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                    +----------------------------------+
                    |   HOST: An Electronic Bulletin   |
                    |  for the History and Philosophy  |
                    |    of Science and Technology     |
                    |----------------------------------|
                    |        Volume 1, Number 2        |
                    |          Spring/Summer           |
                    |           June, 1993.            |
                    |        ISSN # 1192-084 X.        |
                    +----------------------------------+

       +-----------------------------------------------------------+
       |  Institute for the History  | Produced  by  IHPST through |
       |  and Philosophy of Science  | the HOST BBS on EPAS and    |
       |  and Technology, Room 316,  | E-Mail, through INTERNET at |
       |  73 Queen's Park Crescent,  | JSMITH@EPAS.UTORONTO.CA     |
       |  Toronto, Ontario, Canada.  |  IHPST@EPAS.UTORONTO.CA     |
       |  M5S1K7           [IHPST].  |-----------------------------|
       |  Phone:   (416)  978-5047.  | Editors: Julian A. Smith    |
       |  Fax:     (416)  978-3003.  |          Gordon H. Baker    |
       +-----------------------------------------------------------+


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                              +------------+
                              |  Contents  |
                              +------------+

Subscriber's Information

About our Contributors

Articles/Works in Progress

(1) Peter J. Burkholder:
    Alciun of York's _Propositiones ad Acuendos Juvenes_; ("Propositions
    for Sharpening Youths"); Introduction and Commentary.

(2) Peter J. Burkholder:
    _Propositiones Alcuini Doctoris Caroli Magni Imperatoris ad Acuendes
    Juvenes_; _Propositions of Alciun, A Teacher of Emperor Charlemagne,
    for Sharpening Youths_; Translation.

(3) Sharon Low:
    Richard Goldschmidt and William Bateson: Opposition to the Classical
    Conception of the Gene; Obstructionists or Visionaries?


Electronic Resources

(1) Julian A. Smith:
    LISTSERVER Mailing Lists/Discussion  Groups  on BITNET/INTERNET for
    the Historian and Philosopher of Science and Technology.

(2) Julian A. Smith:
    Using "Newsgroups" through BITNET/INTERNET.


Book Reviews

(1) _Storms of Controversy: The Secret Avro Arrow Files Revealed_, by
    Palmiro Campagna.
(2) _The People's Railway: A History of Canadian National_, by Donald
    MacKay.
(3) _The American Way of Birth_, by Jessica Mitford
(4) _Loss of Eden: A Biography of Charles and Anne Morrow Lindbergh_,
    by Joyce Milton.
(5) _The Art of Medieval Technology_, by Richard W. Ungur.
(6) _Hidden Attraction:  The  Mystery and History of  Magnetism_,  by
    Gerrit L. Verschuur.
(7) _Gates:  How Microsoft's Mogul Reinvented an Industry -- And Made
    Himself the Richest Man in America_,  by  Stephen Manes  and Paul
    Andrews.
(8) _The  Hacker  Crackdown:   Law  and  Disorder  on  the Electronic
     Frontier_, by Bruce Sterling.

Information for Authors

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                       +--------------------------+
                       | Subscriber's Information |
                       +--------------------------+

   HOST:   An Electronic Bulletin for the History and Philosophy of Science
and Technology,is  produced by the Institute for the History and Philosophy
of Science  and Technology  (or IHPST)  at Victoria  College, Room  316, 73
Queen's Park Crescent, University of Toronto, Toronto, Ontario, Canada, M5S
1K7.   HOST appears  2 times  a year,  Spring/Summer and  Fall/Winter,  and
contains articles,  works in  progress,   research  notes,  communications,
book  reviews,   electronic  resources,    and  news  of  interest  to  the
profession.
   The HOST  Bulletin is distributed in several formats.  Copies through E-
Mail (INTERNET  at JSMITH@EPAS.UTORONTO.CA  or GBAKER@EPAS.UTORONTO.CA) are
available   free.   Printed copies  ($8) or  disk copies  ($5) may  also be
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or fax  at 416-978-3003.   Inquiries, subscription orders, submissions, and
review copies of books should be sent to IHPST,  addressed
to the HOST Bulletin editors.

---------------------------------------------------------------------------

                        +------------------------+
                        | About our Contributors |
                        +------------------------+


Gordon H.  Baker is  a B.A.  candidate at  IHPST, and an editor of the HOST
Bulletin. Mr. Baker's research interests include 19th century medicine, and
the history of science in Canada.

Julian A.  Smith is  a Ph.D.  candidate at  IHPST, and a History of Science
Instructor at Ryerson Polytechnical University, Toronto.  He is also one of
the editors  of the  HOST Bulletin.  Mr. Smith's research interests include
medieval physics,   19th  century medicine,  astronomy and  cartography  in
Canada, and the history of mathematics.

Sharon  Low   has  recently  completed  her  undergraduate  degree  at  the
University of Toronto's Trinity College.  She specializes in zoology, has a
psychology major  and is  interested in  Biological Rhythms (the subject of
her thesis).   She  is now taking graduate level studies in neuroscience in
the United States.  Her paper on Goldschmidt and Bateson  (in this journal)
was the winner of the 1992 IHPST Undergraduate Essay Competition.

Peter Burkholder  was recently  a graduate  student (MA)  at IHPST, but has
recently transferred his Doctoral studies to the University of Minnesota in
Minneapolis.   His interests  are in  medieval studies  and the  history of
mathematics.

Steven Walton  is  a  graduate student  (MA) at IHPST.  He has completed an
M.A. in Engineering  at Cornell University.   His interests are in medieval
studies and the history of technology.

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                     +----------------------------+
                     | Articles/Works in Progress |
                     +----------------------------+


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            Alciun of York's _Propositiones ad Acuendos Juvenes_
                  ("Propositions for Sharpening Youths")
                        Introduction and Commentary
                          By Peter J. Burkholder
                            Received May, 1992
                           Revised  March, 1993

Introduction

   In the  year 782,  Alcuin of York (735-804) was summoned to the court of
Charlemagne in  Frankia.  By this point, the Frankish king's domain covered
much of  modern France;  Lombardy had  been subjected;  authority had  been
established on the Spanish March; and Bavaria was soon to be Christianized.
With his  sphere of  influence thus  extended, Charlemagne was able to turn
his interests to the revitalization of education among his peoples.  It was
for this reason that Alcuin's presence was requested on the Continent.
   Alcuin, also known by his Latin name of Albinus, was born in Northumbria
in the  year of  the Venerable  Bede's death.[1]  He spent time studying in
Italy and  taught at the cathedral school of York before assuming his place
at the  court of Charlemagne in 782.  Alcuin played an integral part in the
so-called "Carolingian  Renaissance," founding the palace school at Aix-la-
Chapelle where  the  seven  liberal  arts  were  taught  according  to  the
educational system  of Cassiodorus  (ca.  490-580).    His  most  important
writings were  his revisions  of the Vulgate and his voluminous letters,[2]
the latter  being collated  in the  ninth  century  as  a  model  of  Latin
composition.   Alcuin eventually assumed the position of abbot at the abbey
of St-Martin of Tours where he founded an important library and school, and
where he remained until his death on May 19, 804.
  During the course of his tenure, Alcuin is credited with having written a
set of  mathematical exercises entitled "Propositiones ad acuendos juvenes"
or  "Propositions  for  Sharpening  Youths."    These  problems  and  their
solutions, 53  in number,  serve as  valuable  evidence  of  the  state  of
mathematical education  under the  Carolingian kings.   To  the best  of my
knowledge, a  complete translation  of, and commentary on, the Propositions
has never  been undertaken,  while scholarly  treatment of  them  has  been
cursory at  best.  It is hoped that such an endeavor will shed new light on
our knowledge  of medieval  mathematics and mathematical education.  Before
delving into  the  Propositions  themselves,  however,  discussion  of  the
problem of authorship is offered.

The Problem of Authorship

  The composition of the Propositions can only be tentatively attributed to
Alcuin.   The most  compelling reason  to ascribe them as such is the title
given  at   the  head  of  the  manuscript  used  for  the  Migne  edition:
"Propositiones  Alcuini  doctoris  Caroli  Magni  imperatoris  ad  acuendos
juvenes."[3]   This particular  manuscript is  a codex  from the  monastery
Augia Dives,  known today  as Richenau  near Constance,  Switzerland.   The
monastery was  secularized in  1803, with  the manuscripts  being dispersed
between Karlsruhe, London, Stuttgart, St. Paul in Carinthia, and Zurich.[4]
The manuscript  is described by the editor as being "very old," but this is
by no means conclusive evidence of its origin.[5]
   J.A.  Giles,  who  edited  Bede's  works  in  which  a  version  of  the
Propositions  appears,[6]   judges  that   the  style  of  the  queries  is
sufficiently like  that of  Alcuin to imply that he was indeed the original
author.[7]   Conversely, the  literary manner in which the Propositions are
stated is  very unlike  anything produced  by  Bede,  and  thus  cannot  be
considered his.   Corroborating  evidence that  Alcuin may  have  been  the
author of the Propositions comes from a letter sent to Charlemagne in which
Alcuin states,  "I have  sent to your Excellency...some simple arithmetical
problems for  reason of  pleasure."[8]    Such  testimony  is,  of  course,
tenuous, for  Alcuin's authorship  of the Propositions is in no way assured
simply because he sent a copy of them to his king.
  There  is  other evidence,  though inconclusive, which indicates that the
Propositions  may   have  been   penned  by  Alcuin.    In  an  interesting
mathematical correspondence  which took  place around  1025, two  monks  of
Cologne and Liege make reference to a work entitled _Albinus_ the Latinized
form of  Alcuin's name.[9]   The  context in  which the  work is  used is a
debate over  the relation of a square's side to its diagonal.  Although the
Propositions specifically  treat no  such problems,  there are instances of
geometrical  methods   employed  for  questions  of  land  measurement  and
circumference.   Thus, we  have an instance where Alcuin's Propositions may
have  been  widely   utilized  in  the early eleventh century, and commonly
known as his work.
  As stated,  there is  evidence  suggesting that the Propositions may have
been the  work of the  Venerable Bede  (672-735).   An almost word-for-word
version of  this  treatise  appears  under  the  heading  "Incipiunt  aliae
propositiones ad  acuendos juvenes"  in Bede's  works.[10]   If  this  were
indeed the  case, Alcuin obviously could not have been the original author.
However, it  is worth  noting that  Bede never  makes any  mention  of  the
Propositions, even  in his own listing of his works.  Moreover, Giles cites
a  number   of  scientific  writings  attributed  to  Bede,  including  the
Propositions, which must be considered unauthentic.[11]  For these reasons,
Bede's version  of the  Propositions appears  in Migne  under  the  heading
"Dubious and Spurious Works."
   Based on  a manuscript  at Leyden,[12]  Smith argues  that the  probable
compiler of  the Propositions  was a  monk named  Ademar or  Aymar  of  the
ancient house  of Chabanais,  who lived  from 988  to 1030.   The  problems
contained therein  seem to  be based  on Aesop's  Fables,  begun  by  Aesop
himself in  Samos during  the seventh century B.C., and modified by Babrius
around the  third century.   What  the connection  is between  Aesop's  and
Alcuin's works is not readily apparent, and Smith fails to elaborate on his
point.   However, based  in part  on the  method  of  presentation,  Thiele
believes that Ademar did indeed author the Propositions.[13]
   Except for  Giles, scholars  are reluctant  to give  Alcuin  credit  for
production of  the Propositions,  mainly on the grounds that he contributed
little or nothing of originality to learning, and because the vast majority
of his  writings were  works on theology.  Thus, Alcuin assumes the typical
medieval scholastic  role as  transmitter of knowledge, not producer of new
material.   A comprehensive study of the various manuscripts would no doubt
help determine the actual author of the Propositions.

The Problems Themselves

   The fifty-three  problems which  make up the Propositions follow a basic
general pattern:   a  brief heading,  a statement of the problem, a request
for an  answer to  the problem,  and a solution.  There can be little doubt
that the  problems were  read aloud,[14] possibly with the students copying
them down  on papyrus,  tree bark  or parchment.[15]  A call for a response
was then  elicited of  the form,  "Let him say, he who is able..."  Some of
the problems  such as  those pertaining  to logic exercises could have been
deciphered  with  no  recourse  to  writing;  others  involving  drawn  out
arithmetic computations  could have  taken quite  some effort  to  compute,
particularly when working with clumsy Roman numerals.
   There is  no strict  categorical framework  for the  problems,  although
clusters of  certain types  appear intermittently.   Only two problems (1 &
26) pertain to rates and distances, the first being a very odd hypothetical
situation involving a snail's arduous and drawn out trek to a luncheon; the
second and  more advanced  problem involves  a dog's pursuit of a hare,[16]
and actually  involves two  rates over  differing  distances.    It  is  as
follows:

   There is  a field  which is  150 feet  long.  At one end stood a dog, at
the other,  a hare.  The dog advanced behind the hare, namely, to chase the
hare.   But whereas the dog went nine feet per stride, the hare went [only]
seven.   Let him  say, he  who wishes, How many feet and how many leaps did
the dog take in pursuing the fleeing hare until it was caught?

   Alcuin's solution  is ingenious, though cryptic.  Whereas we might solve
such a  problem by  two equations  and two  unknowns, Alcuin notes that the
differing rates of the animals is the key to the entire problem:

   The length of the field was 150 feet.  Taking half of 150 makes 75.  The
dog was  covering nine  feet per  stride, and nine times 75 makes 675.  The
dog thus  ran this  many feet  in chasing  the rabbit  until it  caught the
rabbit with its tenacious teeth.  And indeed, because the rabbit went seven
feet per  stride, take  75 seven  times.  This is how many feet the fleeing
rabbit travelled before being caught.

