[PDF] Thèmes Problème A (3e/4e année) Problème de la semaine

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Sens du nombre 8N9

Géométrie 8G9

Algèbre 8A9

Gestion des données 8D9

Pensée computationnelle 8C9

ThèmesProblème A 23

e o4 e année4 *Les problèmes dans ce livret sont organisés par thème. Un problème peut apparaître dans plusiers thèmes. (Cliquer sur le nom du thème ci-dessus pour sauter à cette sectio n)

Problème de la semaineProblèmes et solutions2020 - 2021(solutions disponibles en anglais seulement)

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Sens du nombre (N)

Problem ofthe Week

Problem Aand Solution

CupcakeCaper Problem

Mei Xingis having apartyand isdecorating 20cupcakes forher friends.There isi ci ngonall of thecup cakes,andshelinesthem upto addextra decoration s. There isnothing buticing onthe rstcup cake. Thereis ac herryonthesecondcupcak e. There arec hocolatechipsonthe thirdcupcake.Thereare sprinkleson thefourth cupcake.

In fact,she putsa cherry onev erysecondcupcak e.

She alsoputs cho colatechipsonevery thirdcupcake.

She alsoputs sprinkleson every fourthcup cake.

A) Howmany cupcakesha venoextrade corationsonthem? B) Doan yofthecup cakes getall threedecorations?Justifyyouransw er.

Solution

One waytosolve thispr oblemisto make achartkeepingtrac kof thedecorations cupcake1234567891011121314151617181920

cherryXXXXXXXXXX ch.c hipsXXXXXX sprinklesXXXXX A) Fromthisc hartw ecanseethere areatotalof 7cup cakes thatha ve noextra decorations. B) Wecanalso seethat thereis onlyone cupcak ethat hasall threeextra decorations.

Teacher"sNotes

It wasimportan tinthedescriptionofthis problemto includet hefirst bullet pointwhenexplaining thedecorating strategy. Withoutthat startingp oint,the solution isactually unknown. Ifareonlygiv enthe factthat cherries areplaced on everysecondcup cake, withoutindicatinghowthepattern starts,our ch art could looklike this:cupcake1234567891011121314151617181920 cherryXXXXXXXXXX ch.c hipsXXXXXX sprinklesXXXXX In thissolution, onlythree ofthe cupcak esha ve nodecorations,whichis different from theansw erweexp ected.Ingeneral,when wearedescribing asituation that is repeatingsomepattern,providing aclear initialsetup iscrucial.Withoutthat precision, theresults may beunpredictable. Another thingto noticein thec hartof theoriginal solution isthat allof the cupcakesthatha ve noextradecorationsarelab elledwith aprimenum ber (with the exceptionof thefirst one).Note that1 isnot aprime numb er.A natural question toask would be,ifw eextendedtheproblem to alargernum ber of cupcakes,willall cupcak es(after thefirstone)thatha ve nodecorations be labelledwit hprimenum bers? Theanswerisno.Forexample, we wouldfindout prettyquic klythatthe25th cupcak ew ouldha venoextradecorations. However,wecan predictthatthe cupcak eslab elledwith anynum berthatis coprimewith 2and coprimewith 3will notha ve any extradecorations. Tosay twonum bersarecoprime,istosay thatthey donot hav ean ypositive factorsin common exceptfor 1.F orexample, 25isnota primen umb er,and itsp ositive factors are1, 5,and 25.The positiv efactors of2 are1and2,andthe posit ive factors of3are 1and 3.So 2and 25are coprimeand 3and 25a re coprime.Note that wedonot hav eto checkifthen umb eriscoprimewith4,since an umb erwill becoprime with4 exactlywhen itis coprimew ith2. Thatis, ifthe num beris coprime with2, thenit does notha veafactor of2, andsocannothave afactor of

2 or4 (thefactors of4). Ifa num ber isnot coprimewith 2,thenithasafactorof

2, whichautomaticallymeans itshares afactor with4 andso isnot coprime

with 4. Diana écritle numéro 1danslerectangle duhaut dutriangle. Elleécrit le numéro1 dansle premieret dernierrectangle dec haqueautre rangée.P ourles

autres rectangles,les numéros inscritsserontdéterminés encalculan tlasommedes deuxrectangles dela rangéesup érieurequi touchen tceluiàremplir. Par exemple, len uméroinscritdansle rectangled umilieu dela troisièmerangée serala sommedes numéros danslesdeuxrect angles dela deuxièmerangée.

Remplis lesrectangles dutriangle deDiana enresp ectant lesrèglemen tsci-dessus. ?????sAlgèbr?, S???? ??mbr?

