nyuuzyou/wikis huggingface dataset:
Propositiones ad Acuendos Juvenes/Problems/11 - De Duobus Hominibus Singulas Sorores Accipientibus/Variant 2

== by : Problem $\text {11 b}$ ==
;De Patre et Filio et Vidua Eiusque Filia: A father and son and a widow and her daughter
:If a relict or widow and her daughter take a father and son in marriage,
:so that the son marries the mother and the father marries the daughter,
:what is the relationship of their sons?
== Solution ==
Each son is simultaneously the nephew and the uncle of the other.
:Let son $1$ be the son of Father and Daughter.
:Let son $2$ be the son of Son and Mother.
Then:
:$(1): \quad$ Son $1$ and Son are both the sons of Father, which makes son $1$ the uncle of son $2$
and at the same time:
:$(2): \quad$ Son $2$ and Daughter are both the children of Mother, which makes son $1$ the nephew of son $2$
and vice versa.
== Historical Note ==
== Sources ==
* 
* 
Category:Propositiones ad Acuendos Juvenes
Category:Strange Family Problems
Propositiones ad Acuendos Juvenes/Problems/12 - De Quodam Patrefamilias et Tribus Filiis Eius

== by : Problem $12$ ==
;De Quodam Patrefamilias et Tribus Filiis Eius: A father and his three sons
:A father, when dying, gave to his sons $30$ glass flasks,
:of which $10$ were full of oil,
:$10$ were half full,
:and the last $10$ were empty.
:Divide the oil and the flasks so that each of the $3$ sons receives equally of both glass and oil.
== Solution ==
To each son will come $10$ flasks, as follows:
:To one son, give him all the half full flasks.
:To the other sons, give them $5$ full flasks and $5$ empty ones.
== Also presented as ==
Some sources state that the flasks contained wine rather than oil.
== Sources ==
* 
* 
* : 'Propositions to Sharpen Up the Young': $80$
Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/13 - De Rege et de Eius Exercitu

== by : Problem $13$ ==
;De Rege et de Eius Exercitu: A King's Army
:A king ordered his servant to collect an army from $30$ manors,
:in such a way that from each manor he would take the same number of men as he had collected up till then.
:The servant went to the first manor alone;
:to the second he went with one other;
:to the next he took $3$ with him.
:How many were collected in all?
== Solution ==
:$2^{30} = 1 \, 073 \, 741 \, 824$
An extraordinarily large army.
== Proof ==
The assumption is that on visiting the first manor, the servant has collected himself, so to speak.
Otherwise the question never gets started::
:the servant collects no soldiers from the first manor,
:the same number as collected so far (none) from the second, and so on,
:until at the $30$th manor he still has collected no soldiers.
So the assumption as made is that on visiting the $1$st manor, the servant has collected $2^0 = 1$ soldier, namely, himself.
After visiting the first manor, the servant has $2^1 = 2$ soldiers.
Let $T_n$ denote the number of soldiers collected after visiting manor $n$.
On visiting manor $n + 1$, the servant has collected $T_n$ soldiers.
Thus on leaving manor $n + 1$, the servant has collected $T_n + T_n$, that is, $2 T_n$ soldiers.
It follows by Principle of Mathematical Induction that after visiting the $n$th manor, he has a total of $2^n$ soldiers.
The result follows.
== Historical Note ==
== Sources ==
* 
* 
* : 'Propositions to Sharpen Up the Young': $78$
Category:Propositiones ad Acuendos Juvenes
Category:Wheat and Chessboard Problem
Propositiones ad Acuendos Juvenes/Problems/13 - De Rege et de Eius Exercitu/Historical Note

== Historical Note on by : Problem $13$: De Rege et de Eius Exercitu ==
This is seen to be an example of the Wheat and Chessboard Problem, taken to not quite as great an extreme.
 and , in their translation of , give 's complete solution, which consists of documenting each of the powers of $2$ in turn:
:After the first stop there were $2$ men; after the second $4$; after the third $8$; after the fourth $16$; ... after the twenty-ninth $536870912$; after the thirtieth $1073741824$.
 goes on to note that in some of the historical manuscripts that propagate this problem, a number of them have incorrect values.
== Sources ==
* 
Category:Historical Notes
Propositiones ad Acuendos Juvenes/Problems/14 - De Bove

