Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
List of prime numbers wikipedia , lookup
Mechanical calculator wikipedia , lookup
Mathematics of radio engineering wikipedia , lookup
Series (mathematics) wikipedia , lookup
Elementary mathematics wikipedia , lookup
Approximations of π wikipedia , lookup
Proofs of Fermat's little theorem wikipedia , lookup
Positional notation wikipedia , lookup
2016 Chapter Competition Solutions Are you wondering how we could have possibly thought that a Mathlete® would be able to answer a particular Sprint Round problem without a calculator? Are you wondering how we could have possibly thought that a Mathlete would be able to answer a particular Target Round problem in less 3 minutes? Are you wondering how we could have possibly thought that a particular Team Round problem would be solved by a team of only four Mathletes? The following pages provide solutions to the Sprint, Target and Team Rounds of the 2016 MATHCOUNTS® Chapter Competition. These solutions provide creative and concise ways of solving the problems from the competition. There are certainly numerous other solutions that also lead to the correct answer, some even more creative and more concise! We encourage you to find a variety of approaches to solving these fun and challenging MATHCOUNTS problems. Special thanks to solutions author Mady Bauer for graciously and voluntarily sharing her solutions with the MATHCOUNTS community for many, many years! SprintRound 1. Given:Countbackwardsfrom155by4. Find:The9thnumber. Thefirstnumberis155. Thesecondnumberis151,or4less. Theninthnumberis4 8 32less. 155–32=123Ans. 2. Given:Thegraph,asdisplayedbelow. Find:Thedifferencebetweenthe maximumandminimumvalues. y 5. Find:thesumofthefirst8termsofthe sequencebeginning‐4,5…whereeach termisthesumoftheprevioustwoterms. Thefirsteighttermsofthesequenceare −4,5,1,6,7,13,20and33.Theirsumis −4+5+1+6+7+13+20+33=81Ans. 6. Given:Thegraphdisplayedbelow. Find:thetotalcosttoship3packages weighing1.8lbs.,2lbs.and4.4lbs. 6 y S R 5 Cost (dollars) 2016ChapterCompetition 4 3 2 1 x 4 0 1 −2 2 4 x −2 −4 Thehighestpointintermsofyisthethird pointfromtherightwithcoordinates (1,3).Thelowestpointintermsofyisthe secondpointfromtherightwith coordinates(−2.5,−4)–atleastitlooks likearound–2.5.Thedifferenceinthe y‐coordinatesis3 4 7Ans. 3. Given: , 3 4 5 Weight (lbs) 2 −4 2 1 Find:a Substituting−1forb,weget1/−1=−1/a. Cross‐multiplying,wefindthat a=(−1)(−1)=1Ans. 4. Find:Thesumofalltwo‐digitmultiplesof 3thathaveaunitsdigitof1. 21isthefirsttwo‐digitmultipleof3 whichhasaunitsdigitof1.Each successivetwo‐digitmultipleof3witha unitsdigitof1is30larger.Thereare threeinall:21,51and81.Theirsumis 21+51+81=153Ans. 6 Itcosts$3toshipthe1.8‐lb.package, $3.50toshipthe2‐lb.packageand$5to shipthe4.4‐lb.package.Thetotalcostto mailallthreepackagesis3 3.50 5 11.50Ans. 7. Given:thegraphbelow. Find:thedegreemeasureoftheportion ofthecookiethatisflour. M’ C D Flour 40% Eggs 4% Butter 16% Sugar 22% Chocolate Chips 18% Thecookieismadeof40%flourand. 40%of360°is360 0.4 144Ans. 8. Given:3zogutsand4gimunscosts$18. 