   The reason  for dividing  the field  in half may not be so clear, but it
simply corresponds  to partitioning  the field  by the  difference  of  the
animals' feet  per stride,  in this  case, two.    Alcuin  then  takes  the
measurement obtained  by thus  dividing and multiplies it by the respective
rates of dog and hare to arrive at the correct answer.
  This method can be generalized as follows.  The dog must always cover the
space between  it and the hare (d1) plus the additional distance covered by
the hare  (d2).   If the dog's rate is r1, then the equation describing the
distance traversed  by the  dog is  given by  d1+d2=r1t. (1)   In a similar
fashion, the  hare's flight  is denoted  by d2=r2t.  (2)   Substituting the
value of d2 in (2) into (1) yields d1+r2t=r1t.  Thus d1=t(r1-r2). (3)  From
(1), we  know that  t=(d1+d2)/r1, and  putting this  value of  t  into  (3)
results  in   d1=(d1+d2)(r1-r2)/r1.    Rearranging  this  equations  yields
d1+d2=r1d1/(r1-r2).   This is  exactly what  Alcuin's method does.  It says
that the  total distance  covered by,  for instance,  the dog is simply the
intervening expanse  divided by  the difference  of the two animals' rates,
times the  dog's rate.  It is easy to see the advantages that such a method
offers in an oral instruction setting.
   A much  larger corpus of problems (e.g. 2, 3, 4) might best be described
as those of an unknown quantity.  In each exercise, the reader is told that
a certain  quantity of  people, animals or objects, if doubled, tripled, or
in some  other way  arithmetically manipulated,  adds up to 100.  A typical
example is problem 36:

   A certain old man greeted a boy, saying to him:  "May you live, boy, may
you live  for as  long as  you have [already] lived, and then another equal
amount of time, and then three times as much.  And may God grant you one of
my years,  and you  shall live  to be 100."  Let him solve, he who can, How
many years old was the boy at that time?

  The answer is a bit trickier than it might appear at first glance, for it
must  be remembered  that  it  is  difficult  to  use base-10 arithmetic in
solving a problem dealing with a 12-month year:

   When [the  old man]  said "may  you live for as long as you have lived,"
[the boy]  had [already]  lived eight  years, three  months.  Another equal
number of  years would  make 16 years, six months, while another equal span
makes 33  years.  Three times this makes 99 years, which with one more year
added makes 100.

   It would  have been  a rather  simple affair for Alcuin to have invented
such an  exercise by  starting with  100 and  working backwards,  and  then
extrapolating the  procedure to  other problems  of  the  same  genre.    A
slightly more  complicated query  of this  type can be found in problem 40,
where portions  of the  original quantity  are doubled,  halved,  and  then
added:

   A certain man saw from a mountain some sheep grazing and said, "O that I
could have  so many,  and then  just as many more, and then half of half of
this [added],  and then  another half  of this  half.  Then I, as the 100th
[member], might head back to my home together."  Let him solve, he who can,
How many sheep did the man see grazing?

   Again, such  a scenario could have easily been derived by beginning with
100, and  then arithmetically manipulating it until the desired problem was
in order:

  36 sheep were first seen by the man when he said, "O that I could have so
many."   Adding an  equal number makes 72, and a half of half of this, that
is, of  36, makes  18.   And again,  a half  of this, that is, of 18, makes
nine.  Therefore add 36 and 36, making 72.  Add to this 18, which makes 90.
Then add nine to 90, making 99.  The man himself added to these will be the
100th one.

   The only  precaution which would be necessary would be to make sure that
fractions do not occur, and this could be easily checked.
   A third  type of problem which Alcuin presents to the student is that of
dividing quantities  amongst various  parties.  This sometimes involves the
division of an inheritance between sons, as in problem 12:

   A certain  father died  and left  as an inheritance to his three sons 30
glass flasks,  of which  10 were  full of  oil; another  10 were half full,
while another  10 were  empty.   Divide, he  who can, the oil and flasks so
that an  equal share  of the  commoditites should  equally come down to the
three sons, both of oil and glass.

  There is little doubt that anyone, whether trained in mathematics or not,
could solve  such a  problem.   One need  only pour  all of  the oil into a
central vat  and divide  the liquid and glass equally from there.  However,
as  an   exercise,  Alcuin  demonstrates  how  such  a  division  might  be
accomplished without recourse to such crude means:

  There are three sons and 30 glass flasks.  However, of the flasks, 10 are
full [of  oil], 10  half full,  and 10  empty.   Take three times 10, which
makes 30,  so each  son shall  receive 10 flasks as his portion.  Divide up
the three portions, that is, give to the first son 10 half [filled] flasks,
to the  second son  five full and five empty [flasks].  Do the same for the
third son, and the brothers' portions of glass and oil shall be the same.

  Questions pertaining to division of an estate are traceable back to Roman
law and  what is  known as  the Testament Problem.[17]  Roman precepts made
definite provisions for the division of property upon a father's death, and
thus we  find problems  like number  35.   Here, a  father leaves  behind a
pregnant wife,  with instructions  for division  of his  inheritance in the
case of  either a  boy or girl being born.  To complicate matters, opposite
sex twins  are produced.   A  long-winded  solution  of  how  the  father's
possessions are to be divided follows.
  These types of problems seem to stress logic more than arithmetic skills.
The exercises  involving distribution  of  corn  by  a  head  of  household
(paterfamilias) to his servants are slightly more complicated, as differing
amounts of corn are allowed for men, women and children:

   A certain  head of household had 30 servants whom he ordered to be given
30 modia  of corn  as follows:   The  men should  receive three  modia; the
women, two;  and the children, a half [modium].  Let him solve, he who can,
How many men, women and children were there?

   As in  the problems  dealing with  an unknown quantity, Alcuin had to be
sure that  his numbers  worked out  evenly in  the end.  Note, too, that he
treats fractional  measurements here,  as each child receives half a modium
of corn:

   If you  take thrice three, you get nine; if you take two five times, you
get 10;  and if  you take half of 22, you get 11.  Thus, three men received
nine modia;  five women  received 10;  and 22  children received  11 modia.
Adding three  and five and 22 makes 30 servants.  Likewise, nine and 11 and
10 makes 30 modia.  Hence there are 30 servants, and 30 modia [of corn].

   Problems of  exactly the same type, but with varying numbers of servants
and corn,  can be  found in exercises 32 and 34, indicating that it was the
procedure which Alcuin wished his pupils to understand.
  Alcuin's logic problems, or slight variations of them, can still be found
today in  textbooks and  on examinations.   The most famous no doubt is the
conundrum of the man, she-goat, wolf and cabbage which needed to be ferried
across a  river. (Problem  18)   As only  two passengers fit in the boat at
once, and  since certain  combinations of  animals and  vegetable cannot be
left alone,  the reader  is left  to solve how a successful transport might
take place.   Alcuin  assumes the role of ferryman and leads us through the
problem step by step:

   ...I would  first take  the she-goat  and leave  behind the wolf and the
cabbage.   When I had returned, I would ferry over the wolf.  With the wolf
unloaded, I  would retrieve  the she-goat and take it back across.  Then, I
would unload  the she-goat and take the cabbage to the other side.  I would
next row back and take the she-goat across.  The crossing should go well by
doing thus, and absent from threat of slaughter.

   Problems of exactly this type appear in exercises 17, 19 and 20 as well.
In each  case, a long explanation of how a successful transnavigation might
be performed is offered.
   Other logic  problems are  more straightforward  and exhibit  a  certain
amount of  humor.  The answer to Alcuin's problem of how many footprints an
ox makes  in the  last furrow  is, of course, none, "because the ox goes in
front of the plow and the plow follows it.  For however many footprints the
ox makes on the ploughed earth by going first, so many the plough following
behind destroys  by ploughing." (Problem 14)  Another question of this type
entails a  man who  wishes to  slaughter 300 pigs in three days, but with a
odd number  being butchered  per day  -- a  problem which Alcuin states "is
indissoluble and composed for rebuking." (Problem 43)
   Alcuin's problems  pertaining to  area are of particular interest.  They
consist of queries as to how many measurements or objects can fit inside of
a larger  confine.   Certain exercises  we might  define  as  dealing  with
acreage, although  such a term is not entirely accurate for measurements of
aripenna, the  standard land  quantity.[18]    Other  problems  are  of  no
immediate practical  value whatsoever, and are thus clearly meant as purely
mathematical exercises.    Take  for  example  the  question  of  how  many
rectangular houses can fit within a circular city (problem 29):

  There is a city which is 8000 feet in circumference.  Let him say, he who
is able, How many houses should the city contain, such that each [house] is
30 feet long, and 20 feet wide?

Solution:

   The city measures 8000 feet around, which is divided into proportions of
one-and-a-half to  one, i.e.  4800 and  3200.   The length and width of the
houses are  [also] of  these [dimensions].   Thus, take half of each of the
above [measurements],  and from  the larger number there shall remain 2400,
while from  the smaller,  1600.   Then, divide 1600 into twenty [parts] and
you will  obtain 80  times 20.   In  a similar fashion, [divide] the larger
number, i.e. 2400, into 30 pieces, deriving 80 times 30.  Take 80 times 80,
making 6400.   This  many houses  can be  built in  the city, following the
above-written proposal.

   Essentially, what  Alcuin does  is to  force the  ratio of the length to
width of each house onto the city.  Thus, as each house measures 30x20, the
ratio of length to width is 3:2.  Alcuin breaks up the circumference of the
town into  two pieces  such that  their ratio  is 3:2 as well.  Having done
this, he  simply straightens  out the pieces and sets them perpendicular to
one another.   This,  however, yields  an unclosed figure.  He thus divides
each side  in two  and rearranges  the four sides in order to make a closed
structure.   The ratio  of 3:2 is preserved since 4800/2:3200/2 equals 3:2.
Now Alcuin  has a  rectangular town  with a circumference of 8000 feet, and
whose dimensions  are proportional  to the  dimensions of the houses.  From
there, the  problem of  the number  of houses  which can fit in the town is
trivial.
  The most obvious shortcoming of Alcuin's method is that the area enclosed
by different  curves of equal length is not the same.  The area of a circle
is given  by pi-r-squared,  whereas the  area for a rectangle is denoted by
length  times  width.    Thus,  the  area  enclosed  by  a  circle  with  a
circumference of  8000 feet  is roughly  5,092,958 square  feet, whereas  a
rectangle measuring  1600 by  2400 encloses only 3,840,000 square feet -- a
difference of  over 1.2 million square feet.  The only conclusion which can
be drawn is that Alcuin was unaware of the consequences of modifying shape,
as he  employs the same methodology in problems 27 and 28.  In addition, we
need only  note the  absence of allowance for streets to realize the purely
hypothetical nature of such a problem.
   There is  further evidence  that Alcuin's  Propositions sought merely to
stir the  minds of  their readers  as opposed  to serving as a handbook for
quotidian problems.  Glaring examples of this are exercises 13 and 41, both
of which  teach the  lesson of  geometric growth.[19]   In  the  former,  a
servant is  ordered by  his king  to assemble  an army  from 30 villages as
follows:

  He should bring back as many men [from each successive village] as he had
taken there.   Thus, [the servant] came to the first village alone; he came
with one other person to the next; three people came to the third, etc...

   Such a gathering can be mathematically modelled by the relation N=2^(v),
where v is each successive village and N is the number of people assembled.
Hence, the total number of villagers conscripted would be given as S 2^(v),
with the  summation beginning  at v=0  and continuing  to v=30  -- a figure
representing an  army which  was far  beyond the  capabilities of  even the
richest or  most ambitious  of kings  to field.   Alcuin's  solution  gives
figures up  to v=15,  while the  edition ascribed to Bede continues all the
way to  v=30, although  with errors beginning at v=22.  Neither attempts to
sum the  figures, nor  is it  expected that a student would be expected to;
rather, it  was the  process which  undoubtedly lay  at the  heart  of  the
problem.
   A similar  hypothetical problem  demonstrates  the  idea  of  arithmetic
progression.   It has  been related that when Gauss (1777-1855) was a young
student, his  mathematics teacher  one day  instructed the class to add the
numbers one through 100.  No sooner had the assignment been made than Gauss
somehow magically produced the correct figure of 5050.  How had he done it?
  The key to the problem is to realize that by adding corresponding low and
high figures,  a simple  multiplication problem  unfolds.  Thus, 1+100=101;
2+99=101; 3+98=101;...;49+52=101; 50+51=101.  It is manifest from this that
one need  only multiply  the constant  sum, 101, by 50, the number of sums.
In this way, the correct response of 5050 is obtained.
  Alcuin's  ladder problem  (42)  shows that this concept was already known
by the ninth century:

   There is  a ladder which has 100 steps.  One dove sat on the first step,
two doves  on the  second, three  on the third, four on the fourth, five on
the fifth,  and so  on up  to the hundredth step.  Let him say, he who can,
How many doves were there in all?

Solution:

   There will  be as  many as  follows:  Take the dove sitting on the first
step and add it to the 99 doves sitting on the 99th step, thus getting 100.
Do the  same with the second and 98th steps and you shall likewise get 100.
By combining  all the steps in this order, that is, one of the higher steps
with one  of the  lower, you shall always get 100.  The 50th step, however,
is alone and without a match; likewise, the 100th stair is alone.  Add them
all and you will have 5050 doves.

   We can  see that  with only  slight  modification,  the  above-described
concept was  in place  almost a  thousand years  before Gauss  dazzled  his
schoolteacher.  Perhaps the young Gauss wasn't so clever after all!


Conclusion and Topics for Further Study

   Whether or  not Alcuin  himself authored  the Propositions  may never be
known, but  this is  not  of  great  consequence.    The  Propositions  are
interesting problems  in their  own right  and reveal the general state and
method of  mathematical instruction  around the  time of  Charlemagne.    A
thread of  continuity with  classical education  can be  discerned in these
puzzles as  well as  the influence of Barbarian values of practical methods
for everyday problems.  However, it must be concluded that the Propositions
sought only  to instill  various simple  methods in  its users,  this being
accomplished by repeated problems of the same genre.
   It should  not be  concluded that the Propositions are indicative of the
general state of mathematics during the eighth or ninth centuries.  We have
prima facie  evidence[20] that  these problems  were utilized primarily for
didactic purposes;  thus, to  argue that the Propositions are an example of
the poor state of mathematics is erroneous.  While such a conclusion may be
justified, it  by no  means is  a necessary  deduction from the evidence at
hand.
   The Propositions  are also  potentially valuable  for the  economic  and
social insight  they offer,  and  a  spreadsheet  of  various  weights  and
measures which  appear throughout  the problems is included as an appendix.
Whether or  not these  values are  consistent with  contemporary conditions
awaits another study.


