MYBESTMATHEMATICAL

ANDLOGICPUZZLES

MartinGardner

DOVERPUBLICATIONS,INC.NewYork

CopyrightCopyright©1994byMartinGardner.

Allrightsreserved.

BibliographicalNoteMyBestMathematicalandLogicPuzzles,firstpublishedbyDoverPublications,Inc.,in1994,isanewandoriginalcollectionofworkpreviouslypublishedinScientificAmerican,Games

magazineandinearliervolumecollections.AnewIntroductionhasbeenwrittenexpresslyforthisDoveredition.

LibraryofCongressCataloging-in-PublicationDataGardner,Martin,1914–Mybestmathematicalandlogicpuzzles/MartinGardner.

p.cm.ISBN-13:978-0-486-28152-0(pbk.)ISBN-10:0-486-28152-3(pbk.)1.Mathematicalrecreations.I.

Title.QA95.G2921994

793.7’4—dc20 94-25660

ManufacturedintheUnitedStatesbyCourierCorporation28152315

www.doverpublications.com

INTRODUCTION

DURING the twenty-five years that I had the great privilege of writing theMathematicalGamescolumninScientificAmerican,mypracticewastodevoteacolumn, about every sixmonths, towhat I called shortproblemsorpuzzles.These puzzles were, of course, mathematical rather than problems involvingwords.IdidmybesttopresentnewandunfamiliarpuzzlesthatwerenottobefoundinclassiccollectionssuchasthebooksbySamLoydandHenryDudeney.Readerswerequick to catchmistakes and to supply in somecases alternate

solutionsorinterestinggeneralizations.Thisvaluablefeedbackwasincorporatedwhenthepuzzlecolumnswerereprintedinbookcollections.Mostoftheproblemsinthisbookareselectedfromthefirstthreecollections.

The last 12 puzzles are selected from two articles I contributed to Gamesmagazine(January/FebruaryandNovember/December,1978).Somehavebeenupdated by adding references to new developments related to the puzzle.Needlesstosay,Iwelcomeanycorrectionsoradditions.Theycanbesenttomeincareofthepublisher,DoverPublications,31East2ndStreet,Mineola,N.Y.11501.

MARTINGARDNER

CONTENTS

PUZZLES1.TheReturningExplorer2.DrawPoker3.TheMutilatedChessboard4.TheForkintheRoad5.ScrambledBoxTops6.CuttingtheCube7.Bronxvs.Brooklyn8.TheEarlyCommuter9.TheCounterfeitCoins10.TheTouchingCigarettes11.TwoFerryboats12.GuesstheDiagonal13.CrosstheNetwork14.The12Matches15.HoleintheSphere16.TheAmorousBugs17.HowManyChildren?18.TheTwiddledBolts19.TheFlightaroundtheWorld20.TheRepetitiousNumber21.TheCollidingMissiles22.TheSlidingPennies23.HandshakesandNetworks24.TheTriangularDuel25.CrossingtheDesert26.LordDunsany’sChessProblem27.TheLonesome828.DividingtheCake29.TheFoldedSheet30.WaterandWine31.TheAbsent-MindedTeller32.AcuteDissection33.HowLongIsa“Lunar”?34.TheGameofGoogol

35.MarchingCadetsandaTrottingDog36.White,BlackandBrown37.ThePlaneintheWind38.WhatPricePets?39.TheGameofHip40.ASwitchingPuzzle41.BeerSignsontheHighway42.TheSlicedCubeandtheSlicedDoughnut43.BisectingYinandYang44.TheBlue-EyedSisters45.HowOldIstheRose-RedCity?46.TrickyTrack47.Termiteand27Cubes48.CollatingtheCoins49.TimetheToast50.AFixed-PointTheorem51.HowDidKantSetHisClock?52.PlayingTwentyQuestionswhenProbabilityValuesAreKnown53.Don’tMateinOne54.FindtheHexahedrons55.OutwiththeOnion56.CutDowntheCuts57.DissectionDilemma58.InterruptedBridge59.DashItAll!60.MovetheQueen61.ReadtheHieroglyphics62.CrazyCut63.FindtheOddball64.BigCross-OutSwindle65.ReversetheDog66.FunnyFold

ANSWERS

PUZZLES

1.TheReturningExplorer

ANOLDRIDDLErunsasfollows.Anexplorerwalksonemileduesouth,turnsandwalks onemile due east, turns again andwalks onemile due north.He findshimself backwhere he started.He shoots a bear.What color is the bear?Thetime-honoredansweris:“White,”becausetheexplorermusthavestartedattheNorthPole.ButnotlongagosomeonemadethediscoverythattheNorthPoleisnot theonlystartingpoint thatsatisfies thegivenconditions!Canyou thinkofanyotherspotontheglobefromwhichonecouldwalkamilesouth,amileeast,amilenorthandfindhimselfbackathisoriginallocation?

2.DrawPoker

TWOMEN PLAY a game of draw poker in the following curiousmanner. Theyspreadadeckof52cardsfaceuponthetablesothattheycanseeallthecards.Thefirstplayerdrawsahandbypickinganyfivecardshechooses.Thesecondplayerdoesthesame.Thefirstplayernowmaykeephisoriginalhandordrawuptofivecards.Hisdiscardsareputasideoutofthegame.Thesecondplayermaynowdrawlikewise.Thepersonwiththehigherhandthenwins.Suitshaveequalvalue,so that twoflushestieunlessoneismadeofhighercards.Afterawhile theplayersdiscover that the first player can alwayswin if hedrawshisfirsthandcorrectly.Whathandmustthisbe?

3.TheMutilatedChessboard

THEPROPSFORthisproblemareachessboardand32dominoes.Eachdominoisof such size that it exactly covers two adjacent squares on the board. The 32dominoesthereforecancoverall64ofthechessboardsquares.Butnowsupposewecutoff twosquaresatdiagonallyoppositecornersof theboardanddiscardoneofthedominoes.Isitpossibletoplacethe31dominoesontheboardsothatalltheremaining62squaresarecovered?Ifso,showhowitcanbedone.Ifnot,proveitimpossible.

4.TheForkintheRoad

HERE’SARECENTtwistonanoldtypeoflogicpuzzle.AlogicianvacationingintheSouthSeasfindshimselfonanislandinhabitedbythetwoproverbialtribesofliarsandtruth-tellers.Membersofonetribealwaystellthetruth,membersofthe other always lie. He comes to a fork in a road and has to ask a nativebystander which branch he should take to reach a village. He has no way oftellingwhetherthenativeisatruth-telleroraliar.Thelogicianthinksamoment,thenasksonequestiononly.Fromthereplyheknowswhichroadtotake.Whatquestiondoesheask?

5.ScrambledBoxTops

IMAGINE THAT YOU have three boxes, one containing two black marbles, onecontaining twowhitemarbles, and the third, one blackmarble and onewhitemarble. The boxes were labeled for their contents—BB, WW and BW—butsomeonehas switched the labels so that every box is now incorrectly labeled.Youareallowed to takeonemarbleata timeoutofanybox,without lookinginside,andbythisprocessofsamplingyouaretodeterminethecontentsofallthreeboxes.Whatisthesmallestnumberofdrawingsneededtodothis?

6.CuttingtheCube

A CARPENTER, working with a buzz saw, wishes to cut a wooden cube, threeincheson a side, into27one-inch cubes.He cando this easilybymaking sixcuts through the cube, keeping the pieces together in the cube shape. Can he

reducethenumberofnecessarycutsbyrearrangingthepiecesaftereachcut?

7.Bronxvs.Brooklyn

AYOUNGMANlivesinManhattannearasubwayexpressstation.Hehastwogirlfriends, one in Brooklyn, one in The Bronx. To visit the girl in Brooklyn hetakesatrainonthedowntownsideoftheplatform;tovisitthegirlinTheBronxhe takes a train on the uptown side of the same platform. Since he likes bothgirlsequallywell,hesimplytakesthefirsttrainthatcomesalong.InthiswayheletschancedeterminewhetherheridestoTheBronxortoBrooklyn.TheyoungmanreachesthesubwayplatformatarandommomenteachSaturdayafternoon.BrooklynandBronxtrainsarriveatthestationequallyoften—every10minutes.YetforsomeobscurereasonhefindshimselfspendingmostofhistimewiththegirlinBrooklyn:infactontheaveragehegoesthereninetimesoutoften.CanyouthinkofagoodreasonwhytheoddssoheavilyfavorBrooklyn?

8.TheEarlyCommuter

A COMMUTER IS in the habit of arriving at his suburban station each eveningexactly at fiveo’clock.Hiswife alwaysmeets the train anddriveshimhome.Onedayhetakesanearliertrain,arrivingatthestationatfour.Theweatherispleasant, so instead of telephoning home he starts walking along the routealwaystakenbyhiswife.Theymeetsomewhereontheway.Hegetsintothecarand they drive home, arriving at their house ten minutes earlier than usual.Assuming that the wife always drives at a constant speed, and that on thisoccasionshe left just in time tomeet thefiveo’clock train,canyoudeterminehowlongthehusbandwalkedbeforehewaspickedup?

9.TheCounterfeitCoins

INRECENTYEARS a numberof clever coin-weighingor ball-weighingproblemshave aroused widespread interest. Here is a new and charmingly simplevariation.Youhave10stacksofcoins,eachconsistingof10half-dollars.Oneentire stack is counterfeit, but you do not knowwhich one.You do know theweightofagenuinehalf-dollarandyouarealsotoldthateachcounterfeitcoinweighs one grammore than it should.Youmayweigh the coins on a pointerscale.What is thesmallestnumberofweighingsnecessary todeterminewhichstackiscounterfeit?

10.TheTouchingCigarettes

FOURGOLFBALLS canbeplaced so that eachball touches theother three.Fivehalf-dollarscanbearrangedsothateachcointouchestheotherfour.Is itpossible toplacesixcigarettesso thateach touches theother five?The

cigarettesmustnotbebentorbroken.

11.TwoFerryboats

TWO FERRYBOATS start at the same instant from opposite sides of a river,travelingacrossthewateronroutesatrightanglestotheshores.Eachtravelsataconstantspeed,butoneisfaster thantheother.Theypassatapoint720yardsfrom the nearest shore.Both boats remain in their slips for 10minutes beforestartingback.Onthereturntripstheymeet400yardsfromtheothershore.Howwideistheriver?

12.GuesstheDiagonal

ARECTANGLE is inscribed in thequadrantof acircleas shown.Given theunitdistancesindicated,canyouaccuratelydeterminethelengthofthediagonalAC?Timelimit:oneminute!

13.CrosstheNetwork

ONEOFTHEoldestoftopologicalpuzzles,familiartomanyaschoolboy,consistsofdrawinga continuous line across the closednetwork shown so that the linecrosses each of the 16 segments of the network only once. The curved lineshownheredoesnotsolvethepuzzlebecauseitleavesonesegmentuncrossed.No “trick” solutions are allowed, such as passing the line through a vertex oralongoneofthesegments,foldingthepaperandsoon.Itisnotdifficulttoprovethatthepuzzlecannotbesolvedonaplanesurface.

Twoquestions:Canitbesolvedonthesurfaceofasphere?Onthesurfaceofatorus(doughnut)?

14.The12Matches

ASSUMINGthatamatchisaunitoflength,itispossibletoplace12matchesonaplane in various ways to form polygons with integral areas. The illustrationshowstwosuchpolygons:asquarewithanareaofninesquareunits,andacrosswithanareaoffive.Theproblemisthis:Useall12matches(theentirelengthofeachmatchmust

beused) to form insimilar fashion theperimeterofapolygonwithanareaofexactlyfoursquareunits.

15.HoleintheSphere

THISISanincredibleproblem—incrediblebecauseitseemstolacksufficientdataforasolution.Acylindricalholesixincheslonghasbeendrilledstraightthroughthecenterofasolidsphere.Whatisthevolumeremaininginthesphere?

16.TheAmorousBugs

FOURBUGS—A,B,CandD—occupythecornersofasquare10inchesonaside.AandCaremale,BandDarefemale.SimultaneouslyAcrawlsdirectlytowardB,BtowardC,CtowardDandDtowardA.Ifallfourbugscrawlatthesameconstantrate,theywilldescribefourcongruentlogarithmicspiralswhichmeetatthecenterofthesquare.Howfardoeseachbug travelbefore theymeet?Theproblemcanbesolved

withoutcalculus.

17.HowManyChildren?

“IHEARsomeyoungstersplayinginthebackyard,”saidJones,agraduatestudentinmathematics.“Aretheyallyours?”“Heavens,no,”exclaimedProfessorSmith,theeminentnumbertheorist.“My

childrenareplayingwithfriendsfromthreeotherfamiliesintheneighborhood,althoughourfamilyhappenstobelargest.TheBrownshaveasmallernumberofchildren,theGreenshaveastillsmallernumber,andtheBlacksthesmallestofall.”“Howmanychildrenaretherealtogether?”askedJones.“Letmeputit thisway,”saidSmith.“Therearefewerthan18children,and

theproductofthenumbersinthefourfamilieshappenstobemyhousenumber

whichyousawwhenyouarrived.”Jones took a notebook and pencil fromhis pocket and started scribbling.A

moment later he looked up and said, “I needmore information. Is theremorethanonechildintheBlackfamily?”As soon as Smith replied, Jones smiled and correctly stated the number of

childrenineachfamily.KnowingthehousenumberandwhethertheBlackshadmorethanonechild,

Jonesfoundtheproblemtrivial.Itisaremarkablefact,however,thatthenumberof children in each family can be determined solely on the basis of theinformationgivenabove!

18.TheTwiddledBolts

TWOIDENTICALBOLTSareplacedtogethersothattheirhelicalgroovesintermesh.If youmove the bolts around each other as you would twiddle your thumbs,holdingeachboltfirmlybytheheadsothatitdoesnotrotateandtwiddlingtheminthedirectionshown,willtheheads(a)moveinward,(b)moveoutward,or(c)remain the same distance from each other? The problem should be solvedwithoutresortingtoactualtest.

19.TheFlightaroundtheWorld

AGROUPofairplanes isbasedonasmall island.The tankofeachplaneholdsjustenoughfueltotakeithalfwayaroundtheworld.Anydesiredamountoffuelcanbe transferred from the tankofoneplane to the tankof anotherwhile theplanesareinflight.Theonlysourceoffuelisontheisland,andforthepurposesoftheproblemitisassumedthatthereisnotimelostinrefuelingeitherintheairorontheground.

Whatisthesmallestnumberofplanesthatwillensuretheflightofoneplanearound the world on a great circle, assuming that the planes have the sameconstant ground speed and rate of fuel consumption and that all planes returnsafelytotheirislandbase?

20.TheRepetitiousNumber

ANUNUSUALparlor trick isperformedas follows.AskspectatorA to jotdownanythree-digitnumber,andthentorepeatthedigitsinthesameordertomakeasix-digitnumber(e.g.,394,394).Withyourback turnedso thatyoucannotseethenumber,askAtopassthesheetofpapertospectatorB,whoisrequestedtodividethenumberby7.“Don’t worry about the remainder,” you tell him, “because there won’t be

any.”Bissurprisedtodiscoverthatyouareright(e.g.,394,394dividedby7is56,342).Withouttellingyoutheresult,hepassesitontospectatorC,whoistoldtodivideitby11.Onceagainyoustatethattherewillbenoremainder,andthisalsoprovescorrect(56,342dividedby11is5,122).With your back still turned, and no knowledge whatever of the figures

obtainedby thesecomputations,youdirect a fourth spectator,D, todivide thelast result by 13. Again the division comes out even (5,122 divided by 13 is394).Thisfinalresultiswrittenonaslipofpaperwhichisfoldedandhandedtoyou.WithoutopeningityoupassitontospectatorA.“Open this,” you tell him, “and you will find your original three-digit

number.”Provethatthetrickcannotfailtoworkregardlessofthedigitschosenbythe

firstspectator.

