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!

2016ChapterCompetition

SprintRound

1. Given:Countbackwardsfrom155by4.Find:The9thnumber.Thefirstnumberis155.Thesecondnumberis151,or4less.Theninthnumberis4 8 32less.155–32=123Ans.

2. Given:Thegraph,asdisplayedbelow.Find:Thedifferencebetweenthemaximumandminimumvalues.

Thehighestpointintermsofyisthethirdpointfromtherightwithcoordinates(1,3).Thelowestpointintermsofyisthesecondpointfromtherightwithcoordinates(−2.5,−4)–atleastitlookslikearound–2.5.Thedifferenceinthey‐coordinatesis3 4 7Ans.

3. Given: , 1

Find:aSubstituting−1forb,weget1/−1=−1/a.Cross‐multiplying,wefindthata=(−1)(−1)=1Ans.

4. Find:Thesumofalltwo‐digitmultiplesof3thathaveaunitsdigitof1.21isthefirsttwo‐digitmultipleof3whichhasaunitsdigitof1.Eachsuccessivetwo‐digitmultipleof3withaunitsdigitof1is30larger.Therearethreeinall:21,51and81.Theirsumis21+51+81=153Ans.

5. Find:thesumofthefirst8termsofthesequencebeginning‐4,5…whereeachtermisthesumoftheprevioustwoterms.Thefirsteighttermsofthesequenceare−4,5,1,6,7,13,20and33.Theirsumis−4+5+1+6+7+13+20+33=81Ans.

6. Given:Thegraphdisplayedbelow.Find:thetotalcosttoship3packagesweighing1.8lbs.,2lbs.and4.4lbs.

Itcosts$3toshipthe1.8‐lb.package,$3.50toshipthe2‐lb.packageand$5toshipthe4.4‐lb.package.Thetotalcosttomailallthreepackagesis3 3.50 511.50Ans.

7. Given:thegraphbelow.Find:thedegreemeasureoftheportionofthecookiethatisflour.

Thecookieismadeof40%flourand.40%of360°is360 0.4 144Ans.

8. Given:3zogutsand4gimunscosts$18.2zogutsand3gimunscosts$13.Find:thecostof1zogutand1gimunLetz=thecostofazogutandg=thecostofagimun.Wecanwritetheequations

x

y

2

4

2 4−4 −2

−4

−2

S������� R��

Weight (lbs)

Co

st (

do

llar

s)

1

2

3

4

0

5

6

1 2 3 4 5 6x

y

40%Flour

22%Sugar

18%

Chocolate Chips

16%Butter

4%Eggs

M�’ C�� � D����

3 4 18(Eq.1)2 3 13(Eq.2)SubtractingEq.2fromEq.1,weget1 1 18 13 5Ans.

9. Given:Arectangularpieceofpaperwitha40‐inchlength.Itisfoldedinhalf.Theratioofthelongsideoftheoriginalsheettotheshortsideoftheoriginalsheetisthesameastheratioofthelongsideofthefoldedsheettotheshortsideofthefoldedsheet.Find:thelengthoftheshortsideoftheoriginalsheet.Lets=thelengthoftheshortsideoftheoriginalsheet,whichisalsothelongsideofthefoldedsheet.Thelengthoftheshortsideofthefoldedsheetis40/2=20.

Wehavetheratio .Cross‐

multiplyingyields 800.Sothelengthoftheshortsideoftheoriginalsheetis

√800 10√8 20√2Ans.

10. Find:thegreatestmultipleof3thatcanbeformedusingoneormoreofthedigits2,4,5and8,usingeachdigitonlyonce.Anumberisamultipleof3ifthesumofitsdigitsisdivisibleby3.Since8+5+4+2=19,whichisnotdivisible3,weknowthatnonumberformedusingallfourdigitswillbeamultipleof3.Thefourcombinationsofthreedigitsthatcanbeusedtoform3‐digitnumbersare8,5,4;8,5,2;8,4,2;and5,4,2.Since8+5+4=17,8+4+2=14and5+4+2=11,noneofwhichisdivisibleby3,weknowthatno3‐digitnumberformedwiththesecombinationsofdigitswillbeamultipleof3.But8+5+2=15,whichisdivisibleby3.Thegreatestmultipleof3formedusingthegivennumbersis852Ans.

11. Given:

Find: Simplifyingtheexpression,weget

So 3, and 4.Therefore,3 4 7Ans.