Problem ofthe Week

Problem Aand Solution

NumberedTriangle

Problem

Diana hasoutlined thefollo wingtria ngleformedof boxes: Diana llsin thetop bo xof thetrianglewitha1.The rstb ox andlast bo xof theremaining rowsof thetriangle arelled witha 1.The restof theb oxes ineac hro ware lledwith the sum ofthe num bersinthetwo boxesin thero wabovethat touchthetopside ofth eb oxDiana is lling.F orexample,themiddle bo xo fthe thirdrowislledwith thesumofthe num bers in the twoboxes inthesecondrow. What doesthetriangle look like afterDianaha slledinall ofthe num bersaccordingtothe rules describedabo ve? Can youndan ypatterns inthenishedtrian gle? Sharethese patternswith othersin your class. Not printingthispage? Youcan lli nthe boxeson ourinteractiveworksheet .

Solution

Here isthe completedtriangle:

Teacher"sNotes

This problemis anexploration ofPascal'sT rianglewhichis namedfor Frenc h mathematician BlaiseP ascal(1623-1 662).There arelots ofpatternstob e found inthis structure.It iseasier todescrib eman yof thepatterns intermsof rowsofthe triangle,in particularw erefer tothe topro wofthetriangle asRow 0 . Hereare afew of thepatterns: The triangleis symmetricalong av erticalline ofsymmetry .Inotherw ords, the leftside matches thetheright side. Starting witha 1in Row1 and movingalongthe diagonal,w esee the whole numbersinorder:1;2;3;4;5;6;7;8. Starting witha 1in Row2 and movingalongthe diagonal,w esee the differences betweenadjacent numbersincreasingby1 ateachstep: (31) =2 ;(63) =3 ;(106) =4 ;(1510) =5 ;(2115) =6 ;(2821) =7 .

The sumof then umb ersinarowistwice thesum ofthe numbers inthe rowabove.Forexample, thesumofthe num bers inRow5 is

1 +5 +10 +10 +5 +1 =32 , andthe sumof the num bersinRow4 is

1 +4 +6 +4 +1 =16 .

In general,the sumo fthe numbers inthe nthrowis 2n.

If youstartat any 1and movealong thediagonal which isnotadiagonalof1s, andthen change directionforthelast num ber, thenthe sumof theentriesalongthe diagonalwill equalthe lastn umb er.F orexample, startingwith a1 inRow3 :1 +4 +10 +20 =35 . Thisis oftenreferred toas the

HockeyStick Identity

, becauseify oucircle thenumb ersin thispath, the shapewill look somewhatlike aho ckeystick. In Algebra,the num bersinthenthrowpro videthecoecien tsfor the expansion ofa binomialexpression raisedto thenthpower.Forexample, x+y)3= 1x3+ 3x2y+ 3xy2+ 1y3. The listab oveisnotcomprehensive,there areman yother patternsinPascal"s Triangle.Hop efullyyourstuden tswereable tofindsomeothers!

Problem ofthe Week

Problem Aand Solution

Collecting CardsProblem

Mitig lovestoplay hoc key.He alsolovestocollecthockey cards.He has169 hockeycardsin total. Mitig'sy oungercousinAkialso lov esto play hockey,and she isjustbeginning her collection ofho ckeycards.Shestarts with9. If Mitiggiv esAki8cards ada y, willthey ever havethesamen umb erof cardsintheir collections? If theydo endup withthe samen umb er ofcards atsomepoint,howmanycards does each haveandho wman ydayswi llittake forthisto happen?

Solution

One waytosolve thisproblem wouldbe tocreate atable thatkeepstrac kofhow manycards Mitigand Akiha ve each day.Day0 representstheda ybeforeMitig starts givingcards toAli. Each day ,Mitig'stotalwill decreaseby8 andAki's total willincreaseby8. DayMitig's TotalAki's Total 01699

116117

215325

314533

413741

512949

612157

711365

810573

99781

108989

So, onthe 10thda y, MitigandAkieachha ve 89ho ck eycards. Another waytosolve thisproblem istoseet hat Mitigand Akiha ve atotal of

169 +9 =178 cards. Theywill hav ethesamenum ber ofcardswheneac hhas

half ofthe total:1782 =89 cards. SoAki needs899 =8 0more cardsto havehalfof them.If Mitig gives 8ca rdseach day,t hatwill take

808 =10 days.S incethereis noremainderinthis division,they willha ve the

same numberofcardsonthe 10thda y.