21.TheCollidingMissiles

TWOMISSILESspeeddirectlytowardeachother,oneat9,000milesperhourandtheotherat21,000milesperhour.Theystart1,317milesapart.Withoutusingpencilandpaper,calculatehowfaraparttheyareoneminutebeforetheycollide.

22.TheSlidingPennies

SIX PENNIES are arranged on a flat surface as shown in the top picture. Theproblemis tomove theminto the formationdepictedatbottomin thesmallestnumberofmoves.Eachmoveconsistsinslidingapenny,withoutdisturbinganyoftheotherpennies,toanewpositioninwhichittouchestwoothers.Thecoins

mustremainflatonthesurfaceatalltimes.

23.HandshakesandNetworks

PROVETHAT at a recentconventionofbiophysicists thenumberof scientists inattendancewhoshookhandsanoddnumberoftimesiseven.Thesameproblemcanbeexpressedgraphicallyasfollows.Putasmanydots(biophysicists)asyouwishonasheetofpaper.Drawasmanylines(handshakes)asyouwishfromanydottoanyotherdot.Adotcan“shakehands”asoftenasyouplease,ornotatall.Provethat thenumberofdotswithanoddnumberof lines joiningthemiseven.

24.TheTriangularDuel

SMITH,BROWNandJonesagreetofightapistolduelunderthefollowingunusualconditions.Afterdrawinglotstodeterminewhofiresfirst,secondandthird,theytake theirplaces at the cornersof an equilateral triangle. It is agreed that theywillfiresingleshots in turnandcontinuein thesamecyclicorderuntil twoof

themaredead.Ateachturnthemanwhoisfiringmayaimwhereverhepleases.All threeduelistsknowthatSmithalwayshitshis target,Brownis80percentaccurateandJonesis50percentaccurate.Assumingthatallthreeadoptthebeststrategy,andthatnooneiskilledbya

wild shot not intended for him, who has the best chance to survive? Amoredifficultquestion:Whataretheexactsurvivalprobabilitiesofthethreemen?

25.CrossingtheDesert

ANUNLIMITEDsupplyofgasolineisavailableatoneedgeofadesert800mileswide, but there is no source on the desert itself. A truck can carry enoughgasolinetogo500miles(thiswillbecalledone“load”),anditcanbuildupitsownrefuelingstationsatanyspotalongtheway.Thesecachesmaybeanysize,anditisassumedthatthereisnoevaporationloss.Whatis theminimumamount(inloads)ofgasolinethetruckwillrequirein

order tocross thedesert?Is therea limit to thewidthofadesert the truckcancross?

26.LordDunsany’sChessProblem

ADMIRERSOF theIrishwriterLordDunsanydonotneedtobe told thathewasfond of chess. (Surely his story “The Three Sailors’ Gambit” is the funniestchess fantasy ever written.) Not generally known is the fact that he liked toinvent bizarre chess problems which, like his fiction, combine humor andfantasy.The problem depicted here was contributed by Dunsany to The Week-End

ProblemsBook,compiledbyHubertPhillips.Itssolutioncallsmoreforlogicalthought than skill at chess, although one does have to know the rules of thegame.White is toplayandmate infourmoves.Theposition isone thatcouldoccurinactualplay.

27.TheLonesome8

THE MOST POPULAR problem ever published in The American MathematicalMonthly,itseditorsrecentlydisclosed,isthefollowing.ItwascontributedbyP.L.ChessinoftheWesting-houseElectricCorporationtotheApril1954issue.“Our good friend and eminent numerologist, Professor Euclide Paracelso

BombastoUmbugio,hasbeenbusily engaged in testingonhisdeskcalculatorthe81X109possible solutions to theproblemof reconstructing the followingexactlongdivisioninwhichthedigitswereindiscriminatelyreplacedbyxsaveinthequotientwheretheywerealmostentirelyomitted:

“DeflatetheProfessor!Thatis,reducethepossibilitiesto(81x109)0.”Becauseanynumberraisedtothepowerofzeroisone,thereader’staskisto

discover theunique reconstructionof theproblem.The8 is incorrectpositionabove the line,making it the thirddigitofa fivedigit-answer.Theproblem is

easierthanitlooks,yieldingreadilytoafewelementaryinsights.

28.DividingtheCake

THEREISasimpleprocedurebywhichtwopeoplecandivideacakesothateachissatisfiedhehasatleasthalf:Onecutsandtheotherchooses.Deviseageneralprocedure so thatn persons can cut a cake inton portions in such away thateveryoneissatisfiedhehasatleast1/nofthecake.

29.TheFoldedSheet

MATHEMATICIANShavenotyetsucceededinfindingaformulaforthenumberofdifferentways a roadmap can be folded, givenn creases in the paper. Somenotion of the complexity of this question can be gained from the followingpuzzleinventedbytheBritishpuzzleexpertHenryErnestDudeney.Divide a rectangular sheet of paper into eight squares and number them on

onesideonly,asshowninthetopdrawing.Thereare40differentwaysthatthis“map”canbefoldedalongtheruledlinestoformasquarepacketwhichhasthe“1”squareface-upontopandallothersquaresbeneath.Theproblemistofoldthissheetsothatthesquaresareinserialorderfrom1to8,withthe1face-upontop.

If you succeed in doing this, try themuchmore difficult task of doing thesamethingwiththesheetnumberedinthemannerpicturedatthebottomoftheillustration.

30.WaterandWine

AFAMILIARchestnutconcernstwobeakers,onecontainingwater,theotherwine.

Acertainamountofwateristransferredtothewine,thenthesameamountofthemixture is transferred back to thewater. Is there nowmorewater in thewinethan there iswine in thewater? The answer is that the two quantities are thesame.Raymond Smullyanwrites to raise the further question:Assume that at the

outset one beaker holds 10 ounces ofwater and the other holds 10 ounces ofwine.Bytransferringthreeouncesbackandforthanynumberoftimes,stirringaftereachtransfer,isitpossibletoreachapointatwhichthepercentageofwineineachmixtureisthesame?

31.TheAbsent-MindedTeller

ANABSENT-MINDEDbanktellerswitchedthedollarsandcentswhenhecashedacheckforMr.Brown,givinghimdollars insteadofcents,andcents insteadofdollars.Afterbuyinga five-centnewspaper,Browndiscovered thathehad leftexactlytwiceasmuchashisoriginalcheck.Whatwastheamountofthecheck?

32.AcuteDissection

GIVENA TRIANGLEwith oneobtuse angle, is it possible to cut the triangle intosmaller triangles, all of themacute? (Anacute triangle is a trianglewith threeacuteangles.Arightangleisofcourseneitheracutenorobtuse.)Ifthiscannotbe done, give a proof of impossibility. If it can be done,what is the smallestnumberofacutetrianglesintowhichanyobtusetrianglecanbedissected?Theillustrationshowsa typicalattempt that leadsnowhere.Thetrianglehas

beendivided into threeacute triangles,but thefourth isobtuse,sonothinghasbeengainedbytheprecedingcuts.

The problem (which came to me by way of Mel Stover of Winnipeg) isamusingbecauseeventhebestmathematicianislikelytobeledastraybyitandcometoawrongconclusion.Mypleasureinworkingonitledmetoaskmyselfarelatedquestion:“Whatisthesmallestnumberofacutetrianglesintowhichasquarecanbedissected?”Fordays Iwasconvinced thatninewas theanswer;thensuddenlyIsawhowtoreduceit toeight.Iwonderhowmanyreaderscandiscoveraneight-trianglesolution,orperhapsanevenbetterone.Iamunabletoprovethateightistheminimum,thoughIstronglysuspectthatitis.

33.HowLongIsa“Lunar”?

INH.G.WELLS’SnovelTheFirstMenintheMoonournaturalsatelliteisfoundto be inhabited by intelligent insect creatures who live in caverns below thesurface.Thesecreatures,letusassume,haveaunitofdistancethatweshallcalla“lunar.”Itwasadoptedbecausethemoon’ssurfacearea,ifexpressedinsquarelunars,exactlyequalsthemoon’svolumeincubiclunars.Themoon’sdiameteris2,160miles.Howmanymileslongisalunar?

34.TheGameofGoogol

IN 1958 JohnH. Fox, Jr., of theMinneapolis-HoneywellRegulatorCompany,andL.GeraldMarnie,oftheMassachusettsInstituteofTechnology,devisedanunusual betting game which they call Googol. It is played as follows: Asksomeonetotakeasmanyslipsofpaperashepleases,andoneachslipwriteadifferentpositivenumber.Thenumbersmayrangefromsmallfractionsof1toanumber the size of a “googol” (1 followed by a hundred 0’s) or even larger.Theseslipsareturnedfacedownandshuffledoverthetopofatable.Oneatatimeyouturntheslipsfaceup.Theaimistostopturningwhenyoucometothenumber thatyouguess tobe the largestof theseries.Youcannotgobackandpickapreviously turnedslip. Ifyou turnoverall theslips, thenofcourseyoumustpickthelastoneturned.Mostpeoplewillsupposetheoddsagainstyourfindingthehighestnumberto

beatleastfivetoone.Actuallyifyouadoptthebeststrategy,yourchancesarealittle better than one in three. Two questions arise. First, what is the beststrategy? (Note that this is not the same as asking for a strategy that willmaximizethevalueoftheselectednumber.)Second,ifyoufollowthisstrategy,howcanyoucalculateyourchancesofwinning?When there are only two slips, your chance of winning is obviously 1/2,

regardless of which slip you pick. As the slips increase in number, the

probabilityofwinning(assuming thatyouuse thebest strategy)decreases,butthecurveflattensquickly,and there isvery littlechangebeyond tenslips.Theprobability never drops below 1/3. Many players will suppose that they canmake the task more difficult by choosing very large numbers, but a littlereflection will show that the sizes of the numbers are irrelevant. It is onlynecessarythattheslipsbearnumbersthatcanbearrangedinincreasingorder.The game hasmany interesting applications. For example, a girl decides to

marrybeforetheendoftheyear.Sheestimatesthatshewillmeettenmenwhocanbepersuadedtopropose,butonceshehasrejectedaproposal,themanwillnot try again. What strategy should she follow to maximize her chances ofaccepting the top man of the ten, and what is the probability that she willsucceed?The strategy consists of rejecting a certain number of slips of paper (or

proposals),thenpickingthenextnumberthatexceedsthehighestnumberamongtherejectedslips.Whatisneededisaformulafordetermininghowmanyslipstoreject,dependingonthetotalnumberofslips.

35.MarchingCadetsandaTrottingDog

ASQUAREFORMATIONofArmycadets,50feetontheside,ismarchingforwardataconstantpace.Thecompanymascot,asmallterrier,startsatthecenteroftherear rank [positionA in the illustration], trots forward in a straight line to thecenterofthefrontrank[positionB],thentrotsbackagaininastraightlinetothecenter of the rear. At the instant he returns to position A, the cadets haveadvancedexactly50 feet.Assuming that thedog trots at a constant speed andlosesnotimeinturning,howmanyfeetdoeshetravel?

If you solve this problem, which calls for no more than a knowledge ofelementary algebra, you may wish to tackle a much more difficult versionproposedbythefamouspuzzlistSamLoyd.(SeeMoreMathematicalPuzzlesofSamLoyd,Dover,1960,page103.)Insteadofmovingforwardandbackthroughthemarchingcadets,themascottrotswithconstantspeedaroundtheoutsideofthe square, keeping as close as possible to the square at all times. (For theproblemweassumethathetrotsalongtheperimeterof thesquare.)Asbefore,theformationhasmarched50feetbythetimethedogreturnstopointA.Howlongisthedog’spath?

36.White,BlackandBrown

PROFESSORMERLEWHITEofthemathematicsdepartment,ProfessorLeslieBlackof philosophy, and Jean Brown, a young stenographer who worked in theuniversity’sofficeofadmissions,werelunchingtogether.“Isn’t it remarkable,” observed the lady, “that our last names are Black,

Brown andWhite and that one of us has black hair, one brown hair and onewhite.”“Itisindeed,”repliedthepersonwithblackhair,“andhaveyounoticedthat

notoneofushashairthatmatcheshisorhername?”“Bygolly,you’reright!”exclaimedProfessorWhite.Ifthelady’shairisn’tbrown,whatisthecolorofProfessorBlack’shair?

37.ThePlaneintheWind

ANAIRPLANEFLIES inastraight linefromairportAtoairportB, thenbackinastraightlinefromBtoA.Ittravelswithaconstantenginespeedandthereisnowind.Willitstraveltimeforthesameroundtripbegreater,lessorthesameif,throughoutbothflights,atthesameenginespeed,aconstantwindblowsfromAtoB?

38.WhatPricePets?

THE OWNER of a pet shop bought a certain number of hamsters and half thatmany pairs of parakeets. He paid $2 each for the hamsters and $1 for eachparakeet.Oneverypetheplacedaretailpricethatwasanadvanceof10percentoverwhathepaidforit.Afterallbutsevenofthecreatureshadbeensold,theownerfoundthathehad

taken infor themanamountofmoneyexactlyequal towhathehadoriginallypaid for all of them. His potential profit, therefore, was represented by thecombinedretailvalueofthesevenremaininganimals.Whatwasthisvalue?

39.TheGameofHip

THE GAME of “Hip,” so named because of the hipster’s reputed disdain for“squares,”isplayedonasix-by-sixcheckerboardasfollows:One player holds eighteen red counters; his opponent holds eighteen black

counters. They take turns placing a single counter on any vacant cell of theboard. Each tries to avoid placing his counters so that four of themmark thecornersofasquare.Thesquaremaybeanysizeandtippedatanyangle.Thereare105possiblesquares,fourofwhichareshownintheillustration.Aplayerwinswhenhisopponentbecomesa“square”byformingoneofthe

105squares.Thegamecanbeplayedonaboardwithactualcounters,orwithpencil andpaper.Simplydraw theboard, then registermovesbymarkingX’sandO’sonthecells.

FormonthsafterIhaddevisedthisgameIbelievedthatitwasimpossiblefora draw to occur in it. Then C.M. McLaury, a mathematics student at theUniversityofOklahoma,demonstratedthat thegamecouldendinadraw.Theproblemistoshowhowthegamecanbedrawnbydividingthe36cellsintotwosetsofeighteeneachsothatnofourcellsofthesamesetmarkthecornersofasquare.

