nyuuzyou/wikis huggingface dataset:
Propositiones ad Acuendos Juvenes/Problems/42 - De Scala Habente Gradus Centum

== by : Problem $42$ ==
;De Scala Habente Gradus Centum: One Hundred Steps
:A stairway consists of $100$ steps.
:On the first step stands a pigeon;
:on the second, two pigeons;
:on the third, three;
:on the fourth, four;
:on the fifth, five;
:and so on every step up to the hundredth.
:How many pigeons are there altogether?
== Solution ==
We count them as follows.
Take the single one on the first step and add it to the $99$ on the $99$th step, making $100$.
Taking the $2$nd with the $98$th likewise gives $100$.
So for each step, one of the higher steps combined with one of the lower steps, in this manner, will always give $100$ for the two steps.
However the $50$th step is alone, not having a pair.
Similarly the $100$th step is alone.
Join all together and get $5050$ pigeons.
== Historical Note ==
== Sources ==
* 
* 
* : 'Propositions to Sharpen Up the Young': $85$
Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/42 - De Scala Habente Gradus Centum/Historical Note

== Historical Note on by : Problem $42$: De Scala Habente Gradus Centum ==
 calls to mind the apocryphal story (quite possibly untrue) about , where at the age of $8$ he solved the problem of adding the numbers from $1$ to $100$ in a few seconds.
 remarks that the knowledge of how to sum an arithmetic series like this seems to have been known to the ancient Babylonians and ancient Egyptians.
It was also known to the ancient Greeks, but interestingly does not appear in .
's is the earliest text which dresses the problem up in fancy clothes, and many European texts follow his example.
In such renditions, the number of elements to add is almost always $100$.
== Sources ==
* 
* : 'Propositions to Sharpen Up the Young': $85$
Category:Historical Notes
Propositiones ad Acuendos Juvenes/Problems/43 - De Porcis

== by : Problem $43$ ==
;De Porcis: A puzzle of pigs
:A man has $300$ pigs, and orders that the pigs must be killed,
:an odd number each day,
:in $3$ days.
:How many pigs should be killed each day?
== Solution ==
In the words of :
:This is a fable.
:No-one can solve how to kill $300$ pigs in $3$ days, an odd number each day.
:This fable is given to children.
Modern interpretations can produce answers along the lines:
:$1$ on the first day, $1$ on the second day, and $298$ on the third day.
:$298$ is a very odd number of pigs to kill in one day.
As such, this is a fallacy of equivocation:
:odd meaning either an odd number as opposed to an even number
:odd meaning unusual or strange.
== Also presented as ==
== Historical Note ==
== Sources ==
* 
* 
* : 'Propositions to Sharpen Up the Young': $76$
Category:Propositiones ad Acuendos Juvenes
Category:Trick Puzzles
Propositiones ad Acuendos Juvenes/Problems/43 - De Porcis/Also presented as

== by : Problem $43$: Also presented as ==
;De Porcis: A puzzle of pigs
Some sources present this problem with the number of pigs being set to $30$.
The answer is the same.
You cannot add $3$ odd numbers and reach $30$.
== Historical Note ==
== Sources ==
* 
* 
Category:Propositiones ad Acuendos Juvenes
Category:Trick Puzzles
Propositiones ad Acuendos Juvenes/Problems/43 - De Porcis/Historical Note

== Historical Note on by : Problem $43$: De Porcis ==
The question is raised in some sources as to whether this problem can be construed as cruelty to children.
Some suggest that may have used this problem to distract the more rambunctious of his pupils.
In any event, this is an early recognition that there exist problems which have no solution.
A more subtle variant on the problem has arisen in more recent years: dividing an odd number of pigs into pens into an even number of pens so that is an odd number of pigs in each pen.
The problem is solved by housing them in concentric pens, for example: $3$ pigs can be partitioned into $2$ pens with one pig in the inside pen, and $2$ in the outside pen.
== Sources ==
* 
* 
* : 'Propositions to Sharpen Up the Young': $76$
Category:Historical Notes
Propositiones ad Acuendos Juvenes/Problems/44 - De Salutatione Pueri ad Patrem

== by : Problem $44$ ==
;De Salutatione Pueri ad Patrem: A Son's Greeting to his Father
:A son greeted his father: "Hello Father;"
:to which the father replied: "Hello Son.
:May you live long, as much as you have lived.
:If you triple that number and add one of my years,
:you will have lived $100$ years."
:How old was the boy at the time?
== Solution ==
:$16$ years and $6$ months.
== Proof ==
Let $x$ years be the age of the boy at the time.
Then we have:
That is, $16$ years and $6$ months.
== Sources ==
* 
* 
Category:Propositiones ad Acuendos Juvenes
Category:Heap Problems
Propositiones ad Acuendos Juvenes/Problems/45 - De Columba

== by : Problem $45$ ==
;De Columba: A Pigeon
:A pigeon sitting in a tree saw others flying, and said to them:
:"I wish there were others with you --
:the same number again and a third time.
:Then with one of me, there would be $100$."
:How many pigeons were there at first?
== Solution ==
:$33$
== Proof ==
Let $x$ years be the number of pigeons there at first.
Then we have:
== Sources ==
* 
* 
Category:Propositiones ad Acuendos Juvenes
Category:Heap Problems
Propositiones ad Acuendos Juvenes/Problems/46 - De Sacculo ab Homine Invento

== by : Problem $46$ ==
;De Sacculo ab Homine Invento: A Man Finding a Purse
:A man walking along a road, found a purse containing $2$ talents.
:Others saw this and said: "Friend, give us a portion of your find."
:He refused to give any to them.
:So they set upon him and took the bag from him,
:and each one took $50$ shillings.
:Seeing that he could not stop them, he reached out and snatched $50$ shillings for himself.
:How many men were there?
== Solution ==
:$216$
== Proof ==
There are $75$ (troy) pounds in a talent.
Apparently, in this context, there are also $72$ gold shillings in a troy pound.
Hence there are $75 \times 72 = 5400$ shillings in a talent.
Let $x$ be the number of men that were there.
Then we have:
== Historical Note ==
== Sources ==
* 
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Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/46 - De Sacculo ab Homine Invento/Historical Note

== Historical Note on by : Problem $46$: De Sacculo ab Homine Invento ==
 refers to similar problems by and .
He also references The Bloom of Thymarides, given by both and , still to be investigated by .
He also suggests that it is related to $16$: De Duobus Hominibus Boves Ducentibus, classified as a Donkey and Mule problem, but the structure of that and this are considerably different.
== Sources ==
* 
Category:Historical Notes
Propositiones ad Acuendos Juvenes/Problems/47 - De Episcopo qui Jussit XII Panes in Clero Dividi

== by : Problem $47$ ==
;De Episcopo qui Jussit $\text {XII}$ Panes in Clero Dividi: A Bishop Dividing $12$ Loaves Among his Clergy
:A certain bishop ordered that $12$ loaves be divided among his clergy.
:He ordered that:
::each priest should receive $2$ loaves,
::each deacon one half
::and each reader one quarter.
:There were as many loaves as clergy.
:How many priests, deacons and readers must there be?
== Solution ==
:$5$ priests
:$1$ deacon
:$6$ readers.
== Proof ==
Let $p$, $d$ and $r$ denote the number of priests, deacons and readers respectively.
We are to solve for $p, d, r\in \N$:
Thus:
:$d = 36 - 7 p$
Inspecting possible contenders for $p$ and $d$ individually, and calculating $r$:
Only $2$ of these satisfie the condition that $12 - \paren {p + 2} \ge 0$.
:$p = 4$, $d = 8$, $r = 0$
:$p = 5$, $d = 1$, $r = 6$
It is understood that there is at least one reader.
Hence the result:
:$p = 5$, $d = 1$, $r = 6$
Thus:
:the priests get $10$ loaves between them;
:the deacon gets half a loaf;
:the readers get $1 \frac 1 2$ loaves divided between them.
== Sources ==
* 
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Category:Propositiones ad Acuendos Juvenes
Category:One Hundred Fowls Problems
Propositiones ad Acuendos Juvenes/Problems/48 - De Homine qui Obviavit Scolaris