An Introduction to the Translations

   In translating  the 53  problems and answers of the Propositions, I have
utilized the  Migne edition  of Alcuin's works.  I have annotated this text
and supplied alternate or additional versions of problems as they appear in
Bede's supposed  previous work.   Where  differences occur,  a footnote  is
provided beginning  with "Bede,"  hence  referring  the  reader  to  Bede's
edition.   A further comparison with Heruagius's edition of Bede's writings
revealed only  trivial discrepancies, and thus alternate readings from this
work have been omitted.
   A quality translation must be true both to the original language and the
language into  which the  material is converted.  With this is mind, I have
tried to  keep verb  tenses consistent  according to  English usage despite
Alcuin's variations  within a given problem.  English words which have been
read into  the Latin  are contained  within square  brackets [   ]  and are
either interpretive  or corroborated  by Bede's  edition.    Certain  words
referring to  weights and measures (e.g. aripennum, denarius, solidus) have
been left in the original.  Though aripennum might be rendered "arpent" and
solidus a  "sous," such  translations either  do little in helping us grasp
what is involved in the usage, or are modernly deceitful.


                                References


[1] Since  the Propositions  to be  discussed cannot  be ascribed to Alcuin
with certainty,  I will offer only a very brief biographical account of the
man.  Secondary literature on Alcuin is plentiful.  See for example Stephen
Allott,  _Alcuin   of  York_,   (York,  1974);   L.  Wallach,  _Alcuin  and
Charlemagne_, (New  York, 1959);  Eleanor Duckett,  _Alcuin,  A  Friend  of
Charlemagne_, (New  York, 1951);  C.J.B. Gaskoin,  _Alcuin:   His Life  and
Works_, (Cambridge,  1904); Andrew  West,  _Alcuin  and  the  Rise  of  the
Christian Schools_,  (London, 1893);  and Frederick  Lorenz, _The  Life  of
Alcuin_, Jane  Slee, trans.,  (London, 1837).   For  a listing  of Alcuin's
texts and  translations, see George Sarton, _Introduction to the History of
Science_, 3 vols., (Baltimore, 1927), vol. 1, part 1, pp. 528-529.

[2] See Rolph Page, _The Letters of Alcuin_, (New York, 1909).

[3] _Alcuini  opera omnia_,  J.P. Migne, ed., vol. 2, found in _Patrologiae
latinae cursus completus..._, vol. 101, (Paris, 1863).

[4] Frederick  Hall, _A  Companion to Classical Texts_, (Oxford, 1913), pp.
294 & 342.

[5] D.E.  Smith tells  us that  the oldest manuscript of the problems dates
from the  eleventh century.   _History of Mathematics_, 2 vols., (New York,
1923; reprint, 1951), vol. 1, p. 186.

[6] One  such manuscript  ascribing the  Propositions  to  Bede  is  _Codex
Latinus Monacensis_,  no. 14272.   Its  origin is  either tenth or eleventh
century.  See _Catalogus codicum Latinorum bibliothecae regiae monacensis_,
(Hildesheim, 1975), vol. 2,2, pp. 152-153.  (This is the only ascription to
Bede noted by Lynn Thorndike, _A Catalog of Incipits of Medieval Scientific
Writings_, (Cambridge, MA, 1963).)

[7] Giles, ed., _The Miscellaneous Works of Venerable Bede, in the Original
Latin_, 6 vols., (London, 1843), vol. 6, p. xiv.

[8] "Misi excellentiae vestrae...aliquas figuras arithmeticae subtilitatis,
laetitiae causa."   Migne,  op. cit., vol. 100, letter 101, col. 314, dated
anno 800.

[9] An  edition of the correspondence, along with scholarly commentary, can
be found  in Paul  Tannery's _Memoires  scientifiques_, vol. 5 of _Sciences
exactes au  moyen-age-, (Paris,  1922), pp.  264-288.   An earlier  partial
edition is  contained in  Jules Clerval's _Les Ecoles de Chartres au moyen-
age, du  ve au xvie siecle_, (Paris, 1895; reprint, Geneva, 1977), pp. 459-
464.   I have  studied these  letters anew  and hope  to make  my  findings
available in  the near  future in  a paper  entitled "Speculum  geometricae
undecimo saeculo:   The Mathematical Correspondence of Ragimbold of Cologne
and Radulf of Liege, ca. 1025."

[10] Migne,  op. cit., vol. 90, cols. 667-676.  These also appear in volume
one of  Joannes Hervagius's edition of Bede's _Opera Bedae Venerabilis..._,
8 vols. bound in 4, (Basil, 1563), but are not included in Giles's edition.
The most  notable difference  between Bede's  version and that of Alcuin is
the lack of solutions for problems 36-53 in the former.

[11] Giles, op. cit., vol. 6, pp. ix-xv.

[12]  Georg  Thiele,  ed.,  _Der  Illustrierte  lateinische  Aesop  in  der
Handschrift des  Ademar_, Codex  Vossianus  Lat.  Oct.  15,  Fol.  195-205,
(Leiden, 1905).

[13] Ibid., pp. 23-25.

[14] Vera  Sanford specifically  places  Alcuin's  Propositions  under  the
rubric "Verbal  Problems."   _A Short  History  of  Mathematics_,  (Boston,
1930), pp. 212-213.

[15] For  the materials  available to  schoolchildren,  see  Pierre  Riche,
_Education  and  Culture  in  the  Barbarian  West,  Sixth  through  Eighth
Centuries_, trans.  from the third edition by John Contreni, (Columbia, SC,
1976), pp. 458-462.

[16] Smith  describes this  problem as  being "already ancient" by Alcuin's
time, but  fails to  cite any  precedents.   Op. cit.,  187.  Sanford dates
pursuit problems  to Roman  legionaries, whose  stride was  so uniform that
time schedules  could be  worked out for marching from place to place.  Op.
cit., pp. 217-218.

[17] See Sanford, op. cit., pp. 218-219.

[18] The  use of aripenna and the smaller perticae, of course, implies that
such measurements were standard and well-known to all.  From problem 25, we
can deduce that one aripennum equals 184.53 perticae.

[19] Sanford  regards problems  of geometric  progression as  some  of  the
oldest types of mathematical endeavors, and cites extant Babylonian tablets
from ca. 2000 b.c. to this effect.  Op. cit., pp. 174-176.

[20] See problem 43.

---------------------------------------------------------------------------

               _Propositiones Alcuini Doctoris Caroli Magni
                   Imperatoris ad Acuendes Juvenes_ [1]
              _Propositions of Alciun,  A Teacher of Emperor
                   Charlemagne, for Sharpening Youths_
                                Translation
                          By Peter J. Burkholder
                            Received May, 1992
                           Revised March, 1993.


I.  propositio de limace.
Limax fuit  ab hierundine  invitatus ad prandium infra leucam unam.  In die
autem non  potuit plus  quam unam unciam pedis ambulare.  Dicat, qui velit,
in quot diebus [2] ad idem prandium ipse limax perambulabat?
1.  proposition concerning the snail.
A snail was invited by a swallow to lunch a league away.  However, it could
not walk  further than  one inch  per day.  Let him say, he who wishes, How
many [years and] days did it take for the snail to walk to that lunch?

Sequitur solutio de limace.
In leuca  una sunt  mille quingenti  passus; vii  d pedes  xc unciae.  Quot
unciae, tot dies fuerunt, qui faciunt annos ccxlvi, et dies ccx.
Here follows the solution of the snail.
In one  league, there  are 1500  passus [3].   7500  feet  [equals]  90,000
inches. There are as many days as there are inches, that is, 246 years, 210
days.


II.  propositio de viro ambulante in via. [4]
Quidam vir  ambulans per  viam vidit sibi alios homines obviantes, et dixit
eis:   Volebam [5],  ut fuissetis  alii tantum,  quanti estis;  et medietas
medietatis; et  hujus numeri  medietas [et  rursum de  medietate medietas];
tunc una mecum c fuissetis.  Dicat, qui velit, quanti fuerunt, qui in prima
ab illo visi sunt?
2.  proposition of the man walking in the street.
A certain  man walking  in the street saw other men coming towards him, and
he said  to them:   "O  that there  were so  many [more]  of you as you are
[now]; and  then half  of half  of this [were added]; and then half of this
number [were  added], and  again, a  half of [this] half.  Then, along with
me, you  would number 100 [men]."  Let him say, he who wishes, How many men
were first seen by the man?

Solutio de eadem propositione.
Qui imprimis ab illo visi sunt, fuerunt xxxvi.  Alii tantum lxxii. Medietas
medietatis xviii.   Et  hujus numeri  medietas sunt  viiii.   Dic ergo sic:
lxxii et  xviii fiunt  xc.   Adde viiii, fiunt xcviiii.  Adde loquentem, et
habebis c.[6]
Solution of the same proposition.
Those who  were first  seen by the man were 36 in number; double this would
be 72.   A  half of  half of this is 18, and a half of this number makes 9.
Therefore, say  this:   72 and  18 makes  90.   Adding 9  to this makes 99.
Include the speaker and you shall have 100.


III.  propositio de duobus proficiscentibus.[7]
Duo homines  ambulantes per  viam, videntesque ciconias, dixerunt inter se:
Quot sunt?   Qui  conferentes numerum dixerunt:  Si essent aliae tantae; et
ter tantae,  et medietas  tertii, adjectis  duobus, c  essent.   Dicat, qui
potest, quantae fuerunt, quae imprimis ab illis visae sunt?
3.  proposition concerning the two travellers.
Two men  were walking  in the  street when  they noticed some storks.  They
asked each  other, "How many are there?"  Discussing the matter, they said:
"If [the storks] were doubled, then taken three times, and then half of the
third [were  taken] and  with two more added, there would be 100."  Let him
say, he who is able, How many [storks] were first seen by the men?

Solutio de ciconiis.
xxviii et  xviii,[8] et  tertio sic:   fiunt  lxxxiiii.  Et medietas tertii
fiunt xiiii.  Sunt in totum xcviii.  Adjectis duobus, c apparent.

Solution concerning the storks.
28 taken  three times  makes 84.  Half of a third makes 14.  Thus, in total
there are 98.  By adding two, there are 100.


IV.  propositio de homine et equis.[9]
Quidam homo vidit equos pascentes in campo, optavit dicens:  Utinam essetis
mei, et  essetis alii tantum, et medietas medietatis; certe gloriarer super
equos c.   Discernat,  qui  vult,  quot  equos  imprimis  vidit  ille  homo
pascentes?
4.  proposition concerning the man and the horses.
A certain  man saw  some horses  grazing in a field and said longingly:  "O
that you  were mine, and that you were double in number, and then a half of
half of this [were added].  Surely, I might boast about 100 horses."    Let
him discern,  he who  wishes, How  many horses  did the  man originally see
grazing?

Solutio de equis.
xl equi erant, qui pascebant.  Alii tantum fiunt lxxx.  Medietas medietatis
hujus, id est, xx, si addatur, fiunt c.
Solution concerning the horses.
There were  40 horses  grazing; double  this makes  80.   A half of half of
this, i.e. 20, if added, makes 100.


V.  propositio de emptore denariorum.[10]
Dixit quidam  emptor:[11]   Volo de  centum denariis  c porcos  emere;  sic
tamen, ut  verres x  denariis ematur;  scrofa autem  v denariis;  duo  vero
porcelli denario  uno.   Dicat, qui  intelligit, quot verres, quot scrofae,
quotve porcelli  esse debeant,  ut in neutris numerus nec superabundet, nec
minuatur?
5.  proposition concerning the buyer and his denarii.
A certain  buyer said:   "I want to buy 100 pigs with 100 denarii in such a
way that  a mature  boar is  bought for 10 denarii; a sow for five denarii;
and two  small female  pigs for  one denarius."  Let him say, he who knows,
How many  boars, sows,  and small female pigs should there be so that there
are neither too many nor too few of either [pigs or denarii]?

Solutio de emptore.
Fac viiii  scrofas et unum verrem in quinquaginta quinque denariis; et lxxx
porcellos in  xl.  Ecce porci xc.  In residuis v denariis, fac porcellos x,
et habebis centenarium numerum in utrisque.
Solution concerning the buyer.
Buy nine  sows and  one boar with 55 denarii, and 80 small female pigs with
40; behold, 90 pigs.  With the remaining five denarii, buy ten small female
pigs, and you shall have 100 pigs for 100 denarii.


VI.  propositio de duobus negotiatoribus c solidos habentis.
Fuerunt duo  negotiatores, habentes  c  solidos  communes,  quibus  emerent
porcos.   Emerunt autem  in solidis duobus porcos v, volentes eos saginare,
atque iterum  venundare, et  in solidis  lucrum facere.   Cumque  vidissent
tempus non  esse ad  saginandos porcos,  et ipsi eos non valuissent tempore
hiemali pascere,  tentavere venundando,  si potuissent,  lucrum facere, sed
non potuerunt;  quia non  valebant eos  amplius venundare,  nisi  ut  empti
fuerant, id  est, ut  de  v  porcis  duos  solidos  acciperent.    Cum  hoc
conspexissent, dixerunt  ad invicem:   Dividamus  eos.  Dividentes autem et
vendentes, sicut emerant, fecerunt lucrum.  Dicat, qui valet, imprimis quot
porci fuerunt;  et dividat ac vendat et lucrum faciat, quod facere de simul
venditis non valuit.
6.  proposition of the two businessmen who had 100 solidi.
There were two businessmen who had 100 solidi between them, with which they
bought some pigs.  For two solidi, they bought five pigs, wishing to fatten
them and  to sell  them again at a profit.  But when they saw that the time
was not right to fatten the pigs, and being unable to pasture them over the
winter, they  tried to  make a  profit by selling them.  However, they were
unsuccessful because  they could  only sell the pigs for what they had paid
(i.e., five  pigs for  two solidi).   When they realized this, they said to
each other,   "We  shall divide the pigs."  But by dividing and selling the
pigs for as much as they had paid, they made a profit.  Let him say, he who
can, How many pigs were there at first, and how did the men divide and sell
for a profit that which they could not do together?

