Almost all the mathematical cuneiform texts in the Schøyen Collection were published in Friberg's

70 Pages·2010·15.57 MB·English

SIXMOREMATHEMATICALCUNEIFORMTEXTSINTHESCHØYENCOLLECTION
JÖRANFRIBERGANDANDREWGEORGE
Almost all the mathematical cuneiform texts in the Schøyen Collection were published in
Friberg’sARemarkableCollection=MSCT1(2007).However,sixmathematicalcuneiformproblem
textsinthecollectionwereidentifiedtoolatetobeincluded.Thepresentarticlewilltakecareofthat
omission. It is a pleasure to be able place this additional material in a volume honouring Martin
Schøyen,whoseenthusiasmforseeinghisobjectspublishedinacademiceditionsismatchedonlyby
hisbenevolencetowardsthosescholarswhosharethisaimwithhim1.
Catalogue
1 MS3299 2problemsforrectangularprisms;2problemsforsquares
Tablet,portraitformat,leftcornersmissing,25+1+11lines,unruled 94x62x22mm
2 MS3976 Alinearequation(findinganoriginalamountofbarleycorn)
Tablet,landscapeformat,complete,11+1lines,unruled 62x75x25mm
3 MS3895 Amarketrateproblem(fineoil),leadingtoaquadraticequation
Tablet,landscapeformat,complete,obversepoor,13+2+11lines,unruled 63x76x28mm
4 MS3928 2quadraticequations(squares).Nosolutionprocedureoranswer
Tablet,portraitformat,complete,10+10lines,ruled 65x46x27mm
5 MS2833 3notveryclearquestionsconcerningaditch.Nosolutionprocedureoranswer
Tablet,portraitformat,damagetorightside,13lines,unruled,reverseuninscribed75x55x24mm
6 MS4905 2notveryclearproblems,oneforarectangle,theotherforarectangularprism
Solutionprocedurebutnoanswer.Tablet,landscapeformat,5+6lines,unruled 40x47x23mm
Introductoryremarks
ThescriptsofthesesixtabletsareconventionalexamplesofOldBabyloniancursive.Theortho-
graphicstyleismainlysyllabic.Thespellingconventionsshowthatfourofthetabletswerecertainly
writteninsouthernBabylonia:e-pe-ši-ka(no.1:3,12,20;no.2:4;no.3:8);as-sú-u‹(no.1:9);sà-
1 TheAssyriologicalpartofthisarticleistheworkofGeorge,themathematicalparttheworkofFriberg,buteach
alsocontributedtotheother’sshare.ThecuneiformcopiesarebyGeorge,unlessotherwisestated.–Georgerecordswith
pleasurehisdebttoChristopherWalker,MarkRonanandotherparticipantsintheLondonCuneiforum,wherethetablets
publishedherewerereadintheautumnof2008.
124 JÖRANFRIBERG–ANDREWGEORGE
ri-iq(no.2:1);ta-na-as-sà-a‹(no.2:7);sà-ni-iq(no.6:5,11).Itisprobablethattextsnos.1-5stem
fromthesamelocation.ThemajorityofOldBabylonianarchivaldocumentsintheSchøyenCollec-
tion came originally from Larsa, and Larsa is thus a plausible provenance for these mathematical
tablets.Aprovenanceelsewhereinthesouthisexpectedfortextno.6,foritenteredthecollectionwith
a sixteenth-century archive dated to the first Sealand dynasty. Strictly speaking, this makes no. 6 a
post-OldBabyloniantablet,butitstextisclearlyintheOldBabyloniantradition.
Thetexts’spelling,terminologyandprovenancearediscussedinmoredetailinthemathemati-
calcommentarybelow.Noteworthyfeaturesofspellingandlanguageare:
(a)ip-se-e=ipsê(no.1:7),theconstructstateofipsûm“squareroot”,clearlyborrowedintoAkkadianfromib.sá
(ib.si ),theSumerianwordofthesamemeaning,onwhichseenowAttinger,ZA98(2008).Becauseitwasonlydis-
8
coveredrecently,AkkadianipsûmdoesnotyetappearinanydictionaryofAkkadian.
(b)GAM.DA(no.1:4,18)fortheverticaldimensionofarectangularpit,presumably=šuplum‘depth’.
(c)ì.GAM(no.5:2,12)foršapālum‘togrowdeep’,probablyalsoforšuplum2.