== by : Problem $48$ ==
;De Homine qui Obviavit Scolaris: A Man Meeting some Scholars
:A man met some scholars and said to them: "How many are there in your school?"
:One of them answered: "I don't want to tell you that.
:Count us twice and multiply by $3$.
:Then divide into $4$ parts.
:If you add me to this $4$th part, you will have $100$."
== Solution ==
:$66$.
== Proof ==
Let $x$ be the number of scholars in the school.
Then we have:
:$\dfrac {3 \paren {x + x} } 4 + 1 = 100$
That is, after algebra:
:$\dfrac {3 x} 2 = 99$
The result follows.
== Sources ==
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Category:Propositiones ad Acuendos Juvenes
Category:Heap Problems
Propositiones ad Acuendos Juvenes/Problems/49 - De Carpentariis

== by : Problem $49$ ==
;De Carpentariis: Carpenters
:$7$ carpenters made $7$ wheels each.
:How many carts were fitted?
== Solution ==
:$12$, with one wheel left over.
== Proof ==
Let $x$ be the number of carts fitted.
Let $y$ be the number of wheels made.
It is presumed that each cart has $4$ wheels.
Thus we have:
That is, we have that:
:$49 = 4 \times 12 + 1$
The result follows.
== Sources ==
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* 
Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/4 - De Homine et Equis in Campo Pascentibus

== by : Problem $4$ ==
;De Homine et Equis in Campo Pascentibus: Some Horses Grazing
:A man saw some horses at pasture and wished they were his,
:and that there were others with them that were his,
:the same number again,
:plus a quarter of the sum that would result,
:for then he would glory in $100$ horses.
:How many did the man see at pasture?
== Solution ==
:$40$.
== Proof ==
Let $x$ be the number seen at pasture.
Then we have:
:$x + x + \dfrac {x + x} 4 = 100$
That is, after algebra:
:$\dfrac {10 x} 4 = 100$
Hence the result.
== Sources ==
* 
* 
Category:Propositiones ad Acuendos Juvenes
Category:Heap Problems
Propositiones ad Acuendos Juvenes/Problems/50 - De Vino in Vasculum

== by : Problem $50$ ==
;De Vino in Vasculum: Wine in Flasks
:How many sextarii are there in $100$ metri of wine,
:and how many meri are there in the same $100$ metri?
== Solution ==
:$4800$ sextarii
which is:
:$28 \, 800$ meri.
== Proof ==
There are $48$ sextarii to the metra.
There are $6$ meri to the sextarius.
Thus the number of sextarii in $100$ metri is:
:$48 \times 100 = 4800$
and the meri in $100$ metri is:
:$6 \times 4800 = 28 \, 800$
== Historical Note ==
== Sources ==
* 
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Category:Propositiones ad Acuendos Juvenes
Category:Heap Problems
Propositiones ad Acuendos Juvenes/Problems/50 - De Vino in Vasculum/Historical Note

== Historical Note on by : Problem $50$: De Vino in Vasculum ==
The original text from gives that there are $289$ meri to the metra, rather than the $288$ that are obtained by straightforward multiplication.
It is not clear whether this is what the exact number of them is, or whether made a mistake in his arithmetic when multiplying $48$ by $6$.
== Sources ==
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Category:Historical Notes
Propositiones ad Acuendos Juvenes/Problems/51 - De Vino in Vasculis a Quodam Patre Distributo

== by : Problem $50$ ==
;De Vino in Vasculis a Quodam Patre Distributo: A Father Dividing Flasks of Wine
:A dying man left his sons four flasks of wine.
:In the first flask there were $40$ measures,
:in the second $30$,
:in the third $20$,
:and in the fourth $10$.
:Calling his steward, he said:
:"Divide these $4$ flasks among my $4$ sons so that each of their portions shall be equal,
:both in wine and in flasks."
== Solution ==
Add all the wine together to get $100$ measures.
Take half of that to get $50$ measures.
Add the $10$ measures to the $40$ to make $50$ measures, and divide that up between $2$ of the sons, so they get $25$ measures each.
Then add the $20$ measures to the $30$ to again make $50$ measures, and divide that up between the other $2$ sons, so they also get $25$ measures each.
== Historical Note ==
== Sources ==
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Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/51 - De Vino in Vasculis a Quodam Patre Distributo/Historical Note