== by : Problem $14$ ==
;De Bove: An ox
:An ox ploughs a field all day.
:How many hoofprints does he leave in the last furrow?
== Solution ==
None.
== Proof ==
Trick question.
As the ox walks, he pulls behind him a plough.
A plough turns over the earth as it works, thus obliterating all the hoofprints made by the ox.
== Sources ==
* 
* 
* : 'Propositions to Sharpen Up the Young': $76$
Category:Propositiones ad Acuendos Juvenes
Category:Trick Puzzles
Propositiones ad Acuendos Juvenes/Problems/15 - De Homine

== by : Problem $15$ ==
;De Homine: A Man Ploughing
:How many furrows has a man made,
:when he has made $3$ turnings at each end of the field?
== Solution ==
:$7$.
== Proof ==
Let the ends of the field be $A$ and $B$.
:He starts at $A$ and ploughs the first furrow.
:He turns at $B$ and ploughs the second furrow.
:He turns at $A$ and ploughs the third furrow.
:He turns at $B$ and ploughs the fourth furrow.
:He turns at $A$ and ploughs the fifth furrow.
:He turns at $B$ and ploughs the sixth furrow.
:He turns at $A$ and ploughs the seventh furrow.
== Historical Note ==
== Sources ==
* 
* 
Category:Propositiones ad Acuendos Juvenes
Category:Fencepost Mistakes
Propositiones ad Acuendos Juvenes/Problems/15 - De Homine/Historical Note

== Historical Note on by : Problem $15$: De Homine ==
 himself gives $6$ as the answer to this.
 correctly gives $7$, as do most translators and interpreters.
It is assumed that has made a Fencepost Mistake.
Careful counting, as has been demonstrated here, indicates that the answer is indeed $7$.
== Sources ==
* 
Category:Historical Notes
Propositiones ad Acuendos Juvenes/Problems/16 - De Duobus Hominibus Boves Ducentibus

== by : Problem $16$ ==
;De Duobus Hominibus Boves Ducentibus: Two Men Leading Oxen
:Two men were leading oxen down the road, and one said to the other:
::""Give me two oxen and I'll have as many as you have."
:Then the other said,
::"Now you give me two oxen, and I'll have double the number you have."
:How many oxen were there, and how many did each have?
== Solution ==
:There were $12$ oxen:
::The first man had $4$
::The second man had $8$.
== Proof ==
Let $a$ and $b$ be the number of oxen owned at the start by the first and second man respectively.
It is assumed, by use of the word now in the question, that the suggested transaction actually went ahead, and the $2$ oxen actually changed hands.
We have:
== Historical Note ==
== Sources ==
* 
* 
Category:Propositiones ad Acuendos Juvenes
Category:Donkey and Mule Problems
Propositiones ad Acuendos Juvenes/Problems/16 - De Duobus Hominibus Boves Ducentibus/Historical Note

== Historical Note on by : Problem $16$: De Duobus Hominibus Boves Ducentibus ==
This is a less common variant of the traditional Donkey and Mule problem, in which the suggested exchange is not actually perceived as going ahead.
Hence, in the usual case, when the second man makes his suggested exchange, the assumption is that they are starting with their original distribution of goods.
In this case, the second suggestion happens after the first exchange has actually been done.
== Sources ==
* 
Category:Historical Notes
Propositiones ad Acuendos Juvenes/Problems/17 - De Tribus Fratribus Singulas Habentibus Sorores

== by : Problem $17$ ==
;De Tribus Fratribus Singulas Habentibus Sorores: Three Friends and their Sisters
:Three friends, each with a sister, needed to cross a river.
:Each one of them coveted the sister of another.
:At the river, they found only a small boat, in which only two of them could cross at a time.
:How did they cross the river, without any of the women being defiled by the men?
== Solution ==
The constraint here is that at no time may a man be on the same bank as (or in the boat with) the sister of another, without the brother of that sister being present as well.
Let $M_1$, $M_2$ and $M_3$ denote the men, and $S_1$, $S_2$ and $S_3$ denote their respective sisters.
Let them be crossing from $A$ to $B$.
As follows:
=== Solution 1 ===
=== Solution 2 ===
== Also presented as ==
This can also be presented as men with their wives, rather than sisters.
== Sources ==
* 
* 
* : 'Propositions to Sharpen Up the Young': $81$
Category:Propositiones ad Acuendos Juvenes
Category:River Crossing Problems
Category:Propositiones ad Acuendos Juvenes/Problems/17 - De Tribus Fratribus Singulas Habentibus Sorores
Propositiones ad Acuendos Juvenes/Problems/17 - De Tribus Fratribus Singulas Habentibus Sorores/Solution 1