40.ASwitchingPuzzle

THEEFFICIENTswitchingofrailroadcarsoftenposesfrustratingproblemsinthefield of operations research. The switching puzzle shown is one that has themeritofcombiningsimplicitywithsurprisingdifficulty.The tunnel is wide enough to accommodate the locomotive but not wide

enough for either car. The problem is to use the locomotive for switching thepositionsofcarsAandB, thenreturnthelocomotiveto itsoriginalspot.Eachendofthelocomotivecanbeusedforpushingorpulling,andthetwocarsmay,ifdesired,becoupledtoeachother.Thebestsolutionistheonerequiringthefewestoperations.An“operation”is

heredefinedasanymovementofthelocomotivebetweenstops,assumingthatitstopswhenitreversesdirection,meetsacartopushitorunhooksfromacarithasbeenpulling.Movementsofthetwoswitchesarenotcountedasoperations.A convenientway toworkon the puzzle is to place a penny, a dime and a

nickelontheillustrationandslidethemalongthetracks,rememberingthatonlythe coin representing the locomotive can pass through the tunnel. In theillustration, the carswere drawn in positions too close to the switches.Whileworking on the problem, assume that both cars are far enough east along thetracksothatthereisamplespacebetweeneachcarandswitchtoaccommodateboththelocomotiveandtheothercar.No “flying switch” maneuvers are permitted. For example, you are not

permitted to turn the switch quickly just after the engine has pushed anunattached car past it, so that the car goes one way and the engine, withoutstopping,goesanotherway.

41.BeerSignsontheHighway

SMITHDROVEatasteadyclipalongthehighway,hiswifebesidehim.“Haveyounoticed,”hesaid,“thatthoseannoyingsignsforFlatzbeerseemtoberegularlyspacedalongtheroad?Iwonderhowfaraparttheyare.”Mrs.Smithglancedatherwristwatch,thencountedthenumberofFlatzbeer

signstheypassedinoneminute.“What an odd coincidence!” exclaimed Smith. “When you multiply that

numberbyten,itexactlyequalsthespeedofourcarinmilesperhour.”Assuming that the car’s speed is constant, that the signs are equally spaced

andthatMrs.Smith’sminutebeganandendedwiththecarmidwaybetweentwosigns,howfarisitbetweenonesignandthenext?

42.TheSlicedCubeandtheSlicedDoughnut

ANENGINEER,notedforhisabilitytovisualizethree-dimensionalstructure,washavingcoffeeanddoughnuts.Beforehedroppedasugarcube intohiscup,heplacedthecubeonthetableandthought:IfIpassahorizontalplanethroughthecube’scenter,thecrosssectionwillofcoursebeasquare.IfIpassitverticallythrough the center and four corners of the cube, the cross section will be anoblongrectangle.NowsupposeIcutthecubethiswaywiththeplane....Tohissurprise,hismentalimageofthecrosssectionwasaregularhexagon.Howwastheslicemade?Ifthecube’ssideishalfaninch,whatisthesideof

thehexagon?Afterdroppingthecubeintohiscoffee,theengineerturnedhisattentiontoa

doughnutlyingflatonaplate.“IfIpassaplanehorizontallythroughthecenter,”hesaidtohimself,“thecrosssectionwillbetwoconcentriccircles.IfIpasstheplanevertically throughthecenter, thesectionwillbe twocirclesseparatedbythe width of the hole. But if I turn the plane so. . . . ” He whistled withastonishment.Thesectionconsistedoftwoperfectcirclesthatintersected!Howwas this slicemade? If thedoughnut isaperfect torus, three inches in

outsidediameterandwithaholeoneinchacross,whatarethediametersoftheintersectingcircles?

43.BisectingYinandYang

TWOMATHEMATICIANSwerediningattheYinandYang,aChineserestaurantonWest Third Street in Manhattan. They chatted about the symbol on therestaurant’smenu,shownintheillustration.

“Isupposeit’soneoftheworld’soldestreligioussymbols,”oneofthemsaid.“Itwouldbehardtofindamoreattractivewaytosymbolizethegreatpolaritiesof nature: good and evil,male and female, inflation and deflation, integrationanddifferentiation.”“Isn’titalsothesymboloftheNorthernPacificRailway?”“Yes. I understand that one of the chief engineers of the railroad saw the

emblem on aKorean flag at theChicagoWorld’s Fair in 1893 and urged hiscompany toadopt it.Hesaid it symbolized theextremesof fireandwater thatdrovethesteamengine.”“Doyousupposeitinspiredtheconstructionofthemodernbaseball?”“Iwouldn’tbesurprised.By theway,didyouknowthat there isanelegant

methodofdrawingonestraightlineacrossthecirclesothatitexactlybisectstheareasoftheYinandYang?”AssumingthattheYinandYangareseparatedbytwosemicircles,showhow

eachcanbesimultaneouslybisectedbythesamestraightline.

44.TheBlue-EyedSisters

IF YOU HAPPEN tomeet two of the Jones sisters (this assumes that the two arerandom selections from the set of all the Jones sisters), it is an exactly even-moneybet thatbothgirlswillbeblue-eyed.What isyourbestguess as to thetotalnumberofblue-eyedJonessisters?

45.HowOldIstheRose-RedCity?

TWOPROFESSORS,oneofEnglishandoneofmathematics,werehavingdrinksinthefacultyclubbar.“It is curious,” said the English professor, “how some poets can write one

immortal line and nothing else of lasting value. John William Burgon, forexample.Hispoemsaresomediocrethatnoonereadsthemnow,yethewroteoneofthemostmarvelouslinesinEnglishpoetry:‘Arose-redcityhalfasoldasTime.’”The mathematician, who liked to annoy his friends with improvised

brainteasers,thoughtforamomentortwo,thenraisedhisglassandrecited:

“Arose-redcityhalfasoldasTime.Onebillionyearsagothecity’sageWasjusttwo-fifthsofwhatTime’sagewillbeAbillionyearsfromnow.Canyoucompute

Howoldthecrimsoncityistoday?”

The English professor had long ago forgotten his algebra, so he quicklyshiftedtheconversationtoanothertopic,butreadersofthisbookshouldhavenodifficultywiththeproblem.

46.TrickyTrack

THREE HIGH SCHOOLS—Washington, Lincoln and Roosevelt— competed in atrackmeet.Eachschoolenteredoneman,andoneonly,ineachevent.Susan,astudentatLincolnHigh,satinthebleacherstocheerherboyfriend,theschool’sshot-putchampion.WhenSusanreturnedhomelaterintheday,herfatheraskedhowherschool

haddone.“We won the shot-put all right,” she said, “butWashington High won the

trackmeet.Theyhadafinalscoreof22.Wefinishedwith9.SodidRooseveltHigh.”“Howweretheeventsscored?”herfatherasked.“Idon’trememberexactly,”Susanreplied,“buttherewasacertainnumberof

pointsforthewinnerofeachevent,asmallernumberforsecondplaceandastillsmallernumberforthirdplace.Thenumberswerethesameforallevents.”(By“number”Susanofcoursemeantapositiveinteger.)“Howmanyeventsweretherealtogether?”“Gosh,Idon’tknow,Dad.AllIwatchedwastheshot-put.”“Wasthereahighjump?”askedSusan’sbrother.Susannodded.“Whowonit?”Susandidn’tnow.Incredible as itmay seem, this last question can be answeredwith only the

informationgiven.Whichschoolwonthehighjump?

47.Termiteand27Cubes

IMAGINE a large cube formed by gluing together 27 smaller wooden cubes ofuniformsizeasshown.Atermitestartsatthecenterofthefaceofanyoneoftheoutside cubes and bores a path that takes him once through every cube. Hismovementisalwaysparalleltoasideofthelargecube,neverdiagonal.Isitpossibleforthetermitetoborethrougheachofthe26outsidecubesonce

andonlyonce,thenfinishhistripbyenteringthecentralcubeforthefirsttime?Ifpossible,showhowitcanbedone;ifimpossible,proveit.

It isassumedthat the termite,once ithasbored intoasmallcube, followsapathentirelywithinthelargecube.Otherwise,itcouldcrawloutonthesurfaceofthelargecubeandmovealongthesurfacetoanewspotofentry.Ifthiswerepermitted,therewould,ofcourse,benoproblem.

48.CollatingtheCoins

ARRANGETHREEpenniesandtwodimesinarow,alternatingthecoinsasshown.The problem is to change their positions to those shown at the bottom of theillustrationintheshortestpossiblenumberofmoves.Amoveconsistsofplacingthetipsofthefirstandsecondfingersonanytwo

touchingcoins,oneofwhichmustbeapennyandtheotheradime,thenslidingthepair toanotherspotalong the imaginary lineshownin the illustration.Thetwocoins in thepairmust touchat all times.Thecoinat left in thepairmustremain at left; the coin at right must remain at right. Gaps in the chain areallowed at the end of anymove except the final one.After the lastmove thecoinsneednotbeatthesamespotontheimaginarylinethattheyoccupiedatthestart.

Ifitwerepermissibletoshifttwocoinsofthesamekind,thepuzzlecouldbesolvedeasilyinthreemoves:slide1,2toleft,fillthegapwith4,5,thenmove5,3fromrighttoleftend.Butwiththeprovisothateachshiftedpairmustincludeadimeandpennyitisabafflingandprettyproblem.

49.TimetheToast

EVEN THE SIMPLEST of household tasks can present complicated problems inoperational research. Consider the preparation of three slices of hot butteredtoast.Thetoasteristheold-fashionedtype,withhingeddoorsonitstwosides.Itholdstwopiecesofbreadatoncebuttoastseachofthemononesideonly.Totoastbothsidesitisnecessarytoopenthedoorsandreversetheslices.Ittakesthreesecondstoputasliceofbreadintothetoaster,threesecondsto

takeitoutandthreesecondstoreverseaslicewithoutremovingit.Bothhandsarerequiredforeachoftheseoperations,whichmeansthatitisnotpossibletoput in, takeoutor turn twoslicessimultaneously.Nor is itpossible tobutteraslicewhile another slice isbeingput into the toaster, turnedor takenout.Thetoasting time foronesideofapieceofbread is thirtyseconds. It takes twelvesecondstobutteraslice.Eachslice isbutteredononesideonly.Nosidemaybebuttereduntil ithas

been toasted.A slice toasted and buttered on one sidemay be returned to thetoasterfortoastingonitsotherside.Thetoasteriswarmedupatthestart.Inhowshortatimecanthreeslicesofbreadbetoastedonbothsidesandbuttered?

50.AFixed-PointTheorem

ONE MORNING, exactly at sunrise, a Buddhist monk began to climb a tallmountain.Thenarrowpath,nomorethanafootortwowide,spiraledaroundthe

mountaintoaglitteringtempleatthesummit.Themonkascended thepathatvaryingratesofspeed,stoppingmany times

alongthewaytorestandtoeatthedriedfruithecarriedwithhim.Hereachedthetempleshortlybeforesunset.Afterseveraldaysoffastingandmeditationhebegan his journey back along the same path, starting at sunrise and againwalkingatvariablespeedswithmanypausesalongtheway.Hisaveragespeeddescendingwas,ofcourse,greaterthanhisaverageclimbingspeed.Prove that there is a spot along thepath that themonkwill occupyonboth

tripsatpreciselythesametimeofday.

51.HowDidKantSetHisClock?

ITISSAIDthatImmanuelKantwasabachelorofsuchregularhabitsthatthegoodpeopleofKönigsbergwouldadjust their clockswhen they sawhimstrollpastcertainlandmarks.One evening Kant was dismayed to discover that his clock had run down.

Evidentlyhismanservant,whohadtakenthedayoff,hadforgotten towindit.The great philosopher did not reset the hands because his watch was beingrepairedandhehadnowayofknowingthecorrecttime.HewalkedtothehomeofhisfriendSchmidt,amerchantwholivedamileorsoaway,glancingattheclockinSchmidt’shallwayasheenteredthehouse.AftervisitingSchmidtforseveralhoursKantleftandwalkedhomealongthe

routebywhichhecame.Asalways,hewalkedwithaslow,steadygaitthathadnotvariedin twentyyears.Hehadnonotionofhowlongthisreturntrip took.(SchmidthadrecentlymovedintotheareaandKanthadnotyet timedhimselfonthiswalk.)Nevertheless,whenKantenteredhishouse,heimmediatelysethisclockcorrectly.HowdidKantknowthecorrecttime?

52.PlayingTwentyQuestionswhenProbabilityValuesAreKnown

INTHEwell-knowngameTwentyQuestionsonepersonthinksofanobject,suchastheLibertyBellinPhiladelphiaorLawrenceWelk’sleftlittletoe,andanotherpersontries toguess theobjectbyaskingnomore than twentyquestions,eachanswerablebyyesorno.Thebestquestionsareusuallythosethatdividethesetofpossibleobjectsintotwosubsetsasnearlyequalinnumberaspossible.Thusif a person has chosen as his “object” a number from 1 through 9, it can beguessed by this procedure in no more than four questions—possibly less. In

twentyquestionsonecanguessanynumberfrom1through220(or1,048,576).Suppose that each of the possible objects can be given a different value to

represent the probability that it has been chosen. For example, assume that adeckofcardsconsistsofoneaceofspades,twodeucesofspades,threethrees,and on up to nine nines,making 45 spade cards in all. The deck is shuffled;someonedrawsacard.Youaretoguessitbyaskingyes-noquestions.Howcanyouminimizethenumberofquestionsthatyouwillprobablyhavetoask?

53.Don’tMateinOne

KARL FABEL, a German chess problemist, is responsible for this outrageousproblem.

Youareasked to findamove forwhite thatwillnot result inan immediatecheckmateoftheblackking.

54.FindtheHexahedrons

APOLYHEDRONisasolidboundedbyplanepolygonsknownasthefacesofthesolid.Thesimplestpolyhedronisthetetrahedron,consistingoffourfaces,eachatriangle[leftfigure].Atetrahedroncanhaveanendlessvarietyofshapes,butifweregarditsnetworkofedgesasatopologicalinvariant(that is,wemayalterthelengthofanyedgeandtheanglesatwhichedgesmeetbutwemustpreserve

thestructureofthenetwork),thenthereisonlyonebasictypeoftetrahedron.Itisnotpossible,inotherwords,foratetrahedrontohavesidesthatareanythingbuttriangles.Thefive-sidedpolyhedronhastwobasicvarieties[middleandrightfigures].

One is represented by the Great Pyramid of Egypt (four triangles on aquadrilateral base). The other is represented by a tetrahedronwith one cornerslicedoff;threeofitsfacesarequadrilaterals,twoaretriangles.JohnMcClellan,anartistinWoodstock,NewYork,asksthisquestion:How

many basic varieties of convex hexahedrons, or six-sided solids, are therealtogether? (A solid is convex if each of its sides can be placed flat against atabletop.)Thecubeis,ofcourse,themostfamiliarexample.Ifyousearchforhexahedronsbychoppingcorners fromsimplersolids,you

mustbecareful toavoidduplication.Forexample, theGreatPyramid,with itsapexslicedoff,hasaskeletonthatistopologicallyequivalenttothatofthecube.Becarefulalsotoavoidmodelsthatcannotexistwithoutwarpedfaces.

55.OutwiththeOnion

ARRANGEFOURpapermatchesonthetableasshownintheleft-handfigure.Theyrepresent amartini glass.Amatchheadgoes inside to indicate the onionof aGibson cocktail. The puzzle is to move just two matches so that the glass isreformed, but the onion—which must stay where it is—winds up outside theglass.Atthefinish,theglassmaybeturnedtotheleftorright,orevenbeupsidedown,butitmustbeexactlythesameshapeasbefore.Theupperrightfigureisnot a solution because the onion is still inside. The lower right figure doesn’tworkbecausethreematcheshavebeenmoved.

56.CutDowntheCuts

YOUHAVEtakenupthehobbyofjewelrymakingandyouwanttojointhefourpiecesofsilverchainshownatthelefttoformthecircularbraceletshownattheright.Sinceittakesabitofdoingtocutopenalinkandweldittogetheragain,younaturallywanttocutasfewlinksaspossible.Whatistheminimumnumberoflinksyoumustcuttodothejob?

