Download 2016 Chapter Competition Solutions

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: