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PROBLEM SOLVING VIA THE AMC - amtt.com.au

The AusT r AliA n MATheMATics TrusT enrichMenT series PROBLEM SOLVING VIA THE AMC W ATKINSW ATKINS5 PROBLEM SOLVINGVIA THE AMCThis book consists of a development of techniques for SOLVING approximately 150 problems which have been set in the Australian Mathematics Competition for the Westpac Awards (AMC). These problems have been selected from the topics: Geometry, Motion, diophantine Equations and Counting Techniques. Students find these areas difficult and in the case of the latter two, the topics are not often covered in standard school AMC is one of the largest competitions in the world with around 490 000 entries yearly or approximately one third of all Australian secondary school Atkins graduated with a BA from Sydney University and an MA from Macquarie University. Warren worked as a Mathematics teacher in NSW high schools, then as a Senior Lecturer in the Faculty of Education at the University of Canberra.

Problem Solving via the AMC Chapter 1 Diophantine Equations Diophantine equations are named after the prominent Greek math- ematician Diophantus of Alexandria (c. 275) who published three

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Transcription of PROBLEM SOLVING VIA THE AMC - amtt.com.au

1 The AusT r AliA n MATheMATics TrusT enrichMenT series PROBLEM SOLVING VIA THE AMC W ATKINSW ATKINS5 PROBLEM SOLVINGVIA THE AMCThis book consists of a development of techniques for SOLVING approximately 150 problems which have been set in the Australian Mathematics Competition for the Westpac Awards (AMC). These problems have been selected from the topics: Geometry, Motion, diophantine Equations and Counting Techniques. Students find these areas difficult and in the case of the latter two, the topics are not often covered in standard school AMC is one of the largest competitions in the world with around 490 000 entries yearly or approximately one third of all Australian secondary school Atkins graduated with a BA from Sydney University and an MA from Macquarie University. Warren worked as a Mathematics teacher in NSW high schools, then as a Senior Lecturer in the Faculty of Education at the University of Canberra.

2 He is now retired but still holds the position of Chairman of the Problems Committee for the AMC and is Editor of Mathematics Competition, the journal of the World Federation of National Mathematics 978-1-876420-07-9 AMT PublishingPublished byAMT PublishingAustralian Mathematics TrustUniversity of Canberra ACT 2601 AUSTRALIAC opyright 1992 AMT Publishing1st Reprint 19962nd Reprint 19993rd Reprint 20034th Reprint 2007 Telephone: +61 2 6201 Limited ACN 083 950 341 National Library of Australia Card Number and ISSNA ustralian Mathematics Trust Enrichment Series ISSN 1326-0170 PROBLEM SOLVING Via the AMCISBN 978-1-876420-07-9 FOREWORD 3 PREFACE 5 1. diophantine EQUATIONS 9 Solutions 21 2. COUNTING TECHNIQUES 31 Solutions 64 3. WHEN THINGS MOVE 89 Solutions 112 4.

3 GEOMETRY 129 Solutions 179 REFERENCES 213 CONTENTSP roblem SOLVING via the AMCF orewardWith over half a million 1992 entries, the Australian MathematicsCompetition for the Westpac Awards is one of the world s greatmathematics competitions. There are many reasons for this. Certainlya significant factor in its extraordinary success has been its annualcollection of motivating and exciting annual infusion ofrich ideas into thousands of schools in Australia, New Zealand and theSouth West Pacific. These questions are intelligent, well put and havethe property of being challenging and questions comeas close to making the point as can be done. said the eminent Americanmathematician, Paul with two important basic ingredients, a rich source of beautiful ques-tions and the world s largest database of achievement in mathematics,the author has taken up the challenge to counter the claim that perhapscertain topics should not be included in the school curriculum as sixteenyear old students achievement is no better than that of thirteen year oldstudents.

