nyuuzyou/wikis huggingface dataset:
Greek Anthology Book XIV: 3. - Problem

== Problem ==
:Cypris thus addressed Love, who was looking downcast:
::"How, my child, has sorrow fallen on thee?"
:And he answered:
::"The Muses stole and divided among themselves, in different proportions, the apples I was bringing from Helicon, snatching them from my bosom.
::Clio got the fifth part,
::and Euterpe the twelfth,
::but divine Thalia the eighth.
::Melpomene carried off the twentieth part,
::and Terpsichore the fourth,
::and Erato the seventh;
::Polyhymnia robbed me of thirty apples,
::and Urania of a hundred and twenty,
::and Calliope went off with a load of three hundred apples.
:So I come to thee with lighter hands, bringing these fifty apples that the goddesses left me."
== Solution ==
Let $n$ be the number of apples Love started out with.
Then:
:$\dfrac n 5$ were stolen by Clio
:$\dfrac n {12}$ were stolen by Euterpe
:$\dfrac n 8$ were stolen by Thalia
:$\dfrac n {20}$ were stolen by Melpomene
:$\dfrac n 4$ were stolen by Terpsichore
:$\dfrac n 7$ were stolen by Erato
:$30$ were stolen by Polyhymnia
:$120$ were stolen by Urania
:$300$ were stolen by Calliope
:$50$ were salvaged by Love.
Thus:
Thus
:$672$ apples were stolen by Clea
:$280$ apples were stolen by Euterpe
:$420$ apples were stolen by Thalia
:$168$ apples were stolen by Melpomene
:$840$ apples were stolen by Terpsichore
:$480$ apples were stolen by Erato
in addition to:
:the $30$ that were stolen by Polyhymnia
:the $120$ that were stolen by Urania
:the $300$ that were stolen by Calliope
:the $50$ that were salvaged by Love
and it is seen that:
:$672 + 280 + 420 + 168 + 840 + 480 + 30 + 120 + 300 + 50 = 3360$
== Sources ==
* : $3$. -- 
Category:Greek Anthology Book XIV
Greek Anthology Book XIV: 4. - Problem

== Problem ==
:On the Dung of Augeas
::Heracles the mighty was questioning Augeas, seeking to learn the number of his herds, and Augeas replied:
::"About the streams of Alpheius, my friend, are the half of them;
::the eighth part pasture around the hill of Cronos,
::the twelfth part far away by the precinct of Taraxippus;
::the twentieth part feed in holy Elis,
::and I left the thirtieth part in Arcadia;
::but here you see the remaining fifty herds."
== Solution ==
Let $n$ be the number of herds owned by Augeas.
Then:
:$\dfrac n 2$ are about the streams of Alpheius
:$\dfrac n 8$ pasture around the hill of Cronos
:$\dfrac n {12}$ pasture far away by the precinct of Taraxippus
:$\dfrac n {20}$ feed in holy Elis
:$\dfrac n {30}$ were left in Arcadia
:$50$ remain to be seen by Heracles.
Thus:
Thus:
:$120$ are about the streams of Alpheius
:$30$ pasture around the hill of Cronos
:$20$ pasture far away by the precinct of Taraxippus
:$12$ feed in holy Elis
:$8$ were left in Arcadia
in addition to:
:the $50$ remain to be seen by Heracles
and it is seen that:
:$120 + 30 + 20 + 12 + 8 + 50 = 240$
== Sources ==
* : $4$. -- 
Category:Greek Anthology Book XIV
Greek Anthology Book XIV: 48. - Problem

== Problem ==
:The Graces were carrying baskets of apples, and in each was the same number.
:The nine Muses met them and asked them for apples, and they gave the same number to each Muse,
::and the nine and three had each of them the same number.
:Tell me how many they gave and how they all had the same number.
== Solution ==
According to tradition, there are .
Let $n$ be the number of apples owned by each party at the end of the transaction.
At the end of the transaction, there are $12$ parties each with the same number $n$.
Each of the then had $4 n$ apples at the start.
In total there were $12 n$ apples.
Any (strictly) positive integer $n$ satisfies the conditions.
== Sources ==
* : $48$. -- 
Category:Greek Anthology Book XIV
Greek Anthology Book XIV: 49. - Problem