(d)nam=ana(no.2:6,9;no.5:12),asoccasionallyinOldBabylonianlegaldocuments.
¢
(e)½(a)du‹=mišil(a)patārum(no.3:12)‘tocalculate(lit.solve)halfofx’.
¢ ¢ ¢
(f)Thequestionkīmasi‘howmuch?’isabbreviatedinno.1:11tokīmas(writtenma-as).
(g)Theunambiguousspelling[tu-u]š-ta-ak-kal-ma(no.1:23)withplene/kk/resolvesthedisagreementover
thederivationofthiswordinfavourofšutākulum‘toconsumeoneself/eachother’<akālum,against*šutakūlum‘to
makeholdoneself/eachother’<kullum,mostrecentlyadvocatedbyHøyrupinLWS(2002),23,andChV(2001),159,
fn.8.Unnoticedbefore,thesameword,withthesamespelling,appearsalsointheLarsatextYBC4675=MCTB,
obv.12,rev.14.–Theconfirmationofthereadingtuštakkalhasconsequencesalsofortheproperunderstandingof
thedebatedtermtakīltum<kullum,whichoccursinseveralOldBabyloniansolutionproceduresforquadraticequa-
tions.Itcannolongerbeexplainedasderivedfromapreceding*tuštakāl<*šutakūlum,asinHøyrup,op.cit.,199-
200.The(presumably) correctmeaningofthetermisexplainedinFriberg,MSCT1(2007), 337, inspiredbythe
discussioninMuroi,HSc12(2003),as‘thatwhichholds(yourmind)’,inotherwords,somethingthatyouhavemade
amentalnoteof.Indeed,normallyinOldBabylonianmetricalgebratexts,thetermstandsforthecoefficientinfront
ofthelinearterminaquadraticequation,multipliedby½.Inthesolutionofaquadraticequationby(geometric)
“completionofthesquare”,thishalvedcoefficientisfirstsquaredandmadeamentalnoteof,tobeusedagainina
laterstageofthesolutionprocedure.Inoneuniquecase,thetermoccursinanothercontext,butitisthenexplicitly
statedtorefertoaprecedingre-eš-kali-ki-il‘mayitholdyourhead’,inthefollowingphrase:a-na640ta-ki-il-tiša
re-eš-kaú-ki-lui-ší‘raise(multiply)by640,thetakīltumthatheldyourhead’.
1.MS3299.Twoproblemsforrectangularprisms;twoproblemsforsquares
1.1.Presentationofthetext
MS3299isanOldBabyloniantabletinscribedinasinglecolumnwith37linesofcuneiformtext.
Thetabletlacksthetwoleft-handcorners,whichwerepatchedupwithclayandsmallfragmentsof
oldtabletsinmoderntimes,presumablytoimprovethevalueofthetabletinthemarketplace.These
pieces,20innumber,cameadriftwhenthetabletwasfiredwhileinconservationattheSchøyenCol-
lection.Uponcloseinspection,atleast7ofthesesmallfragmentsturnedouttoholdtextfromtheEpic
ofGilgamesh.TheyweregiventhenumberMS 3263andarepublishedas textno.6inBabylonian
LiteraryTextsintheSchøyenCollection(George,MSCT4(2009)).
2NoteGAMforšupluminOldBabylonianmathematicaltextsnowintheLouvre,refs.CADŠ/3:325.Ontheuses
ofthesignGAMinOBmathematicaltextsseeRobsonMMTC(1999),132,whoproposesthatinthemeaning‘depth’itcan
bereadBÙR.BÙR,however,isavalueofthesignU,notofGAM;nevertheless,theSumerian‘deep’is/burud/,written
bùr(u)-(d),sothatGAM.DAshouldprobablybereadburudx.da.
SIXMOREMATHEMATICALCUNEIFORMTEXTSINTHESCHØYENCOLLECTION 125
Fig.1.1.MS3299.Conformtransliterationofthecuneiformtext.Scale1:1.
