2021年大聯(lián)盟(Math League)國際夏季五年級數(shù)學挑戰(zhàn)活動一（含答案）

PAGE

11

2021美國"大聯(lián)盟"(MathLeague)國際夏季數(shù)學挑戰(zhàn)活動2021MathLeagueInternationalSummerChallenge

Grade5,IndividualQuestions&Solutions

Question1:

Sarahisusingpopsiclestickstobuildsomegridsforhercityplanningproject.Sheneeds4popsiclestickstomakea1by1grid,and12stickstomakea2by2gridasshownbelow.Howmanysticksdoessheneedtomakea10by20grid?

Answer:430Solution:

Analyzethe2by2grid.TocreateitSarahused3rowsof2stickseachand3columnsof2stickseach.Similartoit,therewouldbe11rowsandthereshouldbe20sticksineachrow.So,thereare20×11=220sticks(placedhorizontally).

Therewouldalsobe21columnsandthereshouldbe10sticksineach.Therefore,thereare21×10=210sticksplacedvertically.

Total=220+210=430popsiclesticks

Question2:

Thesidesofthelargerectangleare20mand16m,figurebelow,notdrawntoscale.Allsixshadedrectanglesareidentical.Whatisthetotalareaofalltheshadedregions,insquaremeters?

Answer:192Solution:

Findthedimensionofthesmallrectangle.Thelength:16÷2=8m.Thewidth:20–8–8=4m.Thetotalareaofalltheshadedregions:6×(8×4)=192m2.

Question3:

Iftwosidesofasquarefieldwereincreasedbyfivefeet,asseeninthediagrambelow,notdrawntoscale,theareaofthefieldwouldincreaseby245squarefeet.Findtheareaoftheoriginalsquare.

Answer:484Solution:

Theareaoftheshadedsquareis5×5=25,figurebelow.Theareaofeachofthetwo“new”rectanglesis(245–25)÷2=110.Thusthelengthofthe“new”rectangle,orthesidelengthoftheoriginalsquare,is110÷5=22.Sotheareaoftheoriginalsquareis22×22=484.

Question4:

Duringpractice,Dana’ssixarrowslandonthetargetshown.Eacharrowisinsideoneoftheregionsofthetarget.Whichofthefollowingtotalscoresispossible:44,31,26,16?(Ifanarrowlandsinsidethesmallestcircleinthecenter,thescoreDanagetsis7.Ifanarrowlandsinsidethesecondcircle,butnotinsidethesmallestcircle,thescoreDanagetsis5.Ifanarrowlandsinsidethelargestcircle,butnotinsidethesecondcircleorthesmallestcircle,thescoreDanagetsis3.Assumenoarrowlandsontheboundaryofanyofthethreecircles.)

44

31

26

16Answer:(c)Solution:

Assumexarrowslandinsidethesmallestcircle,andyarrowslandinsidethesecondcircle.Then6–x–yarrowslandinsidethelargestcircle.Andthetotalscoreis:

7x+5y+3(6–x–y)=4x+2y+18

Thetotalscoreisanevennumber,so(b)isnotgood.

Thetotalscoreisgreaterthanorequalto18,so(d)isnotgood.

Ifthetotalscoreis44,thenwehave4x+2y+18=44.4x+2y=26,2x+y=13.Butitisnotpossiblethat2x+y=13,as0≤x≤6,0≤y≤6,andx+y≤6.So(a)isnotgood.

Whenx=0andy=4orx=1andy=2,orx=2andy=0,4x+2y+18=26.

Question5:

Thereare1000peopleinaroom.Eachpaireithershakeshandsordoesnot.

Isitalwaystruethatsometwopeopleshookthesamenumberofhands?(Pleaseenter1ifyouranswerisYes,and0ifyouranswerisNo.)

Isitalwaystruethatsomethreepeopleshookthesamenumberofhands?(Pleaseenter1ifyouranswerisYes,and0ifyouranswerisNo.)