2zogutsand3gimunscosts$13. Find:thecostof1zogutand1gimun Letz=thecostofazogutandg=thecost ofagimun.Wecanwritetheequations 3 4 18(Eq.1) 2 3 13(Eq.2) SubtractingEq.2fromEq.1,weget 1 1 18 13 5Ans. 9. Given:Arectangularpieceofpaperwitha 40‐inchlength.Itisfoldedinhalf.The ratioofthelongsideoftheoriginalsheet totheshortsideoftheoriginalsheetis thesameastheratioofthelongsideof thefoldedsheettotheshortsideofthe foldedsheet. Find:thelengthoftheshortsideofthe originalsheet. Lets=thelengthoftheshortsideofthe originalsheet,whichisalsothelongside ofthefoldedsheet.Thelengthoftheshort sideofthefoldedsheetis40/2=20. Wehavetheratio .Cross‐ multiplyingyields 800.Sothelength oftheshortsideoftheoriginalsheetis √800 10√8 20√2Ans. 10. Find:thegreatestmultipleof3thatcan beformedusingoneormoreofthedigits 2,4,5and8,usingeachdigitonlyonce. Anumberisamultipleof3ifthesumof itsdigitsisdivisibleby3.Since8+5+4+ 2=19,whichisnotdivisible3,weknow thatnonumberformedusingallfour digitswillbeamultipleof3.Thefour combinationsofthreedigitsthatcanbe usedtoform3‐digitnumbersare8,5,4; 8,5,2;8,4,2;and5,4,2.Since8+5+4= 17,8+4+2=14and5+4+2=11,none ofwhichisdivisibleby3,weknowthat no3‐digitnumberformedwiththese combinationsofdigitswillbeamultiple of3.But8+5+2=15,whichisdivisible by3.Thegreatestmultipleof3formed usingthegivennumbersis852Ans. 11. Given: Find: Simplifyingtheexpression,weget So 3, and 3 4 4.Therefore, 7Ans. 12. Given:8%ofx%of200=4 Find:x Rewritingthepercentsasfractions,the problemstatementcanbeexpressed algebraicallyas 8 200 4 100 100 Simplifyingandsolvingforx,weget 16 400and 25Ans. 13. Given:Thereare1929th‐gradersand 13610th‐gradersenteredinadrawing. Find:theprobabilitya10th‐graderwins thedrawing. Theprobabilityis Ans. 14. Given:2 ∙ 4 2 Find:k 2 2 ∙4 2 ∙ 2 2 ∙2 2 . Therefore, 10Ans. 15. Given:AcirclewithcenterP(5,10) intersectsthex‐axisatQ(5,0). Find:theareaofthecircle. P(5, 10) Q(5, 0) TheradiusofthecircledrawnfromPto Q,asshown,haslength.Thisistheradius 10–0=10.Theareaisπ 100πAns. Nowlet’strytorotatethefirstdiesoitis positionedexactlyliketheseconddieso thatcorrespondingsidesonthedicehave thesamenumberofdotsandarealigned thesameway.Ifwerotatethefirstdieso thatthefrontfaceisnowpositionedon thetop,weobtainthearrangement showninFigure2. 16. Given:Subtract3from2xanddividethe differenceby5.Theresultis7. Find:x Theproblemstatementcanbeexpressed algebraicallyas 7.Solvingforx,we seethat2 3 35and2 38.So, 19Ans. 17. Given:Melina’sratioof2‐pointshot attemptsto3‐pointshotattemptsis4:1. Find:thepercentofMelina’sattempted shotsthatare3‐pointshots. Ofevery5shotattempts,1isa3‐point shotsand4are2‐pointshots.So,the 3‐pointshotsaccountfor Figure1 Ifwerotatethefirstdieagainsothatthe currentfrontfaceisnowontheright,we getthearrangementinFigure3. 