126 JÖRANFRIBERG–ANDREWGEORGE
Thephotographof MS 3299, publishedbelow as pl.1, shows thebaked tabletwith two of the
smallpiecesreattachedtothereverse.Thesmallerisupsidedownandclearlydoesnotbelongwhere
placed;norcoulditbeattachedelsewhere.Itisnotreproducedinthehandcopyofthereverse.After
MS3299hadbeenbakedandthemodernpatchinghadbeendiscovered,itwasgiventhenumberMS
3263/15. The larger piece, a flake containing parts of six lines of text, was given the number MS
3263/9.Itsitsquitewellasgluedbutdoesnotmakeatightjoinwiththebrokenclaybehindit.Be-
causeitisofmathematicalcontent,itwasassumedtobelongtoMS3299,whereitcanonlyfitinthe
position it occupies on the photograph and in the hand copy. However, this assumption is not well-
grounded,forthetextofthispartofthetabletisdifficulttoreconstructiftheflakeistakenintoac-
count. In addition, the flake shows no sign of the tool mark that mars the adjacent surface of the
reverseofMS3299.Therefore,itislikelythattheflakebelongstosomeothertabletentirely.
The tablet contains four mathematical problems.They are (# 1) to find the length and breadth
ofapitexcavatedintheformofarectangularprism,wherevolumeanddepthareknownquantities
andthebreadthisone-fifthofthelength(ll.1-8);(#2)tofindthelength,breadthanddepthofapit
excavatedintheformofcube,wheresurfaceareaandvolumeareknownquantities(ll.9-18);(#3)
tofindthesidesoftwosquarefieldswhenonesideishalfthelengthoftheotherandwhenthecom-
binedareaofthefieldsisaknownquantity(ll.19-28);(#4)tofindthesideofasquarefieldwhena
squarefieldwithhalfthesidehasagivenarea(ll.29-38).Allfourproblemsareworkedout,butonly
thelastproblemisconcludedwithanexplicitlygivenanswer.
1.2a.MS3299#1.Aproblemforarectangularprism(kalakkum‘excavatedpit’)
Thetextofthisexerciseisfairlywellpreserved.Mostofthemissingtext,intheupperandlower
leftcornersoftheobverseofthetablet,canbereconstructedwithouttrouble.
MS3299#1.Transliteration3
1 [x x x x k]i?7½mušar(sar)ka-la-ak-kum‹a-°am-ša¿-atšiddim(uš)pūtum([s]ag.ki)
2 [6šuplum(GAM.DA)]šiddum([u]š)°ù¿pūtum(sag.ki)mi-[n]u
3 [at-ta]i-nae-pe-ši-kaaš-[šum]‹a-am-°ša¿-atšiddim(uš)pūtum(sag.ki)qá-bu-ú
¢¢
4 [1ù1]2ta-la-ap-pa-atigi(igi)šuplim(GAM.DA)tapattar(du‹)-maa-na730eperī(sa‹ar)°tu¿-ub-<ba>-al-ma
5 °115¿i-na-ad-d[i-i]k-kum1°ù¿[1]2tuštakkal(°gu ¿)kal12i-na-ad-di-ik-kum
¢¢ 7
6 igi(igi)1[2]tapattar([d]u‹)-maa-na115tanašši(íl)-[ma]°615¿i-na-ad-di-ik-[k]um
7 ip-se-e615°te-le¿-eq-°qé¿-ma230i-na-ad-di-ik-kum
8 230a-na1ù12tu-ub-°ba¿-al-mašiddam(°uš¿)ùpūtam(sag.ki)i-na-ad-di-ik-kum
Translation
1 [x x x x x](thevolume)ofanexcavatedpitwas7½mušar,thefrontwasafifthofthelength,
2 [thedepthwas6cubits].Whatwerethelengthandthefront?
3 When[you]workitout,since“thefrontwasafifthofthelength”wasstated,
4 you shall note down [1 and 1]2, you shall solve (calculate) the reciprocal of the depth, and you shall carry
(multiply)itby730,the(volumeof)earth:
5 itwillgiveyou115.Youshalllet1and12eateachother(multiplythem),itwillgiveyou12.
6 Youshallsolve(calculate)thereciprocalof12,andyoushallraise(multiply)itby115:itwillgiveyou615.
7 Youshalltaketheequalside(squaresideorsquareroot)of615:itwillgiveyou230.
3AccordingtotheusualAssyriologicalconvention,transliterationsofAkkadianwordsareherewritteninItalicstyle,
whiletransliterationsofSumerianwordsareinRomanstyle.