Isitalwaystruethatsomefourpeopleshookthesamenumberofhands?(Pleaseenter1ifyouranswerisYes,and0ifyouranswerisNo.)

Isitalwaystruethatsomefivepeopleshookthesamenumberofhands?(Pleaseenter1ifyouranswerisYes,and0ifyouranswerisNo.)

Note:

Onedoesn’tshakehis/herownhand.

Onedoesn’tshakethesameperson’shandmorethanonce.

Answer:

(1)1

(2)0

(3)0

(4)0

Solution:

Ifsomeoneshook999times,theneveryoneshookatleastonetime.Sothepossiblenumbersare1to999.Butthereare1000people,fromPigeonholePrinciple,atleasttwopeopleshookthesamenumberofhands.

Nottrue.

Nottrue.

Nottrue.

Question6:

Thenumbers1through10arewritteninarow.Canthesigns“+”and“–”beplacedbetweenthem,sothatthevalueoftheresultingexpressionis0?(Note:Thereare10numbers,andthereare9placestoplacesigns.)

Answer:No(Yes/No)Solution:

1+2+3+4+5+6+7+8+9+10=55

Changingany“+”to“–”,theresultingexpressionisalsoanoddnumber.Sotheresultingexpressionisalwaysanoddnumber.

Question7:

AlbertandBernardjustbecamefriendswithCheryl,andtheywanttoknowwhenherbirthdayis.Cherylgivesthemalistof10possibledates,figurebelow.CherylthentellsAlbertandBernardseparatelythemonthandthedayofherbirthdayrespectively.(ThatisCheryltellsAlbertthemonthofherbirthday,andtellsBernardthedayofherbirthday.)

Conversation:

Albert:Idon’tknowwhenCheryl’sbirthdayis,butIknowthatBernardalsodoesn’tknow.

Bernard:AtfirstIdidn’tknowwhenCheryl’sbirthdayis,butIknownow.

Albert:ThenIalsoknowwhenCheryl’sbirthdayis.WhenisCheryl’sbirthday?

Pleaseenterthemonthfirst,followedbytheday.Whenyouenterthemonth,pleaseenter5forMay,6forJune,7forJuly,and8forAugust.

Answer:

7

16

Solution:

Albert:Idon’tknowwhenCheryl’sbirthdayis,butIknowthatBernardalsodoesn’tknow.

SoBernardknowsthatitisnotMayorJune.IfitisMay,thenitispossiblethatitisMay19.CherylwilltellBernardthatthedayis19,thenBernardwillknowherbirthdayisMay19.SoitisnotMay.ItisnotJuneeither.

Bernard:AtfirstIdidn’tknowwhenCheryl’sbirthdayis,butIknownow.

SoherbirthdayisJuly16,August15,orAugust17.ThenumberthatCheryltellsBernardis15,16,or17.

Albert:ThenIalsoknowwhenCheryl’sbirthdayis.

IfitisAugust,thenAlbertwillnotbeabletotellCheryl’sbirthday.SoithastobeJuly.SotheanswerisJuly16.

Question8:

Goal:

Arrangethesixdifferentjuicesintothefollowingorder.

Theorderis:1–red,2–orange,3–yellow,4–green,5–blue,6–violet.

Rules:

Youmayonlypourafilledcupintoanemptycup.

Youmaynotswitchthepositionsofanycups.

Theemptycupmustendupontheright.

Examples:

2Poursarenecessarytosolvethefollowingexamplepuzzle,figurebelow.

PourOne,figurebelow.

PourTwo,figurebelow:

Foranotherexamplebelow,thisexampleneedsfourpours.

PourOne,figurebelow.

PourTwo,figurebelow.

PourThree,figurebelow.

PourFour,figurebelow.

For6differentjuicesandoneemptycup,thereare5040(=7!=7×6×5×4×3×2×1)possiblepermutations.Twoofthemwerelistedabove.Oneneeds2pours.Theotherneeds4pours.