57.DissectionDilemma

THE TWO TOP figures show how each of two shapes can be divided into fourparts,allexactlyalike.Yourtaskistodividetheblanksquareintofiveparts,allidenticalinsizeandshape.

58.InterruptedBridge

AFTERYOUhavedealtabouthalfthecardsforabridgegame,thetelephonerings.Youputdowntheundealtcards toanswer thephone.Afteryoureturn,neitheryounor anyone else can rememberwhere the last cardwasdealt.Noonehastouchedanyof thedealtcards.Withoutcounting thecards inanyhand,or thenumber of cards yet to be dealt, how can you finish the deal rapidly andaccurately,givingeachplayerexactlythesamecardshewouldhavereceivedifyouhadn’tbeeninterrupted?

59.DashItAll!

SAULANDSALraceeachotherfor100yards.Salwinsby10yards.Theydecideto race again, but this time, to even thingsup,Salbegins10yardsbehind thestartline.Assumingthatbothrunwiththesameconstantspeedasbefore,whowins?

60.MovetheQueen

PLACE a chess queen on a square next to a corner square of a chessboard asshown in the left-hand figure. The problem is to move the queen four times,makingstandardqueenmoves,sothatshepassesthroughallnineoftheshadedsquares. (A queen may move any number of squares in any direction:horizontally,verticallyordiagonally.)Thetopfigureshowsonewaytodoitinsixmoves,butthat’stwotoomany.

61.ReadtheHieroglyphics

THE SEVEN symbols shown below look like some kind of ancientwriting.Butthereisameaningforeachsymbol,andifyoucanpuzzlethemout,youshouldhavenotroubledrawinginthesquarethenextsymbolofthiscurioussequence.

62.CrazyCut

THISONElooksmucheasierthanitis.Youaretomakeonecut(ordrawoneline)—ofcourseitneedn’tbestraight—thatwilldividethefigureintotwoidenticalparts.

63.FindtheOddball

IF YOU HAVE two identical balls, one heavier than the other, you can easilydeterminewhichisheavierbyputtingthemonoppositepansofabalancescale.If therearefourballs,all thesameweightexceptforoneheavierone,youcanfindtheheavieroneintwoweighings.Supposeyouhavenineidenticalballs,oneofwhichisheavierthantheeight

others. What is the smallest number of weighings needed for positivelyidentifyingtheoddball?

64.BigCross-OutSwindle

CROSS OUT nine letters in such away that the remaining letters spell a singleword.

65.ReversetheDog

ARRANGE13papermatchestomakeadogthatfacestotheleftasinthediagram.Byloweringthedog’stailtothetopdottedline,thenmovingthebottommatchofthedog’sheadtotheotherdottedline,youhavechangedthepicturesothatthe dog is looking the opposite way. Unfortunately, this leaves the dog’s tail(nowontheleft)slantingdowninsteadofup.Canyoumovejusttwomatchestomakethedogfacetotheright,butwithhis

tailpointingupwardasbefore?

66.FunnyFold

SCOTTKIM,whotellsmeheinventedthispuzzleinsixthgrade,hascutalargecapitalletterfromasheetofpaperandgivenitasinglefold.ItiseasytoseethatthelettercouldbeanL,butthatisnottheletterKimcut

out.Whatletterisit?

ANSWERS

1.Isthereanyotherpointontheglobe,besidestheNorthPole,fromwhichyoucouldwalkamilesouth,amileeast,andamilenorthandfindyourselfbackatthestartingpoint?Yesindeed;notjustonepointbutaninfinitenumberofthem!You could start from any point on a circle drawn around the South Pole at adistanceslightlymorethan1+1/2πmiles(about1.16miles)fromthePole—thedistanceis“slightlymore”totakeintoaccountthecurvatureoftheearth.Afterwalking a mile south, your next walk of one mile east will take you on acomplete circle around the Pole, and thewalk onemile north from therewillthenreturnyoutothestartingpoint.Thusyourstartingpointcouldbeanyoneoftheinfinitenumberofpointsonthecirclewitharadiusofabout1.16milesfromtheSouthPole.But this is not all.You could also start at points closer to thePole,sothatthewalkeastwouldcarryyoujusttwicearoundthePole,orthreetimes,andsoon.

2.Thereare88winningfirsthands.Theyfallintotwocategories:(1)fourtensandanyothercard(48hands);(2)threetensandanyofthefollowingpairsfromthesuitnotrepresentedbyaten:A-9,K-9,Q-9,J-9,K-8,Q-8,J-8,Q-7,J-7,J-6(40hands).

3. It is impossible tocover themutilatedchessboard(with twooppositecornersquares cut off)with 31 dominoes, and the proof is easy. The two diagonallyoppositecornersareof thesamecolor.Therefore their removal leavesaboardwithtwomoresquaresofonecolorthanoftheother.Eachdominocoverstwosquaresofoppositecolor,sinceonlyoppositecolorsareadjacent.Afteryouhavecovered60squareswith30dominoes,youareleftwithtwouncoveredsquaresof the same color. These two cannot be adjacent, therefore they cannot becoveredbythelastdomino.Suppose two cells ofopposite color are removed from the chessboard.Can

youprovethatnomatterwhichtwocellsareremoved,theboardcanalwaysbecovered with 31 dominoes? For a simple, elegant proof that this is always

possible,seemybookAha!Insight(W.H.Freeman,1978,page19).

4. If we require that the question be answerable by “yes” or “no,” there areseveral solutions, all exploiting the same basic gimmick. For example, thelogicianpointstooneoftheroadsandsaystothenative,“IfIweretoaskyouifthisroadleadstothevillage,wouldyousay‘yes’?”Thenativeisforcedtogivetherightanswer,evenifheisaliar!Iftheroaddoesleadtothevillage,theliarwouldsay“no”tothedirectquestion,butasthequestionisput,heliesandsayshewouldrespond“yes.”Thusthelogiciancanbecertainthattheroaddoesleadtothevillage,whethertherespondentisatruth-telleroraliar.Ontheotherhand,iftheroadactuallydoesnotgotothevillage,theliarisforcedinthesamewaytoreply“no”totheinquirer’squestion.Asimilarquestionwouldbe,“IfIaskedamemberoftheothertribewhether

thisroadleadstothevillage,wouldhesay‘yes’?”Toavoidthecloudinessthatresults from a questionwithin a question, perhaps this phrasing (suggested byWarrenC.Haggstrom,ofAnnArbor,Michigan)isbest:“Ofthetwostatements,‘Youarealiar’and‘Thisroadleadstothevillage,’isoneandonlyoneofthemtrue?”Hereagain,a“yes”answerindicatesitistheroad,a“no”answerthatitisn’t,regardlessofwhetherthenativeliesortellsthetruth.Dennis Sciama, Cambridge University cosmologist, and John McCarthy of

Hanover,NewHampshire,calledmyattentiontoadelightfuladditionaltwistontheproblem.“Suppose,”Mr.McCarthywrote(inaletterpublishedinScientificAmerican,April1957),“thelogicianknowsthat‘pish’and‘tush’arethenativewordsfor‘yes’and‘no’buthasforgottenwhichiswhich,thoughotherwisehecan speak the native language.He can still determinewhich road leads to thevillage.

“Hepointstooneoftheroadsandasks,‘IfIaskedyouwhethertheroadIampointingtoistheroadtothevillagewouldyousaypish?’Ifthenativereplies,‘Pish,’thelogiciancanconcludethattheroadpointedtoistheroadtothevillageeven though hewill still be in the dark as towhether the native is a liar or atruth-tellerandastowhether‘pish’meansyesorno.Ifthenativesays,‘Tush,’hemaydrawtheoppositeconclusion.”Forhundredsof ingeniousproblems involving truth-tellersand liars, see the

puzzlebooksbymathematician-logicianRaymondM.Smullyan.

5.Youcanlearnthecontentsofallthreeboxesbydrawingjustonemarble.Thekeytothesolutionisyourknowledgethatthelabelsonallthreeoftheboxesareincorrect.Youmustdrawamarblefromtheboxlabeled“black-white.”Assume

thatthemarbledrawnisblack.Youknowthenthattheothermarbleinthisboxmustbeblackalso,otherwise the labelwouldbecorrect.Sinceyouhavenowidentifiedtheboxcontainingtwoblackmarbles,youcanatoncetellthecontentsof the box marked “white-white”: you know it cannot contain two whitemarbles,becauseitslabelhastobewrong;itcannotcontaintwoblackmarbles,for you have identified that box; therefore itmust contain one black and onewhitemarble.Thethirdbox,ofcourse,mustthenbetheoneholdingtwowhitemarbles.Youcansolvethepuzzlebythesamereasoningifthemarbleyoudrawfromthe“black-white”boxhappenstobewhiteinsteadofblack.

6.Thereisnowaytoreducethecutstofewerthansix.Thisisatonceapparentwhenyoufocusonthefactthatacubehassixsides.Thesawcutsstraight—oneside at a time. To cut the one-inch cube at the center (the one which has noexposedsurfacestostartwith)musttakesixpassesofthesaw.

7. The answer to this puzzle is a simplematter of train schedules.While theBrooklyn and Bronx trains arrive equally often—at 10-minute intervals—ithappensthattheirschedulesaresuchthattheBronxtrainalwayscomestothisplatformoneminuteafter theBrooklyn train.Thus theBronx trainwillbe thefirst to arriveonly if theyoungmanhappens to come to the subwayplatformduring thisone-minute interval. Ifheenters thestationatanyother time—i.e.,during a nine-minute interval—the Brooklyn train will come first. Since theyoungman’sarrivalisrandom,theoddsareninetooneforBrooklyn.

8.Thecommuterhaswalkedfor55minutesbeforehiswifepickshimup.Sincethey arrive home 10 minutes earlier than usual, this means that the wife haschopped10minutesfromherusual travel time toandfromthestation,or fiveminutesfromhertraveltimetothestation.Itfollowsthatshemetherhusbandfiveminutesbeforehisusualpick-uptimeoffiveo’clock,orat4:55.Hestartedwalking at four, therefore he walked for 55 minutes. The man’s speed ofwalking,thewife’sspeedofdrivingandthedistancebetweenhomeandstationare not needed for solving the problem. If you tried to solve it by jugglingfiguresforthesevariables,youprobablyfoundtheproblemexasperating.

9.Thecounterfeitstackcanbeidentifiedbyasingleweighingofcoins.Youtakeonecoinfromthefirststack,twofromthesecond,threefromthethirdandsoonto the entire 10 coins of the tenth stack. You then weigh the whole samplecollectiononthepointerscale.Theexcessweightof thiscollection, innumberofgrams,correspondstothenumberofthecounterfeitstack.Forexample,ifthe

group of coins weighs seven grams more than it should, then the counterfeitstackmustbetheseventhone,fromwhichyoutooksevencoins(eachweighingonegrammore thanagenuinehalf-dollar).Evenif therehadbeenaneleventhstackof tencoins, theprocedure justdescribedwouldstillwork,fornoexcessweightwouldindicatethattheoneremainingstackwascounterfeit.

10.Thereareseveraldifferentwaysofplacingthesixcigarettes.Thefigureontheleftshowsthetraditionalsolutionasitisgiveninseveraloldpuzzlebooks.

To my vast surprise, about fifteen readers discovered that seven cigarettescouldalsobeplacedsothateachtouchedalloftheothers!Thisofcoursemakestheolderpuzzleobsolete.Thefigureontheright,senttomebyGeorgeRybickiand John Reynolds, graduate students in physics at Harvard, shows how it isdone.“Thediagramhasbeendrawn,”theywrite,“forthecriticalcasewheretheratiooflengthtodiameterofthecigarettesis .Herethepointsofcontactoccurrightattheendsofthecigarettes.Thesolutionobviouslywillworkforanylength-to-diameter ratio greater than . Some observations on actual‘regular’sizecigarettesgivearatioofabout8to1,whichis,infact,greaterthan

, so this is an acceptable solution.” Note that if the center cigarette,pointing directly toward you in the diagram, is withdrawn, the remaining sixprovideaneatsymmetricalsolutionoftheoriginalproblem.

11.Whentheferryboatsmeetfor thefirst time[topillustration], thecombineddistancetraveledbytheboatsisequaltothewidthoftheriver.Whentheyreachthe opposite shore, the combined distance is twice thewidth of the river; andwhentheymeetthesecondtime[bottomfigure],thetotaldistanceisthreetimestheriver’swidth.Sincetheboatshavebeenmovingataconstantspeedforthesameperiodoftime,itfollowsthateachboathasgonethreetimesasfaraswhentheyfirstmetandhadtraveledacombineddistanceofoneriver-width.Sincethe

white boat had traveled 720 yards when the first meeting occurred, its totaldistanceatthetimeofthesecondmeetingmustbe3X720,or2,160yards.Thebottom illustration shows clearly that this distance is 400 yardsmore than theriver’swidth,sowesubtract400from2,160toobtain1,760yards,oronemile,asthewidthoftheriver.Thetimetheboatsremainedattheirlandingsdoesnotenterintotheproblem.

The problem can be approached in other ways. Many readers solved it asfollows. Let x equal the river-width. On the first trip the ratio of distancestraveledbythetwoboatsisx–720:720.Onthesecondtripitis2x–400:x+400.Theseratiosareequal,soitiseasytosolveforx.

12.LineACisonediagonalof therectangle.Theotherdiagonal isclearly the10-unit radiusof the circle.Since thediagonals are equal, lineAC is 10unitslong.

13.Acontinuouslinethatentersandleavesoneoftherectangularspacesmustofnecessitycrosstwolinesegments.SincethespaceslabeledA,BandCinthetop illustration are each surroundedby anoddnumberof segments, it followsthat an end of a line must be inside each if all segments of the network arecrossed.Butacontinuouslinehasonlytwoends,sothepuzzleisinsolubleonaplanesurface.Thissamereasoningappliesifthenetworkisonasphereoronthesideofa torus[drawingat lowerleft].However, thenetworkcanbedrawnonthetorus[drawingatlowerright]sothattheholeofthetorusisinsideoneofthethreespaces,A,BandC.Whenthisisdone,thepuzzleiseasilysolved.

14.Twelvematchescanbeusedtoformarighttrianglewithsidesofthree,fourandfiveunits,asshowninthefirstillustration.Thistrianglewillhaveanareaofsixsquareunits.Byaltering thepositionof threematchesasshownat right intheillustration,weremovetwosquareunits, leavingapolygonwithanareaoffour.Thissolutionistheonetobefoundinmanypuzzlebooks.Therearehundreds

of other solutions. EltonM. Palmer, Oakmont, Pennsylvania, pointed out thateach of the five tetrominoes (figuresmadewith four squares) can provide thebase for a large number of solutions. We simply add and subtract the sameamountintriangularareastoaccommodateall12matches.Theseconddrawingdepictssomerepresentativesamples,eachrowbasedonadifferenttetromino.

Eugene J. Putzer, staff scientist with the General Dynamics Corporation;CharlesShapiro,Oswego,NewYork;andHughJ.Metz,OakRidge,Tennessee,suggestedthestarsolutionshowninthethirddrawing.Byadjustingthewidthofthe star’s points you can produce any desired area between 0 and 11.196, theareaofaregulardodecagon,thelargestareapossiblewiththe12matches.

