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1
Elementary Problems
Let’s begin with some relatively easy problems. The challenges become gradually more
difficultas yougothroughthebook.The problemsinthischaptercan be solvedwithout
advancedmathematics.Knowledgeofbasicarithmetic,algebra,andgeometrywillbehelp-
ful,aswellasyourowncreativethinking.Irecommendthatyouattemptalltheproblems,
evenifyoualreadyknowtheanswers, because youmay discovernewandinterestingas-
pectsofthesolutions.Abonusaftereach solutiondiscussesarelatedmathematicaltopic.
Remember,eachproblemhasanaha!solution.
1.1 Arithmetic
Fair Division
AbbyhasfifteencookiesandBettyhasninecookies.Carly,whohasnocookies,paysAbby
andBetty24centstosharetheircookies.Eachgirleatsone-thirdofthecookies.
Bettysays thatshe and Abbyshoulddividethe 24 cents evenly, each taking12 cents.
AbbysaysthatsinceshesuppliedfifteencookiesandBettyonlynine,sheshouldtake15
centsandBetty9cents.
Whatisthefairdivisionofthe24centsbetweenAbbyandBetty?
Solution
Thekeyistodeterminetheworthofonecookie.Eachgirleatseightcookies.SinceCarly
pays24centsforeightcookies,eachcookieisworth3cents.ThusAbby,whostartswith
fifteencookiesandsellsseventoCarly,shouldreceive21cents,andBetty,whostartswith
ninecookiesandsellsonetoCarly,shouldreceive3cents.
1
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2 1 ElementaryProblems
Bonus: Cookie Jar Division
Abby, Betty, and Carly have 21 cookie jars (all the same size). Seven are full,seven are
half-full,andseven areempty.The girlswishtodividethecookiejarsamongthemselves
sothateach girlgetsthesame numberofjarsandthesame amountofcookies. Howcan
theydothiswithoutopeningthejars?
Therearetwodifferentways,asshownbelow.
Abby F F F H E E E Abby F F H H H E E
Betty F F F H E E E Betty F F H H H E E
Carly F H H H H H E Carly F F F H E E E
Wehave designatedthefullcookiejarsbyF,thehalf-fullonesbyH,andtheemptyones
by E. In bothsolutions, each girlreceives seven jars and 31 jars’ worthof cookies. We
2
havenotcountedasdifferentthesamesolutionwiththegirls’namespermuted.
As we willsee later(in“Integer Triangles” onp. 165),each solutioncorresponds toa
trianglewithinteger sides and perimeter 7, as shown below. In a given triangle,the side
lengthsequal thenumberoffullcookiejars foreach girlinthecorrespondingcookiejar
solution.
(cid:0)@
h(h(h(h(h(3 h(h(h(h( 1 (cid:0)(cid:0)2 (cid:0)(cid:0) @@@2@
3
3
A Mere Fraction
(a) Findanintegralfractionbetween1=4and1=3suchthatthedenominatorisapositive
integerlessthan10.
(b) Findanintegralfractionbetween7=10and5=7suchthatthedenominatorisapositive
integerlessthan20.
Solution
(a) Since3<31 <4,wehave,upontakingreciprocals,theinequalities
2
1 2 1
< < :
4 7 3
Wedouble-check,bycross-multiplication:
1 2
< since 1 7<4 2
4 7 (cid:2) (cid:2)
and
2 1
< since 2 3<7 1:
7 3 (cid:2) (cid:2)
Onecancheckbyexhaustionthat2=7istheonlysolution.
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1.1 Arithmetic 3
(b) Notice that the fractionfound in (a) can be obtained by addingthe numerators and
denominatorsof1=4and1=3:
1 1 2
C :
4 3 D 7
C
Wouldthesametrickworkfor7=10and5=7?Let’strytheplausibleanswer
7 5 12
C :
10 7 D 17
C
Weverifytheinequalities
7 12 5
< <
10 17 7
bycross-multiplication:
7 12
< since 7 17<10 12
10 17 (cid:2) (cid:2)
and
12 5
< since 12 7<17 5:
17 7 (cid:2) (cid:2)
Onecancheckbyexhaustionthat12=17istheonlysolution.