12. Given:8%ofx%of200=4

Find:xRewritingthepercentsasfractions,theproblemstatementcanbeexpressedalgebraicallyas8100 100

200 4

Simplifyingandsolvingforx,weget16 400and 25Ans.

13. Given:Thereare1929th‐gradersand13610th‐gradersenteredinadrawing.Find:theprobabilitya10th‐graderwinsthedrawing.

Theprobabilityis Ans.

14. Given:2 ∙ 4 2 Find:k2 2 ∙ 4 2 ∙ 2 2 ∙ 2 2 .Therefore, 10Ans.

15. Given:AcirclewithcenterP(5,10)intersectsthex‐axisatQ(5,0).Find:theareaofthecircle.

TheradiusofthecircledrawnfromPtoQ,asshown,haslength.Thisistheradius10–0=10.Theareaisπ 100πAns.

P(5, 10)

Q(5, 0)

16. Given:Subtract3from2xanddividethedifferenceby5.Theresultis7.Find:xTheproblemstatementcanbeexpressed

algebraicallyas 7.Solvingforx,we

seethat2 3 35and2 38.So,19Ans.

17. Given:Melina’sratioof2‐pointshot

attemptsto3‐pointshotattemptsis4:1.Find:thepercentofMelina’sattemptedshotsthatare3‐pointshots.Ofevery5shotattempts,1isa3‐pointshotsand4are2‐pointshots.So,the

3‐pointshotsaccountfor 20%Ans.

18. Find:thesumofthedistinctprime

factorsof2016.Theprimefactorizationof2016is2 7 3 .Thedistinctprimefactors2,3and7havethesum2 3 7 12Ans.

19. Given:Theoppositefacesofasix‐sideddieaddupto7.Twoidenticalsix‐sideddiceareplacedasshown.Find:thesumofthenumberofdotsonthetwofacesthattoucheachother.

Figure1

Sincethedotsonoppositefacesofadieaddto7,itfollowsthattheopposingsidesmustbe6and1,5and2,4and3.AsFigure1shows,thefirstdieispositionedsothatthetopfacehas5dotsandthebottomfacehas2dots.Therightfacehas6dots,andtheleftfacehas1dot.Thefrontfacehas3dots,andthebackface,whichtouchesthefrontfaceoftheseconddie,has4dots.

Nowlet’strytorotatethefirstdiesoitispositionedexactlyliketheseconddiesothatcorrespondingsidesonthedicehavethesamenumberofdotsandarealignedthesameway.Ifwerotatethefirstdiesothatthefrontfaceisnowpositionedonthetop,weobtainthearrangementshowninFigure2.

Figure2

Ifwerotatethefirstdieagainsothatthecurrentfrontfaceisnowontheright,wegetthearrangementinFigure3.

Figure3

Nowthatallcorrespondingfacesonthedicearethesame,wecanseethatontheseconddie,thefrontface,whichtouchesthebackfaceofthefirstdie,has1dot.Thesumofthesetwovalues,then,is4 1 5Ans.

20. Find:howmanysetsoftwoorthreedistinctpositiveintegershaveasumof8?Thereare3setsoftwointegersthathaveasumof8:{1,7},{2,6},{3,5}.Thepair{4,4}doesn’tmakeitbecausethevaluesaren’tdistinct.Thereare2setsofthreeintegersthathaveasumof8:{1,2,5},{1,3,4}.Thatmakesthetotalnumberofsets3 2 5Ans.

21. Given: 3, 4, 5

Find: Thethreegivenequationscanberewrittenas

6810.

Addingthesethreeequationsyields2 2 2 24.Dividingbothsidesby2,weget 12Ans.

22. Given:Themeanofasetof5numbersis3k.Addasixthnumbertothesetandthemeanincreasesbyk.Find:theratioofthesixthnumbertothesumofthefirst5numbers.LetSbethesumofthefirstfivenumbers.

Thenwehave 3kandS=15k.Letxbe

thesixthnumber.Then 3k+kand

S+x=24k.Substituting15kforSinS+x=24kyields15k+x=24k.So,x=9k.

Therefore,theratio is Ans.

23. Given:arectanglecomposedof4squares.

Theareaoftherectangleis240.Find:theperimeter.

Letx=thesidelengthofthesmallestsquare.Thesidelengthofthemediumsquareis2x,andthesidelengthofthelargestsquareis3x.Therectanglehaslengthx+x+3x=5xandwidth3x.Theareais(5x)(3x)=240.Solvingforx,weget15x2=240,sox2=16andx=4(xcan’tbe−4).Thus,therectanglehaslength5(4)=20,width3(4)=12.Theperimeter,then,is2(20+12)=64Ans.