15.Withoutresortingtocalculus,theproblemcanbesolvedasfollows.LetRbethe radius of the sphere. As the first illustration indicates, the radius of thecylindrical holewill thenbe the square root ofR2 – 9, and the altitudeof thesphericalcapsateachendofthecylinderwillbeR–3.Todeterminetheresidueafter the cylinder and caps have been removed, we add the volume of thecylinder,6π(R2–9),totwicethevolumeofthesphericalcap,andsubtractthetotalfromthevolumeofthesphere,4πR3/3.Thevolumeofthecapisobtainedbythefollowingformula,inwhichAstandsforitsaltitudeandrforitsradius:πA(3r2+A2)/6.Whenthiscomputationismade,alltermsobliginglycanceloutexcept36π—

thevolumeoftheresidueincubicinches.Inotherwords,theresidueisconstant

regardlessofthehole’sdiameterorthesizeofthesphere!TheearliestreferenceIhavefoundforthisbeautifulproblemisonpage86of

Samuel I. Jones’sMathematicalNuts, self-published,Nashville, 1932.A two-dimensional analog of the problem appears on page 93 of the same volume.Giventhelongestpossiblestraightlinethatcanbedrawnonacirculartrackofanydimensions[seesecondfigure],theareaofthetrackwillequaltheareaofacirclehavingthestraightlineasadiameter.John W. Campbell, Jr., editor of Astounding Science Fiction, was one of

severalreaderswhosolvedthesphereproblemquicklybyreasoningadroitlyasfollows:Theproblemwouldnotbegivenunlessithasauniquesolution.Ifithasauniquesolution,thevolumemustbeaconstantwhichwouldholdevenwhentheholeisreducedtozeroradius.Thereforetheresiduemustequalthevolumeofaspherewithadiameterofsixinches,namely36π.

16.Atanygiveninstantthefourbugsformthecornersofasquarewhichshrinksand rotates as the bugs move closer together. The path of each pursuer willthereforeat all timesbeperpendicular to thepathof thepursued.This tellsusthat as A, for example, approaches B, there is no component in B’s motionwhichcarriesBtowardorawayfromA.ConsequentlyAwillcaptureBinthesame time that itwould take ifBhad remained stationary.The lengthof eachspiralpathwillbethesameasthesideofthesquare:10inches.If three bugs start from the corners of an equilateral triangle, each bug’s

motionwillhaveacomponentof1/2(thecosineofa60-degreeangleis1/2)itsvelocity that will carry it toward its pursuer. Two bugs will therefore have amutual approach speed of 3/2 velocity. The bugs meet at the center of thetriangle after a time interval equal to twice the side of the triangle divided bythreetimesthevelocity,eachtracingapaththatis2/3thelengthofthetriangle’sside.For a generalization of this problem to n bugs at the corners of n-sided

polygons see Chapter 24 of my Sixth Book of Mathematical Games from

ScientificAmerican(W.H.Freeman,1971).

17.WhenJonesbegantoworkontheprofessor’sproblemheknewthateachofthe four familieshadadifferentnumberofchildren,and that the totalnumberwaslessthan18.Hefurtherknewthattheproductofthefournumbersgavetheprofessor’s house number. Therefore his obvious first step was to factor thehousenumber intofourdifferentnumberswhich togetherwould total less than18.Iftherehadbeenonlyonewaytodothis,hewouldhaveimmediatelysolvedthe problem. Since he could not solve it without further information, weconclude that theremusthavebeenmore thanonewayof factoring thehousenumber.Our next step is to write down all possible combinations of four different

numberswhichtotallessthan18,andobtaintheproductsofeachgroup.Wefindthat there are many cases where more than one combination gives the sameproduct.Howdowedecidewhichproductisthehousenumber?TheclueliesinthefactthatJonesaskediftherewasmorethanonechildin

thesmallestfamily.Thisquestionismeaningfulonlyifthehousenumberis120,whichcanbefactoredas1×3×5×8,1×4×5×6,or2×3×4×5.HadSmith answered “No,” the problem would remain unsolved. Since Jones didsolveit,weknowtheanswerwas“Yes.”Thefamiliesthereforecontained2,3,4and5children.ThisproblemwasoriginatedbyLesterR.FordandpublishedintheAmerican

MathematicalMonthly,March1948,asProblemE776.

18. The heads of the twiddled bolts move neither inward nor outward. Thesituationiscomparable to thatofapersonwalkingupanescalatorat thesameratethatitismovingdown.

19.Threeairplanesarequitesufficienttoensuretheflightofoneplanearoundtheworld.Therearemanywaysthiscanbedone,butthefollowingseemstobethemostefficient.Itusesonlyfivetanksoffuel,allowsthepilotsoftwoplanessufficient timeforacupofcoffeeandasandwichbeforerefuelingat thebase,andthereisapleasingsymmetryintheprocedure.Planes A, B and C take off together. After going 1/8 of the distance, C

transfers1/4tanktoAand1/4toB.ThisleavesCwith1/4tank,justenoughtogetbackhome.PlanesAandBcontinueanother1/8oftheway,thenBtransfers1/4tankto

A.Bnowhas1/2tankleft,whichissufficienttogethimbacktothebasewhere

hearriveswithanemptytank.PlaneA,with a full tank, continues until it runs out of fuel 1/4 of theway

fromthebase.ItismetbyCwhichhasbeenrefueledatthebase.Ctransfers1/4tanktoA,andbothplanesheadforhome.Thetwoplanesrunoutoffuel1/8ofthewayfromthebase,wheretheyare

met by refueled plane B. Plane B transfers 1/4 tank to each of the other twoplanes.Thethreeplanesnowhavejustenoughfueltoreachthebasewithemptytanks.The entire procedure can be diagrammed as shown, where distance is the

horizontalaxisand time theverticalaxis.The rightand leftedgesof thechartshould,ofcourse,beregardedasjoined.

20.Writing a three-digit number twice is the sameasmultiplying it by1,001.Thisnumberhasthefactors7,11and13,sowritingthechosennumbertwiceisequivalent to multiplying it by 7, 11 and 13. Naturally when the product issuccessivelydividedby these same threenumbers, the final remainderwill betheoriginalnumber.I presented this classical trick before hand and desk calculators became

omnipresent.Thetrickismosteffectivewhenasinglespectatorusesacalculatorto perform his divisions while your back is turned, only to see his originallychosennumbermysteriouslyappearinthereadoutafterthefinaldivision.

21.Thetwomissilesapproacheachotherwithcombinedspeedsof30,000milesperhour,or500milesperminute.By running thescenebackward in timeweseethatoneminutebeforethecollisionthemissileswouldhavetobe500milesapart.

22.Numberthetopcoininthepyramid1,thecoinsinthenextrow2and3,andthose in the bottom row 4, 5 and 6. The following fourmoves are typical ofmany possible solutions:Move 1 to touch 2 and 4,move 4 to touch 5 and 6,move5totouch1and2below,move1totouch4and5.

23. Because two people are involved in every handshake, the total score foreveryoneat theconventionwillbeevenlydivisibleby twoand thereforeeven.The total score for themenwho shook hands an even number of times is, ofcourse,alsoeven.Ifwesubtractthisevenscorefromtheeventotalscoreoftheconvention,wegetaneventotalscorefor thosemenwhoshookhandsanoddnumber of times. Only an even number of odd numbers will total an evennumber, so we conclude that an even number of men shook hands an oddnumberoftimes.Thereareotherwaystoprovethetheorem;oneofthebestwassenttomeby

GeraldK. Schoenfeld, amedical officer in theU.S.Navy.At the start of theconvention,before anyhandshakeshaveoccurred, thenumberofpersonswhohave shaken hands an odd number of times will be 0. The first handshakeproduces two “odd persons.” From now on, handshakes are of three types:between twoevenpersons, twooddpersons,oroneoddandoneevenperson.Eacheven-evenshakeincreasesthenumberofoddpersonsby2.Eachodd-oddshakedecreasesthenumberofoddpersonsby2.Eachodd-evenshakechangesan odd person to even and an even person to odd, leaving the number of oddpersons unchanged. There is no way, therefore, that the even number of oddpersonscanshiftitsparity;itmustremainatalltimesanevennumber.Bothproofsapplytoagraphofdotsonwhichlinesaredrawntoconnectpairs

of dots.The lines forma networkonwhich the number of dots thatmark themeetingofanoddnumberoflinesiseven.

24. In the triangularpistolduel thepoorest shot, Jones,has thebest chance tosurvive.Smith,whonevermisses,hasthesecondbestchance.BecauseJones’stwo opponents will aim at each other when their turns come, Jones’s beststrategyistofireintotheairuntiloneopponentisdead.Hewillthengetthefirstshotatthesurvivor,whichgiveshimastrongadvantage.Smith’ssurvivalchancesaretheeasiesttodetermine.Thereisachanceof1/2

thathewillgetthefirstshotinhisduelwithBrown,inwhichcasehekillshim.There is a chance of 1/2 that Brownwill shoot first, and since Brown is 4/5accurate,Smithhasa1/5chanceof surviving.SoSmith’schanceof survivingBrownis1/2addedto1/2x1/5=3/5.Jones,whoisaccuratehalfthetime,now

gets a crack at Smith. If hemisses, Smith kills him, so Smith has a survivalchanceof1/2againstJones.Smith’sover-allchanceofsurvivingistherefore3/5×1/2=3/10.Brown’s case ismore complicated becausewe run into an infinite series of

possibilities.His chance of surviving against Smith is 2/5 (we saw above thatSmith’ssurvivalchanceagainstBrownwas3/5,andsinceoneof the twomenmust survive, we subtract 3/5 from 1 to obtain Brown’s chance of survivingagainstSmith).BrownnowfacesfirefromJones.Thereisachanceof1/2thatJoneswillmiss, inwhichcaseBrownhasa4/5chanceofkillingJones.Up tothispoint,then,hischanceofkillingJonesis1/2×4/5=4/10.Butthereisa1/5chance that Brown may miss, giving Jones another shot. Brown’s chance ofsurvivingis1/2;thenhehasa4/5chanceofkillingJonesagain,sohischanceofsurvivingonthesecondroundis1/2×1/5×1/2×4/5=4/100.IfBrownmissesagain,hischanceofkillingJonesonthethirdroundwillbe

4/1,000;ifhemissesoncemore,hischanceonthefourthroundwillbe4/10,000,and so on.Brown’s total survival chance against Jones is thus the sumof theinfiniteseries:

Thiscanbewrittenastherepeatingdecimal.444444...,whichisthedecimalexpansionof4/9.Aswesawearlier,Brownhada2/5chanceofsurvivingSmith;nowwesee

that he has a 4/9 chance to survive Jones. His over-all survival chance istherefore2/5×4/9=8/45.Jones’ssurvivalchancecanbedeterminedinsimilarfashion,butofcoursewe

cangetitimmediatelybysubtractingSmith’schance,3/10,andBrown’schance,8/45,from1.ThisgivesJonesanover-allsurvivalchanceof47/90.Theentireduelcanbeconvenientlygraphedbyusingthetreediagramshown.

Itbeginswithonly twobranchesbecause Jonespasses ifhehas the first shot,leavingonlytwoequalpossibilities:SmithshootingfirstorBrownshootingfirst,with intent to kill. One branch of the tree goes on endlessly. The over-allsurvivalchanceofanindividualiscomputedasfollows:

1.Markalltheendsofbranchesatwhichthepersonissolesurvivor.2.Traceeachendbacktothebaseofthetree,multiplyingtheprobabilitiesof

eachsegmentasyougoalong.Theproductwillbetheprobabilityoftheeventattheendofthebranch.3.Addtogethertheprobabilitiesofallthemarkedend-pointevents.Thesum

willbetheover-allsurvivalprobabilityforthatperson.IncomputingthesurvivalchancesofBrownandJones,aninfinitenumberof

end-points are involved, but it is not difficult to see from the diagramhow toformulatetheinfiniteseriesthatisinvolvedineachcase.

25.Thefollowinganalysisof thedesert-crossingproblemappeared inan issueof Eureka, a publication of mathematics students at the University of

Cambridge. Five hundred miles will be called a “unit”; gasoline sufficient totakethetruck500mileswillbecalleda“load”;anda“trip”isajourneyofthetruckineitherdirectionfromonestoppingpointtothenext.Twoloadswillcarrythetruckamaximumdistanceof1and1/3units.Thisis

doneinfourtripsbyfirstsettingupacacheataspot1/3unitfromthestart.Thetruckbeginswithafullload,goestothecache,leaves1/3load,returns,picksupanotherfullload,arrivesatthecacheandpicksupthecache’s1/3load.Itnowhasafullload,sufficienttotakeittheremainingdistancetooneunit.Threeloadswillcarrythetruck1and1/3plus1/5unitsinatotalofninetrips.

Thefirstcacheis1/5unitfromthestart.Threetripsput6/5loadsinthecache.Thetruckreturns,picksuptheremainingfullloadandarrivesatthefirstcachewith4/5loadinitstank.This,togetherwiththefuelinthecache,makestwofullloads,sufficienttocarrythetrucktheremaining1and1/3units,asexplainedintheprecedingparagraph.Weareaskedfortheminimumamountoffuelrequiredtotakethetruck800

miles.Threeloadswilltakeit766and2/3miles(1and1/3plus1/5units),soweneedathirdcacheatadistanceof33and1/3miles(1/15unit)fromthestart.Infive trips the truck can build up this cache so thatwhen the truck reaches thecacheattheendoftheseventhtrip,thecombinedfueloftruckandcachewillbethree loads.Aswehave seen, this is sufficient to take the truck the remainingdistanceof766and2/3miles.Seventripsaremadebetweenstartingpointandfirstcache,using7/15loadofgasoline.Thethreeloadsoffuel thatremainarejustsufficientfortherestoftheway,sothetotalamountofgasolineconsumedwillbe3and7/15,oralittlemorethan3.46loads.Sixteentripsarerequired.Proceedingalongsimilar lines, four loadswill take the truckadistanceof1

and1/3plus1/5plus1/7units,with three caches located at theboundariesofthesedistances.Thesumof this infiniteseriesdivergesas thenumberof loadsincreases; therefore the truck can cross a desert of anywidth. If the desert is1,000 miles across, seven caches, 64 trips and 7.673 loads of gasoline arerequired.Hundreds of letterswere received on this problem, giving general solutions

and interesting sidelights. Cecil G. Phipps, professor of mathematics at theUniversityofFlorida,summedmattersupsuccinctlyasfollows:“Thegeneralsolutionisgivenbytheformula:

d=m(1+1/3+1/5+1/7+...),

wheredisthedistancetobetraversedandmisthenumberofmilesperloadof

gasoline.Thenumberofdepotstobeestablishedisonelessthanthenumberoftermsintheseriesneededtoexceedthevalueofd.Oneloadofgasolineisusedin the travel between each pair of stations. Since the series is divergent, anydistancecanbereachedbythismethodalthoughtheamountofgasolineneededincreasesexponentially.“Ifthetruckistoreturneventuallytoitshomestation,theformulabecomes:

d=m(1/2+1/4+1/6+1/8+...)

Thisseriesisalsodivergentandthesolutionhaspropertiessimilartothosefortheone-waytrip.”Manyreaderscalledattentiontothreepreviouslypublisheddiscussionsofthe

problem:“TheJeepProblem:AMoreGeneralSolution.”C.G.PhippsintheAmerican

MathematicalMonthly,Vol.54,No.8,pages458–462,October1947.“CrossingtheDesert.”G.G.AlwayintheMathematicalGazette,Vol.41,No.