== Problem ==
:Make me a crown weighing sixty , mixing gold and brass, and with them tin and much-wrought iron.
::Let the gold and bronze together form two-thirds,
::the gold and tin together three-fourths,
::and the gold and iron three-fifths.
:Tell me how much gold you must put in, how much brass, how much tin, and how much iron, so as to make the whole crown weigh sixty .
== Solution ==
Let $g, b, t, i$ be the quantities in of each of gold, brass, tin and iron.
We have:
So the required quantities are:
:gold: $30 \frac 1 2$ 
:brass: $9 \frac 1 2$ 
:tin: $14 \frac 1 2$ 
:iron: $5 \frac 1 2$ .
== Sources ==
* : $49$. -- 
Category:Greek Anthology Book XIV
Greek Anthology Book XIV: 50. - Problem

== Problem ==
:Throw me in, silversmith, besides the bowl itself,
::the third of its weight,
::and the fourth,
::and the twelfth;
:and casting them into the furnace stir them, and mixing them all up take out, please, the mass, and let it weigh one .
== Solution ==
Let $w$ be the weight in of the bowl.
We have:
Thus the bowl weighs $\dfrac 3 5$ of a .
Assuming $100$ to the , that is, $60$ .
== Sources ==
* : $50$. -- 
Category:Greek Anthology Book XIV
Greek Anthology Book XIV: 51. - Problem

== Problem ==
:$A$. I have what the second has and the third of what the third has.
:$B$. I have what the third has and the third of what the first has.
:$C$. And I have ten and the third of what the second has.
== Solution ==
Let $a$, $b$ and $c$ (respectively) be what $A$, $B$ and $C$ have in .
We have:
So:
:$A$ has $45$ 
:$B$ has $37 \frac 1 2$ 
:$C$ has $22 \frac 1 2$ .
== Sources ==
* : $51$. -- 
Category:Greek Anthology Book XIV
Greek Anthology Book XIV: 6. - Problem

== Problem ==
:"Best of clocks, how much of the day is past?"
:There remain twice two-thirds of what is gone.
== Solution ==
Let $h$ be the number of hours that have passed.
A day of $12$ hours is assumed.
Then:
So $5 \frac 1 7$ hours have passed, and $6 \frac 6 7$ remain.
== Sources ==
* : $6$. -- 
* : Metrodorus and the Greek Anthology: $34$
Category:Greek Anthology Book XIV
Greek Anthology Book XIV: 7. - Problem

== Problem ==
:I am a brazen lion; my spouts are my two eyes, my mouth, and the flat of my right foot.
::My right eye fills a jar in two days,
::my left eye in three,
::and my foot in four.
::My mouth is capable of filling it in six hours;
:tell me how long all four together will take to fill it.
== Solution ==
Let $t$ be the number of hours it takes to fill a jar.
Let $r, l, m, f$ be the flow rate in numbers of jars per day of (respectively) right eye, left eye, mouth and foot.
In $t$ hours, the various contributions of each of the spouts is:
:Right eye: $r t$
:Left eye: $l t$
:Right foot: $f t$
:Mouth: $m t$
So for the total contribution to be $1$ jar, we have:
There are $24$ hours in a day.
Hence, we have:
and so:
So the jar will be filled in $4 \frac {44} {61}$ hours, or $4$ hours, $43$ minutes and $17$ seconds, approximately.
== Variant ==
This is the prototype of a whole class of problems of this type which schoolchildren have been set over the centuries, for example:
== Historical Note ==
== Sources ==
* : $7$. -- 
* : Chapter $\text V$: Verbal Problems: History of Certain Standard Problems: The Cistern Problem
* : Light Reflected off a Mirror: $19$
Category:Greek Anthology Book XIV
Category:Cistern Problems
Greek Anthology Book XIV: 7. - Problem/Historical Note