SIXMOREMATHEMATICALCUNEIFORMTEXTSINTHESCHØYENCOLLECTION 127
8 Youshallcarry(multiply)230by1and12:itwillgiveyouthelengthandthefront.
MS3299#1.Mathematicalcommentary
Thestatementoftheproblem(the“question”)inthisexerciseoriginallybeganwithaphrasethat
isnowtotallylost.Itislikelythatthephrasewaspurelydescriptive.Inanycase,itcannothavespec-
ifiedanynumericalparametersessentialforwhatfollows,sincewhatseemstobeacompleteques-
tioniscontainedintheremainderofll.1-2.Thus,withtheexceptionofthelostinitialphrase,thetext
oftheexerciseisfairlywellpreserved,andthefewmissingpartsareeasilyreconstructed.
TheobjectconsideredinMS3299#1,isarectangularprism,calledan“excavatedpit”or“un-
derground store” (kalakkum < Sum. ki.lá.(k)).The stated problem can be explained as follows (ex-
pressed for the readers’convenience in quasi-modern symbolic terms): Let V be the volume of the
prism,letgbethedepth(GAM.DA)oftheprism,andletuandsbethelength(uš)andfrontorwidth
(sag),respectively,oftherectangularcrosssectionoftheprism.Then4
u(cid:0)s·g=V=7½sar, s=u/5, g=6(cubits) ll.1-2
Itis,ofcourse,supposedtobeknownthatthenindan(=c.6meters),thecubit(=¹/12nindan=c.50
cm), and the sar or mušar (= 1 square nindan · 1 cubit) are the Old Babylonian basic units for hori-
zontallengthmeasure,verticallengthmeasure,andvolume,respectively.
Thetasksetintheexerciseistofindthevaluesofthelengthuandthefronts:
u=?, s=? l.3
Thesolutionprocedurebeginsbyrecallingtheconditionthats=u/5(l.3,right).Thenthenumbers
1and12,for1and¹/5,arenoteddown(l.4,left),probablyonapieceofclayservingasscratchpad.
Next,theareaoftherectangularcrosssectionoftheprism(callitA)iscomputedasthevolume
dividedbytheheight,thatis,asfollows:
A=1/g·V=¹/6·7;30=1;15(sq.nindan) l.4,rightandl.5,left
Notetheusehere,intheinterpretationofthetext,ofsemicolonstospecifythe“absolute”valuesof
the sexagesimal numbers, which in MS 3299 #1, as in all Old Babylonian mathematical texts, only
have“relative”or“floating”values.Thus,thenumber730inthetextisinterpretedhereas7;30=7
½,andthenumber115inthetextisinterpretedas1;15=1¼.
Thecomputedvaluefortheareaoftherectangularcrosssectionoftheprismisinthetextcom-
paredwiththeareaofasimilar,butnormalized,“referencerectangle”.(Thetermisborrowedfrom
Proust,TMN(2007),230).Inquasi-modernnotations,ifthesidesofthereferencerectanglearecalled
u'ands',thensayingthatthereferencerectangleisnormalizedmeansthatu'=1(nindan),andsay-
ingthatthereferencerectangleissimilartotherectangularcrosssectionoftheprismmeansthats'
=u'/5=¹/5=;12(nindan).Consequently,theareaofthenormalizedreferencerectanglecanbecom-
putedas
4AsfirstobservedbyHøyrup(seeLWS(2002),23),thegeometricmultiplicationoftwosidesofarectangle(conve-
(cid:0)
nientlyinsymbolicnotationreplacedby )isinOldBabylonianmetricalgebratextsoftendenotedbytheverbšutākulum,
theŠtstemofakālum(=gu7),whilearithmeticmultiplication,forinstanceofageometricentitybyareciprocalorbyasca-
lingfactor(convenientlyinsymbolicnotationreplacedby·)isdenotedbyuseoftheverbsnašûmandwabālum.Alsomul-
tiplicationofarectangularbasebyaheight(ordepth)istreatedasanarithmeticmultiplication!