== by : Problem $17$ ==
== Solution ==
The constraint here is that at no time may a man be on the same bank as (or in the boat with) the sister of another, without the brother of that sister being present as well.
Let $M_1$, $M_2$ and $M_3$ denote the men, and $S_1$, $S_2$ and $S_3$ denote their respective sisters.
Let them be crossing from $A$ to $B$.
As follows:
:$M_1$ and $S_1$ cross from $A$ to $B$.
:$M_1$ crosses back alone from $B$ to $A$.
:$S_2$ and $S_3$ cross from $A$ to $B$.
:$S_1$ crosses back alone from $B$ to $A$.
:$M_2$ and $M_3$ cross from $A$ to $B$.
:$M_2$ and $S_2$ cross back from $B$ to $A$.
:$M_1$ and $M_2$ cross from $A$ to $B$.
:$S_3$ crosses back alone from $B$ to $A$.
:$S_1$ and $S_3$ cross from $A$ to $B$.
:$M_3$ crosses back alone from $B$ to $A$.
:$M_3$ and $S_3$ cross from $A$ to $B$.
:"And that would complete the crossing without anything untoward happening."
This was the original solution given in by .
It is not optimal.
== Sources ==
* 
* 
* : 'Propositions to Sharpen Up the Young': $81$
Category:Propositiones ad Acuendos Juvenes/Problems/17 - De Tribus Fratribus Singulas Habentibus Sorores
Propositiones ad Acuendos Juvenes/Problems/17 - De Tribus Fratribus Singulas Habentibus Sorores/Solution 2

== by : Problem $17$ ==
== Solution ==
The constraint here is that at no time may a man be on the same bank as (or in the boat with) the sister of another, without the brother of that sister being present as well.
Let $M_1$, $M_2$ and $M_3$ denote the men, and $S_1$, $S_2$ and $S_3$ denote their respective sisters.
Let them be crossing from $A$ to $B$.
As follows:
:$M_1$ and $S_1$ cross from $A$ to $B$.
:$M_1$ crosses back alone from $B$ to $A$.
:$S_2$ and $S_3$ cross from $A$ to $B$.
:$S_1$ crosses back alone from $B$ to $A$.
:$M_1$ and $S_1$ cross from $A$ to $B$.
:$M_1$ crosses back alone from $B$ to $A$.
:$M_2$ and $M_3$ cross from $A$ to $B$.
:$S_1$ crosses back alone from $B$ to $A$.
:$M_1$ and $S_1$ cross from $A$ to $B$.
== Sources ==
* : 'Propositions to Sharpen Up the Young': $81$
Category:Propositiones ad Acuendos Juvenes/Problems/17 - De Tribus Fratribus Singulas Habentibus Sorores
Propositiones ad Acuendos Juvenes/Problems/18 - De Lupo et Capra et Fasciculo Cauli

== by : Problem $18$ ==
;De Lupo et Capra et Fasciculo Cauli: A Wolf, a Goat and a Bunch of Cabbages
:A man takes a wolf, a goat and a bunch of cabbages across the river.
:The only boat he could find could take only two of them at a time.
:But he had been ordered to transfer all of these to the other side in good condition.
:How could this be done?
== Solution ==
The constraints here are that:
:$(1): \quad$ The wolf, if left alone with the goat, would eat it, so these two may not be left alone together.
:$(2): \quad$ The goat, if left alone with the bunch of cabbages, would eat it, so these two may not be left alone together.
Let $M$, $W$, $G$ and $C$ denote the man, wolf, goat and cabbages respectively.
Let them be crossing from $A$ to $B$.
As follows:
=== Solution 1 ===
=== Solution 2 ===
== Also presented as ==
The objects being transported can be seen presented in various ways, for example:
:fox, duck and grain
:African leopard, goat and cassava leaves
and so on.
$4$ or more objects may be transported in this way, each being incompatible with its two neighbours, but in this case one of the objects needs to be able to row.
== Sources ==
* 
* 
* : 'Propositions to Sharpen Up the Young': $82$
Category:Propositiones ad Acuendos Juvenes
Category:River Crossing Problems
Category:Propositiones ad Acuendos Juvenes/Problems/18 - De Lupo et Capra et Fasciculo Cauli
Propositiones ad Acuendos Juvenes/Problems/18 - De Lupo et Capra et Fasciculo Cauli/Solution 1