== Historical Note on Problem $7$, Book $\text {XIV}$ of The Greek Anthology ==
In 's $1918$ translation of , he gives:
:The scholia propose several, two of which, by not counting fractions, reach the result of four hours; but the strict sum is $3 \frac {33} {37}$ hours.
The above is correct if it is assumed there are $12$ hours in a day.
It would also need to be assumed that, in order to fulfil the conditions of the statement of the problem, the spouts are turned off at night.
This problem was apparently first presented by in his Metrika.
It was still being taught in classrooms up until the middle of the $20$th Century, and was considered the epitome of "useless" mathematics.
This, as points out in his of $1992$, is a shame, because the idea behind it is far from useless.
== Proof ==
Assuming a $12$-hour day, the following can be calculated:
and so:
So the jar will be filled in $3 \frac {33} {37}$ hours.
== Sources ==
* : $7$. -- 
* : Light Reflected off a Mirror: $19$
Category:Historical Notes
Greek Anthology Book XIV: 7. - Problem/Mistake

== Source Work ==
: 
:The Puzzles:
::Light Reflected off a Mirror: $19$
:::Solution
== Mistake ==
:My right eye fills $1/8$ jar in $6$ hours (taking a day to be $24$ hours, where the Greeks might have taken it to be $12$), and my left eye fills $1/12$ in $6$ hours, and my foot $1/16$. Thus all four fill the jar $1 + 1/8 + 1/12 + 1/16 = \frac {13} {48}$ times in $6$ hours. So the jar will be filled once in $6 \times 48/61$ hours, or $47$ minutes and $13$ seconds, approximately.
::(Sandford, $1930$, p. $216$)
== Correction ==
While the arithmetic has been correctly worked out to be $6 \times 48/61$ hours, this does not work out to be $47$ minutes and $13$ seconds.
In fact it is $4 \frac {44} {61}$ hours, or $4$ hours, $43$ minutes and $17$ seconds, approximately.
Also, the author is , not Sandford.
== Sources ==
* : Light Reflected off a Mirror: $19$
Category:Mistakes/Curious and Interesting Puzzles
Greek Anthology Book XIV: 7. - Problem/Variant

== Problem ==
:$A$ and $B$ together can do a piece of work in $6$ days.
:$B$ and $C$ together can do it in $20$ days.
:$C$ and $A$ together can do it in $7 \frac 1 2$ days.
How many days will each require to do the job separately?
== Solution ==
It takes:
:$8$ days for $A$ to do the job
:$24$ days for $B$ to do the job
:$120$ days for $C$ to do the job.
== Proof ==
Let $a$, $b$ and $c$ denote the number of days it takes $A$, $B$ and $C$ to complete the job respectively.
Then each completes $\dfrac 1 a$, $\dfrac 1 b$ and $\dfrac 1 c$ of the job each day, respectively.
Thus:
Thus we have:
== Sources ==
* 
* : Light Reflected off a Mirror: $20$
Category:Greek Anthology Book XIV
Category:Cistern Problems
Greek Anthology Book XIV: 7. - Problem/Variant/Mistake

== Source Work ==
: 
:The Puzzles:
::Light Reflected off a Mirror: $20$
:::Solution
== Mistake ==
:$A$ and $B$ together can do a piece of work in $6$ days, $B$ and $C$ together in $20$ days, $C$ and $A$ together in $7 \frac 1 2$ days. How long will each require separately to do the same work?
The solution is given as:
:$A$, $B$ and $C$ do the whole job in $10$, $15$ and $30$ days respectively.
== Correction ==
With the numbers given, the results are actually:
:$A$, $B$ and $C$ do the whole job in $8$, $24$ and $120$ days respectively.
For the given solution to be correct, it would be necessary for the question to have been:
:$A$ and $B$ together can do a piece of work in $6$ days, $B$ and $C$ together in $10$ days, $C$ and $A$ together in $7 \frac 1 2$ days. How long will each require separately to do the same work?
(Emphasis added.)
One of the following things has happened:
:$(1): \quad$ There is an error in the source work, and has reported it as it has been printed.
:$(2): \quad$ The source work is correct, but a typo has been introduced into the question when it was entered into .
It has not been possible to check the source work by to determine which is the case.
== Sources ==
* : Light Reflected off a Mirror: $20$
Category:Mistakes/Curious and Interesting Puzzles
Greek Anthology Book XIV: 9. - Riddle

== Riddle ==
:My father-in-law killed my husband and my husband killed my father-in-law;
:my brother-in-law killed my father-in-law, and my father-in-law my father.
== Solution ==
 was the wife of , and the daughter of .
, who was the father of her second husband , killed both and .
, who had a reputation for being somewhat brutal, killed , the father of , and took as a wife (technically, a concubine).
Finally, , the brother of and therefore the brother-in-law of , killed her second father-in-law .
== Sources ==
* : $9$. -- 
* : Metrodorus and the Greek Anthology
Category:Greek Anthology Book XIV
Category:Strange Family Problems
Greek Anthology Book XIV: Metrodorus: 116

== Arithmetical Epigram of Metrodorus ==
:Mother, why dost thou pursue me with blows on account of the walnuts? Pretty girls divided them all among themselves.
::For Melission took two-sevenths of them from me,
::and Titane took the twelfth.
::Playful Astyoche and Philinna have the sixth and third.
::Thetis seized and carried off twenty,
::and Thisbe twelve,
::and look there at Glauce smiling sweetly with eleven in her hand.
:This one nut is all that is left to me.
== Solution ==
Let $n$ be the total number of walnuts.
:$\dfrac {2 n} 7$ were taken by Melission.
:$\dfrac n {12}$ were taken by Titane.
:$\dfrac n 6$ were taken by Astyoche.
:$\dfrac n 3$ were taken by Philinna.
:$20$ were taken by Thetis.
:$12$ were taken by Thisbe.
:$11$ were taken by Glauce.
:$1$ remains to the poor narrator.
Hence:
So the narrator started with $336$ walnuts, of which:
:$96$ were taken by Melission
:$28$ were taken by Titane
:$56$ were taken by Astyoche
:$112$ were taken by Philinna
and as we know:
:$20$ were taken by Thetis
:$12$ were taken by Thisbe
:$11$ were taken by Glauce
and $1$ remains to the narrator.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $116$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 117

== Arithmetical Epigram of Metrodorus ==
:$A$. Where are thy apples gone, my child?
:$B$. Ino has two-sixths
::and Semele one-eighth,
::and Autonoe went off with one-fourth,
::while Agave snatched from my bosom and carried away a fifth.
:For thee ten apples are left,
::but I, yes I swear it by dear Cypris, have only this one.
== Solution ==
Let $n$ be the total number of apples.
:$\dfrac {2 n} 6$ were taken by Ino.
:$\dfrac n 8$ were taken by Semele.
:$\dfrac n 4$ were taken by Autonoe.
:$\dfrac n 5$ were taken by Agave.
:$10$ remained for the child's parent.
:$1$ remained for the child.
Hence:
So the narrator started with $120$ apples, of which:
:$40$ were taken by Ino
:$15$ were taken by Semele
:$30$ were taken by Autonoe
:$24$ were taken by Agave
and as we know:
:$10$ remained for the child's parent
and $1$ remained for the child.
It is understood that $\dfrac {2 n} 6$ is not expressed in canonical form,that is: $\dfrac n 3$, but such is the nature of the epigram.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $117$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 118

== Arithmetical Epigram of Metrodorus ==
:Myrto once picked apples and divided them among her friends;
::she gave the fifth part to Chrysis,
::the fourth to Hero,
::the nineteenth to Psamathe,
::and the tenth to Cleopatra,
::but she presented the twentieth part to Parthenope
::and gave only twelve to Evadne.
:Of the whole number a hundred and twenty fell to herself.
== Solution ==
Let $n$ be the total number of apples.
:$\dfrac n 5$ were given to Chrysis.
:$\dfrac n 4$ were given to Hero.
:$\dfrac n {19}$ were given to Psamathe.
:$\dfrac n {10}$ were given to Cleopatra.
:$\dfrac n {20}$ were given to Parthenope.
:$12$ were given to Evadne.
:$120$ were kept by Myrto.
Hence:
So the narrator started with $380$ apples, of which:
:$76$ were given to Chrysis
:$95$ were given to Hero
:$20$ were given to Psamathe
:$38$ were given to Cleopatra
:$19$ were given to Parthenope
and as we know:
:$12$ were given to Evadne
and $120$ were kept by Myrto.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $118$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 119