Teacher"sNotes

If thisquestion was askedina highschool mathematicsclass, students might use a verydifferent approachtofind theanswer. Eachp erson"scollectiontotalcould be described witha mathematicalequation that describesthe relationshipbet weenthen umberofdaysthatha ve passedand the totaln umberofhockey cardseac hpersonhas.

Letxrepresentthe num berofdaysoftrading.

Letyrepresentthe num berofhockeycards.

ForMitig, theequation showin ghis numberofho ck eycardsovertimeis: y= 1698x ForAki, theequation showin gher numberofho ck eycardsovertimeis: y= 9+ 8x Wecan draw straightlines onagrap hthat represent theseequations. Weknow that anytw olinesthatarenotparallel willin tersectat ap oint. Solvingt his problem isthe samea sfinding thepoint ofin tersectionof thesetwolines.dndGdadQdeddendeGdeadeQdY X en G ra sQ ied eeen e

Aki's Collection

Point of

intersection

Mitig's CollectionIf wepreciselydra weac hoftheselines, wecannd thep oint ofin tersectiongraphically.W ecouldals ouse algebraictechniquesthatwould usethe equationsto solvetheproblem. Inmathematics thereare oftenma ny wa ysto solve thesame problem.

par paquetpar paquet crayons100,50 $ eaces121,00 $ cartables255,00 $ feuilles lignées1001,50 $ marquers eaçables44,00 $ crayonsdecouleur 241,00 $ M. Camerona 20é lèv esdanssaclasse.Chaqueélèvea besoin de5 cray ons,2 eaces, 5cartables, 50f euilles depapier,1marqueureaçableet 1paquet de crayonsdecouleur. M.Cameron aaussi be soin de8 marqueurseaçables pour lui-même. A) M.Cameron a-t-ilsusamm ent d'argentavec100$ pouracheter toutesles fournitures scolairesdon tilab esoin?Justie tarép onse. B) S'ila susamment d'argent,combien enaura-t-ildesurplus ?Sinon,combien d'argentlui manque-t-il??????S??? ? ??mbr?

Problem ofthe Week

Problem Aand Solution

SchoolSupplySp ending Problem

Mr. Cameronhas $100to spend onsc hoolsuppliesfor hisclassro om.Hisstudents need pencils,erasers, binders,lined paper, whiteboard markers,andcra yons.Whenhe looks through thecatalogue, hedisco vers thattheitemsheneedsare onlya vailable inpac kages. The table belowshowstheprices.SuppliesAmountCost perP ackageperP ackage pencils10$0.50 erasers12$1.00 binders25$5.00 lined paper100$1.50 whiteboardma rkers4$4.00 crayons24$1.00 Mr. Cameronhas 20studen tsin hisclass.Each studen tneeds 5p encils,2erasers,5 binders,

50 sheetsof linedpap er,1 whiteboardmarker, and1 entirepack ageofcray ons.Mr. Cameron

needs 8whiteb oardmarkershimself. A) DoesMr.Cameron hav eenough moneytobuyallofhis classroom supplies?Justi fyy our answer. B) Ifhe does haveenough money,howmuc his leftoverfrom the$100?Ifhe doesnothav e enough money,how muchmore wouldheneed?

Solution

A) Tosolve thisprob lem,w eneedtodeterminehowmany packagesof each supplyare needed in orderto hav eenoughforall20studen tsin Mr.Cameron's class. Since thereare 10p encilsp erpackage andeachstuden trequires5p encils,then 1package of pencilsis enoughfor 105 =2 students.Since thereare 20studen tsin theclass,

Mr. Cameronneeds 202 =1 0packagesofp encils.

Alternatively,sincethere are20 students, andeac hstud entneeds 5p encils,thenMr.Cameron needs205 =100 pencils.Since each pack agehas10pencils,heneeds10010 =10 packages of pencils.

Similarly,w ecancalculate:

1 packageofbindersis enoughfo r255 =5 students,so herequires 205 =4 packages

of binders. OR Mr. Cameronneeds atotal of205 =100 binders, whichmeansherequires

10025 =4 packages.

1 packageoflinedpap eris enoughfor 10050 =2 students,so herequires

202 =10 packagesoflined paper.

OR Mr. Cameronneeds atotal of2050 =1000 sheets ofpap er,whichmeans herequires

1000100 =10 packages.

Also, 1pac kageoferasersis enoughf or122 =6 students.This means weneed206 =3 packageswitha remainderof 2.So 3pac kages woul dnot hav eenougherasersfortheclass, andquotesdbs_dbs47.pdfusesText_47

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