== by : Problem $18$ ==
== Solution ==
The constraints here are that:
:$(1): \quad$ The wolf, if left alone with the goat, would eat it, so these two may not be left alone together.
:$(2): \quad$ The goat, if left alone with the bunch of cabbages, would eat it, so these two may not be left alone together.
Let $M$, $W$, $G$ and $C$ denote the man, wolf, goat and cabbages respectively.
Let them be crossing from $A$ to $B$.
As follows:
:$M$ takes $G$ from $A$ to $B$.
:$M$ returns to $A$.
:$M$ takes $W$ from $A$ to $B$.
:$M$ returns to $A$ with $G$.
:$M$ takes $C$ from $A$ to $B$.
:$M$ returns to $A$.
:$M$ takes $G$ from $A$ to $B$.
:"By this procedure there would be some healthy rowing, but no lacerating catastrophe."
== Sources ==
* 
* 
* : 'Propositions to Sharpen Up the Young': $82$
Category:Propositiones ad Acuendos Juvenes/Problems/18 - De Lupo et Capra et Fasciculo Cauli
Propositiones ad Acuendos Juvenes/Problems/18 - De Lupo et Capra et Fasciculo Cauli/Solution 2

== by : Problem $18$ ==
== Solution ==
The constraints here are that:
:$(1): \quad$ The wolf, if left alone with the goat, would eat it, so these two may not be left alone together.
:$(2): \quad$ The goat, if left alone with the bunch of cabbages, would eat it, so these two may not be left alone together.
Let $M$, $W$, $G$ and $C$ denote the man, wolf, goat and cabbages respectively.
Let them be crossing from $A$ to $B$.
As follows:
:$M$ takes $G$ from $A$ to $B$.
:$M$ returns to $A$.
:$M$ takes $C$ from $A$ to $B$.
:$M$ returns to $A$ with $G$.
:$M$ takes $W$ from $A$ to $B$.
:$M$ returns to $A$.
:$M$ takes $G$ from $A$ to $B$.
== Sources ==
* 
Category:Propositiones ad Acuendos Juvenes/Problems/18 - De Lupo et Capra et Fasciculo Cauli
Propositiones ad Acuendos Juvenes/Problems/19 - De Viro et Muliere Ponderantibus Plaustrum

== by : Problem $19$ ==
;De Viro et Muliere Ponderantibus Plaustrum: A Very Heavy Man and Woman
:A man and a woman, each the weight of a cartload,
:with two children who together weigh as much as a cartload,
:must cross a river.
:They find a boat which can only take one cartload.
:Make the transfer if you can, without sinking the boat.
== Solution ==
In order:
:The two children cross the river.
:One child returns with the boat.
:One of the parents crosses.
:The other child takes the boat back.
:Both children cross the river again.
:One child returns with the boat.
:The other parent crosses.
:The other child returns with the boat.
:The two children cross the river one final time.
:"With such ingenious rowing the navigation may be completed without shipwreck."
== Also presented as ==
Another version of this problem is that there are two unrelated children that the man and woman meet on the shore.
The two children wish to remain on their side of the river, with the boat.
The solution works in the same way as this.
== Sources ==
* 
* 
* : 'Propositions to Sharpen Up the Young': $83$
Category:Propositiones ad Acuendos Juvenes
Category:River Crossing Problems
Propositiones ad Acuendos Juvenes/Problems/1 - De Limace