== Arithmetical Epigram of Metrodorus ==
:Ino and Semele once divided apples among twelve girl friends who begged for them.
:Semele gave them each an even number and her sister an odd number,
::but the latter had more apples.
:Ino gave to three of her friends three-sevenths,
::and to two of them one-fifth of the whole number.
::Astynome took eleven away from her
::and left her only two apples to take to the sisters.
:Semele gave two quarters of the apples to four girls,
::and to the fifth one sixth part,
::to Eurychore she made a gift of four;
::she remained herself rejoicing in the possession of the other four apples.
== Solution ==
The wording of the epigram is ambiguous and unclear.
It is assumed that Ino and Semele are sisters, and the sentence:
:Semele gave them each an even number and her sister an odd number
means:
:Semele gave them each an even number and her sister (Ino) (gave them each) an odd number
Let $i$ be the number of apples that Ino distributed.
Let $s$ be the number of apples that Semele distributed.
We interpret the epigram as stating that:
:$\dfrac 3 7$ of Ino's apples were given to be shared between $3$ of her friends
:$\dfrac 1 5$ of Ino's apples were given to be shared between $2$ of her friends
:$11$ of Ino's apples went to Astynome
and:
:$2$ of Ino's apples remained to Ino and Semele.
:$\dfrac 2 4$ of Semele's apples were shared between $4$ of her friends
:$\dfrac 1 6$ of Semele's apples were given to a $5$th friend
:$4$ of Semele's apples were given to Eurychore.
:$4$ of Semele's apples were kept by Semele.
First we investigate Ino's share:
So Ino started with $35$ apples, of which:
:$15$ of Ino's apples were given to be shared between $3$ of her friends
:$7$ of Ino's apples were given to be shared between $2$ of her friends
and as we know:
:$11$ of Ino's apples went to Astynome
and $2$ of Ino's apples remained to Ino and Semele.
Then we investigate Semele's share:
So Semele started with $24$ apples, of which:
:$12$ of Semele's apples were shared between $4$ of her friends
:$4$ of Semele's apples were given to a $5$th friend
and as we know:
:$4$ of Semele's apples were given to Eurychore.
and $4$ of Semele's apples were kept by Semele.
The fact that Semele gave them an even number and Ino gave them an odd number is unnecessary information, and in fact misleadingly inaccurate.
In particular, if Ino gave $7$ apples to be shared between $2$ friends, one of them definitely received an even number of apples.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $119$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 120

== Arithmetical Epigram of Metrodorus ==
:The walnut-tree was loaded with many nuts, but now someone has suddenly stripped it.
:But what does he say?
::"Parthenopea had from me the fifth part of the nuts,
::to Philinna fell the eighth part,
::Aganippe had the fourth,
::and Orithyia rejoices in the seventh,
::while Eurynome plucked the sixth part of the nuts.
::The three Graces divided a hundred and six,
::and the Muses got nine times nine from me.
:The remaining seven you will find still attached to the farthest branches."
== Solution ==
Let $n$ be the number of walnuts that were originally on the tree.
:$\dfrac n 5$ went to Parthenopea
:$\dfrac n 8$ went to Philinna
:$\dfrac n 4$ went to Aganippe
:$\dfrac n 7$ went to Orithyia
:$\dfrac n 6$ went to Eurynome
:$106$ went to the three Graces
:$9 \times 9$ went to the Muses
:$7$ remain unpicked.
So we have:
So there were $1680$ walnuts on the tree, of which:
:$336$ went to Parthenopea
:$210$ went to Philinna
:$420$ went to Aganippe
:$240$ went to Orithyia
:$280$ went to Eurynome
and as we know:
:$106$ went to the three Graces
:$81$ went to the Muses
and $7$ remain unpicked.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $120$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 121

== Arithmetical Epigram of Metrodorus ==
:From Cadiz to the city of the seven hills
::the sixth of the road is to the banks of Baetis, loud with the lowing of herds,
::and hence a fifth to the Phocian soil of Pylades -- the land is Vaccaean, its name derived from the abundance of cows.
::Thence to the precipitous Pyrenees is one-eighth and the twelfth part of one-tenth.
::Between the Pyrenees and the lofty Alps lies one-fourth of the road.
::Now begins Italy and straight after one-twelfth appears the amber of the Po.
:O blessed am I who have accomplished two thousand and five hundred journeying thence!
:For the Palace on the Tarpeian rock is my journey's object.
== Solution ==
The city of the seven hills, and the location of the Palace on the Tarpeian rock, is of course Rome.
Let $n$ be the number of between Cadiz and Rome.
:$\dfrac n 6$ is the distance from Cadiz to the Baetis
:$\dfrac n 5$ is the distance from the Baetis to Pylades
:$\dfrac n 8 + \dfrac n {120}$ is the distance from Pylades to the Pyrenees
:$\dfrac n 4$ is the distance from the Pyrenees to the Alps
:$\dfrac n {12}$ is the distance from the Alps to the Po
:$2500$ is the distance from the Po to Rome itself.
So we have:
So it is $15 \, 000$ from Cadiz and Rome, of which:
:$2500$ is the distance from Cadiz to the Baetis
:$3000$ is the distance from the Baetis to Pylades
:$2000$ is the distance from Pylades to the Pyrenees
:$3750$ is the distance from the Pyrenees to the Alps
:$1250$ is the distance from the Alps to the Po
and as we know:
:$2500$ is the distance from the Po to Rome itself.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $121$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 122

== Arithmetical Epigram of Metrodorus ==
:After staining the holy chaplet of fair-eyed Justice that I might see thee, all-subduing gold, grow so much, I have nothing;
::for I gave forty talents under evil auspices to my friends in vain,
::while, O ye varied mischances of men, I see my enemy in possession of the half, the third, and the eighth of my fortune.
== Solution ==
Let $n$ be the value in of the narrator's fortune.
We have:
So the narrator started with $960$ .
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $122$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams
Greek Anthology Book XIV: Metrodorus: 123

== Arithmetical Epigram of Metrodorus ==
:Take, my son, the fifth part of my inheritance,
::and thou, wife, receive the twelfth;
::and ye four sons of my departed son and my two brothers, and thou my grieving mother, take each an eleventh part of the property.
::But ye, my cousins, receive twelve talents,
::and let my friend Eubulus have five talents.
:To my most faithful servants I give their freedom and these recompenses in payment of their service.
:Let them receive as follows.
::Let Onesimus have twenty-five minae
::and Davus twenty minae,
::Syrus fifty,
::Synete ten
::and Tibius eight,
::and I give seven minae to the son of Syrus, Synetus.
:Spend thirty talents on adorning my tomb and sacrifice to Infernal Zeus.
:From two talents let the expense be met of my funeral pyre, the funeral cakes, and grave-clothes,
::and from two let my corpse receive a gift.
== Solution ==
Let $n$ be the value in of the narrator's inheritance.
:To friends and relatives:
::$\dfrac n 5$ goes to the son
::$\dfrac n {12}$ goes to the wife
::$\dfrac {7 n} {11}$ goes to $4$ grandsons, $2$ brothers and mother
::$12$ go to the cousins
::$5$ go to Eubulus
:To the various servants:
::$25$ go to Onesimus
::$20$ go to Davus
::$50$ go to Syrus
::$10$ go to Synete
::$8$ go to Tibius
::$7$ go to Synetus
:In addition
::$30$ are spent on adorning the tomb and the sacrifice to Zeus
::$2$ are spent on the funeral pyre, party food and gladrags
::$2$ are spent on the gift for the corpse ("probably precious ointment", suggests ).
First let us add up the specific amounts.
The servants get:
Thus the total of the specific allocations:
Finally:
So the narrator's inheritance is valued at $660$ , of which:
:$132$ go to the son
:$55$ go to the wife
:$420$ go to the $4$ grandsons, $2$ brothers and mother
and the remaining $53$ are specifically allocated.
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $123$
Category:Greek Anthology Book XIV
Category:Metrodorus' Arithmetical Epigrams