337,page209,October1947.Problem inLogistics:TheJeepProblem.OlafHelmer.ProjectRandReport

No.RA-15015,December1,1946.

26.ThekeytoLordDunsany’schessproblemisthefactthattheblackqueenisnotonablacksquareasshemustbeatthestartofagame.Thismeansthattheblackkingandqueenhavemoved,and thiscouldhavehappenedonly ifsomeblack pawns havemoved. Pawns cannotmove backward, sowe are forced toconcludethattheblackpawnsreachedtheirpresentpositionsfromtheothersideoftheboard!Withthisinmind,itiseasytodiscoverthatthewhiteknightontherighthasaneasymateinfourmoves.White’sfirstmoveistojumphisknightatthelowerrightcorneroftheboard

to the square just above his king. If blackmoves the upper left knight to therook’sfile,whitematesintwomoremoves.Blackcan,however,delaythemateonemoveby firstmovinghisknight to thebishop’s file insteadof the rook’s.Whitejumpshisknightforwardandrighttothebishop’sfile,threateningmateon the next move. Blackmoves his knight forward to block themate.Whitetakestheknightwithhisqueen,thenmateswithhisknightonthefourthmove.

27. In long division, when two digits are brought down instead of one, theremustbeazero in thequotient.Thisoccurs twice,soweknowatonce that thequotientisx080x.Whenthedivisorismultipliedbythequotient’slastdigit,the

product is a four-digit number. The quotient’s last digit must therefore be 9,becauseeighttimesthedivisorisathree-digitnumber.Thedivisormustbe less than125becauseeight times125 is1,000, a four-

digit number.Wenowcan deduce that the quotient’s first digitmust bemorethan7,forseventimesadivisorlessthan125wouldgiveaproductthatwouldleavemorethantwodigitsafteritwassubtractedfromthefirstfourdigitsinthedividend.Thisfirstdigitcannotbe9(whichgivesafour-digitnumberwhenthedivisorismultipliedbyit),soitmustbe8,makingthefullquotient80809.Thedivisormustbemorethan123because80809times123isaseven-digit

number and our dividend has eight digits. The only number between 123 and125is124.Wecannowreconstructtheentireproblemasfollows:

28.Severalprocedureshavebeendevisedbywhichnpersonscandivideacakein n pieces so that each is satisfied that he has at least 1/n of the cake. Thefollowingsystemhasthemeritofleavingnoexcessbitsofcake.Supposetherearefivepersons:A,B,C,D,E.Acutsoffwhatheregardsas

1/5 of the cake and what he is content to keep as his share. B now has theprivilege,ifhethinksA’ssliceismorethan1/5,ofreducingittowhathethinksis1/5bycuttingoffaportion.Ofcourseifhethinksitis1/5orless,hedoesnottouch it. C,D andE in turn now have the same privilege. The last person totouchtheslicekeepsitashisshare.Anyonewhothinksthatthispersongotlessthan1/5 is naturally pleasedbecause itmeans, in his eyes, thatmore than4/5remains.Theremainderofthecake,includinganycut-offpieces,isnowdividedamong the remaining fourpersons in thesamemanner, thenamong three.Thefinal division is made by one person cutting and the other choosing. Theprocedureisclearlyapplicabletoanynumberofpersons.For a discussionof this andother solutions, see the section “GamesofFair

Division,” pages 363–368, inGames and Decisions, by R. Duncan Luce andHowardRaiffa,JohnWileyandSons,Inc.,1957.The problem of dividing a cake between n persons so that each person is

convinced he has his fair share has been the topic of many papers. Here are

three:“HowtoCutaCakeFairly,”byL.E.DubinsandE.H.Spanier,inAmerican

MathematicalMonthly,Vol.68,January1961,pages1–17.“Preferred Shares,” by Dominic Olivastro, in The Sciences, March/ April

1992,pages52–54.“AnEnvy-FreeCakeDivisionAlgorithm,”byStevenJ.BramsandAlanD.

Taylor,preprint.Morethanfiftyreferencesarecitedinthebibliography.

29.Thefirstsheetisfoldedasfollows.Holditfacedownsothatwhenyoulookdownonitthenumberedsquaresareinthisposition:

Foldtherighthalfontheleftsothat5goeson2,6on3,4on1and7on8.Fold the bottom half up so that 4 goes on 5 and 7 on 6. Now tuck 4 and 5between6and3,andfold1and2underthepacket.Thesecondsheetisfirstfoldedinhalfthelongway,thenumbersoutside,and

heldsothat4536isuppermost.Fold4on5.Therightendofthestrip(squares6and7)ispushedbetween1and4,thenbentaroundthefoldededgeof4sothat6and7gobetween8and5,and3and2gobetween1and4.Formore puzzles involving paper folding see “TheCombinatorics of Paper

Folding,” in my Wheels, Life and Other Mathematical Amusements (W. H.Freeman,1983).

30.Regardlessofhowmuchwineisinonebeakerandhowmuchwaterisintheother, and regardlessofhowmuch liquid is transferredbackand forth at eachstep(provideditisnotalloftheliquidinonebeaker),itisimpossibletoreachapointatwhichthepercentageofwineineachmixtureisthesame.Thiscanbeshown by a simple inductive argument. If beaker A contains a higherconcentrationofwine thanbeakerB, thena transfer fromAtoBwill leaveAwiththehigherconcentration.SimilarlyatransferfromBtoA—fromaweakertoastrongermixture—issuretoleaveBweaker.Sinceeverytransferisoneofthesetwocases, itfollowsthatbeakerAmustalwayscontainamixturewithahigherpercentageofwinethanB.Theonlywaytoequalizetheconcentrationsisbypouringallofonebeakerintotheother.

31.TodeterminethevalueofBrown’scheck,letxstandforthedollarsandyforthecents.Theproblemcannowbeexpressedbythefollowingequation:l00y+x

–5=2(l00x+y).Thisreducesto98y–199x–5,aDiophantineequationwithan infinite numberof integral solutions.A solutionby the standardmethodofcontinuedfractionsgivesasthelowestvaluesinpositiveintegers:x=31andy=63, making Browns check $31.63. This is a unique answer to the problembecause the next lowest values are:x = 129,y = 262,which fails tomeet therequirementthatybelessthan100.Thereisamuchsimplerapproachtotheproblemandmanyreaderswroteto

tellme about it. As before, let x stand for the dollars on the check, y for thecents.Afterbuyinghisnewspaper,Brownhas left2x+2y.Thechange thathehasleft,fromthexcentsgivenhimbythecashier,willbex–5.Weknowthatyislessthan100,butwedon’tknowyetwhetheritislessthan

50cents.Ifitislessthan50cents,wecanwritethefollowingequations:

2x=y2y=x–5

Ifyis50centsormore,thenBrownwillbeleftwithanamountofcents(2y)that is a dollar ormore.We therefore have tomodify the above equations bytaking100from2yandadding1to2x.Theequationsbecome:

2x+1=y2y–100=x–5

Each set of simultaneous equations is easily solved. The first set gives x aminusvalue,whichisruledout.Thesecondsetgivesthecorrectvalues.

32.Anumberofreaderssent“proofs”thatanobtusetrianglecannotbedissectedinto acute triangles, but of course it can. The illustration shows a seven-piecepatternthatappliestoanyobtusetriangle.

Itiseasytoseethatsevenisminimal.Theobtuseanglemustbedividedbyaline.This line cannot go all theway to the other side, for then itwould formanotherobtusetriangle,whichinturnwouldhavetobedissected,consequentlythe pattern for the large trianglewould not beminimal. The line dividing theobtuse angle must, therefore, terminate at a point inside the triangle. At thisvertex,atleastfivelinesmustmeet,otherwisetheanglesatthisvertexwouldnotallbeacute.Thiscreatestheinnerpentagonoffivetriangles,makingatotalofseventrianglesinall.WallaceManheimer,aBrooklynhighschoolteacheratthetime, gave this proof as his solution to problem E1406 in AmericanMathematical Monthly, November 1960, page 923. He also showed how toconstructthepatternforanyobtusetriangle.The question arises: Can any obtuse triangle be dissected into seven acute

isoscelestriangles?Theanswerisno.VernerE.Hoggatt,Jr.,andRussDenman(AmericanMathematicalMonthly,November1961,pages912–913)provedthateight such triangles are sufficient for all obtuse triangles, and Free Jamison(ibid., June-July 1962, pages 550–552) proved that eight are also necessary.Thesearticlescanbeconsultedfordetailsastoconditionsunderwhichlessthaneight-piece patterns are possible. A right triangle and an acute nonisoscelestrianglecaneachbecutintonineacuteisoscelestriangles,andanacuteisoscelestriangle can be cut into four congruent acute isosceles triangles similar to theoriginal.Asquare,Idiscovered,canbecut intoeightacutetrianglesasshown.If the

dissection has bilateral symmetry, points P and P’must liewithin the shadedareadeterminedbythefoursemicircles.DonaldL.Vanderpoolpointedoutinaletter that asymmetric distortions of the pattern are possible with point Panywhere outside the shaded area provided it remains outside the two largesemicircles.

About 25 readers sent proofs, with varying degrees of formality, that theeight-piece dissection is minimal. One, by Harry Lindgren, appeared inAustralianMathematics Teacher, Vol. 18, pages 14–15, 1962. His proof alsoshowsthatthepattern,asidefromtheshiftingofpointsPandP’asnotedabove,isunique.H.S.M.Coxeterpointedout thesurprisingfact that foranyrectangle,even

thoughitssidesdifferinlengthbyanarbitrarilysmallamount,thelinesegmentPP’canbeshiftedto thecenter togive thepatternbothhorizontalandverticalsymmetry.Free Jamison found in 1968 that a square can be divided into ten acute

isosceles triangles. SeeTheFibonacciQuarterly (December 1968) for a proofthatasquarecanbedissectedintoanynumberofacuteisoscelestrianglesequalorgreaterthan10.

33.Thevolumeofasphereis4π/3timesthecubeoftheradius.Itssurfaceis4πtimesthesquareoftheradius.Ifweexpressthemoon’sradiusin“lunars”andassumethatitssurfaceinsquarelunarsequalsitsvolumeincubiclunars,wecandetermine the length of the radius simply by equating the two formulas andsolvingforthevalueoftheradius.Picancelsoutonbothsides,andwefindthattheradiusisthreelunars.Themoon’sradiusis1,080miles,soalunarmustbe360miles.

34. Regardless of the number of slips involved in the game of Googol, the

probabilityof picking the slipwith the largest number (assuming that thebeststrategyisused)neverdropsbelow.367879.Thisisthereciprocalofe,andthelimitoftheprobabilityofwinningasthenumberofslipsapproachesinfinity.If there are ten slips (a convenient number to use in playing the game), the

probabilityofpickingthetopnumberis.398.Thestrategyistoturnthreeslips,note the largest number among them, then pick the next slip that exceeds thisnumber.Inthelongrunyoustandtowinabouttwooutofeveryfivegames.Whatfollowsisacompressedaccountofacompleteanalysisofthegameby

LeoMoserandJ.R.PounderoftheUniversityofAlberta.Letnbethenumberofslipsandp thenumber rejectedbeforepickinganumber larger thananyonthe/)slips.Numbertheslipsseriallyfrom1ton.Letk+1bethenumberoftheslipbearingthe largestnumber.Thetopnumberwillnotbechosenunlessk isequaltoorgreaterthanp(otherwiseitwillberejectedamongthefirstpslips),andthenonlyifthehighestnumberfrom1tokisalsothehighestnumberfrom1top(otherwisethisnumberwillbechosenbeforethetopnumberisreached).Theprobabilityoffindingthetopnumberincaseit isonthek+1slip isp/k,andtheprobabilitythatthetopnumberactuallyisonthek+1slipis1/n.Sincethelargestnumbercanbeononlyoneslip,wecanwritethefollowingformulafortheprobabilityoffindingit:

Givenavalueforn(thenumberofslips)wecandetermine/)(thenumbertoreject) by picking a value for p that gives the greatest value to the aboveexpression.Asnapproachesinfinity,p/napproachesl/e,soagoodestimateofpis simply the nearest integer to n/e. The general strategy, therefore, when thegame is played with n slips, is to let n/e numbers go by, then pick the nextnumberlargerthanthelargestnumberonthen/eslipspassedup.Thisassumes,ofcourse, thataplayerhasnoknowledgeof the rangeof the

numbers on the slips and therefore no basis for knowing whether a singlenumberishighorlowwithintherange.Ifonehassuchknowledge,theanalysisdoesnotapply.Forexample,ifthegameisplayedwiththenumbersontenone-dollarbills,andyourfirstdrawisabillwithanumberthatbeginswith9,yourbest strategy is to keep the bill. For similar reasons, the strategy in playingGoogolisnotstrictlyapplicabletotheunmarriedgirlproblem,asmanyreaderspointed out, because the girl presumably has a fair knowledge of the range invalue of her suitors, and has certain standards in mind. If the first man who

proposescomesveryclose toher ideal,wroteJosephP.Robinson,“shewouldhaverocksinherheadifshedidnotacceptatonce.”FoxandMarnieapparentlyhitindependentlyonaproblemthathadoccurred

to others a few years before. A number of readers said they had heard theproblembefore1958—onerecalledworkingonitin1955—butIwasunabletofindanypublishedreferencetoit.Theproblemofmaximizingthevalueoftheselected object (rather than the chance of getting the object of highest value)seemsfirst tohavebeenproposedbythefamousmathematicianArthurCayleyin1875. (SeeLeoMoser,“OnaProblemofCayley,” inScriptaMathematica,September-December1956,pages289–292.)

35.Let1bethelengthofthesquareofcadetsandalsothetimeittakesthemtomarchthislength.Theirspeedwillalsobe1.Letxbethetotaldistancetraveledbythedogandalsoitsspeed.Onthedog’sforwardtriphisspeedrelativetothecadetswillbex–1.Onthereturntriphisspeedrelativetothecadetswillbex+1. Each trip is a distance of 1 (relative to the cadets), and the two trips arecompletedinunittime,sothefollowingequationcanbewritten:

Thiscanbeexpressedas thequadratic:x2–2x–1=0, forwhichxhas the

positivevalueof .Multiplythisby50togetthefinalanswer:120.7+feet. Inotherwords, thedog travels a total distance equal to the lengthof thesquareofcadetsplusthatsamelengthtimesthesquarerootof2.Loyd’s version of the problem, in which the dog trots around the moving

square,canbeapproached inexactly the sameway. Iparaphraseaclear,briefsolutionsentbyRobertF.JacksonoftheComputingCenterattheUniversityofDelaware.Asbefore,let1bethesideofthesquareandalsothetimeittakesthecadets

togo50feet.Theirspeedwillthenalsobe1.Letxbethedistancetraveledbythe dog and also his speed. The dog’s speedwith respect to the speed of thesquarewillbex–1onhisforwardtrip, oneachofhistwotransversetrips,andx+1onhisbackwardtrip.Thecircuitiscompletedinunittime,sowecanwritethisequation:

Thiscanbeexpressedas thequartic equation:x4 –4x3 –2x2+4x+5=0.Onlyonepositiverealrootisnotextraneous:4.18112+.Wemultiplythisby50togetthedesiredanswer:209.056+feet.TheodoreW.Gibson,oftheUniversityofVirginia,foundthatthefirstform

oftheaboveequationcanbewrittenasfollows,simplybytakingthesquarerootofeachside:

whichisremarkablysimilartotheequationforthefirstversionoftheproblem.Manyreaderssentanalysesofvariationsofthisproblem:asquareformation

marching inadirectionparallel to thesquare’sdiagonal, formationsof regularpolygonswithmorethanfoursides,circularformations,rotatingformations,andsoon.ThomasJ.MeehanandDavidSalsburgeachpointedoutthattheproblemis the same as that of a destroyer making a square search pattern around amovingship,andshowedhoweasily itcouldbesolvedbyvectordiagramsonwhattheNavycallsa“maneuveringboard.”