Bonus: Mediant Fractions
Theanswersto(a)and(b)arecalledmediantfractions.1Themediantfractionofa=band
c=d is.a c/=.b d/.Thus,themediantfractionof1=4and1=3is2=7,andthemediant
C C
fractionof7=10and5=7is12=17.Ifa=b <c=d (withb andd positive),then
a a c c
< C < :
b b d d
C
Iencouragethereadertoprovetheseinequalitiesusingcross-multiplication.
Here isanaha!proofofthemediantfractioninequalities.Assumingthata,b,c andd
are allpositive,wewillinterpretthefractionsas concentrationsofsaltinwater. Suppose
thatwe have twosolutionsofsaltwater, thefirstwitha teaspoonsofsaltinb gallonsof
water, and the second with c teaspoons of salt in d gallons of water. The concentration
ofsalt inthe firstsolutionisa=b teaspoons/gallon,whilethe concentrationof saltinthe
second solutionisc=d teaspoons/gallon.Suppose thatthe firstsolutionisless salty than
thesecond,i.e., a=b < c=d.Now, ifwe combinethetwosolutions,weobtainasolution
with a c teaspoons of salt in b d gallons of water, so that the salinity of the new
C C
solutionis.a c/=.b d/teaspoons/gallon.Certainly,thenewsolutionissaltierthanthe
C C
firstsolutionandlesssaltythanthesecond.Thatistosay,
a a c c
< C < :
b b d d
C
Ourproofcantrulybecalledasalinesolution!
1MediantfractionsariseinthestudyofFareysequencesandinthe“solution”ofSimpson’sParadox.
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4 1 ElementaryProblems
A Long Sum
Whatisthesumofthefirst100integers,
1 2 3 100‹
C C C(cid:1)(cid:1)(cid:1)C
Ofcourse,wecouldlaboriouslyaddupthenumbers.Butweseekinsteadanaha!solu-
tion,asimplecalculationthatimmediatelygivestheanswerandinsightintotheproblem.
Solution
Observethatthenumbersmaybepairedasfollows:
1and100,
2and99,
3and98,
...,
50and51.
Wehave50pairs,eachpairaddingupto101,sooursumis50 101 5050.
(cid:2) D
Thissolutionworksingeneralforthesum
1 2 3 n;
C C C(cid:1)(cid:1)(cid:1)C
wherenisanevennumber.Wepairthenumbersasbefore:
1andn,
2andn 1,
(cid:0)
3andn 2,
(cid:0)
...,
n=2andn=2 1.
C
Wehaven=2pairs,eachaddingupton 1,sooursumisn=2 .n 1/ n.n 1/=2.
C (cid:2) C D C
What if n is odd? We can no longer pair all the numbers (as the middle term has no
mate).However,throwingina0doesn’tchangethetotal:
0 1 2 3 n:
C C C C(cid:1)(cid:1)(cid:1)C
Nowwehaveanevennumberofterms,andtheymaybepairedas
0andn,
1andn 1,
(cid:0)
2andn 2,
(cid:0)
...,
.n 1/=2and.n 1/=2 1.
(cid:0) (cid:0) C
Wehave.n 1/=2pairs,eachaddingupton,sooursumisn.n 1/=2(again).
C C
A“duplicationmethod”worksforbothnevenandnodd.LetS bethesum,andintro-
duceaduplicateofS,writtenbackwards:
S 1 2 3 n:
D C C C (cid:1)(cid:1)(cid:1) C
S n .n 1/ .n 2/ 1:
D C (cid:0) C (cid:0) C (cid:1)(cid:1)(cid:1) C
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1.1 Arithmetic 5
AddthetwoexpressionsforS,summingthefirstterms,thenthesecondterms,andsoon:
2S .n 1/ .n 1/ .n 1/ .n 1/:
D C C C C C C(cid:1)(cid:1)(cid:1)C C
(Thetermn 1occursntimes.)Thissimplifiesto
C
2S n.n 1/;
D C
or
n.n 1/
S C :
D 2
Bonus: Sum of an ArithmeticProgression
CarlFriedrichGauss (1777–1855),oneofthegreatestmathematicians ofalltime,issaid
tohavesolvedourproblemforn 100whenhewasa10-year-oldschoolpupil.However,
D
accordingtoE.T.Bell[2],theproblemthatGausssolvedwasactuallymoredifficult:
The problem was of the followingsort, 81297 81495 81693 ::: 100899,
C C C C
where the step from one number tothe next is the same all along(here 198),and a
givennumberofterms(here100)aretobeadded.
