2021年大聯(lián)盟(Math League)國際夏季四年級數(shù)學(xué)挑戰(zhàn)活動(dòng)一（含答案）

PAGEPAGE82021美國"大聯(lián)盟"(MathLeague)國際夏季數(shù)學(xué)挑戰(zhàn)活動(dòng)2021MathLeagueInternationalSummerChallengeGrade4,IndividualQuestions&SolutionsQuestion1:Thereisatableinfrontofyouwithonehundredquartersonit.Youhavebeenblindfoldedandarewearingathickpairofgloves.Youarenotabletoseewhetherthebecauseyouareblindfolded.Andyouarenotabletofeelwhetherthequartersareheadsortailsbecauseofthethickgloves.Yourfriendtellsyouthattwentyofthesequartersaretailsandremainingeightyareheads,butyoudonotknowwhicharewhich.Hetellsyouthatifyouareabletosplitthequartersintotwopileswherethenumberoftailsquartersisthesameineachpile,youwillwinallofthequarters.Youarefreetomovethequarters,flipthemover,andarrangethemintotwopilesofanynumber.Foryoutowinallofthequarters,howmanyquartersareineachofthetwopiles?(Pleaseenteryouranswersinascendingorder.)Note:heads:thefrontsideofacoin.tails:thebacksideofacoin.Figurebelow,headsandtailsofaquarter.Answer:2080Solution:Youcanwinthisgamewithjustoneelegantmove:Takeanytwentyquartersfromthehundred,putthemintoaseparatepile,andflipthemover.Nomatterhowmanyofthequartersaretailsinthegroupflipover,youwillalwayswindupwithtwogroupsofquarterswiththesamenumberoftails.Ifyouhappenedtoselectalltwentyofthequartersthatwerealreadytails,youwouldseparatethemandflipthem,resultingintwogroupsofquarterswithzerotailsinthem.Ifyouhappenedtoselecttwentyquarterswithonlysixtailsamongthem,youwouldleavefourteentailsintheoriginalpile,andafterflippingthetwentyquartersyouselected,youwouldhavefourteentailsandsixheadsinthenewpile.Thiswillalwaysworkaslongasyouonlytaketwentyquarters,putthemintotheirownpile,andflipthemallover.Question2:Thepagesofabookareconsecutivelynumberedfrom1through384.Howmanydigit‘8’appearinthisnumbering?Answer:73Solution:Oneway:Wecanlookattheonesandtensdigitsseparately:Theonesdigitof8from1through384:sinceitoccursonetimeineverysetof10consecutivenumbers,thereare38completesetsof10consecutivenumbers.So,thedigit‘8’appears38timesasaonesdigit.Thetensdigitsof8from1through384:sinceitoccurs10timesineverysetof100consecutivenumbers,thereare3completesetsof100(1-100,101-199,200-299).Thedigit8appears30timesasatensdigit.Inaddition,thenumbers380-384contain5moretensdigitof8.Inall,thedigit‘8appearsatotalof38+30+5=73times.Orifyouarelookingatthefirsthundrednumberstherewillbe20digits8inthem:8,18,28,38,48,58,68,78,80,81,82,83,84,85,86,87,88,89,98.Sincewehave384pagesitmeans20×3+13=73timesdigit8willbehappeninginthepagenumeration1-384.Question3:Annabellaisvisitinghergrandmawhoresides295milesawayfromher.Shestartedherdrivingtripataspeedof65mph(milesperhour)forthefirst3hours.Fortherestofthetrip,shedroveataspeedof50mph.Howmanyhoursdidshetaketodrivetohergrandma?Answer:5Solution:Forthefirst3hours,shecompletes65mph×3hours=195miles.Shethenhas295–195=100milestogo.Sinceherspeedis50mph,itwilltakeher100÷50=2hourstocompletetherestofherjourney.Totalhours:2+3=5hours.Question4:Sarahisusingpopsiclestickstobuildsomegridsforhercityplanningproject.Sheneeds4popsiclestickstomakea1by1grid,and12stickstomakea2by2gridasshownbelow.Howmanysticksdoessheneedtomakea10by20grid?Answer:430Solution:Analyzethe2by2grid.TocreateitSarahused3rowsof2stickseachand3columnsof2stickseach.Similartoit,therewouldbe11rowsandthereshouldbe20sticksineachrow.So,thereare20×11=220sticks(placedhorizontally).Therewouldalsobe21columnsandthereshouldbe10sticksineach.Therefore,thereare10×21=210sticksplacedvertically.Total=220+210=430popsiclesticksQuestion5:Thesidesofthelargerectangleare20mand16m,figurebelow,notdrawntoscale.Answer:192Allsixshadedrectanglesareidentical.Whatisthetotalareaofalltheshadedregions,insquareAnswer:192Answer:192Solution:Findthedimensionofthesmallrectangle.Thelength:16÷2=8m.Thewidth:20–8–8=4m.Thetotalareaofalltheshadedregions:6×(8×4)=192m2.Question6:Alice,Bibi,andClaryarethreeintelligentgirls.Oneday,theirteacherdecidestoplayasimplegameandawardsomeprizetothefirstgirlwhosolvesit:head,andeachgirlcanseetheothergirls’hats,butnotherown.Theyareinstructedtoraisetheirhandsiftheyseeatleastoneblackhat.Andthefirstgirltotelltheteacherwhatcolorhatsheiswearingandhowoutwillgettheprize.AfterthehatsareplacedonAlice,Bibi,andAfterafewseconds,Alicesays:“Ihavefigureditout!”wearing?Threechoices:BlackWhiteAnswer:(a)Solution:Aliceknowsthatsheiswearingablackhatandthisishowsheisabletodeduceit:Itisclear,becauseallthreepersonsraisedtheirhands,thatthereareatleasttwoblackhats.However,ifthereweretwoblackhatsandonewhitehat,eitherofthepeoplewiththeblackhatswouldseethewhitehat,andseethateveryonehadtheirhandraisedwhichwouldallowhertoinstantlydeducethecoloroftheirhatisblack.Afterafewsecondshadpassedwithoutanyonespeakingaboutthecoloroftheirhat,itbecamecleartoAlice,thebrightestofthebunch,thattheymustallbewearingblackhats,otherwiseeitherBibiorClary,whoareintelligent,wouldhavesaidsomethingbynow.Question7:Youhaveabalancescaleandsixweights.Therearetworedweights,twoorangeweights,andtwoblueweights.Ineachpairofcoloredweights,oneweightisslightlyheavierthantheother,butisotherwiseidentical.Thethreeheavierweightsallweighthesame,andthethreelighterweightsallweighthesame.Whatisthefewestnumberoftimesyouneedtousethebalancescaleinordertopositivelyidentifytheheavierweightineachpair?Answer:2Solution:Tofindtheheavierweightofeachcolor,youonlyneedtousethebalancescaletwice.First,youmustweigharedandanorangeweightagainstablueandanorangeweight.Ifthepansbalance:Youcanbesurethatthereisaheavyandalightweightoneachpan.Takeboththeredandblueweightsoffthescaleandleavetheorangeweightsoneachside.Thiswillshowyouwhichoftheorangeweightsistheheavierone,whichwillthenshowyouwhethertheredorblueweightthatwasjustonthepanwastheheavierone,whichwillthenallowyouto,byprocessofelimination,determinetheweightoftheredandblueweightthatwereneveronthepan.Ifthepansdonotbalance:Youcanbesurethattheorangeweightonthesideofthescalethatwentdownistheheavierofthetwoorangeweights.Tofindtheheavierweightsoftheredandbluepair,youmusttaketheredweightthatyoujustweighedandweighitagainsttheblueweightthathasnotyetbeenonthescale.Seeingwhathappenshere,plusrememberingwhathappenedwhenyoumadethefirstweighing,willallowyoutocorrectlylabeleachweight.Question8:Inaclassof5thgradestudents,15havepetcats,12havepetdogs,5havebothpetcatsandpetdogs,and8haveneitherpetcatsnorpetdogs.Howmanytotalstudentsareintheclass?Answer:30Solution:5havebothcatsanddogs.15–5=10havecatsonly.12–5=7havedogsonly.8haveneithercatsordogs.5+10+7+8=30.Question9:Whatisthe5-digitmysterynumberthatfitsalltheconditionsbelow?Thesumofthefirsttwodigits(countingfromlefttoright,samebelow)isonesmallerthanthethirddigit.Thethirddigitisdoublethefourthdigit.Thefourthdigitisdoublethelastdigit.Thethirddigitistheproductofthefourthandfifthdigits.Theseconddigitisfivemorethanthefirstdigit.Thefirstdigitisone-eighthofthethirddigitandalsoone-fourthofthefourthdigit.Answer:16842Solution:“Thefirstdigitisone-eight