128 JÖRANFRIBERG–ANDREWGEORGE
A'=u'·s'=1·;12=;12(sq.nindan) l.5,right
Theratiooftheareaoftherectangularcrosssectiontotheareaofthereferencerectangleiscomputed
as
1/A'·A=1/;12·1;15=5·1;15=6;15 l.6
Sincethetworectanglesaresimilar(thatis,ofthesameform),thisarearatioisthesquareofthelin-
ear“scalingfactor”f(aconvenientnamefortheratioofthelengthsofthesidesofthefirstrectangle
tothelengthsofthecorrespondingsidesofthereferencerectangle).Thevalueofthescalingfactor
itself is computed as the ‘equalside’(meaning the square root or, more correctly, in the geometric
sense,thesquareside)ofthearearatio,thatis,as
f=sqs.6;15=2;30 l.7
Thesidesoftherectangularcrosssectionoftheprismcannow,finally,becomputedas
u=f·u'=2;30·1(nindan) and s=f·s'=2;30·;12(nindan) l.8
Theactualresultsoftheseeasycomputationsarenotgiven,butitisclearthat
u=2;30(nindan)=2½nindan and s=;30(nindan)=½nindan.
Itiseasytocheckthecorrectnessofthesecomputedvaluesforthelengthuandthefronts,although
thereisnosuchcheckinthetextoftheexercise.Indeed,withthecomputedvalues,
u(cid:0)s·g=2;30nindan·;30nindan·6cubits=7;30sq.nindan·1cubit=7½volume-sar,
and
u/5=2;30nindan/5=;30nindan=s.
1.2b.MS3299#2.Aproblemforacubicalprism(anotherkalakkum‘excavatedpit’)
Thetextofexercise#2ismoredamagedthanthetextofexercise#1.Luckily,atleastthestate-
mentoftheproblem(thequestion)islargelyintact,andthereforethetextofthewholeexercisecould
bereconstructedwithoutanygreatereffort.
MS3299#2.Transliteration
9 i-na15šiqil(gín)qaqqarim(ki)°1¿½mušar(sar)eperī(sa‹ar)°as¿-sú-u‹
10 [k]a-°la¿-ak-kumma-l[i]im-°ta‹¿-ru°iš¿-pil
¢
11 [ki-i]aim-°ta¿-[‹ar]ù°ki¿ma-asiš-pil
12 [at-ta]i-nae-pe-ši-°ka¿aš-šum!ma-liim-ta‹-ru!(IM)iš-pi-lu
¢¢
13 [1ù1]2ta-la-ap-pa-atigi(igi)15šiqil(gín)qaqqarim(ki)tapattar(du‹)-ma
14 [a-na130eperī(sa‹a]r)tu-ub-ba-°al-ma¿6i-na-ad-di-ik-kum
15 [1a-na12ta-n]a-aš-ši-ma°12ta-am¿-ma-ar
¢¢
16 [igi(igi)12tapattar(du‹)-maa-n]a6tu-ub-ba-al-ma30i-na-ad-di-ik-kum
17 [30a-na1ù]°12¿ta-na-aš-ši-°ma¿mi-it-°‹ar¿-ta-ka
18 (vacat) ùšuplam(GAM.DA)i-na-ad-di-ik-kum
Translation
9 From15shekels(=¼mušar)ofground(area)Iremoved1½mušar(volumeof)earth.
10 Theexcavatedpitwasasdeepasitwasequalsided(square).
11 Howmuchwasitequalsided(square)(eachway),andhowdeepwasit?
12 When[you]workitout,sinceitwasasdeepasitwasequalsided(square),
SIXMOREMATHEMATICALCUNEIFORMTEXTSINTHESCHØYENCOLLECTION 129
13 youshallnotedown[1and12].Youshallsolve(calculate)thereciprocalof15shekelsofground(area),and
14 youshallcarry(multiply)it[by130,theearth(volume)]:itwillgiveyou6.
15 Youshallraise(multiply)[1by12];youwillsee12.
16 [Youshallsolve(calculate)thereciprocalof12,and]youshallcarry(multiply)itby6:itwillgiveyou30.
17 Youshallraise(multiply)[30by1and]12:yourequalside(squareside)
18 (emptyspace) anddepthitwillgiveyou.
MS3299#2.Mathematicalcommentary
The object considered in the exercise MS 3299 # 2 is an excavated rectangular prism with its
depthequaltothesideofitssquarecrosssection.Inmodern(ormorespecifically,Greek)termsthat
isacube.However,justliketheobjectinexercise#1,theexcavatedcubeinexercise#2iscalleda
kalakkum.Thetaskoftheexerciseistocomputethesquaresidem(calledmit‹artu,meaningsome-
thinglike‘equalside’),andthedepthg(called,unspecifically,‘asmuchasthepitisdeep’).