4 The author has the conviction that this lack of improvement inachievement is mainly due to the lack of such experiences in the presentschool a result he has produced a book which will provide what I considerwill be a great resource for the enrichment of mathematics learning insecondary schools everywhere. I congratulate the author, Warren Atkins,for a splendid book and I wish him every success with its J. OHalloran,Executive Director,Australian Mathematics Competition,September 1992.**3 PROBLEM SOLVING via the AMCF orewardWith over half a million 1992 entries, the Australian MathematicsCompetition for the Westpac Awards is one of the world s greatmathematics competitions. There are many reasons for this. Certainlya significant factor in its extraordinary success has been its annualcollection of motivating and exciting annual infusion ofrich ideas into thousands of schools in Australia, New Zealand and theSouth West Pacific.

5 These questions are intelligent, well put and havethe property of being challenging and questions comeas close to making the point as can be done. said the eminent Americanmathematician, Paul with two important basic ingredients, a rich source of beautiful ques-tions and the world s largest database of achievement in mathematics,the author has taken up the challenge to counter the claim that perhapscertain topics should not be included in the school curriculum as sixteenyear old students achievement is no better than that of thirteen year oldstudents. The author has the conviction that this lack of improvement inachievement is mainly due to the lack of such experiences in the presentschool a result he has produced a book which will provide what I considerwill be a great resource for the enrichment of mathematics learning insecondary schools everywhere. I congratulate the author, Warren Atkins,for a splendid book and I wish him every success with its J.

6 OHalloran,Executive Director,Australian Mathematics Competition,September 1992.**3 PROBLEM SOLVING via the ,CountingTechniques,MotionandRatesofChan ge, , ,whoseoriginwasduetothefore-sight,energy andadministrativeskillsofmycolleaguePete rO HalloranoftheUniversityofCanberra,begani n1978andhasexpandedsuchthatin1992itisone ofthelargestintheworldwithover500000stud ententries, ,Juniorforyears7and8, ,inthreesectionsof10questions, ,itismyviewthatthelackofimprovementinmos toftheseskillareasisnotduetotheinappropr iatenessoftheskillfortheparticularageorg radelevel,butratherthattheappropriatepro blemsolvingskillsarenotemphasised5 PROBLEM SOLVING via the ,GeometryandWhen Things Move( ),astheywereperceivedtobedifficultbymost students, diophantine EquationsandCounting Techniquesasthesetopicswerealsoseentobed ifficultand,inaddition, , ,PeterO Halloran,RobinThornelyandJohnCarty,fello wmembersoftheProblemsCommittee,andGraham Pollard,TanyaFordandSallyBakkeroftheAust ralianMathematicsCompetition, ,oftheAustralianMathematicsTrust, Atkins,Faculty of Education,University of Canberra,March SOLVING via the AMCC hapter 1 diophantine EquationsDiophantineequationsarenamedaft ertheprominentGreekmath-ematicianDiophan tusofAlexandria( )whopublishedthreeworks,mainlypertaining totheareaswenowcallalgebraandtheso-lutio n(s) ,probablydatingfromthe5thcentury.

7 Hisboyhoodlasted16ofhislife,hisbeardgrew after112more,after17morehemarried,5years laterhissonwasborn,thesonlivedtohalfhisf ather sage, (seebelow), equationisusedtorefertoanyequation,usual lyinseveralvariables,whicharisesinaprobl emwherethesolutionsarerequiredtobeintege rs(positiveintegersinmanycases), +y= , ( ),wehavethesixsolutionsfor(x,y):(0,5);(1 ,4);(2,3);(3,2);(4,1)and(5,0).Notealsoth atthepairofvaluesx=0andy=5,forexample, ,theprocedureisusually9 PROBLEM SOLVING via the ,forexample, ,however,developamethod, (1985J23): Findthesmallestpositiveintegerwhich,when dividedby6,givesaremainderof1andwhendivi dedby11, ,itisoftheform6m+ ,7,13,19,25,31,37,43,49,55,61,..Similarl y,thepositiveintegerswhichhavearemainder of6whendividedby11are6,17,28,39,50,61,.. :ThiscorrespondstothesolutionoftheDiopha ntineequation6m+1=11n+ 11n=5whichhasmanysolutions,withm=10andn= 5correspondingtotheabovesolutionof61.

8 [1985 Percentage Correct: Junior 26%]10 PROBLEM SOLVING via the AMCN owtryD2(1979I27,S21): Whatistheleastpositiveintegerwhichhasrem aindersof1,1and5whendividedby3,5,and7res pectively?[Solution Page 21]Analysing diophantine (1982J26,I21): Ifxandyarepositiveintegersandx+y+xy=34,f indx+ 1:Lookingatthelefthandsideoftheequationx +y+xy=34,wemayseethatx+y+xy=(x+1)(y+1) 1sothattheequationmaybewrittenas(x+1)(y+ 1)=35 Now,sincexandyareintegers,x+1andy+1areal sointegers, ,5and7(otherthan35and1,whichisnotpossibl einthiscase).Thistellsusthatx+1=7y+1=5}o rx+1=5y+1=7}11 PROBLEM SOLVING via the AMCW hilewecannotdifferentiatebetweenthetwoca sesanddeterminetheactualvaluesforxandy,w ecansayfromthesymmetry(orbyaddingtheequa tionsineithercase)thatx+y=6+4= 2:Ifwedidrecognisethefactorisationgivena bove, , , +y+xy=34x+xy=34 yx(1+y)=34 yx=34 y1+yRememberingthatxandyareintegral, x332323314305=6296287=4278 Therearenomoreintegralvaluesuntilxbecome snegative,hencex+y=6+4(or4+6)= <5,wecouldhavestoppedattheentryfory=5in1 2 PROBLEM SOLVING via the AMCthetable,astheothervaluesmaybeobtaine dbyinterchangingthexandthey.

9 [1982 Percentage Correct: Junior 31%, Intermediate 54%]AndconsiderD4(1984J22,I20,S12): ,20cand50ccoinsandthetotalvalueofthecoin sis$ ,howmany10ccoinshasshe? (x+y) $5=500c,wethenobtaintheequation50x+20y+1 0(20 (x+y))=5005x+2y+20 x y=504x+y=30 HerewehaveatypicalDiophantineequation,wh ereweareonlyinterestedinpositiveintegers olutionsasxandyarerepresentingnumbersoft hings(coinsinthiscase).Expressingyinterm sofx,wehavey=30 4xandrememberingthatthereare20coinsinall , ,2and3,asasubstitutionshowsthatthenumber ofcoinstomakeup$ SOLVING via the AMCL istingthepossibilities,weget50c(x)20c(y) 10c4142 51056687211 Again,wecanseetherearenosolutionsforx> $5with20coinsthatare50c, ,inthisproblem, givesustheonlysolutionof: four50c,fourteen20candtwo10ccoins.[1984 Percentage Correct: Junior 22%, Intermediate 28%, Senior 45%]NowtryD5(1991J27I22): Ashopbuys40pensofthreedifferenttypesatac ostof$ ,$1and$5each,andtherearemore$1pensthan$5 pens,howmany25cpenswerebought?

10 [Solution Page 22]AndtryD6(1984I29): ArticlesX, $ $1eachandarticleZcosts$ $ [Solution Page 23]Further SOLVING via the AMCC onsiderD7(1983I29,S27): Oneofthesolutionstotheequation19x+83y=19 83inpositiveintegersxandyisobviouslygive nby(x,y)=(100,1).Itturnsoutthatthereisex actlyonemorepairofpositiveintegers(x,y) ,sometimesshorter,inparticularcases(onew illbedevelopedlaterinAlternative 2forthisquestion), : Alternative 1:Fromthegivenequation19x+83y=1983weget1 9x=1983 83yx=1983 83y19=104 4y+7 4y+uwhereumustbeintegralandu=7 7y19andhence7y+19u=7 ThislastequationisanotherDiophantineequa tion,similartothefirst.


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