Ofallthe5040possiblepermutations,howmanypermutationsareimpossibletosolve?Thatisnomatterhowyoupourthejuices,youcan’tarrangethejuicesinorder.

Answer:0Solution:

Everypermutationissolvable.Inthecaseof6differentjuices,ifwelookateverycupfrom

theleftmosttotheonerightbeforetherightmost,thereareatmost6cupsthatarenotfillewiththecorrectjuices.Let’scallthesecups“incorrectcups.”Foreveryincorrectcup,weneedatmosttwopourstomakeitcorrect,thatistofillthiscupwiththecorrectjuice.Soeverypermutationissolvable.

<endofsolution>

Whatisthelargestnumberofpoursyouneedtosolveanysolvablepermutation?

Answer:9Solution:

Let’sfirstlookatthecasewhenthereareonly2differentjuices,figurebelow.Thelargestnumberofpoursis3.AndithappenswhenthepositionsofJuice1andJuice2areswapped,andtheemptycupisontheright.

Thenlet’slookatthecasewhenthereare4differentcups,figurebelow.Thelargestnumberofpoursis6.Andithappenswhenthepositionsof2juicesareswapped,andthepositionsoftheother2juicesareswappedtoo,andtheemptycupisontheright.

Sowhenthenumberofjuicesisn,nisanevennumber,thelargestnumberofpoursisn3.

2

<endofsolution>

For1000001differentjuicesandoneemptycup,thereare1000002!possiblepermutations.Thereisapre-definedorderofthe1000001juices,justastheorderforthe6juicesabove.

Ofallthe1000002!possiblepermutations,howmanypermutationsareimpossibletosolve?Thatisnomatterhowyoupourthejuices,youcan’tarrangethejuicesinorder.

Answer:0Solution:

Everypermutationissolvable.Inthecaseof1000001differentjuices,ifwelookatevery

cupfromtheleftmosttotheonerightbeforetherightmost,thereareatmost1000001cupsthatarenotfillewiththecorrectjuices.Let’scallthesecups“incorrectcups.”Foreveryincorrectcup,weneedatmosttwopourstomakeitcorrect,thatistofillthiscupwiththecorrectjuice.Soeverypermutationissolvable.

<endofsolution>

Whatisthelargestnumberofpoursyouneedtosolveanysolvablepermutation(for1000001differentjuicesandoneemptycup)?

Answer:1500001Solution:

Whenthenumberofjuicesisn,nisanoddnumber,thelargestnumberofpoursis

n131.

2

<endofsolution>

Question9:

Twelvestraightjacketedprisonersareonthedeathrow.Tomorrowtheywillbearrangedinasingle-fileline,positionsdeterminedrandomly,allfacingthesamedirection.Thepersoninthebackoftheline,whowillbereferredtoasthetwelfthperson,canseetheelevenpeoplestandinginfrontofhim;theeleventhpersoncanseethetenpeoplestandinginfrontofhim,butnotthetwelfthpersonwhoisstandingbehindhim;soonandsoforthuntilyougettothefirstpersoninlinewhocannotseeanyone.Thewardenofthisprisonlikestoplaygameswithhisprisonersandhaslinedeveryoneupinthisfashionsothathecangivethemanopportunitytogaintheirfreedom.

Thewardenwillplace,atrandom,ablackorwhitehatoneachprisoner’shead.Theprisonerscannotseethecolorofthehattheyarewearing,buttheycanseethecolorofthehatsofeachprisonerstandinginfrontoftheminline.Afterallofthehatshavebeenplacedontheprisoners’heads,thewardenasksthepersoninthebackoftheline,thetwelfthperson,“whatisthecolorofthehatonyourhead?”Theprisonercanreplybysayingeither“black”or“white”,butnothingelse.Whentheprisoneranswersthewarden,alltheprisonersinlinecanhearthereply.Iftheprisonerrepliedwiththecorrectcolorofhishat,hewillbefreedfromtheprison.Ifheiswrong,hewillbeexecutedthatnight.Thisgamecontinuesuntilalltwelveoftheprisonersinlinehavebeenaskedaboutthecoloroftheirhat.Whilethegameisinprogress,noprisonerisallowedtospeak,move,ordoanythinguntilitistheirturntoanswerthequestion.Thewardenallowstheprisonerstogettogetherinthecourtyardthedaybeforethisgametocomeupwithaplantomaximizethenumberoflivestheycansave.Theprisonerscomeupwithaplanthatguaranteesxprisoner(s)willsurvive,and12–xprisoner(s)willeachhavea50%chancetosurvive.Whatisthelargestpossiblevalueofx?