== by : Problem $1$ ==
;De Limace: A Slug
:A leech invited a slug for lunch a leuca away.
:But he could only walk an inch a day.
:How many days will he have to walk for his meal?
== Solution ==
:$90 \, 000$ days, or $246$ years and $210$ days.
(This does not take into account leap years, as it assumes a year of strictly $365$ days.)
== Proof ==
It is understood that there are:
:$1500$ paces to a leuca
:$60$ inches to a pace.
Hence there are $90 \, 000$ inches to a leuca.
Thus at $1$ inch per day, it will take the slug $90 \, 000$ days to arrive for his meal.
== Sources ==
* 
* 
Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/20 - De Ericiis

== by : Problem $20$ ==
;De Ericiis: Hedgehogs
:About a male and female hedgehog with two young,
:having weight,
:wanting to cross a river.
The Latin text appears to be defective, and the question does not lend itself to rational interpretation.
However, it appears to be another example of $19$: De Viro et Muliere Ponderantibus Plaustrum (A Very Heavy Man and Woman).
== Solution ==
In order:
:The two younglings cross the river.
:One child returns with the boat.
:One of the parents crosses.
:The other child takes the boat back.
:Both children cross the river again.
:One child returns with the boat.
:The other parent crosses.
:The other child returns with the boat.
:The two children cross the river one final time.
== Sources ==
* 
* 
Category:Propositiones ad Acuendos Juvenes
Category:River Crossing Problems
Propositiones ad Acuendos Juvenes/Problems/21 - De Campo et Ovibus in Eo Locandis

== by : Problem $21$ ==
;De Campo et Ovibus in Eo Locandis: Sheep in a Field
:There is a field $200$ feet long and $100$ feet wide.
:I want to put sheep in it, so that each sheep has $5$ feet by $4$.
:How many sheep can I put in there?
== Solution ==
:$1000$.
== Proof ==
We have that:
:$5$ goes into $200$ a total of $40$ times
:$4$ goes into $100$ also a total of $25$ times.
Thus we can put $40 \times 25 = 1000$ sheep into that field.
== Sources ==
* 
* 
Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/22 - De Campo Fastigioso

== by : Problem $22$ ==
;De Campo Fastigioso: An Irregular Field
:There is an irregular field,
:measuring $100$ perches along each side,
:and $50$ perches along each end,
:but $60$ perches across the middle.
:How many acres does it contain?
== Solution ==
The area of an acre used here is $144$ square perches.
The length of the field is $100$ perches.
The length of the edges of the field is $50$ perches.
At the middle it is $60$ perches.
Join the length of the enda with the middle, which is $160$.
Take the $3$rd part of that, which is $53$.
Multiply it by $100$, which is $5300$.
Divide this into $12$ equal parts, giving $37$.
This gives the number of acres in the field.
== Historical Note ==
== Sources ==
* 
* 
Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/22 - De Campo Fastigioso/Historical Note

== Historical Note on by : Problem $22$: De Campo Fastigioso ==
's solution to this is suspect in a number of places.
:$(1): \quad$ Fractions are rounded inconsistently:
:::At one point $\dfrac {160} 3 = 53 \tfrac 1 3$ is rounded to $53$.
:::At another point $\dfrac {5200} {12} = 441 \tfrac 2 3$ is truncated to $441$.
:::At yet another point $\dfrac {441} {12} = 36 \tfrac 3 4$ is rounded up to $37$.
:::However, the actual value of $\dfrac {16 \, 000} {3 \times 12 \times 12} = 37.037 \ldots$ is actually remarkably accurate.
:$(2): \quad$ The method of averaging the $50$s and $60$s to get $\dfrac {160} 3$ is dubious.
:::If the field is a double trapezium, then the mean width is $55$.
:$(3): \quad$ The length given is measured along the side, which may not be the true length.
== Sources ==
* 
Category:Historical Notes
Propositiones ad Acuendos Juvenes/Problems/23 - De Campo Quadrangulo

== by : Problem $23$ ==
;De Campo Quadrangulo: A Four-Sided Field
:A four-sided field measures $30$ perches down one side
:and $32$ down the other;
:it is $34$ perches across the top
:and $32$ across the bottom.
:How many acres are included in the field?
== Solution ==
The sum of the lengths of the sides is $62$ perches, half of which is $31$.
The sum of the widths is $66$, half of which is $33$.
$31$ times this is $1023$.
Divide this into $12$ parts, as in $22$: De Campo Fastigioso, and there are $85$.
Again divide $85$ by $12$, and this gives $7$. 
Therefore there are $7$ acres here.
== Historical Note ==
== Sources ==
* 
* 
Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/23 - De Campo Quadrangulo/Historical Note

