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.