24. Find:theleastpositiveintegersuchthat !isdivisibleby1000.

Since1000=23×53,wearelookingfornsuchthatn!=2×2×2×5×5×5×k.Withfactorsof2and4wehavethree2s.Withthefactorsof5,10and15,wehavethree5s.Theleastvalueisn!=1×2 ×3×···×14×15.So,n=15Ans.

25. Given:△ABCisanisoscelestriangle.AB=AC,m∠A=32°.TrianglesABCandPQRarecongruent,andm∠PXC 114°.Find:thedegreemeasureof∠PYC.

Figure4

SinceisoscelestrianglesABCandPQRarecongruent,AB=AC=PQ=PR,m∠P m∠A 32° andm∠B m∠C m∠Q

m∠R 74°.Sincem∠PXC

114°,itfollowsthatm∠AXZ 66°.

Figure5

Nowlet’slookat△RXWand△WCV,showninFigure5.Bypropertiesofverticalangles,m∠RXW m∠AXZ.Wealsoknowthatm∠RWX 180– 74 66 180–140 40°.Again,bypropertiesofverticalangles,m∠VWC m∠RWX 40°.

2x3x

3xxx x

A

B

XZ

W

V

RP

QY

C

32˚

114˚

?

A

B

XZ

W

V

RP

QY

C

32˚

114˚

?

Sincem∠RWX 40°andm∠C 74°,itfollowsthat△RXW~△WCV(Angle‐Angle)andm∠CVW 66°.

Figure6

Nowlet’slookatand△YVQ,showninFigure6.Onceagain,bypropertiesofverticalanglesm∠YVQ m∠CVW=66°.Sincem∠YVQ 66°andm∠Q 74°,itfollowsthat△WCV~△YVQandm∠VYQ 40°.SinceanglesPYCandVYQaresupplementary,weknowthatm∠PYC 180–40 140°Ans.

26. Given:Theintegers1‐66arearrangedasshown.Find:sumofthenumbersincolumnD.

ThenumbersincolumnDalldifferby6andrangefrom5to65.Thesumofthesenumbersis5 11 ⋯ 59 65

70 35 11 385Ans.

27. Given:OnMonday,asingleworker

paintedafencealonefortwohours.Thentwomorepainterscameand,together,theyfinishedthejob1.5hourslater.OnTuesday,asingleworkerbeganpaintinganidenticalfenceat8:00a.m.Later2moreworkersshowedupand,together,the3workersfinishedthefenceat10:54.Find:thenumberofminutesthefirstworkerpaintedaloneonTuesday.

Allthreeworkerspaintatthesamerate;wewillcallthisRunitsoffencesperhour.ForMonday,whentheentirejobwascompletedin3.5hours,wehaveR×2+3R×1.5 6.5R.NowletTbethetimeworker1paintedalone.ForTuesday,whentheentirejobwascompletedin2.9hours,wehaveR×T+3R×(2.9–T) 8.7R–2RT.SettingthesetwoequationsequaltoeachotherandsolvingforT,weget6.5R=8.74R–2RTand2.2R=2RTsoT=1.1hours.Convertingtominutes,1.1×60 66Ans.

28. Given:△LMNhasaltitudeMH.CirclesareinscribedintrianglesMNHandMLH,tangenttoaltitudeMH.MA:AT:TH=4:2:1Find:theratioofthesmallercircle’sareatothelargercircle’sarea.

Letrbetheradiusofthesmallercircle.SoTH=r,AT=2randMA=4r.Itfollows,then,thattheradiusofthelargercirclehaslengthAH=2r+r=3r.Theareasof

thetwocirclesareπx2andπ(3x)2=9πx2,

andtheratiooftheareasis

Ans.

A

B

XZ

W

V

RP

QY

C

32˚

114˚

?

A1

7

B2

8

C3

6

9

12

D

5

11

E

4

10

�����

M

NLH

T

A2r

3r

4r

r

29. Given:Abugcrawlsnmilesatn+1mi/hononeday.Thenextdayitcrawls2n+1milesatn2+nmi/h.Thetotaltimeforthetripis6hours.Find:thebug’saveragespeed.Onthefirstdaythebugspends

hourscrawling.Onthesecondday,

thebugspends hourscrawling.Since

theentiretriptookatotalof6hoursover

thetwodays,wehave 6.

Solvingforn,weget

12 1

16

2 16 1

2 1 6 6 5 4 1 05 1 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‐digitpalindromewithleadingdigit9oftheform9______9.Since9×7=63,thelastdigitinthe6‐digitproductwillbea3,butthereinnowaytohavetheleadingdigitbea3.Nextwecheckthosewithleadingdigit8oftheform8______8.Since8×7=56,thelastdigitinthe6‐digitproductwillbea6,andtheleadingdigitcouldpossiblybea6dependingonwhattheotherdigitsare.

Thegreatestpalindromewecanformis89998,but7×89998=629986,whichisnotapalindrome.Wecontinuelookingatpalindromesoftheform89__98indescendingorder.Noticethatwitheachsuccessivepalindromeoflessvalue,theresulting6‐digitproductof7andthepalindromedecreasesby700.Also,theseconddigitintheproductwillalwaysremaina2andthefifthdigitwillalwaysbean8.Next,lookingat88__88,thegreatestpalindromewecanformis88988,but7×88988=622916,whichisnotapalindrome.Thefifthdigitwillalwaysbea1,sowearelookingtobringtheseconddigitdowntoa1.Wearesubtracting700eachtime.Welookatthesecond,thirdandfourthdigittofigureoutwhatmultipleof3makestheseconddigita1andthethirdandfourthdigitequal.229−63=166.Wecangettothe6‐digitpalindrome616616bymultiplying7and88088Ans.

TargetRound

1. Given:Aseasonskipasscosts$395.Adayticketcosts$40.Find:Howmuchmoneyissavedwithaseasonskipasscomparedto38daytickets?38dayticketscost$40×38=$1520$1520–$395=$1125Ans.

2. Given:Astringofnumbersconsistsofone1,two2s,three3s,etc.Find:the50thdigit.Thesumof1through9is45.Therefore,startingwiththe46thdigitwehave10101.The50thdigitis1Ans.

3. Given:Eunicetakes2minutes30secondstorunaroundaquarter‐miletrack.Find:Howmanyminutesittakeshertorun1mile.1mileis4quarter‐milelapswitheachlaptaking2minutes30seconds.2lapstake5minutes.4lapstake10minutesAns.

4. Given:RectangleABCDiscomposedof11congruentrectangles.ThelengthofABis33.Find:theareaofrectangleABCD.

Letx=thewidthofoneoftherectangles.Lety=thelengthoftherectangle.IFABhaslength33,2 33ThelengthofEFisthesameasthelength

ofAD.Therefore,4 3 and

Substitutingbackintothefirstequation:

234

33

114

33

11 13212

3 3636 33 1188units2Ans.

5. Given:Sumofthefirst5termsofasequenceis90lessthanthesumofthenext5terms.Find:theabsolutedifferencebetweentwoconsecutivetermsofthissequence.Termsofanarithmeticsequencedifferbythesameamount.Leta=thefirsttermanddbethecommondifferencebetweenconsecutiveterms.Thesumofthefirst5termsisa+a+d+a+2d+a+3d+a+4d=5a+10d.Thesumofthenext5terms isa+5d+a+6d+a+7d+a+8d+a+9d=5a+35d.Thedifferencebetweenthesetwovaluesis(5a+35d)–(5a+10d)=90.Simplifyingandsolvingfordyields

25d=90andd Ans.

6. Given:thehourandminutehandsofa

clockcreatea60°angleat2:00.Find:howmanysecondslateristhenexttimewhenthehourandminutehandscreatea60°angle.Theminutehandtakes1hour,3600seconds,togoaroundtheclock.

Therefore,theminutehandmoves

°everysecond.Thehourhandtakes12

hours,12×3600=43200seconds,togoaroundtheclock.Therefore,thehour

handmoves °everysecond.

Nowlet’slookataclockshowing2:00.

A B

CD

Theminutehandisat0°andthehourhandisat60°. Everysecondtheminute

handmoves °andthehourhandmoves

°.Thismeansthateachsecondthe

minutehandcomes

°closertothehourhand.

Letx=thenumberofsecondsitwilltakeforthetwohandstocoincide.Then60

,and 654. 54seconds.

Atthispointthehandsareexactlyontopofeachother.Itwilltakethesameamountoftimefortheminutehandtogopastthehourhandtoforma60°angle.654. 54 2 1309. 09 1309Ans.

7. Given:Abagcontainssomenumberofblueandexactly11yellowmarbles.When5bluemarblesareadded,theprobabilityofrandomlydrawingabluemarbleexceeds70%.Find:theleastpossiblenumberofbluemarblesinthebagoriginally.Letb=theoriginalnumberofbluemarbles.Wehavethefollowing:

55 11

710

516

710

Cross‐multiplying,weget10 50 7 112.Solvingforbyields

3 62,so 20 .Substituting,the

leastintegergreaterthan20 ,weget

21 521 16

2637

70.207%

whenb=21Ans.

8. Given:Aroomhaseightlightswitches.Initially,exactly5ofthelightsareon.Threepeopleentertheroomsequentiallyandindependentlyfliponeswitchrandomly.Find:theprobabilitythatafterthethirdpersonhasexitedtheroom,exactlysixofthelightsareon.Inordertoendupwithexactly6lightson,thethreepeoplemustturntwoonandoneoff,insomeorder.Otheroptionsthatwouldnotresultinsixlightsonareturningonthreelights,turningoffthreelightsorturningoneonandtwooff.Thereare3×2×1=6waystoturnthreelightson.Thisisbecausewhenthefirstpersonenterstheroomthenumberoflightsoffis3,whenthesecondpersonenterstheroomthereare2lightsoffandwhenthethirdpersonenterstheroomthereis1lightoff.Similarly,thereare5×4×3=60waystoturnthreelightsoff.Thereare5×4×5=100waystoturntwolightsoffthenoneon,inthatorder.Thereare5×4×5=100waystoturnonelightoff,onelightonandonelightoff.Thereare3×6×5=90waystoturnonelightonthentwolightsoff.Thereare3×2×7=42waystoturntwolightsonthenoneoff.Thereare3×6×3=54waystoturnonelighton,onelightoffthenonelighton.Thereare5×4×3=60waystoturnonelightoffthentwolightson.Thereareatotalof6+60+100+100+90+42+54+60=512waysforthreeoftheeightswitchestobeflipped.Ofthese512ways,42+54+60=156waystoendwithexactly6lightson.Theprobabilityofexactly6lightsbeingonafterthethirdpersonexitstheroomis

Ans.

12

6

398

7 54

2111

10

TeamRound

1. Given:Apackageof8frankfurterscosts$5.Apackageof12bunscosts$3.Find:thepercentofthecostofahotdogsandwichthatcomesfromabun.

Afrankfurtercosts $0.625

Abuncosts $0.25

0.250.625 0.25

0.250.875

0.285714

28.5714% 29%Ans.

2. Given:TrianglesABC,ACD,ADEandAEFareisoscelesrighttriangles.

Find:thevalueof .

Letx=thelengthofAB.ACisthehypotenuseoftriangleABC.Sinceallthetrianglesare45‐45‐90,weknowtheratioofthesidetothehypotenuseis1to√2.SoAC=x√2,AD=2x,AE=2√2xandAF=4x.

Ans.

3. Find:the6‐digitnumberthathasthese

properties:‐Noneofthedigitsrepeat.‐Theonesdigitisaprimenumber.‐Theonesdigitisthesumofthetensandhundredsdigits.‐Thethousandsdigitisthesumofthehundredsandten‐thousandsdigits.‐Thehundred‐thousandsdigitisneitherprimenorcomposite.‐Thesumofthedigitsis21.Thehundred‐thousandsdigitisnotprime,soitcannotbe2,3,5or7.Itisalso

notcomposite.Soitcan’tbe4,6,8or9.Therefore,itmustbe1.Theonesdigitisprime.Therefore,itcanbe2,3,5,or7.Butitisalsothesumofthetensandhundredsdigit.Sotheonesdigitcannotbe2becausethatwouldmakethetensandhundredsdigitbothbe1.Theonesdigitalsocannotbe3becausethenoneofthetensorhundredsdigitswouldbe1andthatisalreadytakenbythehundredthousandsdigit.Sotheonesdigitcanbe5or7.Supposetheonesdigitis5.Thenthesumofthetensandhundredsdigitsis5.Thatcouldbe1and4or2and3.1isalreadyused.Solet’snowsupposethatthetensdigitis3andthehundredsdigitis2.Thesumofallthedigitsis21.Wehavealreadyused1,2,3,and5.Sotheothertwovaluesmustaddupto10.Thatcouldbe1and9,2and8,3and7,or4and6.Only4and6haven’tbeenused.Ifthethousandsdigitis6,itwouldhavetobethesumofthehundredsdigit(2)andtheten‐thousandsdigit,whichwouldhavetobe4.Thenumber:146,235Ans.