20%Ans. 18. Find:thesumofthedistinctprime factorsof2016. Theprimefactorizationof2016is 2 7 3 .Thedistinctprimefactors2, 3and7havethesum2 3 7 12Ans. 19. Given:Theoppositefacesofasix‐sided dieaddupto7.Twoidenticalsix‐sided diceareplacedasshown. Find:thesumofthenumberofdotson thetwofacesthattoucheachother. Sincethedotsonoppositefacesofadie addto7,itfollowsthattheopposingsides mustbe6and1,5and2,4and3.As Figure1shows,thefirstdieispositioned sothatthetopfacehas5dotsandthe bottomfacehas2dots.Therightfacehas 6dots,andtheleftfacehas1dot.The frontfacehas3dots,andthebackface, whichtouchesthefrontfaceofthesecond die,has4dots. Figure2 Figure3 Nowthatallcorrespondingfacesonthe dicearethesame,wecanseethatonthe seconddie,thefrontface,whichtouches thebackfaceofthefirstdie,has1dot. Thesumofthesetwovalues,then,is 4 1 5Ans. 20. Find:howmanysetsoftwoorthree distinctpositiveintegershaveasumof8? Thereare3setsoftwointegersthathave asumof8:{1,7},{2,6},{3,5}.Thepair {4,4}doesn’tmakeitbecausethevalues aren’tdistinct.Thereare2setsofthree integersthathaveasumof8:{1,2,5}, {1,3,4}.Thatmakesthetotalnumberof sets3 2 5Ans. 21. Given: 3, 4, 5 Find: Thethreegivenequationscanbe rewrittenas 6 8 10. Addingthesethreeequationsyields 2 2 2 24.Dividingbothsides by2,weget 12Ans. 22. Given:Themeanofasetof5numbersis 3k.Addasixthnumbertothesetandthe meanincreasesbyk. Find:theratioofthesixthnumbertothe sumofthefirst5numbers. LetSbethesumofthefirstfivenumbers. Thenwehave 24. Find:theleastpositiveinteger suchthat !isdivisibleby1000. Since1000=23×53,wearelookingforn suchthatn!=2×2×2×5×5×5×k. Withfactorsof2and4wehavethree2s. Withthefactorsof5,10and15,wehave three5s.Theleastvalueisn!=1×2 × 3×···×14×15.So,n=15Ans. 25. Given:△ABCisanisoscelestriangle. AB=AC,m∠A=32°.TrianglesABCand PQRarecongruent,andm∠PXC 114°. Find:thedegreemeasureof∠PYC. A 32˚ 3kandS=15k.Letxbe thesixthnumber.Then Therefore,theratio is Ans. 23. Given:arectanglecomposedof4squares. Theareaoftherectangleis240. Find:theperimeter. R W 114˚ 3k+kand S+x=24k.Substituting15kforSin S+x=24kyields15k+x=24k.So,x=9k. X Z P B VC ? Y Q Figure4 SinceisoscelestrianglesABCandPQRare congruent,AB=AC=PQ=PR,m∠P m∠A 32° andm∠B m∠C m∠Q m∠R 74°.Sincem∠PXC 114°,itfollowsthatm∠AXZ 66°. A 2x 3x 32˚ x x x 3x Letx=thesidelengthofthesmallest square.Thesidelengthofthemedium squareis2x,andthesidelengthofthe largestsquareis3x.Therectanglehas lengthx+x+3x=5xandwidth3x.The areais(5x)(3x)=240.Solvingforx,we get15x2=240,sox2=16andx=4 (xcan’tbe−4).Thus,therectanglehas length5(4)=20,width3(4)=12.The perimeter,then,is2(20+12)=64Ans. P 114˚ B Figure5 X Z R W VC ? Y Q Nowlet’slookat△RXWand△WCV, showninFigure5.Bypropertiesof verticalangles,m∠RXW m∠AXZ.Wealso knowthatm∠RWX 180– 74 66 180–140 40°.Again,bypropertiesof verticalangles,m∠VWC m∠RWX 40°. Allthreeworkerspaintatthesamerate; wewillcallthisRunitsoffencesperhour. ForMonday,whentheentirejobwas completedin3.5hours,wehave R×2+3R×1.5 6.5R. NowletTbethetimeworker1painted alone.ForTuesday,whentheentirejob wascompletedin2.9hours,wehave R×T+3R×(2.9–T) 8.7R–2RT. Settingthesetwoequationsequaltoeach otherandsolvingforT,weget 6.5R=8.74R–2RTand2.2R=2RTso T=1.1hours.Convertingtominutes, 1.1×60 66Ans. Sincem∠RWX 40°andm∠C 74°, itfollowsthat△RXW~△WCV (Angle‐Angle)andm∠CVW 66°. A 32˚ X Z P 114˚ B Figure6 ? Y R W V Q C Nowlet’slookatand△YVQ,shownin Figure6.Onceagain,bypropertiesof