Solutio de porcis.
Imprimis ccl  porci erant, qui c solidis sunt comparati, sicut supra dictum
est, in  duobus solidis  v porcos:   quia  sive  quinquagies  quinos,  sive
quinquies l  dixeris, ccl numerabis.  Quibus divisis unus tulit cxxv, alter
similiter.   Unus vendidit deteriores tres semper in solido; alter meliores
duos in  solido.   Sic evenit, ut is qui deteriores vendidit, de cxx porcis
xl  solidos  est  consecutus.[12]    Qui  vero  meliores,  lx  solidos  est
consecutus; quia  de inferioribus  xxx semper  in x  solidis; de melioribus
viginti autem  in x  solidis sunt  venundati:   et remanserunt  utrisque  v
porci, ex quibus ad lucrum iiii solidos et duos denarios facere potuerunt.

Solution concerning the pigs.
There were  250 pigs  to begin  with.  These were bought for 100 solidi, as
stated above,  at the  price of  two solidi per five pigs.  Because whether
you say  "50 times  five" or  "five times  50," you arrive at 250.  One man
sold three  inferior pigs  at a  price of one solidi; the other, two better
pigs per  solidi.   Thus it  happened that  he who  sold the  inferior pigs
obtained 40  solidi for  120 pigs,  whereas the  better pigs  brought in 60
solidi. This  is because it was always 30 inferior pigs for ten solidi, and
20 better  pigs for  ten solidi.   For  each man, there remained five pigs,
from which they could make four solidi and two denarii in profit.


VII.  propositio de disco pensante libras xxx.
Est discus  qui pensat  libras xxx  sive solidos  dc, habens  in se  aurum,
argentum, aurichalcum,  et stannum.   Quantum  habet auri, ter tantum habet
argenti.   Quantum argenti, ter tantum aurichalci.  Quantum aurichalci, ter
tantum stanni.  Dicat, qui potest, quantum in unaquaque specie pensat?
7.  proposition concerning the plate weighing 30 pounds.
There is  a plate  weighing 30 pounds or 600 solidi.  In it, there is gold,
silver, brass  and tin.   It has three times are much silver as gold, three
times as  much brass  as silver, and three times as much tin as brass.  Let
him say, he who can, How much does each type of metal weigh?

Solutio.
Aurum pensat  uncias novem:  argentum ter incias viiii, id est, libras duas
et tres uncias.  Aurichalcum pensat ter libras duas et [ter] iii uncias, id
est, libras  vi et  viiii uncias.   Stannum  pensat ter  libras vi,  et ter
uncias viiii,  hoc est,  libras xx,  et iii  uncias.   viiii unciae,  et ii
librae cum  iii unciis:   et  vi librae cum viiii unciis:  et xx librae cum
iii unciis adunatae, xxx libras efficiunt.
Solution.
The gold  weighs nine ounces.  The silver weighs three times this, i.e. two
pounds, three  ounces.   The brass  weighs three  times two  pounds,  three
ounces, i.e.  six pounds,  nine ounces.   The  tin weighs  three times  six
pounds, nine  ounces, i.e.  20 pounds,  three ounces.  Nine ounces, and two
pounds, three  ounces, and  six pounds,  nine ounces,  and 20 pounds, three
ounces, taken together, make 30 pounds.

Item aliter  ad solidum.   Aurum pensat solidos argenteos xv.  Argentum ter
xv, id  est, xlv.   Aurichalcum  ter xlv, id est, cxxv.  Stannum ter cxxxv,
hoc est,  ccccv.   Junge ccccv,  et cxxxv:  et xlv:  et xv; et invenies dc,
qui sunt librae xxx.
Another method.     The gold  weighs 15 silver solidi.  The silver is three
times the gold, i.e. 45.  The brass is three times 45, i.e. 125 [sic].  The
tin is  three times  135, i.e. 405.  Add 405 and 135 and 45 and 15, and you
will get 600 [solidi], which equals 30 pounds.


VIII.  propositio de cupa.
Est cupa  una, quae  c metretis  impletur capientibus  singulis modia tria;
habens fistulas  iii.   Ex numero  modiorum tertia  pars  et  vi  per  unam
fistulam currit:   per alteram tertia pars sola:  per tertiam sexta tantum.
Dicat nunc, qui vult, quot sextarii per unamquamque fistulam cucurrissent.
8.  proposition concerning the cask.
There is  a cask  which has  three cracks  in it.   It  is filled  with 100
metretae, each  holding three  modia.  Of the modia, a third and sixth part
run out through one crack.  Through another [crack], only a third part runs
out.   Only a  sixth part runs out of the third crack.  Let him say now, he
who wishes, how many sextarii ran out through each crack.

Solutio.
Per primam fistulam iii dc sextarii cucurrerunt.  Per secundum ii cccc.[13]
Per tertiam i cc.
Solution.
3600 sextarii  run through  the first  crack; 2400  through the second; and
1200 through the third.


IX.  propositio de sago.
Habeo sagum habentem in longitudine cubitos c, et in latitudine lxxx.  Volo
exinde per  portiones sagulos  facere, ita  ut unaquaeque  portio habeat in
longitudine cubitos  v, et in latitudine cubitos iiii.  Dic, rogo, sapiens,
quot saguli exinde fieri possint?
9.  proposition concerning the cloak material.
I have  a material  for cloaks  which is  100 cubits  long, 80 cubits wide.
From it,  I wish  to make  smaller cloaks  from portions in such a way that
each portion  is five  cubits in length and four cubits wide.  I ask you to
tell me, wise one, How many smaller cloaks can be made from [the material]?

Solutio.
De quadrigentis  octogesima pars  v sunt;  et centesima  iiii.   Sive  ergo
octuagies v,  sive centies  iiii duxeris,  semper cccc  invenies.  Tot sagi
erunt.[14]
Solution.
An eightieth  part of  400 is five, and a hundredth part, four.  Therefore,
whether you  measure off 80 [lengths] of five [cubits], or 100 of four, you
shall always arrive at 400.  There shall be this many cloaks.


X.  propositio de linteo.[15]
Habeo linteamen  unum longum  cubitorum lx, latum cubitorum xl.  Volo ex eo
portiones facere,  ita ut  unaquaeque portio  habeat in longitudine cubitos
senos, et  in latitudine  quaternos, ut  sufficiat ad  tunicam  consuendam.
Dicat, qui vult, quot tunicae exinde fieri possint?
10.  proposition concerning the linen cloth.
I have  a single  linen cloth  which is  60 cubits long, 40 cubits wide.  I
wish to  make it  into smaller  portions, each  being six cubits in length,
four cubits  in width, so that each piece is ample for making a tunic.  Let
him say,  he who  wishes, How  many tunics  can be  made [from  the  larger
piece]?

Solutio. [16]
Decima pars sexagenarii vi sunt.  Decima vero quadragenarii iiii sunt. Sive
ergo decimam sexagenarii, sive decimam quadragenarii decies miseris, centum
portiones vi cubitorum longas; et iiii cubitorum latas invenies.
Solution.
One tenth  of 60 is six, and a tenth of 40 is four.  Therefore, whether you
shall have  taken ten  times a tenth of 60 [cubits] or ten times a tenth of
40, you  will arrive  at 100  portions of  six cubits  in length,  and four
cubits wide.


XI.  propositio de duobus hominibus sorores accipientibus.
Si duo  homines ad  invicem, alter alterius sororem in conjugium sumpserit;
dic, rogo, qua propinquitate filii eorum sibi pertineant?
11.  proposition concerning the two men marrying [one another's] sister.
If two  men should marry one another's sister, tell me, I ask, What will be
the sons' relations to each other?

Solutio ejusdem. [17]
Verbi gratia:  si ego accipiam sororem socii mei, et ille meam, et ex nobis
procreentur filii;  ego denique  sum patruus  filii sororis  meae; et  illa
amita filii mei.  Et ea propinquitate sibi invicem pertinent.
Solution of the same [proposition].
As stated,  if I should marry my friend's sister, and he should marry mine,
sons would  be produced  by us.   Thus, I shall be the paternal uncle of my
sister's son, and she shall be my son's maternal aunt.  The relation of the
two men [to the sons] shall be the same.


XII.  propositio de quodam patrefamilias et tribus filiis ejus.
Quidam paterfamilias  moriens dimisit [18] haereditatem tribus filiis suis,
xxx ampullas  vitreas, quarum  decem fuerunt  plenae  oleo.    Aliae  decem
dimidiae.   Tertiae decem  vacuae.  Dividat, qui potest, oleum et ampullas,
ut unicuique  eorum de tribus filiis aequaliter obveniat tam de vitro, quam
de oleo.
12.  proposition concerning a certain father and his three sons.
A certain father died and left as an inheritance to his three sons 30 glass
flasks, of  which 10  were full  of oil;  another 10  were half full, while
another 10  were empty.   Let him divide, he who can, the oil and flasks so
that an  equal share  of the  commodities should  equally come  down to the
three sons, both of oil and glass.

Solutio.
Tres igitur  sunt filii,  et xxx ampullae.  Ampullarum autem quaedam x sunt
plenae, et  x mediae,  et x  vacuae.  Duc ter decies; fiunt xxx.  Unicuique
filio veniunt  x ampullae  in portionem.   Divide autem per tertiam partem,
hoc est,  da primo filio x semis ampullas, ac deinde da secundo v plenas et
v vacuas.   Similiter  dabis tertio, et erit trium aequa germanorum divisio
tam in oleo, quam in vitro.
Solution.
There are  three sons  and 30 glass flasks.  However, of the flasks, 10 are
full [of  oil], 10  half full,  and 10  empty.   Take three times 10, which
makes 30,  so each  son shall  receive 10 flasks as his portion.  Divide up
the three portions, that is, give to the first son 10 half [filled] flasks,
to the  second son  five full and five empty [flasks].  Do the same for the
third son, and the brothers' portions of glass and oil shall be the same.


XIII.  propositio de rege.
Quidam rex  jussit famulo suo colligere de xxx villis exercitum, eo modo ut
ex unaquaque  villa tot  homines sumeret  quotquot illuc  adduxisset.  Ipse
tamen ad  villam primam solus venit; ad secundam cum altero; jam ad tertiam
tres venerunt.   Dicat,  qui potest,  quot homines fuissent collecti de xxx
villis.
13.  proposition concerning the king.
A certain  king ordered  his servant  to gather an army from 30 villages as
follows:   He should  bring back as many men [from each successive village]
as he  had taken  there.   Thus, [the  servant] came  to the  first village
alone; he  came with one other person to the next; three people came to the
third.   Let him  say, he who is able, how many men were collected from the
30 villages.

Solutio. [19]
In prima igitur mansione duo fuerunt; [20] in secunda iiii, in tertia viii,
in quarta  xvi, in  quinta xxxii,  in sexta  lxiiii, in septima cxxviii, in
octava cclvi,  in nona  dxii, in decima i xxiiii, in undecima ii xlviii, in
duodecima iiii  xcvi, in  quarta decima  xvi ccclxxxiiii.  In quinta decima
xxxii dcclxviii, etc.

Solution.
In the  first village, there were two [people]; in the second, four; in the
third, eight; in the fourth, 16; in the fifth, 32; in the sixth, 64; in the
seventh, 128;  in the eighth, 256; in the ninth, 512; in the 10th, 1024; in
the 11th,  2048; in  the 12th, 4096; in the 14th, 16, 384; in the 15th, 32,
768; etc.


XIV.  propositio de bove.
Bos qui tota die arat, quot vestigia faciat in ultima riga?
14.  proposition concerning the ox.
How many  footprints in  the last  furrow does  an ox  make which  has been
plowing all day?

Solutio.
Nullum omnino  vestigium facit  bos in  ultima riga, eo quod ipse praecedit
aratrum, et  hunc aratrum  sequitur.   Quotquot  enim  hic  praecedendo  in
exculta terra  vestigia figit,[21]  tot ille subsequens excolendo resolvit.
Propterea illius nullum reperitur vestigium in ultima riga.
Solution.
An ox  makes no  footprints whatsoever in the last furrow.  This is because
the ox  goes in  front of  the plow,  and the plow follows it.  For however
many footprints  the ox makes on the ploughed earth by going first, so many
the plough  following behind destroys by ploughing.  On account of this, no
footprints appear in the last furrow.


XV.  propositio de homine.
Quaero a te ut dicas mihi quot rigas factas habeat homo in agro suo, quando
de utroque capite campi tres versuras factas habuerit?
15.  proposition concerning the man.
I ask  you in  order that  you might  tell me, How many furrows might a man
have in  his field  if he  shall have  made three turns at each head of the
field?

Solutio.
Ex uno capite campi iii.  Ex altero iii, quae faciunt rigas versuras vi.
Solution.
Three [furrows]  from one  head of  the field,  and three  from the  other,
making six plowed furrows.[22]


XVI.  propositio de duobus hominibus boves ducentibus.
Duo homines  ducebant boves  per viam, e quibus unus alteri dixit:  Da mihi
boves duos;  et habeo tot boves quot et tu habes.  At ille ait:  Da mihi et
tu duos  boves, et habeo duplum quam tu habes.  Dicat, qui vult, quot boves
fuerunt, quot unusquisque habuit.
16.  proposition concerning the two men leading oxen.
Two men  were leading oxen along the road when one said to the other, "Give
me two  oxen, and  I shall have as many oxen as you."  Then the other said,
"You give  me two  oxen, and  I shall  have twice as many as you."  Let him
say, he who wishes, how many oxen there were, and how many each man had.

Solutio.
Prior, qui  dari sibi  duos rogavit,  boves habebat  iiii.   At  vero,  qui
rogabatur,  habebat  viii.    Dedit  quippe  rogatus  postulanti  duos,  et
habuerunt uterque  sex.   Qui enim  prius acceperat,  reddidit  duos  danti
priori, qui  habebat sex, et habuit viii, quod est duplum a quator, et illi
remanserunt iiii, quod est simplum ab viii.
Solution.
At first,  the man who asked for two to be given to him had four oxen.  But
indeed, the  man who was asked had eight.  Of course, having been asked, he
gave two  to the  one asking, and each of the two had six.  For the man who
first asked  returned two to the one first giving (who now had six), and he
had eight,  which is  double four,  and four remained to that one, which is
half of eight.


XVII.  propositio de tribus fratribus singulas habentibus sorores.
Tres fratres  erant qui  singulas sorores  habebant,  et  fluvium  transire
debebant (erat  enim unicuique  illorum concupiscentia  in  sorore  proximi
sui), qui venientes ad fluvium non invenerunt nisi parvam naviculam, in qua
non potuerunt  amplius nisi  duo ex  illis transire.   Dicat,  qui  potest,
qualiter fluvium transierunt, ne una quidem earum ex ipsis maculata sit?
17.  proposition concerning the men [23] who had unmarried  sisters.
There were  three men, each having an unmarried sister, who needed to cross
a river.   Each  man was  desirous of  his friend's  sister.  Coming to the
river, they  found only  a small boat in which only two persons could cross
at a  time.   Let him say, he who is able, How did they cross the river, so
that none of the sisters were defiled by the men?

Solutio.
Primo omnino  ego et  soror mea  introissemus in  navem et transfretassemus
ultra;  transfretatoque   fluvio  dimisissem   sororem  meam  de  nave,  et
reduxissem navem  ad ripam.   Tunc vero introissent sorores duorum virorum,
illorum videlicet,  qui ad  littus remanserant.   Illis igitur feminis navi
egressis, soror  mea [quae  prima transierat], intraret, navemque reduceret
ad nos.   Illa  egrediente foras, duo in navem fratres intrassent, ultraque
venissent.   Tunc unus  ex illis  una cum  sorore sua navem ingressi ad nos
transfretassent.   Ego autem et ille, qui navigaverat, sorore mea remanente
foras, ultra  venissemus.   Nosque ad  littora vectos,  una ex illis duabus
quaelibet mulieribus,  ultra navem  reduceret, sororeque  mea secum recepta
pariter ad  nos ultra  venissent.   Et ille,  cujus soror ultra remanserat,
navem ingressus  eam secum  reduceret.  Et fieret expleta transvectio nullo
maculante contagio. [24]

Solution.
First of  all, my  sister and  I got  into the  boat and  crossed.   Having
crossed the  river, I  let my sister out and recrossed the river.  Then the
sisters of  the two  men who remained on the bank got in.  When these women
had gotten  out of the boat, my sister, who had already gone across, got in
and brought  the boat  back to  us.  She then got out, and the two brothers
crossed in the boat.  Then, one of the brothers and his sister crossed over
to us.   However,  I and the brother who piloted the boat went across while
my sister remained behind.  When we had been taken to the [other] side, one
of the  other women took the boat back across, and my sister came across to
us with  her at  the same  time.  Then the man whose sister had remained on
the other  side got in the boat and and brought it back with her.  Thus the
crossing was accomplished, with no one being defiled.

XVIII.  propositio de homine et capra et lupo.
Homo quidam  debebat  ultra  fluvium  transferre  [25]  lupum,  capram,  et
fasciculum cauli.   Et  non potuit  aliam navem  invenire, nisi  quae  duos
tantum ex  ipsis ferre  valebat.  Praeceptum itaque ei fuerat ut omnia haec
ultra illaesa transire potuit? [26]
18.  proposition concerning the man, the she-goat, and the wolf.
A certain  man needed  to take  a wolf,  a she-goat  and a  load of cabbage
across a  river.   However, he could only find a boat which would carry two
of these  [at a  time].   Thus, what rule did he employ so as to get all of
them across unharmed?

Solutio.
Simili namque  tenore ducerem  prius capram  et dimitterem  foris lupum  et
caulum.  Tum deinde venirem, lupumque transferrem: [27] lupoque foris misso
capram  navi  receptam  ultra  reducerem;  capramque  foris  missam  caulum
transveherem ultra;  atque iterum  remigassem,  capramque  assumptam  ultra
duxissem.   Sicque faciendo  facta erit remigatio salubris, absque voragine
lacerationis.
Solution.
In a  similar manner,  I would first take the she-goat and leave behind the
wolf and  the cabbage.   When  I had returned, I would ferry over the wolf.
With the  wolf unloaded,  I would  retrieve the  she-goat and  take it back
across.   Then, I  would unload  the she-goat  and take  the cabbage to the
other side.   I  would next  row back,  and take  the she-goat across.  The
crossing should  go well  by doing  thus, and  absent from  the  threat  of
slaughter.


XIX.  propositio de viro et muliere ponderantibus [plaustri pondus onusti].
De viro  et muliere,  quorum uterque  pondus habebat  plaustri onusti, duos
habentes infantes  inter  utrosque  plaustrali  pondere  pensantes  fluvium
transire debuerunt.  Navem invenerunt quae non poterat ferre plus nisi unum
pondus plaustri.   Transfretari  faciat,  qui  se  putat  posse,  ne  navis
mergatur.
19.   proposition concerning  the man and his wife, [each] weighing as much
as a loaded cart.
A man  and his wife, each the weight of a loaded cart, who had two children
each the  weight of  a small  cart, needed  to cross a river.  However, the
boat they  came across  could only  carry the  weight of one cart.  Let him
devise [a way] of crossing in order that the boat should not sink.

Solutio.
Eodem quoque  ordine, ut  superius.    Prius  intrassent  duo  infantes  et
transissent unusque  ex illis  reduceret navem.   Tunc mater navem ingressa
transisset.   Deinde filius  ejus reduceret  navem.   Qua transvecta frater
illius navim  ingressus ambo  ultra transissent, rursusque unus ex illis ad
patrem reduceret  navem.  Qua reducta, filio foris stante, pater transiret:
rursusque filius,  qui ante  transierat, ingressus  navim eamque ad fratrem
reduceret:     jamque  reductam  ingrediantur  ambo  et  transeant.    Tali
subremigante ingenio erit expleta navigatio forsitan sine naufragio.
Solution.
Also in  the same  manner, first,  the two  children get  in [the boat] and
cross; one  of them then brings the boat back.  Then the mother gets in the
boat and  crosses; her  son brings  the boat back.  With the boat back, the
brother of  this one  gets in  the boat  and both  cross; one  of them then
brings the  boat back  to the  father.  When the boat has returned and with
the son  on the  bank, the father may cross.  Then the brother who had gone
across before  get in the boat and brings it back to his brother.  Now with
the boat  returned, both  brothers get in and cross.  By such a clever plan
of crossing,  the navigation  can  perhaps  take  place  without  the  boat
sinking.


XX.  propositio de hirtitiis.[28]
De hirtitiis  masculo et femina habentibus duos natos libram ponderantibus,
flumen transire volentibus.
20.  proposition concerning the hirtitii.
A masculine  and feminine [....] who had two children weighing [29] a pound
wished to cross a river.

Solutio.
Similiter, ut  superius, transissent  prius duo  infantes, et unus ex illis
navem reduceret;  in quam pater ingressus ultra transisset; et ille infans,
qui prius  cum fratre  transierat, navim ad ripam reduceret, in quam frater
illius rursus  ingressus ambo  ultra venissent;  unusque propterea ex illis
foras egressus; et alter ad matrem reduceret navim:  in quam mater ingressa
ultra venisset:   qua  egrediente foras,  filius ejus,  qui ante  cum patre
transierat, navim  rursus ingressus eam ad fratrem ultra reduceret; in quam
ambo  ingressi   ultra  venissent,  et  fieret  expleta  transvectio  nullo
formidante naufragio.
Solution.
Again, as above, first the two children go across.  One of them brings back
the boat,  in which the father crosses.  Then, the child who had first gone
across with  his brother  brings the boat back to the river, and he and his
brother both  go across.  One of them gets out on the [opposite] shore; the
other takes  the boat  back to the mother.  The mother gets in and crosses.
When she  has unloaded at the [opposite] shore, her son, who had previously
crossed with  his father,  gets in  the boat again and takes it over to his
brother.   Both brothers  get in  and cross.  A crossing can be carried out
thusly, free from dread of accident.


XXI.  propositio de campo et ovibus in eo locandis.
Est campus  qui habet  in longitudine  pedes cc,  et in latitudine pedes c.
Volo ibidem  mittere oves;  sic tamen  ut unaquaeque  ovis habeat  in longo
pedes v,  et in  lato pedes  iv.   Dicat, rogo,  qui valet,  quot oves [30]
ibidem locari possint?
21.  proposition concerning the field and the sheep to be placed in it.
There is  a field  which is  200 feet  long, 100  feet wide.  I want to put
sheep in  it as  follows:   Each sheep should have [an area] five feet long
and four  feet wide.  Let him say, I ask he who is able, How many sheep can
be put in such a place?

Solutio.
Ipse campus  habet in longitudine pedes cc.  Et in latitudine pedes c.  Duc
bis [31]  quinquenos de cc, fiunt xl.  At deinde c divide per iiii.  Quarta
pars centenarii  xxv.   Sive ergo  xl vicies quinquies; sive xxv quadragies
ducti, [32]  millenarium implent  numerum.   Tot ergo ibidem oves colfocari
[33] possunt.

Solution.
The field  is 200  feet long and 100 feet wide.  Divide 200 by five, making
40.   Then, divide  100 by  four, a  fourth part  of which  is 25.   Hence,
whether 40  times 25,  or 25  times 40,  the number 1000 is obtained.  This
many sheep can inhabit such a place.

XXII.  propositio de campo fastigioso.
Est campus  fastigiosus, qui  habet in  uno latere perticas c, et in altero
latere perticas  c, et in fronte perticas l, et in medio perticas lx, et in
altera fronte  perticas l.  Dicat, qui potest, quot aripennas [34] claudere
debet?
22.  proposition concerning the slanting field.
There is  a slanting  field which is 100 perticae on each side, 50 perticae
on one  front, 60  perticae in  the middle,  and 50  perticae on  the other
front.   Let him  say, he who is able, How many aripennae does [this field]
enclose?

Solutio.
Longitudo hujus campi c perticis, et utriusque frontis latitudo l, medietas
vero lx  includitur.   Junge utriusque  frontis numerum  cum medietate,  et
fiunt clx.   Ex  ipsis assume  tertiam partem,  id est, liii, et multiplica
centies, fiunt  v ccc.   Divide  [35] in  xii aequas partes, et inveniuntur
ccccxli. [36]   Item  eosdem divide  in xii  partes, et reperiuntur xxxvii.
Tot sunt in hoc campo aripenni. [37]

Solution.
The field  is 100  perticae in  length, 50  perticae on  each front, and 60
perticae in  the middle.   Add  the length  of each  front with the middle,
making 160.   Take  one third of this, that is, 53, and multiply it by 100,
making 5300.   Divide  this into  12 equal  parts, and  you arrive  at 441.
Likewise, divide  this into 12 equal parts, and you get 37.  There are this
many aripenni in the field.

XXIII.  propositio de campo quadrangulo.
Est campus  quadrangulus qui  habet in  uno latere perticas xxx, et in alio
perticas xxxii, et in fronte perticas xxxiiii, et in altera perticas xxxii.
Dicat, qui potest, quot aripenni in eo concludi debent?
23.  proposition concerning the quadrangular field.
There is  a field which is 30 perticae on one side, 32 perticae on another,
34 perticae  in the  front, and 32 perticae on the remaining side.  Let him
say, he who can, How many aripenni are contained in such a field?

Solutio.
Duae ejusdem  campi longitudines  faciunt lxii.   Duc  dimidiam lxii, fiunt
xxxi.   Ac duae  ejusdem campi  latitudines junctae  fiunt lxvi.   Duc vero
mediam de  lxvi, fiunt  xxxiii.   Duc vero  [38] terties semel, fiunt i xx.
Divide per duodecimam partem bis sicut superius, hoc est, de mille viginti,
duc per  xii, fiunt lxxxv, rursusque lxxxv divide per xii, fiunt vii.  Sunt
ergo in hoc aripenni numero septem.

Solution.
Two lengths  of this  field make  62 [perticae].  Half of 62 makes 31.  But
[the other]  two sides  of the  field added  together make  66.  Half of 66
makes 33.   Take  [33] 31 times, making 1020.  Divide [1020] twice by 12 as
above, first  getting 85,  then 85  by 12,  making 7.  Thus there are seven
aripenni in this field.

XXIV.  propositio de campo triangulo.
Est campus  triangulus qui  habet in  uno latere  perticas xxx,  et in alio
perticas xxx,  et in  fronte perticas  xviii. [39]  Dicat, qui potest, quot
aripennos concludere debet?
24.  proposition concerning the triangular field.
There is  a field which is 30 perticae on one side, 30 perticae on another,
and 18  perticae in  the front.  Let him say, he who can, How many aripenni
must be contained [in such a field]?

Solutio.
Junge duas longitudines istius campi, et fiunt lx.  Duc mediam de lx, fiunt
xxx, et  quia in  fronte perticas  xviii habet,  duc mediam de xviii, fiunt
viiii.  Duc vero novies triginta, fiunt cclxx.  Fac exinde bis xii, id est,
divide cclxx,  per duodecimam,  fiunt xxii  et semis;  atque iterum xxii et
semis per  duodecimam divide partem....[40]  fit aripennis unus et perticae
x, et dimidia.
Solution.
Adding two  lengths of  the field  makes 60.  Removing half of 60 makes 30.
Because there  are 18  perticae in  front, take  half of  this away, making
nine.   Taking nine  times 30  makes 270.   Then,  divide [270]  by twelve,
making 22-and-a-half.   Again, divide 22-and-a-half by twelve, [making two,
[41] with  four left  over, which  is a  third of  12.   Thus there are two
aripenna in  this amount and three parts of a third aripennum.]  This makes
one aripennum, and 10-and-a-half perticae.


XXV.  propositio de campo rotundo.
Est campus  rotundus, qui  habet in gyro perticas cccc.  Dic quot aripennos
capere debet?
25.  proposition concerning the round field.
There is a round field which contains 400 perticae in its circle.  Tell me,
How many aripenni ought it to hold?

Solutio.
Quarta quidem  pars hujus campi, qui cccc includitur perticis est c, hos si
per semetipsos  [42] multiplicaveris,  id est, si centies duxeris, x millia
fiunt, hos in xii partes dividere debes; etenim de x millibus duodecima est
dcccxxxiii, quam  cum item  in xii  partitus fueris, invenies lxviiii.  Tot
enim aripennis hujusmodi campus includitur. [43]

Solution.
A quarter  of this  field, which  contains 400  perticae, is  100.   If you
multiply [100] by 100, you get 10,000, which you must divide into 12 parts.
For indeed,  a twelfth  of 10,000 is 833, which when again partitioned into
twelfths gives 69. [44]  This many aripenni are included in the field.

XXVI.  propositio de cursu cbnks. bc. fvgb. lfp:rks. [45]
Est campus  qui habet in longitudine pedes cl.  In uno capite stabat canis,
et in  alio stabat  lepus.   Promovit namque  canis ille  post illum,  [46]
scilicet leporem  currere.   Ast ubi ille canis faciebat in uno saltu pedes
viiii, lepus  transmittebat vii.   Dicat,  qui velit,  quot  pedes  quotque
saltus canis  persequendo, et  lepus fugiendo, quoadusque comprehensus est,
fecerunt? [47]
26.   proposition concerning  the chase  of the  dog and  the flight of the
hare.
There is  a field  which is  150 feet long.  At one end stood a dog, at the
other, a  hare.   The dog  advanced behind [the hare], namely, to chase the
hare.   But whereas the dog went nine feet per stride, the hare went [only]
seven.   Let him  say, he  who wishes, How many feet and how many leaps did
the dog take in pursuing the fleeing hare until it was caught?

Solutio.
Longitudo hujus  videlicet campi  habet pedes  cl.  Duc mediam de cl, fiunt
lxxv.   Canis vero  faciebat in  uno saltu  pedes viiii, quippe lxxv novies
ducti fiunt  dclxxv, tot  pedes leporem  consequendo [48]  canis  cucurrit,
quoadusque eum  comprehendit dente  tenaci.   At vero  quia lepus  faciebat
pedes vii, in uno saltu, duc ipsos lxxv septies. [49]  Tot vero pedes lepus
fugiendo peregit, donec consecutus est.

Solution.
The length  of this  field was 150 feet.  Taking half of 150 makes 75.  The
dog was  covering nine  feet per  stride, and nine times 75 makes 675.  The
dog thus  ran this  many feet  in chasing  the rabbit  until it  caught the
rabbit with its tenacious teeth.  And indeed, because the rabbit went seven
feet per  stride, take  75 seven  times.  This is how many feet the fleeing
rabbit travelled before being caught.


XXVII.  propositio de civitate quadrangula.
Est civitas  quadrangula quae habet in uno latere pedes mille centum; et in
alio latere pedes mille; et in fronte pedes dc, et in altera pedes dc. Volo
ibidem tecta  domorum ponere, sic, ut habeat unaquaeque casa in longitudine
pedes xl,  et in latitudine pedes xxx.  Dicat, qui velit, quot casas capere
debet?
27.  proposition concerning the quadrangular city.
There is  a quadrangular city which has one side of 1100 feet, another side
of 1000 feet, a front of 600 feet, and a final side of 600 feet.  I want to
put some  houses there so that each house is 40 feet long and 30 feet wide.
Let him say, he who wishes, How many houses ought the city to contain?

Solutio.
Si fuerunt  duae  hujus  civitatis  longitudines  junctae,  facient  ii  c.
Similiter duae,  si fuerunt  latitudines junctae,  faciunt i  cc.  Ergo duc
mediam de  i cc,  faciunt [50] dc, rursusque duc mediam de ii c, fiunt i l.
Et quia  unaquaeque domus  habet in  longitudine [51]  pedes xl, et in lato
xxx: deduc  [52] quadragesimam partem de mille l, fiunt xxvi.  Atque iterum
assume tricesimam de dc, fiunt xx.  Vicies ergo xxvi ducti, fiunt dxx.  Tot
domus capiendae sunt.

Solution.
If the  two lengths  of this  city were joined together, they would measure
2100 [feet].   Likewise,  if the  two sides were joined, they would measure
1200.  Therefore, take half of 1200, i.e. 600, and half of 2100, i.e. 1050.
Because each  house is 40 feet long and 30 feet wide, take a fourtieth part
of 1050,  making 26.  Then, take a thirtieth of 600, which is 20.  20 times
26 is 520, which is the number of houses to be contained in the city.


XXVIII.  propositio de civitate triangula.
Est civitas  triangula quae  in uno habet latere pedes c, et in alio latere
pedes c,  et in  fronte  pedes  xc,  volo  enim  ibidem  aedificia  domorum
construere, [53] sic tamen, ut unaquaeque domus habeat in longitudine pedes
xx, et  in latitudine  pedes x.  Dicat, qui potest, quot domus capi debent?
28.  proposition concerning the triangular city.
There is  a triangular city which has one side of 100 feet, another side of
100 feet,  and a  third of  90 feet.   Inside  of this,  I want  to build a
structure of  houses, however,  in such a way that each house is 20 feet in
length, 10  feet in width.  Let him say, he who can, How many houses should
be contained [within this structure]?

Solutio.
Duo igitur  hujus civitatis latera juncta fiunt cc, atque duc mediam de cc,
fiunt c.   Sed  quia in fronte habet pedes xc, duc mediam de xc, fiunt xlv.
Et quia  longitudo uniuscujusque  domus habet pedes xx, et latitudo ipsarum
pedes x, duc xx partem in [54] c, fiunt v.  Et pars decima quadragenarii iv
sunt.   Duc itaque  quinquies iiii,  fiunt xx.  Tot domos hujusmodi captura
[55] est civitas.

Solution.
Two sides  of the  city joined  together make 200; taking half of 200 makes
100.   But because  the front  is 90 feet, take half of 90, making 45.  And
since the  length of  each house  is 20 feet while the width is 10, take 20
into 100,  making five.   A  tenth part of 40 is four; thus, take four five
times, making 20.  The city is to contain this many houses in this way.


XXVIIII.  propositio de civitate rotunda.
Est civitas  rotunda quae  habet in circuitu pedum viii millia.  Dicat, qui
potest, quot  domos capere  debet, ita  ut unaquaeque habeat in longitudine
pedes xxx, et in latitudine pedes xx?
29.  proposition concerning the round city.
There is  a city  which is 8000 feet in circumference.  Let him say, he who
is able, How many houses should the city contain, such that each [house] is
30 feet long, and 20 feet wide?

Solutio.
In hujus  civitatis ambitu  viii millia  pedum numerantur, qui sesquialtera
proportione dividuntur  in xxxx  dccc, et  in  iii  cc.    In  illis  autem
longitudo domorum;  in istis latitudo versatur.  Subtrahe itaque de utraque
summa medietatem, et remanent de majori ii cccc:  de minore vero i dc.  Hos
igitur i  dc divide  in vicenos  et invenies  octoagies viginti,  rursumque
major  summa,   id  est   ii  cccc,  in  xxx  partiti,  octoagies  triginta
dinumerantur.   Duc octoagies  lxxx, et  fiunt vi  millia  cccc.    Tot  in
hujusmodi civitate  domus, secundum  propositionem supra scriptam, construi
[56] possunt.

Solution.
This city  measures 8000  feet around, which is divided into proportions of
one-and-a-half to  one, i.e.  4800 and  3200.   The length and width of the
houses are  to be  of these  [dimensions].   Thus, take half of each of the
above [measurements],  and from  the larger number there shall remain 2400,
while from  the the  smaller, 1600.   Then, divide 1600 into twenty [parts]
and you will obtain 80 times 20.  In a similar fashion, [divide] the larger
number, i.e. 2400, into 30 pieces, deriving 80 times 30.  Take 80 times 80,
making 6400.   This  many houses  can be  built in  the city, following the
above-written proposal.

XXX.  propositio de basilica.
Est basilica  quae habet  in longitudine  pedes ccxl, et in lato pedes cxx.
Laterculi vero  stratae ejusdem unus laterculus habet in longitudine uncias
xxiii, hoc  est, pedem unum et xi uncias.  Et in latitudine uncias xii, hoc
est, pedem i.  Dicat, qui velit, quot laterculi eamdem debent implere?
30.  proposition concerning the basilica.
There is a basilica which is 240 feet long, 120 feet wide.  One tile of the
tiled basilica is 23 inches long, that is, one foot, 11 inches, while being
12 inches  wide, i.e. one foot.  Let him say, he who wishes, How many tiles
are needed to cover the basilica?

Solutio.
cxl pedes  longitudinis implent  cxxvi laterculi;  et cxx pedes latitudinis
cxx laterculi;  quia unusquisque  laterculus in  latitudine pedis  mensuram
habet.   Multiplica itaque  centum vicies  cxxvi,  in  xv  cxx  [57]  summa
concrescit.     Tot  igitur  in  hujusmodi  basilica  laterculi  pavimentum
contegere possunt.

Solution.
126 tiles  build 140  [sic] feet of length, [58] and 120 tiles, 120 feet of
width, because  each brick measures one foot in length.  Thus, multiply 120
by 126,  obtaining 15,120.  Therefore in this way so many tiles are able to
cover the ground of the basilica.


XXXI.  propositio de canava. [59]
Est canava  quae habet  in longitudine pedes c, et latitudine pedes lxiiii.
Dicat, qui  potest, quot cupas capere debet?  ita tamen, ut unaquaeque cupa
habeat in  longitudine pedes  vii, et in lato, hoc est in medio pedes iiii,
et pervius unus habeat pedes iiii. [60]
31.  proposition concerning the wine cellar.
There is  a wine  cellar which  is 100 feet long and 64 feet wide.  Let him
say, he  who can, How many casks can it hold, given that each cask is seven
feet long  and four  feet wide,  and given that there is an aisle four feet
wide in the middle [of the cellar]?

Solutio.
In centum  autem quaterdecies  vii  numerantur,  in  lxiiii  vero  sedecies
quaterni continentur,  ex quibus  iiii ad  pervium reputantur, [61] quod in
longitudinem ipsius  canavae ducitur.  [62]   Quia ergo  in  lx  quindecies
quaterni sunt;  et in  centum quaterdecies  septeni; duc  quindecies xiiii,
[63] fiunt  ccx.   Tot cupae  juxta suprascriptam magnitudinem in hujusmodi
canava [64] contineri possunt.

Solution.
There are  fourteen sevens  in 100,  and sixteen fours in 64, of which four
are needed  for the  aisle which runs the length of this cellar.  And since
there are  fifteen fours in 60, and since there are fourteen sevens in 100,
take 15 times 14, making 210.  This many casks can be stored in the type of
wine cellar described above. [65]

XXXII.  propositio de quodam patrefamilias.
Quidam paterfamilias habuit familias xx.  Et jussit eis dare [66] de annona
modios xx.   Sic jussit, ut viri acciperent [67] modios ternos, et mulieres
binos, et infantes singula semodia.  Dicat, qui potest, quot viri, aut quot
mulieres, vel quot infantes esse debent? [68]
32.  proposition concerning a certain head of household.
A certain head of household had 20 servants. He ordered them to be given 20
modia  of  corn  as follows: The men should receive three modia; the women,
two; and  the children,  half a  modium.  Let him say, he who can, How many
men, women and children must there have been?

Solutio.
Duc semel  ternos,  fiunt  iii,  hoc  est,  unus  vir  ut  modios  accepit.
Similiter et  quinquies bini, fiunt x, hoc est, quinque mulieres acceperunt
modia [69]  x.  Duc vero septies binos, fiunt xiiii, hoc est xiiii infantes
acceperunt modios  vii.   Junge ergo  i et  v et xiiii, fiunt xx.  Hae sunt
familiae xx.   Ac  deinde junge  iii et vii et x, fiunt xx, haec sunt modia
xx.  Sunt ergo simul familiae xx, et modia [70] xx.
Solution.
Take one  three times  which makes  three; that  is, each man received this
many modia.   Likewise, take five twice, making 10; in this way, five women
received 10  modia.   Then, take  two seven  times,  making  14;  thus,  14
children received seven modia.  Add one and five and 14, making 20; this is
the number  of servants.   Then, add three and seven and 10, this being the
number of modia.  Thus there are 20 servants and 20 modia [of corn].


XXXIII.   propositio de  alio patrefamilias erogante suae familiae annonam.
Quidam paterfamilias  habuit familias  xxx, quibus  jussit dari  de  annona
modios xxx.  Sic vero jussit, ut viri acciperent modios ternos, et mulieres
binos, et infantes singula semodia.  Sovat, qui potest, quot viri, aut quot
mulieres, quotve infantes fuerunt?
33.   proposition concerning another head of household distributing corn to
his servants.
A certain  head of household had 30 servants whom he ordered to be given 30
modia of  corn as  follows:  The men should receive three modia; the women,
two; and  the children, a half modium.  Let him solve, he who can, How many
men, women and children were there?

Solutio.
Si duxeris  ternos ter,  fiunt viiii.  Et si duxeris quinquies binos, fiunt
x, ac  deinde duc  vicies bis semis, fiunt xi, hoc est, viri iii acceperunt
modia viiii,  et quinque mulieres acceperunt x, et xxii infantes acceperunt
xi modia.   Simul juncti iii et v, et xxii faciunt familias xxx.  Rursusque
viiii et  xi, et  x, simul  juncti faciunt  modia xxx.    Quod  sunt  simul
familiae xxx, et modii xxx. [71]

Solution.
If you take thrice three, you get nine; if you take two five times, you get
10; and  if you take half of 22, you get 11.  Thus, three men received nine
modia; five  women received  10; and 22 children received 11 modia.  Adding
three and  five and  22 makes  30 servants.   Likewise,  nine and 11 and 10
makes 30 modia.  Hence there are 30 servants, and 30 modia [of corn].

XXXIV.  propositio altera de patrefamilias partiente familiae suae annonam.
Quidam paterfamilias  habuit familias  c, quibus  praecepit dare  de annona
modios c, eo vero tenore, ut viri acciperent modios ternos, mulieres binos,
et infantes  singula semodia.   Dicat  ergo, qui  valet,  quot  viri,  quot
mulieres, aut quot infantes fuerunt?
34.   another proposition  concerning a head of household distributing corn
to his servants.
A certain  head of  household had  100 servants.   He  ordered that they be
given 100  modia of  corn as  follows:  The men should receive three modia;
the women,  two; and the children, half a modium.  Thus let him say, he who
can, How many men, women, and children were there?

