[PDF] Thèmes Problème C (7e/8e année) Problème de la semaine

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Sens du nombre (N)

Géométrie (G)

Algèbre (A)

Gestion des données (D)

Pensée computationnelle (C)

ThèmesProblème C (7

e /8 e année) *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)

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Problem ofthe Week

Problem Cand Solution

This Productisa Mystery

ProblemThe numberA8is at wo-digitnumber withtensdigitAand units(ones) digit8. Similarly, 3B is at wo-digitnumber withtensdigit3and units digitB. WhenA8is multipliedby 3B, theresult isthe four-digitn umb erC730. IfA,B, andCare eachdifferen tdigitsfrom0to9, determinethe values ofA,B, andC.

Solution

In am ultiplicationquestiontherearethreeparts: themultiplier,multiplicandandproduct. In ourproblem, A8is them ultiplier,3Bis them ultiplicand,andC730is thepro duct. The unitsdigit ofthe product C730is 0.The unitsdigit ofa product isequal tothe units digit ofthe resultobtained by multiplying theunitsdigitsof themultiplier andm ultiplicand. So8Bmustequal an umb erwithunitsdigit0.Theon lyc hoicesfor Bare 0and 5,since no other singledigit multiplied by8pro ducesanum ber endinginzero. However,ifB= 0, theunits digitof thepro ductis 0and theremainingthreedigits ofthe product,C73, arepro ducedbym ultiplying 3A8. But3A8producesa num berendingin4, not 3as required.Therefore B6= 0andBmustequal 5.S othe multiplicandis35. SinceA8is at wo-digitnumber, thelargestpossiblevalueforAis9. Since9835 =3430 , the largest possiblevalueof Cis3. Also,the product C730is afour-digit num bersoC6= 0. Therefore, theonly possible valuesforCare1;2;and3. Wewillexamine each possibilit yfor C. IfC= 1, thenC730becomes1730. Wewan tA835 =1 730. Alternatively,wew ant

173035 =A8. But173035:= 49:4, whichisnot awh ol en umber.Therefore,C6= 1.

IfC= 2, thenC730becomes2730. Wewan tA835 =2730 . Alternatively,wew ant

273035 =A8. Since273035 =78 , whichisa wholen umb er.Th erefore,C= 2producesa

validv alueforA, namelyA= 7. IfC= 3, thenC730becomes3730. Wewan tA835 =3730 . Alternatively,wew ant

373035 =A8. But373035:= 106:6, whichisnot awhole num ber. Therefore,C6= 3.

Weha veexaminedevery validpossibilit yforCand foundonly onesolution. Therefore, A= 7, B= 5andC= 2is theonly valid solution.Wecan easilyv erifythat7835 =2730 .

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Problem ofthe Week

Problem Cand Solution

HomestyleLemon Drink

ProblemArestauranto wnerwasexp erimentingwiththe tasteofherhomestylelemon drink.She started with60 litresof water. Sheremo ved15litres ofthe waterandreplaceditwith 15litres of purelemon juice.After thoroughlystirring thenew mixture,she discov eredthat itw asto o lemony.So shere mov ed10litresofthenewmixtureand replac edit with1 0litres ofw ater. She thoroughlystirred themixture andconcluded thatthe newmixture was justrigh t. Determine theratio ofpure le monjuice towaterinthenal 60litre mixture.

Solution

Weneed todetermine theamoun tof purelemon juiceandthea mount ofw ater in thenal mixture. The ownerstartswith 60litres ofw aterand nolemon juice.After removing15 litres ofw aterandadding15 litresof purelemo njuice, shehas 15litres ofpure lemon juiceand 6015 =45 litres ofw ater.So1560 =14 ofthe newmixture is pure lemonjuice and 4560
=34 ofthe newmixture isw ater.

She thenremo ves10litresofthe newmixture,

14 ofwhichispure lemon juiceand 34
ofwhichis water. Sotheowner remov es 14 10= 104
=52 litresof purelemon juice and 34
10= 304
=152 litresof water.

Before addinganother 10litr esof watershehas 1552

=302 52
=252 litresof pure lemonjuice and45152 =902 152
=752 litresof water. After addingthe nal10 litres ofw ater,shehas10 +752 =202 +752
=952 litresof water.

The nalratio ofpure lemonjuice tow ateris

252
:952 =25 :95 =5 :19 : Therefore, thenal ratioof purelemon juiceto wa ter inthe homestyle lemon drink is5 :19 .

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Problem ofthe Week

Problem Cand Solution

More PowertoYou

Problem555555555555555555555555555555555555555555555555 In mathematicsw eliketo writeexpressionsconcisely. For example,w ewill oftenwritethe expression5555as54. Thelo wernumber 5is calledthe base,the raised4is calledthe exponent,andthe wholeexpression 54is calleda po wer.So53means555and isequal to

125. Whatare the lastthreedig itsin thein tegerequalto52020?

Solution

Let's startb yexaminingthelast threedigits ofv ar iou sp ow ersof5. 5 1=005 5 2=025 5 3=125 5 4=625 5

5= 3125

5

6= 15625

5

7= 78125

5

8= 390625

Notice thatther eisapattern forthe lastthree digitsafter the rstt wo po wers of

5. Forevery oddint egerexponen tgreaterthan2,thelastthreedigits are125.

Forev eryevenin tegerexp onentgreaterthan2,thelastthree digitsare625.If the patterncon tinues,then59will end125 sincethe exponen t9 isoddand510 will end625 sincethe expo nen t10iseven.Thisiseasilyveried since 5

9= 1953125and510= 9765625.

Wecan easilyjustify why thispattern continues.If ap ow erendsin125,then the last3 digitsof the nextp owerare thesame asthelastthreedigitsofthe product1255 =62 5. Thatis, thelast threedigits of thenext po werare625. If ap owerendsin625,thenthe las t3 digitsof thenext po werarethesameas the lastthree digitsof thepro duct6255 =3 125. Thatis, thelast threedigits of thenext po werare125. For52020, theexp onent2020isgreaterthan 2and isan even num ber.

Therefore, thelast threedigits of52020are 625.

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Problem ofthe Week

Problem Cand Solution

A GrapeProblem

ProblemThere aresev eralbowls containingvariousamounts ofgrapeson atable.When12 ofthebo wls eachhad 8more grapes addedto them,themean(a verage) num ber ofgrap esp erbowl increased by6.H ow manybowlsof grapesareon thetable?

Solution

Solution 1:

The meann umberofgrapesper bo wlisequalto thetotalnum berofgrapes dividedb ythe numberofbo wls.S othemeannumb erofgrapes per bo wlincreasesby6ifthetotal num berof grapesadded dividedb ythe number ofb owlsisequalto6. If 12of theb owls eachhad8moregrap esaddedtothem, thenthe total num berofgrap es added is128 =9 6. Thus,96 div idedbythenumb erofb owls equals6.Since9616 =6 , thereare 16b owls.

Solution 2:

This solutionuses variables torepresentth eunkno wnvalues. Weare toldthat themean num ber ofgrap esperbowlincreasedb y6. Wecanwrite thisas follows. old mean+ 6= new mean The meanis equalto the totaln umberof grap esdividedbythenumb erof bowls.Letbbethe numberofbo wls. Letgbethe totaln umb erofgrapesthatwereoriginallyin theb owls.If12 of theb owlseachhad8 moregrapesadded tothem, thenthe totaln umberof grapes addedis

128 =96 . Nowwe canrewriteourequation abo ve using variables.

gb + 6= g+ 96b

Wecan separatethe fraction onthe rightside.

gb + 6= gb +96b

Nowif we subtract

gb from bothsidesof the equation,w eareleftwith anequation thathas only onev ariable.Wecan thensolvethis equation. 6 = 96b

Multiply bothsidesb yb:6b= 96

Divide bothsidesb y6:b=966

b= 16

Therefore, thereare 16b owls.