Answer:11Solution:

Theprisonerscancomeupwithaplanthatwillgivethetwelfthprisonertoacta50/50chanceatsurvivalandwillguaranteethattheotherelevenprisonersinlineallsurvive.Thisishowtheydoit:

Theprisonersdecidethatwhoeverisstandinginthetwelfthpositionandspeaksfirstwillsimplysaywhichevercolorheseesanevenamountof.Thereisnowayforthisprisonertoraisehischancesofsurvivalabove50%,butbysayingwhichevercolorheseesanevenamountof,everyotherprisonercandeducethecoloroftheirownhat.

Forexample:

Ifthetwelfthpersonsays“black”,thentheeleventhpersoninlineknowswhatcolorhatheiswearingandcancorrectlystateittothegroupbasedontheremaininghatsthathecansee.Ifheseesanoddnumberofblackhatsinfrontofhim,thenhecanbesurethathishatisblackbecausethepersoninlinebehindhimsawanevenamount.Now,the

tenthpersoninline,knowingthatthetwelfthpersonsawanevenamountofblackhatsandknowingthattheeleventhpersoninlineiswearingablackhat,candeducethecolorofhishatbasedontheremainingninehats.

Ifthetwelfthpersoninlinesawanevenamountofblackhats,andtheeleventhpersoninlineiswearingablackhat,thenthetenthpersoninlineknowsthatifheseesanoddnumberofblackhats,heiswearingawhitehat,andifheseesanevennumberofblackhats,hemustbewearingablackhathimself.

Thissystemallowsforelevenprisonerstoguaranteetheirsurvival.Nosomuchfunthetwelfthprisoner.

Question10:

Somewhereintheworldthereisasmallneighborhoodfullofveryeccentricandintelligentpeople.Therearefivehousesthatareeachadifferentcolor.Thereisapersonofadifferentnationalitywholivesineachhouse.Eachofthesefivepeopledrinkstheirownspecialdrink,smokestheirownspecialbrandofcigarettes,andhastheirownspecialpet.Noonehasthesamepet,smokesthesamebrandofcigarettes,ordrinksthesamedrink.Oneofthesefivepeopleownsapetfish.Withjustthefollowinginformation,figureoutwhoownsthecatandwhoownsthefish:

TheBritishmanlivesintheredhouse.

TheSwedishmanhasadogforapet.

TheDanishmandrinkstea.

Thegreenhouseandthewhitehousearenexttoeachother,andthegreenhouseistotheleftofthewhitehouse.

Theownerofthegreenhousedrinkscoffee.

ThepersonthatsmokesPallMallcigaretteshasabirdforapet.

TheowneroftheyellowhousesmokesDunhillcigarettes.

Thepersonwholivesinthemiddlehousedrinksmilk.

TheNorwegianlivesinthefirsthouse,countingfromlefttoright.

ThepersonwhosmokesBlendcigaretteslivesnexttotheonethathasacatforapet.

ThepersonwhohasahorseforapetlivesnexttotheonewhosmokesDunhillcigarettes.

TheonewhosmokesBluemastercigarettesdrinksbeer.

TheGermansmokesPrincecigarettes.

TheNorwegianlivesnexttoabluehouse.

ThepersonwhosmokesBlendcigaretteshasaneighborwhodrinkswater.

Thefivehouses