4. Given:abicycletirehasadiameterof22inches.Amotorcycletirehasadiameterof25inches.Find:howmanymorefeetthemotorcycletravelsthanthebicycleaftereachtirehasmade1000revolutions.After1000revolutionsofthetire,thebicycletraveled22,000πinches.Themotorcycletraveled25,000πinches.That’sadifferenceof3000πinches.

Convertingtofeet: 250πAns.

5. Find:thesumofthepositiveintegers

from1through500thataredivisibleby2,3,4,5and6.

AB

C

D E

F

TheLCMof2,3,4,5and6is2×2×3×5=60.Themultiplesof60between1and500are60,120,180,240,300,360,420and480.Theresumis60 120 180240 300 360 420 480 540 4 2160Ans.

6. Given:themean,median,uniquemodeandrangeof10integersareall10.Find:greatestpossibleintegerinthelist.Sincethemeanis10,thesumofthe10integersis100.Themedianis10whichmeansthatthefifthandsixthintegersareboth10ortheaverageofthefifthandsixthintegersis10.However,sincethemodeisalso10,thefifthandsixthintegersmustbe10.Thismeansthelargesttheminimumvaluecouldpossiblybewouldbe10,andsincetherangeis10,thiswouldmakethemaximum20.Thelistofintegerswouldbe10,10,10,10,10,10,10,10,10,20,butthisdoesn’tworksincethissumsto110.Sohowabout19asthemaximum?Wecanstartwith9andendwith19.9 19 28Thatleaves8integersbetween10and18tosumto100 28 72That’snotgoingtowork.Howaboutahighof18?Thatwillwork.Wecanstartwith8andendwith18.8 18 26Thatleaves8integerstosumto100 26 748,8,8,8,10,10,10,10,10,18These10numberssumto100forameanof10.Themedianis10,themodeis10andtherangeis10.Thegreatestpossibleintegeris18Ans.

7. Given:A200millilitersolutionis7%detergent.Find:Howmanymillilitersof100%detergentneedtobeaddedsothesolutionwillbe14%detergent?7%of200millilitersis14millilitersofdetergent.Letx=theamountweneedtoaddtogetasolutionof14%detergent.14200

0.14

14 28 0.14 0.86 14

.16.27906 16.3mlAns.

8. Find:howmany3‐digitnumbersare

theresuchthateachofthedigitsisprimeandthesumofthedigitsisprime?Theprimesingledigitnumbersare2,3,5and7.Consider,first,thateachdigitisdifferent.Forgetusinganyothercombinationwithone2becausetwooddsplusoneevenwillalwaysbeanevennumber.3 5 7 15Thisisn’tprime.Nowconsiderthattwoofthedigitsarethesame.2 2 3 7Thatworksandthereare3permutations.2 2 5 9Thatdoesn’twork.2 2 7 11Thatworks.Thereare3permutations.3 3 5 11Thatworks.Thereare3permutations.3 3 7 13Thatworks.Thereare3permutations.5 5 3 13Thatworks.Thereare3permutations.5 5 7 17Thatworks.Thereare3permutations.7 7 3 17

Thatworks.Thereare3permutations.7 7 5 19Thatworks.Thereare3permutations.Wedonotneedtoconsider3ofthesameprimebecausethesumof3ofthesameprimeisalwaysgoingtobedivisibleby3.Sowenowhave8×3=24numbersAns.

9. Find:HowmanydifferentpathsfromtoptobottomspellALGEBRA?

Fromthefirstline(withA)tothesecondline(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).2 64Ans.

10. Given:ThenumberoftokensDevinhasisasquarenumberlessthan100.IfKevingivesDevinhistokens,Devinwillhaveatotalnumberoftokensthatisstillasquare.IfDevingivesKevinthesamenumberoftokensthatKevinalreadyhas,thenumberoftokensDevinisleftwithisalsosquare.Find:HowmanytokensdoesKevinhave?Listthenumberofsquareslessthan100:1,4,9,16,25,36,49,64,81ForDevintogiveKevinthesamenumberoftokensthatKevinalreadyhas,DevinmusthavemoretokensthanKevin.Sowhatwehaveisthefollowing:

Letd=thenumberoftokensDevinhasLetk=thenumberoftokensKevinhas

isasquareisasquare

dcannotbe1or4,sinceKevinwouldhavemoretokensthanDevin.Suppose 99 7 16; 9 7 2Thiscannotwork.Therearenomorevaluesforkwhere .Suppose 1616 9 25; 16 9 7Thiscannotwork.Therearenomorevaluesforkwhere .Suppose 25.25 11 36; 25 11 1425 24 49; 25 24 1Andwehaveit.

24Ans.