36. The assumption that the “lady” is Jean Brown, the stenographer, quicklyleadstoacontradiction.Heropeningremarkbringsforthareplyfromthepersonwithblackhair,thereforeBrown’shaircannotbeblack.Italsocannotbebrown,forthenitwouldmatchhername.Thereforeitmustbewhite.Thisleavesbrownfor the color of Professor Black’s hair and black for ProfessorWhite. But astatementbythepersonwithblackhairpromptsanexclamationfromWhite,sotheycannotbethesameperson.It is necessary to assume, therefore, that Jean Brown is a man. Professor

White’shaircan’tbewhite(forthenitwouldmatchhisorhername),norcanitbeblackbecausehe(orshe)repliestotheblack-hairedperson.Thereforeitmustbe brown. If the lady’s hair isn’t brown, then ProfessorWhite is not a lady.Brownisaman,soProfessorBlackmustbethelady.Herhaircan’tbeblackorbrown,soshemustbeaplatinumblonde.

37.Sincethewindbooststheplane’sspeedfromAtoBandretardsitfromBtoA,one is tempted to suppose that these forcesbalanceeachother so that totaltravel timefor thecombinedflightswill remainthesame.This isnot thecase,because the timeduringwhich theplane’s speed isboosted is shorter than thetimeduringwhichitisretarded,sotheover-alleffectisoneofretardation.Thetotal travel time in a wind of constant speed and direction, regardless of the

speedordirection,isalwaysgreaterthaniftherewerenowind.

38.Letxbethenumberofhamstersoriginallypurchasedandalsothenumberofparakeets.Lety be thenumberofhamsters among the sevenunsoldpets.Thenumber of parakeets among the seven will then be 7 – y. The number ofhamsterssold(atapriceof$2.20each,whichisamarkupof10percentovercost)willbex–y,andthenumberofparakeetssold(at$1.10each)willbex–7+y.Thecostofthepetsistherefore2xdollarsforthehamstersandxdollarsfor

theparakeets—atotalof3xdollars.Thehamstersthatweresoldbrought2.2(x–3;)dollarsandtheparakeetssoldbrought1.1(x–74-y)dollars—atotalof3.3x–1.1y–7.7dollars.Wearetoldthatthesetwototalsareequal,soweequatethemandsimplifyto

obtainthefollowingDiophantineequationwithtwointegralunknowns:

3x=11y+77

Since x and y are positive integers and j is notmore than 7, it is a simplemattertotryeachoftheeightpossiblevalues(includingzero)forytodeterminewhichof themmakesxalsointegral.Thereareonlytwosuchvalues:5and2.Eachwould lead to a solution of the problemwere it not for the fact that theparakeets were bought in pairs. This eliminates 2 as a value for y because itwould give x (the number of parakeets purchased) the odd value of 33. Weconcludethereforethatyis5.Acompletepicturecannowbedrawn.Theshopownerbought44hamsters

and22pairsofparakeets,payingaltogether$132forthem.Hesold39hamstersand21pairsofparakeetsforatotalof$132.Thereremainedfivehamstersworth$11 retail and two parakeetsworth $2.20 retail—a combined value of $13.20,whichistheanswertotheproblem.

39. The illustration shows the finish of a drawn game ofHip. This beautiful,hard-to-find solution was first discovered by C. M. McLaury, a mathematicsstudent at the University of Oklahoma to whom I had communicated theproblembywayofRichardAndree,oneofhisprofessors.Two readers (William R. Jordan, Scotia, New York, and Donald L.

Vanderpool, Towanda, Pennsylvania) were able to show, by an exhaustiveenumeration of possibilities, that the solution is unique except for slightvariationsin thefourbordercells indicatedbyarrows.Eachcellmaybeeither

color,providedallfourarenotthesamecolor,butsinceeachplayerislimitedinthegametoeighteenpieces,twoofthesecellsmustbeonecolor,twotheothercolor. They are arranged here so that nomatter how the square is turned, thepatternisthesamewheninverted.

Theorder-6boardisthelargestonwhichadrawispossible.Thiswasprovedin1960byRobertI.Jewett,thenagraduatestudentattheUniversityofOregon.He was able to show that a draw is impossible on the order-7, and since allhighersquarescontainaseven-by-sevensubsquare,drawsareclearlyimpossibleonthemalso.Asaplayablegame,Hiponanorder-6boardisstrictlyforthesquares.David

H.Templeton,professorofchemistryattheUniversityofCalifornia’sLawrenceRadiationLaboratoryinBerkeley,pointedoutthatthesecondplayercanalwaysforceadrawbyplayinga simplesymmetrystrategy.Hecaneithermakeeachmovesothatitmatcheshisopponent’slastmovebyreflectionacrossaparallelbisectoroftheboard,orbya90-degreerotationabouttheboard’scenter.(Thelatterstrategycouldleadtothedrawdepicted.)Analternatestrategyistoplayinthe corresponding opposite cell on a line from the opponent’s last move andacrossthecenteroftheboard.Second-playerdrawstrategieswerealsosentbyAllan VV. Dickinson, Richmond Heights, Missouri, and Michael Merritt, astudentatTexasA.&M.College.Thesestrategiesapplytoalleven-orderfields,

and since no draws are possible on such fields higher than 6, the strategyguaranteesawinforthesecondplayeronalleven-orderboardsof8orhigher.Evenontheorder-6,areflectionstrategyacrossaparallelbisectorissuretowin,becausetheuniquedrawpatterndoesnothavethattypeofsymmetry.Symmetry play fails on odd-order fields because of the central cell. Since

nothing is known about strategies on odd-order boards, the order-7 is the bestfield for actual play. It cannot end in a draw, and no one at present knowswhetherthefirstorsecondplayerwinsifbothsidesplayrationally.In 1963 Walter W. Massie, a civil engineering student at Worcester

Polytechnic Institute, devised aHip-playingprogram for the IBM1620digitalcomputer,andwroteatermpaperaboutit.Theprogramallowsthecomputertoplay first or secondonany square fieldoforders4 through10.Thecomputertakes a random cell if it moves first. On other plays, it follows a reflectionstrategy except when a reflectedmove forms a square, then itmakes randomchoicesuntilitfindsasafecell.On all square fields of ordern, the number of different squares that can be

formedbyfourcellsis(n4–n2)/12.Thederivationofthisformula,aswellasaformulafor rectangularboards, isgiven inHarryLangman,PlayMathematics,Hafner,1962,pages36–37.As far as I know, no studies have beenmade of comparable “triangle-free”

coloringsontriangularlatticefields.

40.The locomotivecanswitch thepositionsofcarsAandB,andreturn to itsformerspot,in16operations:

1.Locomotivemovesright,hookstocarA.2.PullsAtobottom.3.PushesAtoleft,unhooks.4.Movesright.5.Makesaclockwisecirclethroughtunnel.6.PushesBtoleft.Allthreearehooked.7.PullsAandBtoright.8.PushesAandBtotop.AisunhookedfromB.9.PullsBtobottom.10.PushesBtoleft,unhooks.11.Circlescounterclockwisethroughtunnel.12.PushesAtobottom.13.Movesleft,hookstoB.

14.PullsBtoright.15.PushesBtotop,unhooks.16.Moveslefttooriginalposition.

Thisprocedurewilldothejobevenwhenthelocomotiveisnotpermittedtopullwithitsfrontend,providedthatatthestartthelocomotiveisplacedwithitsbacktowardthecars.HowardGrossman,NewYorkCity,andMoisesV.Gonzalez,Miami,Florida,

eachpointedout that if the lowersiding iseliminatedcompletely, theproblemcanstillbesolved,althoughtwoadditionalmovesarerequired,making18inall.Canthereaderdiscoverhowitisdone?Manydifferent train-switchingpuzzles canbe found in thepuzzlebooksby

SamLoyd,ErnestDudeneyandothers.Tworecentarticlesdealwiththistypeofpuzzle:A.K.Dewdney’sComputerRecreationscolumninScientificAmerican(June1987)and“ReversingTrains:ATurnoftheCenturySortingProblem,”byNancyAmatoetal.,JournalofAlgorithms,Vol.10,1989,pages413–428.

41.ThecuriousthingabouttheproblemoftheFlatzbeersignsis that it isnotnecessarytoknowthecarsspeedtodeterminethespacingofthesigns.Letxbethenumberofsignspassedinoneminute.Inanhourthecarwillpass60xsigns.Thespeedofthecar,wearetold,is10x:milesperhour.In10xmilesitwillpass60xsigns,soinonemileitwillpass60x/10x,or6,signs.Thesignsthereforeare1/6mile,or880feet,apart.

42.Acube,cutinhalfbyaplanethatpassesthroughthemidpointsofsixsidesasshown,producesacrosssectionthatisaregularhexagon.Ifthecubeishalfaninchontheside,thesideofthehexagonis inch.

Tocutatorussothatthecrosssectionconsistsoftwointersectingcircles,theplanemustpassthroughthecenterandbetangenttothetorusaboveandbelow,asshowninthesecondpicture(nextpage).Ifthetorusandholehavediametersof three inches and one inch, each circle of the section will clearly have adiameteroftwoinches.Thiswayofslicing,andthetwowaysdescribedearlier,aretheonlywaysto

sliceadoughnutso that thecrosssectionsarecircular.EverettA.Emerson, intheelectronicsdivisionofNationalCashRegister,Hawthorne,California,sentafullalgebraicproofthatthereisnofourthway.

43.The illustrationshowshowtoconstructastraight line thatbisectsboth theYin and the Yang. A simple proof is obtained by drawing the two brokensemicircles.CircleK’sdiameter ishalf thatof themonad; therefore its area isone-fourth that of themonad. Take regionG from this circle, addH, and theresultingregionisalsoone-fourththemonad’sarea.ItfollowsthatareaGequalsareaH,andofcoursehalfofGmustequalhalfofH.Thebisecting line takeshalfofGawayfromcircleK,butrestoresthesamearea(halfofH)tothecircle,sotheblackareabelowthebisectinglinemusthavethesameareaascircleK.Thesmallcircle’sareaisone-fourththelargecircle’sarea,thereforetheYinisbisected.ThesameargumentappliestotheYang.TheforegoingproofwasgivenbyHenryDudeneyinhisanswer toproblem

158,AmusementsinMathematics(ThomasNelson&Sons,1917;Doverreprint,

1958).AfteritappearedinScientificAmerican,fourreaders(A.E.Decae,F.J.Hooven, Charles W. Trigg and B. H. K. Willoughby) sent the followingalternative proof, which is much simpler. Draw a horizontal diameter of thesmallcircleK.Thesemicirclebelowthislinehasanareathatisclearly1/8thatof the largecircle.Abovethediameter isa45-degreesectorof the largecircle(boundedbythesmallcircle’shorizontaldiameterandthediagonalline)whichalso is obviously 1/8 the area of the large circle.Together, the semicircle andsectorhaveanareaof1/4thatofthelargecircle,thereforethediagonallinemustbisectbothYinandYang.ForwaysofbisectingtheYinandYangwithcurvedlines, the reader is referred to Dudeney’s problem, cited above, and Trigg’sarticle,“BisectionofYinandofYang,”inMathematicsMagazine,Vol.34,No.2,November-December1960,pages107–108.

The Yin-Yang symbol (called the T’ai-chi-t’u in China and the Tomoye inJapan) is usually drawnwith a small spot ofYin inside theYang and a smallspotofYanginsidetheYin.Thissymbolizesthefactthatthegreatdualitiesoflife are seldom pure; each usually contains a bit of the other. There is anextensive Oriental literature on the symbol. Sam Loyd, who bases severalpuzzlesonthefigure(SamLoyd’sCyclopediaofPuzzles,page26),calls it theGreatMonad.Theterm“monad”isrepeatedbyDudeney,andalsousedbyOlinD.WheelerinabookletentitledWonderland,publishedin1901bytheNorthernPacificRailway.Wheeler’sfirstchapterisdevotedtoahistoryofthetrademark,and is filled with curious information and color reproductions from Orientalsources.For more on the symbol, see Schuyler Cammann, “The Magic Square of

Three inOldChinese Philosophy andReligion,”History of Religions,Vol. 1,

No. 1, Summer 1961, pages 37–80, myNew Ambidextrous Universe (W. H.Freeman,1979),pages219–220,andGeorgeSarton,AHistoryofScience,Vol.1(HarvardUniversityPress,1952;Doverreprint,underthetitleAncientSciencethroughtheGoldenAgeofGreece,1993),page11.CarlGustavJungcitessomeEnglish references on the symbol in his introduction to the book of IChing(1929),andthereisabookcalledTheChineseMonad:ItsHistoryandMeaning,byWilhelmvonHohenzollern,thedateandpublisherofwhichIdonotknow.

44.Thereareprobablythreeblue-eyedJonessistersandfoursistersaltogether.Iftherearengirls,ofwhichbareblue-eyed,theprobabilitythattwochosenatrandomareblue-eyedis:

We are told that this probability is 1/2, so the problem is one of findingintegralvaluesforbandnthatwillgivetheaboveexpressionavalueof1/2.Thesmallestsuchvaluesaren=4,b=3.Thenexthighestvaluesaren=21,b=15,but it is extremelyunlikely that therewouldbe asmany as21 sisters, so foursisters,threeofthemblue-eyed,isthebestguess.

45.Therose-redcity’sageissevenbillionyears.Letxbethecity’spresentage;y, thepresentageofTime.Abillionyearsagothecitywouldhavebeenx–1billionyearsoldandabillionyearsfromnowTime’sagewillbey+1.Thedataintheproblempermittwosimpleequations:

Theseequationsgivex,thecity’spresentage,avalueofsevenbillionyears;and y, Time’s present age, a value of fourteen billion years. The problempresupposesa“BigBang”theoryofthecreationofthecosmos.

46.ThereisspaceonlytosuggesttheprocedurebywhichitcanbeshownthatWashington High won the high jump event in the trackmeet involving threeschools. Three different positive integers provide points for first, second andthirdplaceineachevent.Theintegerforfirstplacemustbeatleast3.Weknowthereareatleasttwoeventsinthetrackmeet,andthatLincolnHigh(whichwontheshot-put)hadafinalscoreof9,sotheintegerforfirstplacecannotbemore

than8.Canitbe8?No,becausethenonlytwoeventscouldtakeplaceandthereisnowaythatWashingtonHighcouldbuildupatotalof22points.Itcannotbe7becausethispermitsnomorethanthreeevents,andthreearestillnotsufficientto enable Washington High to reach a score of 22. Slightly more involvedargumentseliminate6,4and3astheintegerforfirstplace.Only5remainsasapossibility.If5isthevalueforfirstplace,theremustbeatleastfiveeventsinthemeet.

(FewereventsarenotsufficienttogiveWashingtonatotalof22,andmorethanfivewouldraiseLincoln’stotal tomorethan9.)Lincolnscored5fortheshot-put,soitsfourotherscoresmustbe1.Washingtoncannowreach22inonlytwoways: 4, 5, 5, 5, 3 or 2, 5, 5, 5, 5. The first is eliminated because it givesRoosevelt a score of 17, and we know that this score is 9. The remainingpossibility gives Roosevelt a correct final tally, so we have the uniquereconstructionofthescoringshowninthetable.

Washington High won all events except the shot-put, consequently it musthavewonthehighjump.Manyreaderssentshortersolutionsthantheonejustgiven.Tworeaders(Mrs.

Erlys Jedlicka, Saratoga, California, and Albert Zoch, a student at IllinoisInstituteofTechnology)noticedthattherewasashortcuttothesolutionbasedontheassumptionthattheproblemhadauniqueanswer.Mrs.Jedlickaputitthisway:

DearMr.Gardner:Did you know this problem can be solved without any calculation

whatever? The necessary clue is in the last paragraph. The solution to theintegerequationsmustindicatewithoutambiguitywhichschoolwonthehighjump.Thiscanonlybedoneifoneschoolhaswonalltheevents,notcountingtheshot-put;otherwisetheproblemcouldnotbesolvedwiththeinformationgiven, even after calculating the scoring and number of events. Since theschoolthatwontheshot-putwasnottheover-allwinner,itisobviousthatthe

over-allwinnerwontheremainingevents.HencewithoutcalculationitcanbesaidthatWashingtonHighwonthehighjump.

47. It isnotpossible for the termite topassonce through the26outsidecubesandenditsjourneyinthecenterone.Thisiseasilydemonstratedbyimaginingthat the cubes alternate in color like the cells of a three-dimensionalcheckerboard,orthesodiumandchlorineatomsinthecubicalcrystallatticeofordinarysalt.Thelargecubewillthenconsistof13cubesofonecolorand14oftheothercolor.Thetermite’spathisalwaysthroughcubesthatalternateincoloralongtheway;thereforeifthepathistoincludeall27cubes,itmustbeginandendwithacubebelongingtothesetof14.Thecentralcube,however,belongstothe13set;hencethedesiredpathisimpossible.The problem can be generalized as follows:A cube of even order (an even

numberofcellsontheside)hasthesamenumberofcellsofonecolorasithascellsoftheothercolor.Thereisnocentralcube,butcompletepathsmaystartonanycellandendonanycellofoppositecolor.Acubeofoddorderhasonemorecellofonecolor thantheother,soacompletepathmustbeginandendonthecolorthatisusedforthelargerset.Inodd-ordercubesoforders3,7,11,15,19.. . the central cell belongs to the smaller set, so it cannot be the end of anycompletepath.Inodd-ordercubesof1,5,9,13,17...thecentralcellbelongstothelargerset,soitcanbetheendofanypaththatstartsonacellofthesamecolor.No closed path, going through every unit cube, is possible on anyodd-ordercubebecauseoftheextracubeofonecolor.Many two-dimensional puzzles can be solved quickly by similar “parity

checks.” For example, it is not possible for a rook to start at one corner of achessboardandfollowapaththatcarriesitoncethrougheverysquareandendsonthesquareatthediagonallyoppositecorner.

48.Thedimeandpennypuzzlecanbesolvedinfourmovesasfollows.Coinsarenumberedfromlefttoright.1.Move3,4totherightof5butseparatedfrom5byagapequaltothewidth

oftwocoins.2.Move1,2totherightof3,4,withcoins4and1touching.3.Move4,1tothegapbetween5and3.4.Move5,4tothegapbetween3and2.

49.Ioriginallypublishedasolutionshowinghowthethreeslicesofbreadcouldbetoastedintwominutes.Fivereaderssurprisedmebycuttingthetimeto111

seconds.Ihadoverlookedthepossibilityofpartiallytoastingonesideofaslice,removingit,thenreturningitlatertocompletethetoasting.Seconds Operation

1–3 PutinsliceA.3–6 PutinB.6–18 Acompletes15secondsoftoastingononeside.18–21 RemoveA.21–23 PutinC.23–36 Bcompletestoastingononeside.36–39 RemoveB.39–42 PutinA,turned.42–54 ButterB.54–57 RemoveC.57–60 PutinB.60–72 ButterC.72–75 RemoveA.75–78 PutinC.78–90 ButterA.90–93 RemoveB.93–96 Put in A, turned to complete the toasting on its partially

toastedside.96–108 Acompletesitstoasting.108–111

RemoveC.

Allslicesarenowtoastedandbuttered,butsliceAisstillinthetoaster.Evenif Amust be removed to complete the entire operation, the time is only 114seconds.

50.Aman goes up amountain one day, down it another day. Is there a spotalong the path that he occupies at the same time of day on both trips? Thisproblem was called to my attention by psychologist Ray Hyman, of theUniversity of Oregon, who in turn found it in a monograph entitled “OnProblem-Solving,”bytheGermanGestaltpsychologistKarlDuncker.Dunckerwritesofbeingunabletosolveitandofobservingwithsatisfactionthatotherstowhomheputtheproblemhadthesamedifficulty.Thereareseveralwaystogo

about it,hecontinues,“butprobablynone is . . .moredrasticallyevident thanthe following. Let ascent and descent be divided between two persons on thesameday.Theymustmeet.Ergo....Withthis,fromanuncleardimconditionnoteasilysurveyable,thesituationhassuddenlybeenbroughtintofulldaylight.”

51.ImmanuelKantcalculatedtheexacttimeofhisarrivalhomeasfollows.Hehadwoundhisclockbeforeleaving,soaglanceatitsfacetoldhimtheamountoftimethathadelapsedduringhisabsence.Fromthishesubtractedthelengthoftime spent with Schmidt (having checked Schmidt’s hallway clock when hearrivedandagainwhenheleft).Thisgavehimthetotaltimespentinwalking.Sincehe returnedalong the same route, at the samespeed,hehalved the totalwalkingtimetoobtainthelengthoftimeittookhimtowalkhome.Thisaddedto the time of his departure from Schmidt’s house gave him the time of hisarrivalhome.

52.Thefirststepistolistinordertheprobabilityvaluesfortheninecards:1/45,2/45,3/45....Thetwolowestvaluesarecombinedtoformanewelement:1/45plus 2/45 equals 3/45. In other words, the probability that the chosen card iseitheranaceordeuceis3/45.Therearenoweightelements:theace-deuceset,thethree,thefour,andsoonuptonine.Againthetwolowestprobabilitiesarecombined:theace-deucevalueof3/45andthe3/45probabilitythatthecardisathree.Thisnewelement,consistingofaces,deucesandthrees,hasaprobabilityvalue of 6/45. This is greater than the values for either the fours or fives, sowhenthetwolowestvaluesarecombinedagain,wemustpairthefoursandfivestoobtainanelementwiththevalueof9/45.Thisprocedureofpairingthelowestelementsiscontinueduntilonlyoneelementremains.Itwillhavetheprobabilityvalueof45/45,or1.Thischartshowshowtheelementsarecombined.Thestrategyforminimizing

thenumberofquestionsistotakethesepairingsinreverseorder.Thusthefirstquestion could be: Is the card in the set of fours, fives and nines? If not, youknowitisintheothersetsoyouasknext:Isitasevenoreight?Andsoonuntilthecardisguessed.

Note that if thecardshouldbeanaceordeuce itwill takefivequestions topinpoint it. A binary strategy, of simply dividing the elements as nearly aspossible into halves for each question, will ensure that no more than fourquestions need be asked, and you might even guess the card in three.Nevertheless, the previously described procedure will give a slightly lowerexpected minimum number of questions in the long run; in fact, the lowestpossible.Inthiscase,theminimumnumberisthree.Theminimumiscomputedasfollows:Fivequestionsareneededifthecardis

an ace. Five are also needed if the card is a deuce, but there are two deuces,makingtenquestionsinall.Similarly,thethreethreescallforthreetimesfour,or 12, questions. The total number of questions for all 45 cards is 135, or anaverageofthreequestionspercard.This strategy was first discovered by David A. Huffman, an electrical

engineeratM.I.T.,whilehewasagraduatestudentthere.It isexplainedinhispaper “A Method for the Construction of Minimum-Redundancy Codes,”Proceedings of the Institute of Radio Engineers, Vol. 40, pages 1098–1101,September1952.ItwaslaterrediscoveredbySethZimmerman,whodescribeditin his article on “An Optimal Search Procedure,” American MathematicalMonthly, Vol. 66, pages 690–693, October 1959. A good nontechnicalexposition of the procedurewill be found in JohnR. Pierce,Symbols, Signals

andNoise(Harper&Brothers,1961),beginningonpage94.TheeminentmathematicianStanislawUlam,inhisbiographyAdventuresofa

Mathematician (Scribner’s, 1976, page 281), suggested adding the followingrule to theTwentyQuestionsgame.Thepersonwhoanswers ispermittedonelie.Whatis theminimumnumberofquestionsrequiredtodetermineanumberbetween1andonemillion?Whatifheliestwice?Thegeneralcaseisfarfromsolved.Iftherearenolies,theanswerisofcourse

20. If just one lies, 25questions suffice.ThiswasprovedbyAndrzejPelc, in“Solution of Ulam’s Problem on Searching with a Lie,” in Journal ofCombinatorialTheory (SeriesA),Vol. 44, pages 129–140, January1987.Theauthor also gives an algorithm for finding the minimum number of neededquestionsforidentifyinganynumberbetween1andn.Adifferentproofofthe25minimumisgivenbyIvanNivenin“CodingTheoryAppliedtoaProblemofUlam,”MathematicsMagazine,Vol.61,pages275–281,December1988.When two lies are allowed, the answer of 29 questions was established by

Jurek Czyzowicz, Andrzej Pelc and Daniel Mundici in the Journal ofCombinatorialTheory (SeriesA),Vol.49,pages384–388,November1988.Inthe same journal (Vol. 52, pages 62–76, September 1989), the same authorssolved the more general case of two lies and any number between 1 and 2n.WojciechGuziki, ibid.,Vol.54,pages1–19,1990,completelydisposedof thetwo-liecaseforanynumberbetween1andn.Howabout three lies?Thishasbeen answeredonly fornumbersbetween1

andonemillion.ThesolutionisgivenbyAlbertoNegroandMatteoSereno,inthesamejournal,Vol.59,1992.Itis33questions,andthat’snolie.Thefour-liecaseremainsunsolvedevenfornumbersinthe1toonemillion

range.Ofcourseifoneisallowedtolieeverytime,thereisnowaytoguessthenumber.Ulam’sproblemiscloselyrelatedtoerror-correctingcodingtheory.IanStewartsummarizedthelatestresultsin“HowtoPlayTwentyQuestionswithaLiar,”inNewScientist(October17,1992),andBarryCipradidthesamein“AllTheoremsGreatandSmall,”SIAMNews(July1992,page28).

53.Inthechessproblemwhitecanavoidcheckmatingblackonlybymovinghisrookfoursquarestothewest.Thischeckstheblackking,butblackisnowfreetocapturethecheckingbishopwithhisrook.When this problem appeared in Scientific American, dozens of readers

complainedthatthepositionshownisnotapossibleonebecausetherearetwowhitebishopsonthesamecolorsquares.Theyforgotthatapawnonthelastrowcanbe exchanged for anypiece, not just thequeen.Either of the twomissing

whitepawnscouldhavebeenpromotedtoasecondbishop.Therehavebeenmanygamesbymasters inwhichpawnswerepromoted to

knights. Promotions to bishops are admittedly rare, yet one can imaginesituationsinwhichitwouldbedesirable.Forinstance,toavoidstalematingtheopponent.Orwhitemayseethathecanuseeitheranewqueenoranewbishopinasubtlecheckmate.Ifhecallsforaqueen,itwillbetakenbyablackrook,inturncapturedbyawhiteknight.But ifwhite calls for abishop,blackmaybereluctanttotradearookforbishopandsoletthebishopremain.

54. The seven varieties of convex hexahedrons, with topologically distinctskeletons,areshownintheillustration(nextpage).Iknowofnosimplewaytoprovethattherearenoothers.AninformalproofisgivenbyJohnMcClellaninhisarticleon“TheHexahedraProblem,”RecreationalMathematicsMagazine,No.4,August1961,pages34–40.There are 34 topologically distinct convex heptahedra, 257 octahedra and

2,6069-hedra.Therearethreenonconvex(concaveorre-entrant)hexahedra,26nonconvexheptahedraand277nonconvexoctahedra.See thefollowingpapersbyP.J.Federico:“EnumerationofPolyhedra:TheNumberof9-Hedra,”JournalofCombinatorialTheory,Vol.7,September1969,pages155–161;“Polyhedrawith4to8Faces,”GeometriaDedicata,Vol.3,1975,pages469–481;and“TheNumberofPolyhedra,“PhilipsResearchReports,Vol.30,1975,pages220–231.

A formula for calculating the number of topologically distinct convexpolyhedra,giventhenumberoffaces,remainsundiscovered.PaulR.BurnettcalledmyattentiontotheOldTestamentverseZechariah3:9.

InamoderntranslationbyJ.M.PowisSmithitreads:“ForbeholdthestonewhichIhavesetbeforeJoshua;uponasinglestonewith

sevenfacetsIwillengraveitsinscription.”An outline of a formal proof that there are just seven distinct convex

hexahedra isgiven in“Euler’sFormula forPolyhedraandRelatedTopics,”byDonald Crowe, in Excursions into Mathematics, by Anatole Beck, MichaelBleicherandDonaldCrowe(Worth,1969,pages29–30).

55.

55.

56.Itisonlynecessarytocutthethreelinksofonepiece.Theycanthenbeusedtojointheremainingthreepiecesintothecircularbracelet.

57.

58.Continuethedealbytakingcardsfromthebottomofthepacketofundealtcards,dealingfirsttoyourself,thencounterclockwisearoundthetable.

59.Salwinsagain.Inthefirstracesheran100yardsinthetimeittookSaultorun90.Therefore,inthesecondrace,afterSaulhasgone90yards,Salwillhavegone 100, so shewill be alongside him.Bothwill have 10more yards to go.SinceSalisthefasterrunner,shewillfinishbeforeSaul.

60.

60.

61.Thesymbolsarethenumerals1,2,3,4,5,6,7shownalongsidetheirmirrorreflections.Thenextsymbol,therefore,isthedouble8,asshownatthefarright,below.

62.Thefigureiscutintocongruenthalveslikethis:

63. Two weighings will do the job. Divide the nine balls into three sets oftriplets.Weigh one triplet against another. If a pan goes down you know theheavyballisamongthethreeonthatpan.Pickanytwooftheseballsandweighone against theother. If one sidegoesdown,youhave found theball. If theybalance,theheavyballmustbetheoneyouputaside.Ineithercase,youhavefoundtheoddballintwoweighings.Supposethetwotripletsbalanceonthefirstweighing.Youknowthenthatthe

heavyballisintheremainingtriplet.Asdescribedabove,theheavierballofthistripletiseasilyidentifiedbyweighinganyballofthetripletagainstanyother.

64.Crossouteveryother letter, startingwithN.ThiseliminatesNINELETTERS,leavingASINGLEWORD.AfterthisanswerappearedinGames,DonDwyer,Jr.,senttothemagazinea

secondsolution.CrossouttheninelettersAEILNRSTWeverytimetheyappear,toleavethewordGOD.

65.

66.Asthepicturesbelowshow,thefoldedletterisanupside-downandturned-overF.

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