Withthesameuseofquasi-modernsymbolicnotationsasinthecommentaryabovetoexercise
#1,thequestioninexercise#2canbereformulatedasfollows:
A=15area-shekels=;15area-sar, V=1½volume-sar, l.9
m=g l.10
m=?, g=? l.11
Here,asinallsimilarOldBabylonianmathematicaltexts,
1area-sar=1squarenindan, and 1volume-sar=1squarenindan·1cubit.
Thesolutionprocedurestartsbyrecalling(l.12,right)thatthedepthoftheexcavatedpitissup-
posedtobeequaltothesideofthesquarecrosssection,andbynotingdown(l.13,left)thenumbers
1and12,thusstatingthedifferentfactorsforcalculatingthesides(innindan)anddepth(incubits),
factorswhicharethekeynumbersinl.17.(Atleast,thisiswhatissuggestedintheproposedrecon-
structionofthecorrespondingmissingpartofthetext).
Thenthedepth(g)iscomputedasthevolumedividedbytheareaofthecrosssection,asfollows:
g=1/A·V=1/;15·1;30=(4·1;30=)6(cubits) l.13,right-l.14
Next,thevolumeofanormalized“referencecube”iscomputedas
1·12=12(sq.nindan·1cubit) l.15
Theexplanationforthiscomputationisthatifacube-shapedexcavatedpitisnormalizedinthesense
thatitssquarecrosssectionhasthesidem'=1(nindan),thenitsdepthisg'=12cubits.(Recallthat
1nindan=12cubits).Theareaofthesquarecrosssectionofthecubeisthen1(sq.nindan),andthe
volumeofthecubeis1(sq.nindan)·12(cubits)=12(sq.nindan·1cubit).
Theratioofthedepthofthegivencube-shapedexcavatedpittothedepthofthenormalizedref-
erencecubeis,clearly,
1/g'·g=¹/12·6=;30 l.16
Thismeansthatthe“scalingfactor”inthissituation(theratiobetweenthelengthsofthesidesofthe
twocubes)isf=;30.Therefore,thesidesoftheoriginalcube-shapedexcavedpitcanbecomputed
asfollows:
m=f·m'=;30·1(nindan) and g=f·g'=;30·12(cubits) ll.17-18
Consequently,althoughthisisnotstatedexplicitlyinthetext,m=g=½nindan=6cubits.Nevermind
thatitshouldhavebeenobvious,withoutanyfurthercomputations,thatifthedepthofacubeis6cu-
bits, as shown in l. 14, then all sides of the cube must have the common length 6 cubits.And never
mindthatifthesquarecrosssectionofacubeisknowntohavethearea;15sar=¼squarenindan,
130 JÖRANFRIBERG–ANDREWGEORGE
asstatedinthefirstlineoftheexercise,thenthesideofthecubemusthavethelengthsqs.(¼square
nindan)=½nindan=6cubits.
Apparently, the author of this exercise was so intent on giving another example of the method
of using normalized reference objects and scaling factors that he did not bother to find the simplest
solution to the stated problem! Or, maybe, the correct explanation is that this is an example of how
instructioninBabylonianmathematicswasbasedonlearningprocedure.Shortcutswouldrisktheun-
dermining of the rigid procedure, and the student who wrote the tablet was obliged to show that he
knewtheprocedure.
1.2c.MS3299#3.Aproblemfortwosquares(‘equalsides’)
The text of exercise # 3 is extensively damaged. More than half the text is missing or unread-
able.Inaddition,whatremainsofthequestionisawkwardlyformulatedandthereforequiteobscure.
Fortunately, however, only the last few lines of the exercise are completely broken away, and
theircontent,withthelaststepsofthesolutionalgorithm,iseasilyreconstructed.Oncethesolution
procedurehadbeenreconstructed,inthewayshownbelow,itwasnottoodifficulttofindalsoarea-
sonablereconstructionandinterpretationofthequestion.
MS3299#3.Transliteration
¢
19 <mi-it-‹ar-ti>[uš-ta-ki-il]a-na2e‹!(›AR)-peuš-ta-ki-il°a¿-na°libbi(šà)¿eqlim(a.šà)usib(da‹)-ma°2¿05
20 [mi-it-‹ar-tik]i-ia-aim-ta-‹arat-tai-nae-pe-ši-ka
21 [aš-šum2qá]-bu-ú1ù30lu-pu-ut
22 [1a-na2te-‹e-e]p-[p]e-ma30i-na-ad-di-ik-kum
23 [30ù30tu-u]š-ta-ak-kal-ma15i-na-ad-di-ik-kum
¢ ¢
24 [15a-na1tu-u]s-sa-am-ma°115¿ta-am-ma-ar
¢¢
25 [igi(igi)115tapattar(du‹)-ma]°48¿ta-°am¿-ma-ar
26 [a-na205ta-n]a-aš-ši-ma1°40i-na-ad¿-di-i[k-kum]
27 [ip-se-e140te-le-eq-qé-ma10ta-am-ma-ar]
28 [a-na1ta-na-aš-ši-mami-it-‹ar-ta-kata]-°am-ma¿-[ar]
Translation
19 [I made <my equalside> eat itself (I squared it)], I broke it in 2 (and) I made it eat itself (I squared it).
Iaddedit(theareaofthesecondsquare)ontothefield(area)(oftheoriginalsquare):205.
20 Howmuchwas[myequalside]equalsided(whatwasthesquareside)?Whenyouworkitout,
21 [since“2”was]stated,notedown1and30.
22 [Youshallbreak1in2]:itwillgiveyou30.
23 [You]shallmake[30and30]eateachother(youshallmultiplythem):itwillgiveyou15.
24 [You]shalladd[15to1]:youwillsee115.
25 [Youshallsolve(calculate)thereciprocalof115]:youwillsee48.
26 [Youshall]raise(multiply)it[by205]:itwillgive[you]140.
27 [Youshalltaketheequalside(squareroot)of140:itwillgiveyou10.]
28 [Youshallraise(multiply)itby1:you]willsee[yourequalside(squareside)].
MS3299#3.Mathematicalcommentary
Thestatementoftheproblem(thequestion)inthisexerciseisquiteobscureinanumberofways
(inadditiontothefactthattheveryfirstwordinthetextofthequestionismissing).Inparticular,it
is(probably)notsaidexplicitlywhattheobjectoriginallyconsideredshouldbe.Nevertheless,itseems
tobeclearfrombywhatremainsofthequestionthattheobjectoriginallyconsideredinthisexercise
isamit‹artum‘equalside(squareside)’,whichisfirstsquared,thencutinhalf,andthenagainsquared.
SIXMOREMATHEMATICALCUNEIFORMTEXTSINTHESCHØYENCOLLECTION 131
Afterthat,thesumoftheareasofthetwosquaresisgiven.Inquasi-modernterms,ifmistheorigi-
nalsquareside,thestatedproblemcanbeinterpretedasthefollowingequation:
sq.m+sq.(m/2)=205, m=? ll.19-20
The somewhat curiously stated question in l. 20, “[My equalside, how] much is it equalsided?” ob-
viouslyasksforthelengthofthesquaresidem.
Thesolutionprocedurestartsbyrecalling(inl.21)thehalvingoftheoriginalsquareside,and
bynotingdown‘1’(forthewholesquareside)and‘30’(forthehalvedsquareside).
Thenthefirststepoftheactualcomputationistocompute
1·½=;30 and sq.;30=;15 ll.22-23
Inthenextstepofthesolutionprocedureiscomputed
(sq.1+sq.;30=)1+;15=1;15 l.24
Whatisgoingonhereis,ofcourse,thecomputationofthecombinedareaofa“normalizedreference
pairofsquares”,beginningwithasquareofside1.
Next,theratioofthecombinedareaofthefirstpairofsquarestothecombinedareaofthenor-
malizedreferencepairofsquaresiscomputedinthefollowingway:
¹/1;15=;48(4/5), ;48·205=140 ll.25-26
Sincethisratio,obviously,isthesquareofthescalingfactorf,thevalueoffiscomputedas
sqs.140=10(nindan) l.27
Consequently,thelengthsoftheoriginalsquaresidecanbecomputedasfollows:
10(nindan)·1=10nindan l.28
Itisinterestingtoobservethatthisapparentlycomplicatedsolutionprocedureintermsofaref-
erencepairofsquaresandascalingfactorismathematicallyequivalenttothemodernwayofsolv-
ingthestatedproblembyuseofsymbolicnotationandalgebraicmanipulationofequations.Indeed,
themodernsolutionprocedurewouldbeasfollows:
sq.m+sq.(m/2)=125 ↔ (1+¼)sq.m=125 ↔ sq.m=125/1.25=100 ↔ m=√100=10.