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Problem ofthe Week

Problem Cand Solution

In TheirPrime

ProblemAprime numberis anin tegergreaterthan1 withexactly tw odifferen tp ositiv efactors,1and the numberitself.Thereare threechildrenin afamily .Eac hof theiragesisa primen umb er. The sumof theirages is41 andat leastt wo ofthe children hav eagesthatdifferby 16. Determine allp ossibilitiesfortheages ofthe children.

Solution

Wecan startb ylist ingallofthe primenumb ersless than41. Thep os sibleprime ages are2, 3,5, 7,11, 13,17, 19,23, 29,31, and37. We coulda ctua lly eliminate some ofthe largerp rimes fromthislistsincethereare threedieren tprimes in the sum. Noww ewilllook forall primepairsfrom thislist thatdier by 16.The pairs include 3and 19,7 and23, and13 and2 9.W ewill look ateac hofthesepairs and determinethe thirdn umb ersothatthesumofthethree agesis 41. Forthe pair3 and19, thethird agew ouldb e41319 =19 , whichisprime. The agesof thethree children wo uldbe 3,19,and19.Thisisap ossiblesolution. Forthe pair7 and23, thethird agew ouldb e41723 =11 , whichisprime. The agesof thethree children would be7,11 ,and23.This isanotherposs ible solution. Forthe pair13 and29, thes umof theset wo agesis13 +29 =42 . Thissum is already over41,sothis isnot ap ossible solution. Therefore, thereare tw opossibilitiesfortheages ofthechildren. Thec hildren are either7, 11and 23y earsold or3, 19and19y earsold. Forfurther thought: Howwould theproblemchange ifthew ordsprime num ber wereremo vedfromtheproblemand replacedwith positiv ein teger?

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4321 =24

4? =24

161514 321

1? +2? +3? ++ 2019?+ 2020?+ 2021???????Algè???? ?????? ??????

Problem ofthe Week

Problem Cand Solution

HowIt Ends

ProblemThe productofthep ositive integers 1to4is4321 =24 and canb ewritteninan abbreviated forma s4!. Wesay 4factorial. So4! =24 . The productofthepositiv ein tegers1 to16is161514321and canbe written in anabbreviated formas 16!. Wesay 16factorial. Therepresentsthe pro ductofallthemissingin tegersb etw een14 and3. In general,the product ofthepositiv ein tegers1 tonisn!. Notethat 1! =1 . Determine thetens digitand units(ones) dig itof thesum

1! +2! +3! ++ 2019!+ 2020!+ 2021!

Solution

Atrst glan ceseemslikethereisa greatdeal ofworkto do.Ho wev er,b yexamining several factorials, wewilldisco ver otherwise. 1! =1

2! =2 1 =2

3! =3 21 =6

4! =4 321 =2 4

5! =5 4321 =120

Now6! =6 (54321) =6 5! =6(120) =720 ,

7! =7 (654321) =7 6! =7(720) =5040 ,

8! =8 (7654321) =8 7! =8(5040) =40 320,

9! =9 (87654321) =9 8! =9 (40320)=362880 , and

10! =10 (987654321) =10 9! =10(362 880)=3628 800.

An interestingobservation surfaces, 9! =9 8!;10! =10 9!;11! =11 10!;and soon. Furthermore,thelastt wo digitsof 10!are00.Everyfactorial ab ov e10 !will alsoend with00 since multiplyinganin tegerthat endswith00b yanother integer produces anin tegerproduct that endsin 00.So allfactorials abo ve 10!will endwith00andwillnotchange thetens digit or theunits digitin therequired sum .W ecan determinethelasttwo digitsoftherequired sum byadding thelasttw odigits ofeac hofthefactorialsfrom1! to9!.