== Historical Note on by : Problem $51$: De Vino in Vasculis a Quodam Patre Distributo ==
 points out that "the solution is hardly more enlightening than the problem."
It may be intended that some shifting of the wine is to happen.
== Sources ==
* 
* 
Category:Historical Notes
Propositiones ad Acuendos Juvenes/Problems/52 - De Homine Patrefamilias

== by : Problem $52$ ==
;De Homine Patrefamilias: A Lord of the Manor
:A man ordered that $90$ measures of grain were to be moved from one of his houses to another, $30$ leucas away.
:One camel was to carry the grain in $3$ journeys, carrying $30$ measures on each journey.
:The camel eats one measure for each leuca.
:How many measures will remain?
== Solution ==
$20$ measures.
== Proof ==
On the $1$st journey, the camel carries $30$ measures over $20$ leucas, eating $1$ measure for each leuca, leaving $10$ leucas.
On the $2$nd journey, the camel likewise carries $30$ measures over $20$ leucas, eating $1$ measure for each leuca, leaving $10$ leucas.
On the $3$rd journey, the camel again carries $30$ measures over $20$ leucas, eating $1$ measure for each leuca, leaving $10$ leucas.
Now he is at the point $10$ leucas away, with $30$ measures to take over $10$ leucas.
During that journey the camel eats $10$ measures, leaving $20$ measures.
== Historical Note ==
== Sources ==
* 
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Category:Propositiones ad Acuendos Juvenes
Category:Desert Crossing Problems
Propositiones ad Acuendos Juvenes/Problems/52 - De Homine Patrefamilias/Historical Note

== Historical Note on by : Problem $52$: De Homine Patrefamilias ==
 suggests that the camel does in fact make $4$ journeys, not $3$.
However, contend that the $3$rd journey to carry $30$ measures $20$ leucas, then $30$ leucas the remaining $10$ leucas, is in fact one journey, just broken into $2$ parts.
It is suggested that there may be a further unspoken constraint given here: that a camel may not be able to carry more than $30$ measures at one time.
Note also that the camel does not require food when he is not carrying a load.
There exists a better solution if more than $3$ trips are allowed:
:The camel makes $3$ trips carrying $30$ measures to $10$ leucas, there now being $60$ measures at $10$ leucas.
:The camel makes $2$ trips carrying $30$ measures to a point another $15$ leucas further, there now being $30$ measures $25$ leucas on.
:The camel makes $1$ further trip of $5$ leucas carrying those $30$ measures to the destination, eating $5$ and so carrying $25$ measures in total.
This may be the earliest appearance of a desert crossing problem.
== Sources ==
* 
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Category:Historical Notes
Propositiones ad Acuendos Juvenes/Problems/53 - De Homine Patrefamilias Monasterii XII Monachorum

== by : Problem $53$ ==
;De Homine Patrefamilias Monasterii $\text {XII}$ Monachorum: An Abbot with $12$ Monks
:An abbot had $12$ monks in his monastery.
:Calling his steward he gave him $204$ eggs and ordered that he give equal shares to each monk.
:Thus he ordered that he give:
::$85$ eggs to the $5$ priests,
::and $68$ to the $4$ deacons
::and $51$ to the $3$ readers.
:How many eggs went to each monk, so that none had too many or too few,
:but all received equal portions as above?
== Solution ==
$17$.
== Proof ==
Take the $12$th part of $204$.
This $12$th part is $17$, so $204$ is $12$ times $17$ or $17$ times $12$.
Just as $85$ is $5$ times $17$, so is $68$ four times and $51$ three times.
Now $5$ and $4$ and $3$ are $12$.
There are $12$ men.
Again add $85$ and $68$ and $51$ which is $204$.
There are $204$ eggs.
Therefore to each comes $17$ eggs as the $12$th part.
== Historical Note ==
== Sources ==
* 
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Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/53 - De Homine Patrefamilias Monasterii XII Monachorum/Historical Note

== Historical Note on by : Problem $53$: De Homine Patrefamilias Monasterii $\text {XII}$ Monachorum ==
It is suggested that this problem has been corrupted.
It feels as though it ought to be a one hundred fowls problem, but the answer is provided as part of the question.
The answer looks as though the scribe is bewilderedly wondering if it be a trick question.
Compare it with Problem $47$: De Episcopo qui Jussit XII Panes in Clero Dividi.
== Sources ==
* 
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Category:Historical Notes
Propositiones ad Acuendos Juvenes/Problems/5 - De Emptore in C Denariis

== by : Problem $5$ ==
;De Emptore in $\text C$ Denariis: A Merchant and $100$ Pence
:A merchant wanted to buy $100$ pigs at $100$ pence.
:For a boar he would pay $10$ pence;
:and for a sow $5$ pence;
:while he would pay $1$ penny for a couple of piglets.
:How many boars, sows and piglets must there have been for him to have paid exactly $100$ pence for the $100$ animals?
== Solution ==
:$1$ boar
:$9$ sows
:$90$ piglets.
== Proof ==
Let $x$, $y$ and $z$ denote the number of boars, sows and piglets respectively.
We are to solve for $x, y, z \in \N$:
Note that both $x$ and $y$ need to be (strictly) positive.
We need to find possible values of $x$ such that $100 - 19 x$ is divisible by $9$.
Inspecting possible contenders for such an $x$ individually:
Only one of these satisfies the condition, that is:
:$x = 1$
:$y = 9$
which leaves:
:$z = 100 - 1 - 9 = 90$
Hence the result.
== Historical Note ==
== Sources ==
* 
* 
Category:Propositiones ad Acuendos Juvenes
Category:One Hundred Fowls Problems
Propositiones ad Acuendos Juvenes/Problems/6 - De Duobis Negotiatoribus C Solidos Communis Habentibus

== by : Problem $6$ ==
;De Duobis Negotiatoribus $\text C$ Solidos Communis Habentibus: $2$ Wholesalers with $100$ Shillings
:$2$ wholesalers with $100$ shillings between them bought some pigs with the money.
:They bought at the rate of $5$ pigs for $2$ shillings, intending to fatten them up and sell them again, making a profit.
:But when they found that it is not the right time of year for fattening pigs,
:and they were not able to feed them through the winter,
:they tried to sell them again to make a profit.
:But they couldn't, because they could only sell them for the price they had paid for them, $2$ shillings for $5$ pigs.
:When they saw this, they said to each other, "Let's divide them.
:By dividing them, and selling them again at the rate they had bought them for, they made a profit.
:How could they be divided to make a profit, which could not be made by selling them all at once?
== Solution ==
First there were $250$ pigs bought with $100$ shillings at the above mentioned rate, for $5$ fifties are $250$.
On division, each merchant had $125$ pigs.
One sold the poorer quality pigs at $3$ for a shilling.
The other sold the better quality pigs at $2$ for a shilling.
The one who sold the poorer quality pigs received $40$ shillings for $120$ pigs.
The one who sold the better quality pigs received $60$ shillings for $120$ pigs.
Then there remained $5$ of each sort of pig.
From these they could make a profit of $4$ shillings and $2$ pence.
== Historical Note ==
== Sources ==
* 
* 
* : 'Propositions to Sharpen Up the Young': $77$
Category:Propositiones ad Acuendos Juvenes
Category:Applesellers' Problem
Propositiones ad Acuendos Juvenes/Problems/6 - De Duobis Negotiatoribus C Solidos Communis Habentibus/Historical Note

== Historical Note on by : Problem $6$: De Duobis Negotiatoribus ==
According to , this is the first appearance of the type of problem known as the Applesellers' Problem, also known as the missing penny or marketwomen's problem.
It hinges on the fact that a common mean is erroneously being calculated of $2$ for a monetary unit plus $3$ for a monetary unit equals $5$ for $2$ monetary units.
 goes on to remark that, while this problem appears everywhere in European puzzle-books since its appearance here, he has never seen a non-European version.
However, claims in his that it appears in 's , which dates from c. $850$. However, this has not been corroborated.
== Sources ==
* 
Category:Historical Notes
Propositiones ad Acuendos Juvenes/Problems/7 - De Disco Pesante Libras XXX