verticalanglesm∠YVQ m∠CVW=66°. Sincem∠YVQ 66°andm∠Q 74°,it followsthat△WCV~△YVQand m∠VYQ 40°.SinceanglesPYCandVYQ aresupplementary,weknowthat m∠PYC 180–40 140°Ans. 26. Given:Theintegers1‐66arearrangedas shown. Find:sumofthenumbersincolumnD. A 1 B 2 7 8 C 3 6 9 12 D E 5 4 11 10 28. Given:△LMNhasaltitudeMH.Circlesare inscribedintrianglesMNHandMLH, tangenttoaltitudeMH. MA:AT:TH=4:2:1 Find:theratioofthesmallercircle’sarea tothelargercircle’sarea. M 4r A r L ThenumbersincolumnDalldifferby6 andrangefrom5to65.Thesumofthese numbersis5 11 ⋯ 59 65 70 35 11 T 3r N H Letrbetheradiusofthesmallercircle.So TH=r,AT=2randMA=4r.Itfollows, then,thattheradiusofthelargercircle haslengthAH=2r+r=3r.Theareasof 385Ans. 27. Given:OnMonday,asingleworker paintedafencealonefortwohours.Then twomorepainterscameand,together, theyfinishedthejob1.5hourslater.On Tuesday,asingleworkerbeganpainting anidenticalfenceat8:00a.m.Later2 moreworkersshowedupand,together, the3workersfinishedthefenceat10:54. Find:thenumberofminutesthefirst workerpaintedaloneonTuesday. 2r thetwocirclesareπx2andπ(3x)2=9πx2, andtheratiooftheareasis Ans. Thegreatestpalindromewecanformis 89998,but7×89998=629986,whichis notapalindrome.Wecontinuelookingat palindromesoftheform89__98in descendingorder.Noticethatwitheach successivepalindromeoflessvalue,the resulting6‐digitproductof7andthe palindromedecreasesby700.Also,the seconddigitintheproductwillalways remaina2andthefifthdigitwillalways bean8.Next,lookingat88__88,the greatestpalindromewecanformis 88988,but7×88988=622916,whichis notapalindrome.Thefifthdigitwill alwaysbea1,sowearelookingtobring theseconddigitdowntoa1.Weare subtracting700eachtime.Welookatthe second,thirdandfourthdigittofigureout whatmultipleof3makestheseconddigit a1andthethirdandfourthdigitequal. 229−63=166.Wecangettothe6‐digit palindrome616616bymultiplying7and 88088Ans. 29. Given:Abugcrawlsnmilesatn+1mi/h ononeday.Thenextdayitcrawls 2n+1milesatn2+nmi/h.Thetotaltime forthetripis6hours. Find:thebug’saveragespeed. Onthefirstdaythebugspends hourscrawling.Onthesecondday, thebugspends hourscrawling.Since theentiretriptookatotalof6hoursover 6. thetwodays,wehave Solvingforn,weget 2 1 6 1 1 2 1 6 1 5 5 2 4 1 1 6 6 1 0 1 0 (ncan’tbe−1) Substituting,wefindthat,onthefirstday, thebugcrawled mileat 1 1 mi/h.Onthesecondday,itcrawled 2 1 1 milesat mi/h.Thisisatotalof 1 1 milesin6hours.Sotheaveragespeed was1 6 Ans. 30. Find:thegreatest5‐digitpalindromen suchthat7nisa6‐digitpalindrome. Startbylookingata5‐digitpalindrome withleadingdigit9oftheform9______9. Since9×7=63,thelastdigitinthe 6‐digitproductwillbea3,butthereinno waytohavetheleadingdigitbea3. Nextwecheckthosewithleadingdigit8 oftheform8______8. Since8×7=56,thelastdigitinthe 6‐digitproductwillbea6,andtheleading digitcouldpossiblybea6dependingon whattheotherdigitsare. TargetRound 2 1. Given:Aseasonskipasscosts$395.Aday ticketcosts$40. Find:Howmuchmoneyissavedwitha seasonskipasscomparedto38day tickets? 