== Historical Note on by : Problem $23$: De Campo Quadrangulo ==
The formula being used here for a quadrilateral whose sides are $a$, $b$, $c$ and $d$ is the Egyptian Formula for Area of Quadrilateral:
:$\AA = \dfrac {a + b} 2 \times \dfrac {c + d} 2$
which may be a fair approximation if the field is approximately rectangular.
But the area of a quadrilateral depends not only on the lengths of its sides but also its angles.
The maximum area is obtained when the quadrilateral is cyclic is the area, in which case Brahmagupta's Formula can be used:
:$\AA^2 = \paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d}$
which in this case gives approximately $1022$ square units.
So in this case the Egyptian Formula for Area of Quadrilateral works fairly well.
== Sources ==
* 
Category:Historical Notes
Propositiones ad Acuendos Juvenes/Problems/24 - De Campo Triangulo

== by : Problem $24$ ==
;De Campo Triangulo: A Triangular Field
:A triangular field is $30$ perches along $2$ sides
:and $18$ perches along the bottom.
:How many acres must be enclosed?
== Solution ==
Join the $2$ long sides and make $60$.
Half of $60$ is $30$.
Because the bottom is $18$, take half of this which is $9$.
$9$ times $30$ is $270$.
Divide $270$ by $12$ which makes $22 \tfrac 1 2$.
Again divide by $12$, giving $1$ acre and $12$ perches.
== Historical Note ==
== Sources ==
* 
* 
Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/24 - De Campo Triangulo/Historical Note

== Historical Note on by : Problem $24$: De Campo Triangulo ==
When the answer is given in perches, it is clear that square perches is really meant.
Note that the acre as given is defined as $144$ square perches rather than the modern definition as $160$ of them.
As points out, the answer is in fact numerically incorrect: $22 \frac 1 2$ acre is in fact $126$ square perches.
However, using Heron's Formula:
:$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
where:
we obtain:
At $144$ square perches to the acre, this equals $1$ acre and $113.56$ square perches.
== Sources ==
* 
Category:Historical Notes
Propositiones ad Acuendos Juvenes/Problems/25 - De Campo Rotundo

== by : Problem $25$ ==
;De Campo Rotundo: A Round Field
:A round field is $400$ perches in circumference.
:How many acres does it contain?
== Solution ==
The fourth part of the field's $400$ perches in circumference is $100$.
Multiply this by itself to get $10 \, 000$.
Divide this in $12$ parts; the result is $833$.
Divide this into $12$ parts again; the result if $69$.
This is the number of acres in the field.
== Historical Note ==
== Sources ==
* 
* 
Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/25 - De Campo Rotundo/Historical Note

== Historical Note on by : Problem $25$: De Campo Rotundo ==
The area $\AA$ of a circle in terms of its circumference is:
:$\AA = \dfrac {C^2} {4 \pi}$
However, what has done is used a value for $\pi$ of $4$.
The actual value (assuming $144$ square perches to the acre) is approximately $88.4$ acres.
Another translator uses a tortuous method to obtain $\AA = 96$ acres, but this uses a value for $\pi$ of $3$.
== Sources ==
* 
Category:Historical Notes
Propositiones ad Acuendos Juvenes/Problems/26 - De Campo et Cursu Canis ac Fugo Leporis

== by : Problem $26$ ==
;De Campo et Cursu Canis ac Fugo Leporis: A Dog Chasing a Hare
:There is a field $150$ feet long.
:At one end is a dog, and at the other a hare.
:The dog chases when the hare runs.
:The dog leaps $9$ feet at a time,
:while the hare travels $7$ feet.
:How many feet will be travelled by the pursuing dog and the fleeing hare before the hare is seized?
== Solution ==
The dog travels $675$ feet.
The hare travels $525$ feet.
== Proof ==
Let $x$ be the length of time it takes the dog to catch the hare.
The dog travels $9 x$ during the time the hare travels $7 x$.
Thus we have:
:$9 x = 150 + 7 x$
which after algebra gives:
:$x = 75$
Hence the result.
== Sources ==
* 
* 
Category:Propositiones ad Acuendos Juvenes
Category:Hound and Hare Problems