== by : Problem $7$ ==
;De Disco Pesante Libras $\text {XXX}$ : A Dish weighing $30$ Pounds
:A dish weighing $30$ pounds is made of gold, silver, brass, and lead.
:It contains $3$ times as much silver as gold;
:$3$ times as much brass as silver;
:and $3$ times as much lead as brass.
:How much is there of each?
== Solution ==
:Gold: $9$ ounces
:Silver: $2$ (troy) pounds and $3$ ounces
:Brass: $6$ (troy) pounds and $9$ ounces
:Lead: $20$ (troy) pounds and $3$ ounces.
== Proof ==
Let $x$ be the amount in (troy) pounds of gold in the dish.
Then:
:the amount of silver in the dish is $3 x$
:the amount of brass in the dish is $3 \times 3 x = 9 x$
:the amount of lead in the dish is $3 \times 9 x = 27 x$
Thus, knowing the total quantity of metal in the dish:
Given that there are $12$ (troy) ounces in one (troy) pound, we can express this as:
== Sources ==
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Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/8 - De Cupa

== by : Problem $8$ ==
;De Cupa: A Cask and Three Pipes
:A cask is filled with $100$ metretae through $3$ pipes.
:One third plus one sixth of the capacity flows through one pipe,
:one third of the capacity flows through another,
:but only a sixth of the capacity flows through the third.
:How many sextarii flow in through each pipe?
== Solution ==
:$3600$ sextarii flow in through the first pipe;
:$2400$ sextarii flow in through the second pipe;
:$1200$ sextarii flow in through the third pipe.
== Proof ==
First note that there are $72$ sextarii to the metreta.
Hence there are $7200$ sextarii in total in the cask.
Through the first pipe we have:
:$\dfrac 1 3 \times 7200 + \dfrac 1 6 \times 7200 = 3600$
Through the second pipe we have:
:$\dfrac 1 3 \times 7200 = 2400$
Through the third pipe we have:
:$\dfrac 1 6 \times 7200 = 1200$
== Historical Note ==
== Sources ==
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Category:Propositiones ad Acuendos Juvenes
Category:Cistern Problems
Propositiones ad Acuendos Juvenes/Problems/9 - De Sago

== by : Problem $9$ ==
;De Sago: A Cloak
:I have a cloak, $100$ cubits long and $80$ cubits wide.
:I wish to make small cloaks with it, each small cloak $5$ cubits long and $4$ cubits wide.
:How many small cloaks can I make?
== Solution ==
:$400$.
== Proof ==
It is assumed there is no cloth spent on hems, and the making of a small cloak consists of little more than cutting a piece of cloth to the required size.
We have that:
:$5$ goes into $100$ a total of $20$ times
:$4$ goes into $80$ also a total of $20$ times.
Thus the cloth can be cut into $20 \times 20 = 400$ pieces that are $5$ cubits long and $4$ cubits wide.
== Historical Note ==
== Sources ==
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Category:Propositiones ad Acuendos Juvenes
Propositiones ad Acuendos Juvenes/Problems/9 - De Sago/Historical Note

== Historical Note on by : Problem $9$: De Sago ==
According to the translator , the explanation for the answer given in the original manuscript was "not enlightening".
It is supposed that a scribe may have skipped a line when transcribing it.
The text as given by makes little sense:
:The eightieth part of $400$ is $5$ and the hundredth part of $400$ is $4$.
:Both eighty $5$s and one hundred $4$s give the same result, $400$.
:There are that many cloaks.
== Sources ==
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Category:Historical Notes
Prosthaphaeresis Formulas

== Theorem ==
=== Sine plus Sine ===
=== Sine minus Sine ===
=== Cosine plus Cosine ===
=== Cosine minus Cosine ===
=== Hyperbolic Sine plus Hyperbolic Sine ===
=== Hyperbolic Sine minus Hyperbolic Sine ===
=== Hyperbolic Cosine plus Hyperbolic Cosine ===
=== Hyperbolic Cosine minus Hyperbolic Cosine ===
== Also known as ==
== Also see ==
* Simpson's Formulas
== Linguistic Note ==
== Sources ==
* 
* 
* 
* 
Category:Prosthaphaeresis
Category:Trigonometric Identities
Category:Hyperbolic Functions
Category:Compound Angle Formulas
Prosthaphaeresis Formulas/Also known as

== Prosthaphaeresis Formulas: Also known as ==
The Prosthaphaeresis Formulas are also known as:
* the Factor Formulas (or Factor Formulae)
* Simpson's Formulas (or Simpson's Formulae), although this is usually used for similar results.
== Sources ==
* 
* 
* 
Category:Prosthaphaeresis