Solutio.
Undecim terni  fiunt xxxiii.   Et  xv bis  ducti fiunt xxx, [72] id est, xi
viri acceperunt  xxxiii modios;  et xv  mulieres acceperunt  xxx et lxxiiii
infantes acceperunt  xxxvii, qui simul juncti, id est, xi et xv, et lxxiiii
fiunt c,  quae sunt  familiae c.   Similiter junge xxxiii, et xxx et xxxvii
faciunt [73]  c, qui sunt modii c.  His ergo simul junctis habes familias c
et modios c.
Solution.
11 times three makes 33, and twice 15 makes 30; that is, 11 men received 33
modia [of  corn].   15 women  received 30 [modia], and 74 children received
37.   Adding these together, that is, 11 and 15 and 74, makes 100, which is
the number of servants.  Likewise, adding 33 and 30 and 37 makes 100, which
is the  number of  modia.  Thus with these sums, you have 100 servants, and
100 modia [of corn].


XXXV.  propositio de obitu cujusdam patrisfamilias.
Quidam paterfamilias  moriens  reliquit  infantes,  et  in  facultate  sua,
solidorum dcccclx,  [74] et  uxorem praegnantem.    Qui  jussit  ut  si  ei
masculus nasceretur,  acciperet de  omni massa  dodrans,  hoc  est,  uncias
viiii.   Et mater ipsius acciperet quadrans, hoc est, uncias iii.  Si autem
filia nata esset, [75] acciperet septunx, hoc est vii [76] uncias, et mater
ipsius acciperet  quincunx, hoc  est, v  uncias.  Contigit autem ut geminos
parturiret, id est, puerum et puellam.  Solvat, qui potest, quantum accepit
mater, et quantum filius, quantumve filia?
35.  proposition concerning the death of a certain father.
A certain  father died  and left  behind children, a pregnant wife, and 960
solidi from  his estate.  [However, on his deathbed], he stipulated that if
a son  should be  born to her, then the son should receive three fourths of
the inheritance -- that is, nine twelfths.  The mother should get a quarter
[of the  estate], that  is, three  twelfths.   However, if  a daughter were
born, she should receive seven twelfths, and the mother, five twelfths. But
as it  happened, she gave birth to twins -- both a boy and a girl.  Let him
solve, he who can, How much did the mother, son and daughter each receive?

Solutio. [77]
Junge ergo  viiii et  iii, fiunt  xii, xii  namque unciae  libram  faciunt.
Rursusque junge  similiter vii  et v,  fiunt iterum  xii.   Ideoque bis xii
faciunt xxiiii,  xxiiii autem  faciunt duas  libras, id  est,  solidos  xl.
Deinde ergo [duc] per vicesimam quartam partem dcccclx solidos, et vicesima
quarta pars  eorum fiunt  xl.   Deinde duc,  quia facit  [78] dodrans  sive
dodrans, xl  in nonam partem, ideo novies xl accepit filius, hoc est, xviii
libras, quae  faciunt solidos  ccclx.   Et quia mater tertiam partem contra
filium accepit,  et quintam  contra filiam,  iii et  v, fiunt viii.  Itaque
duc, quia  legitur, quod  faciat bis  seu bisse  xl in parte octava; octies
ergo xl  accepit mater,  hoc est,  libras xvi,  quae faciunt solidos cccxx.
Deinde duc,  quia legitur, quod faciat septunx, xl in vii partibus:  postea
duc septies  xl, fiunt xiiii librae, quae faciunt solidos cclxxx, hoc filia
accepit.   Junge ergo  ccclx et  cccxx et  cclxxx, fiunt  dcccclx solidi et
xlviii librae.
Solution.
Add nine and three, making 12.  12 ounces make a pound.  Then add seven and
five which make another 12.  12 taken twice makes 24 [ounces], equaling two
pounds, itself  equal to  40 solidi.  Then take a twenty-fourth part of the
960 solidi  which is  40.   Then, because the son received three fourths or
nine twelfths  [of the  inheritance], take a ninth of 40.  The son received
nine times  40 [ounces],  that is, 18 pounds, which equals 360 solidi.  And
since the  mother received  a third as much as the son received and a fifth
as much  as the  daughter, [she  got] three  and five  which  makes  eight.
Therefore, as  prescribed, take  twice 40  and divide  it into eight parts.
Thus the mother received eight times 40 [ounces], that is, 16 pounds, which
is 320  solidi.   Then, as  stipulated, divide 40 into seven parts so as to
get seven  twelfths.   After this, take seven times 40, that is, 14 pounds,
which equals  280 solidi.  This is what the daughter received.  Add 360 and
320 and 280, making 960 solidi, 48 pounds.


XXXVI.  propositio de salutatione cujusdam senis ad puerum.
Quidam senior salutavit puerum, cui et dixit:  Vivas, filii, vivas, inquit,
quantum vixisti,  et aliud  tantum, et ter tantum.  Addatque tibi Deus unum
de annis  meis, et  impleas annos centum.  Solvat, qui potest, quot annorum
tunc tempore puer erat?
36.  proposition concerning a certain old man's greeting to a boy.
A certain  old man  greeted a  boy, saying to him:  "May you live, boy, may
you live  for as  long as  you have [already] lived, and then another equal
amount of time, and then three times as much.  And may God grant you one of
my years,  and you  shall live  to be 100."  Let him solve, he who can, How
many years old was the boy at that time?

Solutio.
In eo  vero, quod dixit, vivas, quantum vixisti, vixerat ante annos viii et
menses tres:   et  aliud tantum  fiunt anni  xvi et  menses vi,  et alterum
tantum fiunt  anni xxxiii,  qui ter  multiplicati fiunt  anni xcviiii, unum
ipsis additum fiunt c.
Solution.
When [the  old man] said "may you live for as long as you have lived," [the
boy] had  [already] lived  eight years, three months.  Another equal number
of years  make 16  years, six  months, while  another equal  span makes  33
years.   Three times  this makes  99 years,  which with one more year added
makes 100.


XXXVII.  propositio de quodam homine volente aedificare domum.
Homo quidam,  volens aedificare  domum, locavit  artifices vi,  ex quibus v
magistri et  unus discipulus  erat, et  convenit inter  eum, qui aedificare
volebat; et  artificies, ut  per singulos  dies xxv  denarii eis in mercede
darentur, sic tamen, ut discipulus medietatem de eo, quod unus ex magistris
accipiebat, acciperet.  Dicat, qui potest, quantum unusquisque de illis per
unamquamque diem accepit?
37.  proposition concerning a certain man wishing to build a house.
A certain  man, wanting  to build  a house, found six workmen, of whom five
were masters  and one  an apprentice.   It  was agreed  between the man who
wanted to build and the workmen that 25 denarii should be given to them per
day as  pay, and  that the  apprentice should receive half what the masters
receive.   Let him  say, he  who can, How much did each of them receive per
day?

Solutio.
Tolle primum  xxii denarios  et divide  eos in vi partes.  Sic unicuique de
magistris, qui  quinque sunt, iiii denarios; nam quinquies quatuor xx sunt.
Duos, qui  remanserunt, quae est medietas de uno, tolle et da discipulo; et
sunt adhuc  iii denarii  desidui; quos  sic distribues.   Fac  de unoquoque
denario partes  xi, ter  undecim fiunt xxxiii, tolle illas triginta partes,
divide eas  inter magistros  v.   Quinquies seni  fiunt xxx.  Accidunt ergo
unicuique  magistro   partes  vi.    Tolle  tres  partes,  quae  super  xxx
remanserunt, quod est medietas senarii, et da discipulo.
Solution.
First, take  22 denarii  and divide them into six parts.  Give four denarii
to each  of the  five masters,  since five  times four  is 20.    Take  the
remaining two  denarii, which  is half  of [a  share], and give them to the
apprentice.   There are  still three denarii remaining which you distribute
thusly:   Divide each  denarius into  11 parts, making 33.  Take 30 of them
and divide  them amongst  the five  masters, as  five times  six makes  30.
Hence, six parts go to each master.  Take the remaing three parts, that is,
half of  the six  [which the  masters  received],  and  give  them  to  the
apprentice.


XXXVIII.  propositio de quodam emptore in animalibus centum. [79]
Voluit quidam  homo emere  animalia promiscua  c de solidis c, ita ut equus
tribus solidis  emeretur; bos vero in solido i, et xxiiii [80] oves in sol.
i. Dicat, qui valet, quot caballi, vel quot boves, quotve fuerunt oves?
38.  proposition concerning a certain purchaser and [his] 100 animals.
A certain  man wanted to buy 100 various animals for 100 solidi.  He wished
to pay  three solidi per horse, one solidus per cow, and one solidus per 24
sheep.   Let him  say, he  who can,  How many  horses, cows  and sheep were
there?

Solutio.
Duc ter  vicies tria  i, fiunt  lxviiii.   Et duc bis vicies quatuor, fiunt
xlviii.   Sunt ergo  caballi xxiii,  et solidi lxviiii.  Et oves xlviii, et
solidi ii.   Et  boves xxviiii,  in solidis  xxviiii.   Junge ergo xxiii et
xlviii et  xxviiii, fiunt  animalia c.   Ac  deinde junge  lxviiii et ii et
xxviiii, fiunt solidi c.  Sunt ergo simul juncta animalia c, et solidi c.

Solution.
Take three times 23, making 69.  Then, take two times 24, making 48.  There
are thus  23 horses  [which cost] 69 solidi, 48 sheep [costing] two solidi,
and 29  cows [which  cost] 29  solidi.   Therefore, add  23 and  48 and 29,
making 100  animals.  Then, add 69 and two and 29, making 100 solidi.  Thus
there are 100 animals and just as many solidi.


XXXVIIII.  propositio de quodam emptore in oriente.
Quidam homo  voluit de c solidis animalia promiscua emere c in oriente; qui
jussit famulo  suo, ut  camelum v  solidis acciperet;  asinum solido i.  xx
oves in  solido compararet.   Dicat, qui vult, quot cameli, vel asini, sive
oves in negotio c solidorum fuerunt?
39.  proposition concerning a certain purchaser in the east.
A certain  man wished  to buy  100 assorted  animals for  100 solidi in the
East.  He ordered his servant to pay five solidi per camel, one solidus per
ass, and  one solidus  per 20  sheep.  Let him say, he who wishes, How many
camels, asses and sheep were obtained for 100 solidi?

Solutio.
Si duxeris x novies, [et] v fiunt xcv, hoc est, cameli xviiii sunt empti in
solidis xcv.   Adde  cum ipsis  unum, hoc  est, in solido i asinum i, fiunt
xcvi.  Ac deinde duc vicies quater, fiunt lxxx, hoc est, in quatuor solidis
oves lxxx.   Junge  ergo xviiii et i et lxxx, fiunt c.  Haec sunt animalia.
Ac deinde  junge xcv,  et i  et iiii,  fiunt solid.  c.   Simul ergo juncti
faciunt pecora c, et solidos c.

Solution.
If you  take 10 nine times and add five, you get 95; that is, 19 camels are
bought for 95 solidi.  Add to this one solidus for an ass, making 96. Then,
take 20 times four, making 80 -- that is, 20 sheep for four solidi.  Add 19
and one  and 80,  making 100 -- this is the number of animals.  Then add 95
and one  and four,  making 100  solidi.  Hence there are 100 beasts and 100
solidi.

XL.  propositio de homine et ovibus in monte pascentibus.
Quidam homo vidit de monte oves pascentes, et dixit, utinam haberem tantum,
et aliud  tantum et  medietatem de  medietate, et  de hac  medietate  aliam
medietatem, [81]  atque ego centesimus una cum ipsis ingrederer meam domum.
Solvat, qui potest, quot oves vidit ibidem pascentes?
40.  proposition concerning a man and [some] sheep grazing on a mountain.
A certain  man saw  from a  mountain some sheep grazing and said, "O that I
could have  so many,  and then  just as many more, and then half of half of
this [added],  and then  another half  of this  half.  Then I, as the 100th
[member], might  head back  to my home together with them."  Let him solve,
he who can, How many sheep did the man see grazing?

Solutio.
In hoc  ergo, quod  dixit; haberem  tantum; xxxvi oves primum ab illo visae
sunt.   Et aliud  tantum fiunt  lxxii,  atque  medietas  de  hac  videlicet
medietate, hoc  est, de  xxxvi, fiunt  x et viii.  Rursusque de hac secunda
scilicet medietate  assumpta medietas, id est, de xviii fiunt viiii.  Junge
ergo xxxvi  et xxxvi,  fiunt lxxii.   Adde cum ipsis xviii, fiunt xc.  Adde
vero viiii  cum xc,  fiunt xcviiii.   Ipse vero homo cum ipsis additus erit
centesimus.

Solution.
36 sheep  were first  seen by the man when he said, "O that I could have so
many."   Adding an  equal number makes 72, and a half of half of this, that
is, of  36, makes  18.   And again,  a half  of this, that is, of 18, makes
nine.  Therefore add 36 and 36, making 72.  Add to this 18, which makes 90.
Then add nine to 90, making 99.  The man himself added to these will be the
100th one.

XLI.  propositio de sode et scrofa.
Quidam paterfamilias  stabilivit curtem  novam, [82] in qua posuit scrofam,
quae peperit  porcellos vii in media sode, qui83 una cum matre, quae octava
est, pepererunt  igitur unusquisque  in omni angulo vii.  Et ipsa iterum in
media sode  cum omnibus  generatis peperit  vii.   Dicat, qui vult, una cum
matribus quot porci fuerunt?
41.  proposition concerning the pigsty and the sow.
A certain  head of household set up a new [quadrangular] enclosure in which
he placed  a sow.  The sow gave birth to seven piglets in the middle of the
sty.  The offspring, along with the mother, the eighth pig, each gave birth
to another  seven piglets in each corner [of the sty].  Then, in the middle
of the  sty, the  mother and  all her  offspring [each] gave birth to seven
more.   Let him  say, he who wishes, How many pigs were there [in the end],
including the mother?