38dayticketscost$40×38=$1520 $1520–$395=$1125Ans. 2. Given:Astringofnumbersconsistsofone 1,two2s,three3s,etc. Find:the50thdigit. Thesumof1through9is45.Therefore, startingwiththe46thdigitwehave10101. The50thdigitis1Ans. 3. Given:Eunicetakes2minutes30seconds torunaroundaquarter‐miletrack. Find:Howmanyminutesittakesherto run1mile. 1mileis4quarter‐milelapswitheachlap taking2minutes30seconds. 2lapstake5minutes. 4lapstake10minutesAns. 4. Given:RectangleABCDiscomposedof11 congruentrectangles.ThelengthofABis 33. Find:theareaofrectangleABCD. A B D C Letx=thewidthofoneoftherectangles. Lety=thelengthoftherectangle. IFABhaslength33, 2 33 ThelengthofEFisthesameasthelength ofAD.Therefore,4 3 and Substitutingbackintothefirstequation: 11 4 11 3 4 33 33 132 12 3 36 1188units2Ans. 36 33 5. Given:Sumofthefirst5termsofa sequenceis90lessthanthesumofthe next5terms. Find:theabsolutedifferencebetween twoconsecutivetermsofthissequence. Termsofanarithmeticsequencedifferby thesameamount. Leta=thefirsttermanddbethe commondifferencebetweenconsecutive terms.Thesumofthefirst5termsis a+a+d+a+2d+a+3d+a+4d= 5a+10d.Thesumofthenext5terms is a+5d+a+6d+a+7d+a+8d+a+9d= 5a+35d.Thedifferencebetweenthese twovaluesis(5a+35d)–(5a+10d)=90. Simplifyingandsolvingfordyields 25d=90andd Ans. 6. Given:thehourandminutehandsofa clockcreatea60°angleat2:00. Find:howmanysecondslateristhenext timewhenthehourandminutehands createa60°angle. Theminutehandtakes1hour,3600 seconds,togoaroundtheclock. Therefore,theminutehandmoves °everysecond.Thehourhandtakes12 hours,12×3600=43200seconds,togo aroundtheclock.Therefore,thehour handmoves °everysecond. Nowlet’slookataclockshowing2:00. 11 12 1 10 9 8 7 6 5 2 3 4 Theminutehandisat0°andthehour handisat60°. Everysecondtheminute handmoves °andthehourhandmoves °.Thismeansthateachsecondthe minutehandcomes °closertothehourhand. Letx=thenumberofsecondsitwilltake forthetwohandstocoincide.Then60 ,and 654. 54seconds. Atthispointthehandsareexactlyontop ofeachother.Itwilltakethesame amountoftimefortheminutehandtogo pastthehourhandtoforma60°angle. 654. 54 2 1309. 09 1309Ans. 7. Given:Abagcontainssomenumberof blueandexactly11yellowmarbles.When 5bluemarblesareadded,theprobability ofrandomlydrawingabluemarble exceeds70%. Find:theleastpossiblenumberofblue marblesinthebagoriginally. Letb=theoriginalnumberofblue marbles.Wehavethefollowing: 5 7 5 11 10 5 7 16 10 Cross‐multiplying,weget 10 50 7 112.Solvingforbyields 3 62,so 20 .Substituting,the leastintegergreaterthan20 ,weget 26 21 5 70.207% 21 16 37 whenb=21Ans. 8. Given:Aroomhaseightlightswitches. Initially,exactly5ofthelightsareon.Three peopleentertheroomsequentiallyand independentlyfliponeswitchrandomly. Find:theprobabilitythatafterthethird personhasexitedtheroom,exactlysixof thelightsareon. Inordertoendupwithexactly6lights on,thethreepeoplemustturntwoonand oneoff,insomeorder.Otheroptionsthat wouldnotresultinsixlightsonare turningonthreelights,turningoffthree lightsorturningoneonandtwooff. Thereare3×2×1=6waystoturnthree lightson.Thisisbecausewhenthefirst personenterstheroomthenumberof lightsoffis3,whenthesecondperson enterstheroomthereare2lightsoffand whenthethirdpersonenterstheroom thereis1lightoff. Similarly,thereare5×4×3=60waysto turnthreelightsoff. Thereare5×4×5=100waystoturn twolightsoffthenoneon,inthatorder. Thereare5×4×5=100waystoturn onelightoff,onelightonandonelightoff. Thereare3×6×5=90waystoturnone lightonthentwolightsoff. Thereare3×2×7=42waystoturntwo lightsonthenoneoff. Thereare3×6×3=54waystoturnone lighton,onelightoffthenonelighton. Thereare5×4×3=60waystoturnone lightoffthentwolightson. Thereareatotalof6+60+100+100+ 90+42+54+60=512waysforthreeof theeightswitchestobeflipped. Ofthese512ways,42+54+60=156 waystoendwithexactly6lightson. Theprobabilityofexactly6lightsbeing onafterthethirdpersonexitstheroomis Ans. TeamRound 1. Given:Apackageof8frankfurterscosts $5.Apackageof12bunscosts$3. Find:thepercentofthecostofahotdog sandwichthatcomesfromabun. Afrankfurtercosts Abuncosts $0.625 $0.25 0.25 0.25 0.285714 0.625 0.25 0.875 28.5714% 29%Ans. 2. Given:TrianglesABC,ACD,ADEandAEF areisoscelesrighttriangles. Find:thevalueof D . E C B A F Letx=thelengthofAB.ACisthe hypotenuseoftriangleABC. Sinceallthetrianglesare45‐45‐90,we knowtheratioofthesidetothe hypotenuseis1to√2.SoAC=x√2,AD= 2x,AE=2√2xandAF=4x. Ans. 3. Find:the6‐digitnumberthathasthese properties: ‐Noneofthedigitsrepeat. ‐Theonesdigitisaprimenumber. ‐Theonesdigitisthesumofthetensand hundredsdigits. ‐Thethousandsdigitisthesumofthe hundredsandten‐thousandsdigits. ‐Thehundred‐thousandsdigitisneither primenorcomposite. ‐Thesumofthedigitsis21. Thehundred‐thousandsdigitisnot prime,soitcannotbe2,3,5or7.Itisalso notcomposite.Soitcan’tbe4,6,8or9. Therefore,itmustbe1. Theonesdigitisprime.Therefore,itcan be2,3,5,or7.Butitisalsothesumofthe tensandhundredsdigit.Sotheonesdigit cannotbe2becausethatwouldmakethe tensandhundredsdigitbothbe1. Theonesdigitalsocannotbe3because thenoneofthetensorhundredsdigits wouldbe1andthatisalreadytakenby thehundredthousandsdigit. Sotheonesdigitcanbe5or7. Supposetheonesdigitis5. Thenthesumofthetensandhundreds digitsis5.Thatcouldbe1and4or2and 3.1isalreadyused.Solet’snowsuppose thatthetensdigitis3andthehundreds digitis2. Thesumofallthedigitsis21.Wehave alreadyused1,2,3,and5.Sotheother twovaluesmustaddupto10.Thatcould be1and9,2and8,3and7,or4and6. Only4and6haven’tbeenused. Ifthethousandsdigitis6,itwouldhave tobethesumofthehundredsdigit(2) andtheten‐thousandsdigit,whichwould havetobe4.Thenumber:146,235Ans. 4. Given:abicycletirehasadiameterof 22inches.Amotorcycletirehasa diameterof25inches. Find:howmanymorefeetthemotorcycle travelsthanthebicycleaftereachtirehas made1000revolutions. After1000revolutionsofthetire,the bicycletraveled22,000πinches.The motorcycletraveled25,000πinches. That’sadifferenceof3000πinches. Convertingtofeet: 250πAns. 5. Find:thesumofthepositiveintegers from1through500thataredivisibleby 2,3,4,5and6. TheLCMof2,3,4,5and6is2×2×3×5 =60.Themultiplesof60between1and 500are60,120,180,240,300,360,420 and480.Theresumis60 120 180 240 300 360 420 480 540 4 2160Ans. 6. Given:themean,median,uniquemode andrangeof10integersareall10. Find:greatestpossibleintegerinthelist. Sincethemeanis10,thesumofthe10 integersis100.Themedianis10which meansthatthefifthandsixthintegersare both10ortheaverageofthefifthand sixthintegersis10.However,sincethe modeisalso10,thefifthandsixth integersmustbe10.Thismeansthe largesttheminimumvaluecouldpossibly bewouldbe10,andsincetherangeis10, thiswouldmakethemaximum20.The listofintegerswouldbe10,10,10,10,10, 10,10,10,10,20,butthisdoesn’twork sincethissumsto110.Sohowabout19 asthemaximum? Wecanstartwith9andendwith19. 9 19 28 Thatleaves8integersbetween10and18 tosumto100 28 72 That’snotgoingtowork. Howaboutahighof18? Thatwillwork. Wecanstartwith8andendwith18. 8 18 26 Thatleaves8integerstosumto 100 26 74 8,8,8,8,10,10,10,10,10,18 These10numberssumto100foramean of10.Themedianis10,themodeis10 andtherangeis10.Thegreatestpossible integeris18Ans. 7. Given:A200millilitersolutionis7% detergent. Find:Howmanymillilitersof100% detergentneedtobeaddedsothe solutionwillbe14%detergent? 7%of200millilitersis14millilitersof detergent. Letx=theamountweneedtoaddtogeta solutionof14%detergent. 14 0.14 200 14 28 0.14 0.86 14 . 16.27906 16.3mlAns. 8. Find:howmany3‐digitnumbersare theresuchthateachofthedigitsisprime andthesumofthedigitsisprime? Theprimesingledigitnumbersare2,3,5 and7. Consider,first,thateachdigitisdifferent. Forgetusinganyothercombinationwith one2becausetwooddsplusoneeven willalwaysbeanevennumber. 3 5 7 15 Thisisn’tprime. Nowconsiderthattwoofthedigitsare thesame. 2 2 3 7 Thatworksandthereare3permutations. 2 2 5 9 Thatdoesn’twork. 2 2 7 11 Thatworks.Thereare3permutations. 3 3 5 11 Thatworks.Thereare3permutations. 3 3 7 13 Thatworks.Thereare3permutations. 5 5 3 13 Thatworks.Thereare3permutations. 5 5 7 17 Thatworks.Thereare3permutations. 7 7 3 17 Letd=thenumberoftokensDevinhas Letk=thenumberoftokensKevinhas isasquare isasquare dcannotbe1or4,sinceKevinwould havemoretokensthanDevin. Suppose 9 9 7 16; 9 7 2 Thiscannotwork.Therearenomore valuesforkwhere . Suppose 16 16 9 25; 16 9 7 Thiscannotwork.Therearenomore valuesforkwhere . Suppose 25. 25 11 36; 25 11 14 25 24 49; 25 24 1 Andwehaveit. 24Ans. Thatworks.Thereare3permutations. 7 7 5 19 Thatworks.Thereare3permutations. Wedonotneedtoconsider3ofthesame primebecausethesumof3ofthesame primeisalwaysgoingtobedivisibleby3. Sowenowhave8×3=24numbersAns. 9. Find:Howmanydifferentpathsfromtop tobottomspellALGEBRA? Fromthefirstline(withA)tothesecond line(withL)therearetwopaths(e.g.,2 ): AL1andAL2. Fromthesecondlinetothethirdline (withG)thereare4paths: AL1G1,AL1G2,AL2G2,AL2G3(e.g.,2 ). Andwecanseethepattern. Thereare23pathstothefourthline(with E),24pathstothefifthline(withB),25 pathstothesixthline(withR)and26 pathstotheseventhline(withA). 64Ans. 2 10. Given:ThenumberoftokensDevinhasis asquarenumberlessthan100.IfKevin givesDevinhistokens,Devinwillhavea totalnumberoftokensthatisstilla square.IfDevingivesKevinthesame numberoftokensthatKevinalreadyhas, thenumberoftokensDevinisleftwithis alsosquare. Find:HowmanytokensdoesKevinhave? Listthenumberofsquareslessthan100: 1,4,9,16,25,36,49,64,81 ForDevintogiveKevinthesamenumber oftokensthatKevinalreadyhas,Devin musthavemoretokensthanKevin. Sowhatwehaveisthefollowing: