the fourth corner there are $4096 \times 8 = 32 \, 768$.
Multiplying this by $8$ gives $262 \, 144$.
This gives the total number of pigs after the final litters are produced in the centre pigsty.
Hence the result.
== Historical Note ==
== Sources ==
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Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/41 - De Sode et Scrofa/Historical Note

== Historical Note on by : Problem $41$: De Sode et Scrofa ==
First it is noted that all the piglets born must be female in order to have litters of their own.
The method of impregnation has also not been considered, but that is a mere detail.
What is interesting is that makes a mistake in his arithmetic, calculating $32 \, 788$ as the number of pigs in the fourth corner.
This mistake is propagated to the final quantity of pigs, which he gives as $262 \, 304$.
Arithmetic clearly continues to be difficult throughout history, as translations through the ages give several different wrong results.
It is also noted that the word sode is not a standard Latin word, but all the texts have it.
== Sources ==
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Category:Historical Notes