Solutio.
In prima  igitur parturitione,  quae fuit  facta  in  media  sode,  fuerunt
porcelli vii,  et mater  eorum octava.   Octies  igitur  octo  ducti  fiunt
lxiiii.   Tot porcelli  una cum  matribus fuerunt  in i  angulo.  Ac deinde
sexagies quater  octo ducti  fiunt dxii.  Tot cum matribus suis porcelli in
angulo ii.   Rursusque  dxii octies  ducti fiunt  i.ii xcvi.  Tot in tertio
angulo cum  matribus suis  fuerunt.   Qui si  octies multiplicentur,  fiunt
xxxii dcclxxxviii,  tot cum  matribus in quarto fuerunt angulo.  Multiplica
quoque octies  xxxii dcclxxxviii,  fiunt cc  lxii et  ccciiii.    Tot  enim
creverunt, cum in media sode novissime partum fecerunt.
Solution.
In the  first birth,  which took place in the middle of the sty, there were
seven piglets,  with the  mother being  the eighth  [member].   Eight taken
eight times  is 64 -- this many piglets, along with the mother, were in the
first corner.   Then,  64 taken eight times makes 512 -- this many piglets,
including their  mothers, were in the second corner.  512 taken eight times
yields 4096  -- this  many piglets,  along with  their mother,  were in the
third corner.   If  [4096] is multiplied eight times, one gets 32,788 [sic]
[84] -- this many piglets, including the mother, were in the fourth corner.
Taking eight times 32,788 [sic] makes 262,304 [sic]. [85]  There grew to be
this many [pigs] in the last stage in the middle of the sty.


XLII.  propositio de scala habente gradus centum.
Est scala  una habens  gradus c.   In  primo gradu  sedebat columba una; in
secundo duae;  in tertio  tres; in  quarto iiii;  in quinto v.  Sic in omni
gradu usque  ad centesimum.   Dicat,  qui potest,  quot columbae  in  totum
fuerunt?
42.  proposition concerning the ladder having 100 steps.
There is a ladder which has 100 steps.  One dove sat on the first step, two
doves on  the second,  three on  the third, four on the fourth, five on the
fifth, and  so on  up to  the hundredth step.  Let him say, he who can, How
many doves were there in all?

Solutio.
Numerabitur autem  sic:   a primo  gradu in  quo una sedet, tolle illam, et
junge ad  illas xcviiii,  quae nonagesimo [nono] gradu consistunt, et erunt
c.   Sic secundum  ad nonagesimum octavum et invenies similiter c.  Sic per
singulos gradus,  unum de  superioribus gradibus, et alium de inferioribus,
hoc ordine conjunge, et reperies semper in binis gradibus c. Quinquagesimus
autem gradus  solus et  absolutus  est,  non  habens  parem;  similiter  et
centesimus solus remanebit.  Junge ergo omnes et invenies columbas vl.

Solution.
There will  be as many as follows:  Take the dove sitting on the first step
and add  to it the 99 doves sitting on the 99th step, thus getting 100.  Do
the same with the second and 98th steps and you shall likewise get 100.  By
combining all  the steps  in this  order, that  is, one of the higher steps
with one  of the  lower, you shall always get 100.  The 50th step, however,
is alone and without a match; likewise, the 100th stair is alone.  Add them
all and you will find 5050 doves.


XLIII.  propositio de porcis.
Homo quidam habuit ccc porcos, et jussit, ut tot porci numero impari in iii
dies occidi  deberent. [86]   Similis  est et de xxx sententia.  Dicat, qui
potest, quot  porci impares  sive de  ccc sive  de  xxx,  inter  tres  dies
occidendi sunt?  Haec ratio indissolubilis ad increpandum composita est.
43.  proposition concerning the pigs.
A certain  man had  300 pigs.   He ordered all of them slaughtered in three
days, but  with an uneven number being killed each day.  He wished the same
thing to be done with 30 pigs.  Let him say, he who can, What odd number of
pigs out  of 300  or 30  were to  be killed  in three days?  (This ratio is
indissoluble and was composed for rebuking.)

Solutio.
Ecce fabula!  quae a  nemine solvi  potest, ut  ccc porci, sive triginta in
tribus diebus  impari numero  occidantur.  Haec fabula est tantum ad pueros
increpandos.

Solution.
Behold an  impossibility which  is able  to be solved by nobody!, in such a
way that  30 [pigs]  be killed  in three  days by  an odd  number.  Such an
implausible story is only for teasing young boys.


XLIIII.  propositio de salutatione pueri ad patrem.
Quidam puer  salutavit patrem;  Ave, inquit,  pater!   Cui pater:   Valeas,
fili! vivas,  quantum vixisti,  quos annos  geminatos triplicatos;  [87] et
sume unum  de annis  meis; et  habebis annos  c.   Dicat, qui  potest, quot
annorum tunc tempore puer erat?
44.  proposition concerning the boy's greeting to his father.
A certain  boy addressed  his father,  saying, "Greetings,  father!"    The
father responded,  "May you fare well, my son, and may you live three times
twice your  years.   Then, adding  one of my own years, you will live to be
100."  Let him say, he who can, How many years was the boy at the time?

Solutio.
Erat enim  puer annorum xvi, et mensium vi, qui geminati cum mensibus fiunt
anni xxxiii,  qui triplicati  fiunt xcviiii.   Addito  uno  patris  anno  c
apparent.
Solution.
They boy  was 16  years, six  months.   Double this  makes 33  years, which
tripled is 99.  Having added one year of the father, there are 100.


XLV.  propositio.
Columba sedens in arbore vidit alias volantes; dixit eis:  Utinam fuissetis
aliae tantum et ternae tantum, [88] tunc una mecum fuissetis c.  Dicat, qui
potest, quot columbae erant in primis volantes?
45.  proposition.
A dove  sitting in  a tree saw some other doves flying and said to them, "O
that you  were doubled,  and then  tripled.  Then, along with me, you would
number 100."   Let  him say,  he who  can, How  many doves  were  initially
flying?

Solutio.
Triginta iii  erant columbae,  quas prius  conspexit volantes.   Item aliae
tantae fiunt  lxvi.   Et tertiae tantum, fiunt xcviiii.  Adde sedenteni, et
erunt c.
Solution.
There were  33 doves  flying at  first.   Double this makes 66, while three
times [33] makes 99.  Adding in the sitting dove makes 100.


XLVI.  propositio de sacculo ab homine invento.
Quidam homo  ambulans per  viam invenit  sacculum cum talentis duobus.  Hoc
quoque alii  videntes dixerunt  ei:  Frater, da nobis portionem inventionis
tantum. [89]  Qui renuens noluit eis dare.  Ipsi vero irruentes diripuerunt
sacculum, et  tulit sibi  quisque solidos  quinquaginta.   Et ipse postquam
vidit se  resistere non  posse, misit manum et rapuit solidos quinquaginta.
Dicat, qui vult, quot homines fuerunt?
46.  proposition concerning the small bag found by the man.
A certain  man walking  in the  street found  a small  bag  containing  two
talents.   Some other people saw this and said to him:  "Brother, give us a
portion of your discovery."  But the man shook his head and did not want to
give them any.  The others then rushed at him and tore apart the sack, each
obtaining for  himself 50  solidi.   And when  the man saw that he could no
longer resist  [their attack],  he grabbed  50 solidi for himself.  Let him
say, he who wishes, How many men were there?

Solutio.
Apud quosdam  talentum lxxii  vel pondo vel habet libras.  Libra vero habet
solidos aureos  lxxii.   Sexagies quinquies  lxxii ducti  fiunt v cccc, qui
numerus duplicatus  fiunt decies  dccc.   In x millibus et octingentis sunt
quinquagenarii ccxvi.  Tot homines idcirco fuerunt.

Solution.
Each talent  has 72  pounds in  it by  weight, and  a pound  equals 72 gold
solidi.   65 times 72 equals 5400 [sic], [90] twice which makes 10,800.  50
goes into  10,800 216  times, which  is the number of men [in the problem].
[91]

XLVII.  propositio de episcopo qui jussit xii panes dividi.
Quidam episcopus  jussit xii  panes dividi in clero.  Praecepit enim sic ut
singuli  presbyteri  binos  acciperent  panes;  diaconus  dimidium,  lector
quartam partem:   ita  tamen fiat, ut clericorum et panum unus sit numerus.
Dicat, qui vult, quot presbyteri, vel quot diacones, aut quot lectores esse
debent?
47.  proposition concerning the bishop who ordered 12 loaves of bread to be
divided.
A certain bishop ordered 12 loaves of bread divided amongst the clergy.  He
stipulated that  each priest  should receive  two loaves;  a deacon, half a
loaf; and  a lector,  a quarter  part.   Hence, it should turn out that the
number of  clerics and  loaves is  the same.   Let him say, he who can, How
many priests, deacons and lectors must there have been?

Solutio.
Quinquies bini  fiunt x,  id est,  v presbyteri decem panes receperunt:  et
diaconus unus  dimidium panem:   et  inter lectores  vi habuerunt  panem et
dimidium.  Junge v et i et vi in simul, et fiunt xii.  Rursusque junge x et
semis et  unum et  semis, fiunt  xii.   Et illi  sunt xii  panes; qui simul
juncti faciunt  homines xii et panes xii.  Unus est ergo numerus clericorum
et panum.
Solution.
Twice five is 10; that is, five priests received 10 loaves.  The deacon got
half a loaf, and there was a loaf and a half for the six lectors.  Add five
and one  and six,  making 12.   Then  add 10-and-a-half and one-and-a-half,
making 12,  this being  the number  of loaves.   Hence,  there are  12  men
altogether and  12 loaves.   Therefore, the number of clerics and loaves is
the same.


XLVIII.  propositio de homine qui obviavit scholaribus.
Quidam homo  obviavit scholaribus,  [92] et  dixit eis:   Quanti  estis  in
schola? Unus  ex eis respondit dicens:  Nolo hoc tibi dicere, tu numera nos
bis, multiplica  ter; tunc  divide in  quatuor partes.  Quarta pars numeri,
[93] si  me addis cum ipsis, centenarium explet numerum.  Dicat qui potest,
quanti fuerunt, qui pridem obviaverunt ambulanti per viam?
48.  proposition concerning the man who met [some] students.
A certain  man met some students and asked them, "How many of you are there
in school?"   One  of [the  students] responded  to him:  "I do not want to
tell you  [except as  follows]:   double the number of us, then triple that
number; then,  divide that number into four parts.  If you add me to one of
the fourths,  there will  be 100."   Let  him say,  he who  can,  How  many
[students] first met the man?

Solutio.
Terties ter  bini [id est, bis xxxiii] fiunt lxvi:  tanti erant, qui pridem
obviaverunt  ambulanti;   qui  numerus  bis  ductus  cxxxii  reddit.    Hos
multiplica ter, fiunt cccxcvi, horum quarta pars xcviiii sunt.  Adde puerum
respondentem et reperies c.

Solution.
Twice 33  makes 66; this is the number [of students] who first met the man.
Twice this  number yields  132, and  three times  this number  gives 396, a
quarter part  of which  is 99.   Add in the responding boy and you will get
100.


XLVIIII.  propositio de carpentariis.
Septem carpentarii septenas rotas fecerunt.  Dicat, qui potest, quot carrae
rexerunt? [94]
49.  proposition concerning the carpenters.
Seven carpenters  [each] made  seven wheels.   Let him say, he who can, How
many carts did they build?

Solutio.
Duc septies  vii fiunt  xlviiii, tot rotas fecerunt.  xii vero quater ducti
xlviii reddunt.   Super  xl et  viiii rotas  xii carra  sunt erecta, et una
superfuit rota.
Solution.
Take seven  times seven,  making 49,  this being  the number of wheels.  12
taken four  times yields  48.   12 carts were assembled from the 49 wheels,
with one wheel left over.


L.  propositio de vino in vasculis.
Centum metra  vini, rogo,  ut dicat,  qui vult, quot sextarios capiunt? vel
ipsa etiam centum metra quot meros habent?
50.  proposition concerning the wine in small vessels.
I ask so that one who wishes might respond:  How many sextarii do 100 metra
of wine contain, and how many meri do 100 metra have?

Solutio.
Unum metrum  capit sectarios xl et viii.  Duc centies xlviii, fiunt quatuor
millia dccc.   Tot  sextarii sunt.   Similiter  et unum  metrum habet meros
cclxxxviiii, duc centies cclxxxviiii fiunt xxviii dcccc.  Tot sunt meri.

Solution.
One metrum  containes 48 sextarii.  Take 48 a hundred times, making 4800 --
this is  the number  of sextarii  [in 100  metra].   Likewise,  one  metrum
contains 289  meri.   100 times 289 is 28,900 -- this is the number of meri
[in 100 metra].


LI.  propositio de vini in vasculis a quodam patre divisione. [95]
Quidam paterfamilias  moriens dimisit  [96] iiii filiis, iiii vascula vini:
in primo vase erant modia xl, in secundo xxx, in tertio xx, et in quarto x;
qui vocans  dispensatorem domus  suae ait:   Haec  quatuor vascula cum vino
intrinsecus manente  divide  inter  quatuor  filios  meos;  sic  tamen,  ut
unicuique eorum una [97] sit portio tam in vino, quam in vasis.  Dicat, qui
intelligit, quomodo  dividendum est,  ut omnes  aequaliter ex  hoc accipere
possint?
51.   proposition concerning the wine in small vessels divided by a certain
father.
A certain dying father left four small vessels of wine to his four sons. In
the first  vessel, there were 40 modia [of wine]; in the second, 30; in the
third, 20;  and in  the fourth,  10.  Calling his house treasurer, he said:
"Divide these  four vessels  containing wine amongst my four sons in such a
way that  each son  receives an equal portion of wine and vessels." Let him
say, he  who can,  How must  the vessels have been divided so that all [the
sons] received an equal amount from this?

Solutio.
In primo  siquidem vasculo  fuerunt modia xl, in secundo xxx, in tertio xx,
in quarto  x.   Junge igitur  xl et  xxx et  xx et x, fiunt c.  Tunc deinde
centenarium idcirco  numerum per quartam divide partem.  Quarta namque pars
centenarii xxv  reperitur, qui  numerus bis  ductus