1.2d.MS3299#4.Aproblemforasinglesquare(‘equalside’)
ThetextonthereverseofMS3299,includingtheentiretextofexercise#4,ispartlylost,partly
damaged.Nevertheless,withsomeeffortithasbeenpossibletoreconstructtheentiretextofexercise
#4withrelativeconfidence.Thereconstructionisbasedonthereasonableassumptionthatthisex-
erciseiscloselyrelatedtotheprecedingexercise.
MS3299#4.Transliteration
29 [mi-it-‹ar-tia-na2e‹-peuš-ta-ki-i]l
30 [1 aša a.šà-šumi-it-‹ar-tiki-i]aim-ta-[‹]ar
iku 5
31 [at-tai-nae-pe-ši-ka1a-n]a2te-‹[e-ep-pe]-ma30ta-am-ma-ar
32 [30ù30tu-uš-ta-ak-kal]-ma15i-°na-ad-di¿-ik-kum
¢¢
33 [igi(igi)15tapattar(du‹)-ma4ta-a]m-ma-ar
34 [1 aša x-ma140t]a-am-ma-ar
iku 5
35 [4a-na140eqlim(a.šà)t]u-°ub¿-b[a-a]l-°ma!¿640ta-°am¿-ma-ar
36 [ip-se-e640te]-le-eq-q[é-ma2]0ta-am-ma-ar
37 20a-na°1¿[ta-na-aš]-ši-ma20mi-i[t-‹ar]-ta-ka
38 (vacat)ta-am-°ma¿-ar
132 JÖRANFRIBERG–ANDREWGEORGE
Translation
29 Ibroke[myequalside(squareside)in2,Imadeiteatitself(Isquaredit)].
30 [1ikuwasitsfield(area).Howmuchwasmyequalside(squareside)]equalsided(whatwasitssquareside)?
31 [Whenyouworkitout,youshallbreak[1]in2:youwillsee30.
32 [You]shallmake[30and30eat]eachother(multiplythem):itwill[give]you15.
33 [Youshallsolve(calculate)thereciprocalof15:you]willsee[4].
34 [Youshallx(convert)1iku:]youwillsee140.
35 Youshallcarry(multiply)[4by140,thefield(area)]:youwillsee640.
36 [You]shalltake[theequalside(squareroot)of640:]youwillsee[20].
37 [You]shallraise(multiply)20by1:20,yourequalside(squareside),
38 youwillsee.
MS3299#4.Mathematicalcommentary
According to our tentative reconstruction above of the question in ll. 29-30, of which hardly a
traceremains,theobjectconsideredinthisexerciseisasinglesquareside.Thesquaresideishalved,
andthenthehalvedsquaresideismultipliedbyitself.Theresultisasquarewiththegivenarea1iku.
Inquasi-modernsymbolicnotations:
sq.(m/2)=1iku=140sar(sq.nindan), m=? ll.29-30
Thisisareallysimpleequationwiththeobvioussolution
m/2=sqs.140=10(nindan), sothat m=2·10(nindan)=20nindan.
However,hereagaintheauthorofMS3299stickstohispreferredmethodofusingnormalizedref-
erenceobjectsandscalingfactors.Therefore,hestartsbycomputingthelengthofahalvednormal-
izedreferencesquareside,whichis,ofcourse,
(½·1=);30·1=;30 l.31
Theareaofthesquareonthishalvednormalizedsquaresideisthen,justasobviously,
sq.;30=;15 l.32
anditsreciprocalis
1/;15=4 l.33
Thegivenareaofthesquareonthehalvedoriginalsquareside,ontheotherhand,is
1iku=140sar(sq.nindan) l.34
Theratioofthisgivenareatotheareaofthesquareonthehalvedreferencesquaresideis
(1/;15·140=)4·140=640 l.35
Sincethisratioisthesquareofthescalingfactorf,itfollowsthatthevalueoffitselfis
sqs.640=20(nindan) l.36
Consequently,theoriginalsquaresidemustbe,asstatedinthetext,
20(nindan)·1=20nindan ll.37-38
MS3299.Conclusion
MS3299isaquitebrief“recombinationtext”withexercises##1-2borrowedfromonetheme