The sumof thelast tw odigits of1!to9!will equal

1 +2 +6 +24 +20 +20 +40 +20 +80 =213 :

Therefore, forthe sum1!+2!+3!++2019!+2020!+2021!, thetens digitwill be 1and the units (ones)digit willb e3. Fromwhatw eha vedone,w edo notknowthe hundredsdigit. ForF urtherThought: Ifw ehadonlyb eenin terestedin theunitsdigitin therequiredsum , howman yfactorialswould we needtocalculate?If wewa nted tokno wthe lastthree digits, howman ymorefactorialsw ouldb erequired?

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Problem ofthe Week

Problem Cand Solution

PaintingtheW ay

ProblemAlexia,Benito, andCarmen won ateamartcomp etitionat theirlo calpark. Aspartoftheir prize, theycan paint the5km(5000 m)pa ved paththrough thepark any waythey like. They decidedAlexia willpain tthe rst70m,then Benitowi llpain tthe next15 m,then Carmen willpain tthenext35 m.They willk eeprep eatingthis patternun tilthey reachthe end ofthe 5km pa th. Whatpercentageofthepath willeac hpersonpaint?

Solution

After eachperson paintstheirrst section,theywillha ve cov ered

70 +15 +35 =120 m intotal. We willcallthisone cycle.Th en umb erof cycles

in 5000m is5000120 =1253 =41 23

So,there are41 completecycles and

23
ofanother cycle.The totaldistance coveredin41 cyclesis 41120 =4920 m. Thatmeans thereare

50004920 =80 m remaining.Alexia would painttherst 70m,leaving the

last 10m forBenito topain t. Wecan organizethis informationin at able tocalculate thetotal distanceeach personwill paint. DistancePaintedby EachPerson (m)

AlexiaBenitoCarmen

First41 cycles4170= 28704115= 6154135= 1435

Lastpartialcycle70100

Total2870+70= 2940615+ 10= 6251435

We canno wcalculatethep ercentage ofthe patheachpersonwill pai nt.

Alexia:

29405000

100%= 58:8%

Benito:

6255000

100%= 12:5%

Carmen:

14355000

100%= 28:7%

Therefore, Alexiawill paint 58:8%of thepath, Benitowill paint 12:5%of the path, andCarmen willpain t28:7%of thepath.

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Problem ofthe Week

Problem Cand Solution

A PathUsing Math

ProblemA landscaperneedstollapath measuring2 feetb y6 feetwith patios ton es.The patiostones are each1fo otb y2feet,so thelandscapercalculates thats hewill need6 ofthem. Before arranging thepatio stones,the landscaper wan tstoloo katallofheroptions. Shecann otcut or overlapthestones,and theyall must tinside thepath areawithout anygaps.T wo possible arrangementsof thestones aresho wn.Ho wman ydifferentarrangem entsarethere intotal?

Solution

Let"s considerthe wa ysthatthepatiostones canb earranged. Wewillimaginewe arelo oking at thep athfromtheside,just like inthe imagessho wnin thequestion.First,notice tha tthere mustalw aysbean evennum berofpatio stonesthathav eahorizontal orientation, because they mustbe placedinpairs. All patiostones arev ertical(and nonearehorizontal) This canb edoneinonly onew ay .

Fourpatio stonesare vertical andt woarehorizon talThere couldb e0,1,2, 3or 4v erticalstones toth erigh tof thehorizontalstones. Sothere are5 wa ysthatfourpatiostonesare vertical andt wo arehorizon tal.

Twopa tiostonesareverticaland fourare horizontal Weneed toconsider subcases:

Case 1:There areno vertical stonesb etweenthe horizontal stones.There couldb e0,1or 2v erticalstones tothe right ofthehorizontal stones.So

there are3 wa ysthattwopatio stonesa reverticalandfour arehorizon tal whenquotesdbs_dbs29.pdfusesText_35

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