TheproblemmentionedbyBellisthesumofanarithmeticprogression.Wecanevaluate
thesumusingourformulaforthesumofthefirstnintegers.Thecalculationis
81297 81495 100899 81297 100 198 .1 99/
C C(cid:1)(cid:1)(cid:1)C D (cid:2) C (cid:2) C(cid:1)(cid:1)(cid:1)C
99 100
8129700 198 (cid:2)
D C (cid:2) 2
9109800:
D
Sums of Consecutive Integers
Beholdtheidentities
1 2 3
C D
4 5 6 7 8
C C D C
9 10 11 12 13 14 15
C C C D C C
16 17 18 19 20 21 22 23 24
C C C C D C C C
25 26 27 28 29 30 31 32 33 34 35
C C C C C D C C C C
36 37 38 39 40 41 42 43 44 45 46 47 48:
C C C C C C D C C C C C
Whatisthepatternandwhydoesitwork?2
2RogerB.Nelsengivesa“proofwithoutwords”forthisproblemintheFebruary1990issueofMathematics
Magazine.
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6 1 ElementaryProblems
Solution
Considerthethirdidentity:
9 10 11 12 13 14 15:
C C C D C C
Ifweadd4toeachofthefirstthreenumbersontheleft(9,10,and11),thenweobtainthe
threenumbersontheright(13,14,and15).Wehaveadded4 3 12ontheleft,whichis
(cid:1) D
thefourthnumberontheleft.
Thenthidentity(forn 1)is
(cid:21)
n2 .n2 1/ .n2 n/ .n2 n 1/ .n2 n 2/ .n2 n n/:
C C C(cid:1)(cid:1)(cid:1)C C D C C C C C C(cid:1)(cid:1)(cid:1)C C C
There are n 1terms ontheleftandnterms ontheright.Ifwe addn 1toeach term
C C
ontheleftexcept thelastterm, thenwe obtainallthe termsonthe right.We have added
.n 1/n n2 nontheleft,andthisisthelasttermontheleft.
C D C
Bonus: Finding aPolynomial
It’seasy tofindthesum givenbythenthidentity.Theaverage ofthen 1termsonthe
C
leftis.n2 .n2 n//=2,sothesumis
C C
.n 1/.n2 .n2 1//=2 n.n 1/.2n 1/=2:
C C C D C C
Let’ssupposethatwedidn’tknowthisformula,butonlythesums:
3; 15; 42; 90; 165; 273; ::::
How can we find thepolynomialp.n/, whose values for n 1;2;3;:::are these num-
D
bers?Themethod(fromfinitedifferencecalculus)istomakeasequenceofdifferencesof
consecutivevaluesofourstartingsequence:
12; 27; 48; 75; 108; ::::
Werepeatthisprocess,creatingasequenceofsequences:
3; 15; 42; 90; 165; 273; :::
12; 27; 48; 75; 108; :::
15; 21; 27; 33; :::
6; 6; 6; ::::
Having obtained a constant sequence, we stop. Now, the polynomialp.n/ is defined by
multiplyingthefirstcolumnofourarraybysuccessivebinomialcoefficientsandadding:
n n n n .n 1/.n 2/.2n 3/
p.n/ 3 12 15 6 C C C :
D 0!C 1!C 2!C 3!D 2
Thispolynomialgivesthevaluesofoursequencestartingatp.0/.Sincewewanttostartat
p.1/,wesimplyreplacenbyn 1toobtainthepolynomialp.n/ n.n 1/.2n 1/=2.
(cid:0) D C C
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1.1 Arithmetic 7
Pluses and Minuses
Evaluate
1002 992 982 972 962 952 22 12:
(cid:0) C (cid:0) C (cid:0) C(cid:1)(cid:1)(cid:1)C (cid:0)
Solution
Using the formula for the difference of two squares, x2 y2 .x y/.x y/, the
(cid:0) D C (cid:0)
expressioncanbewrittenas
.100 99/.100 99/ .98 97/.98 97/ .96 95/.96 95/ .2 1/.2 1/:
C (cid:0) C C (cid:0) C C (cid:0) C(cid:1)(cid:1)(cid:1)C C (cid:0)
Sinceeveryotherterminparenthesesisequalto1,thisexpressionsimplifiesto
100 99 98 97 96 95 2 1:
C C C C C C(cid:1)(cid:1)(cid:1)C C
In“ALongSum,”wefoundthissumtobe5050.
Bonus: Curious Identities.3
Canyouexplainthepatternfortheidentities
32 42 52
C D
102 112 122 132 142
C C D C
212 222 232 242 252 262 272
C C C D C C
362 372 382 392 402 412 422 432 442‹
C C C C D C C C
Let’sprovethelastoneinawaythatindicateswhat’sgoingoningeneral.Transposingall
termsontheleftexcept362totherightsideoftheequationyields
362 .412 402/ .422 392/ .432 382/ .442 372/:
D (cid:0) C (cid:0) C (cid:0) C (cid:0)
Noticethatwepairedthelargestandsmallestterms,thesecond-largestandsecond-smallest,
etc.Now,usingourdifferenceofsquaresformula,wehave
362 .41 40/.41 40/ .42 39/.42 39/
D C (cid:0) C C (cid:0)
.43 38/.43 38/ .44 37/.44 37/
C C (cid:0) C C (cid:0)
81 1 81 3 81 5 81 7
D (cid:1) C (cid:1) C (cid:1) C (cid:1)
81.1 3 5 7/
D C C C
81 16:
D (cid:1)
And certainly, 362 92 42 81 16. Our computationtells the storyin general. For
D (cid:1) D (cid:1)
n 1,weclaimthat
(cid:21)
Œn.2n 1/(cid:141)2 Œ2n.n 1/(cid:141)2 .2n2 2n 1/2 .2n2 3n/2:
C C(cid:1)(cid:1)(cid:1)C C D C C C(cid:1)(cid:1)(cid:1)C C
3MichaelBoardmangivesa“proofwithoutwords”forthisproblemintheFebruary2000issueofMathematics
Magazine.
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8 1 ElementaryProblems
Transposingtermsaswedidinourtestcase,weobtain
Œn.2n 1/(cid:141)2 Œ.2n2 2n 1/2 .2n.n 1//2(cid:141)
C D C C (cid:0) C C(cid:1)(cid:1)(cid:1)
Œ.2n2 3n/2 .n.2n 1/ 1/2(cid:141)
C C (cid:0) C C
.4n2 4n 1/.1 3 5 7 2n 1/:
D C C C C C C(cid:1)(cid:1)(cid:1)C (cid:0)
Weneedaformulaforthesumofthefirstnoddnumbers,
1 3 5 7 2n 1:
C C C C(cid:1)(cid:1)(cid:1)C (cid:0)
Since this is thesum of an arithmetic progression,we coulduse the method in“A Long
Sum.”However,bythediagrambelow,wecanseethatthesumisequalton2.
n
n
Usingourformulaforthesumofthefirstnoddnumbers,ouridentitybecomes
Œn.2n 1/(cid:141)2 .2n 1/2n2;
C D C
whichisobviouslytrue.Ouridentityisverified!
Which is Greater?
Whichnumberisgreater,
p6 p10 or p5 p12‹
C C
We could compute square roots, but we seek instead an aha! solution that gives the
answerimmediately.
Solution
Thewinningideaistocomparethesquaresofthetwonumbers(thuseliminatingsomeof
thesquareroots).Thelargernumberhasthelargersquare.
Thesquaresofthegivennumbersare
2
p6 p10 6 2p60 10 16 2p60
C D C C D C
(cid:16) (cid:17)
and
2
p5 p12 5 2p60 12 17 2p60:
C D C C D C
Thesecondsquarei(cid:16)slarger.Ther(cid:17)efore,p5 p12isgreaterthanp6 p10.
C C
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1.1 Arithmetic 9
Bonus: ADiophantine Equation
The numbers inthis problem—let’scall the smaller one x and the larger one y—satisfy
theequation
x2 1 y2:
C D
ThisisanexampleofaDiophantineequation,aftertheGreekmathematicianDiophantus
(c.200–c.284).Diophantuscalledforrationalnumbersolutionstosuchequations.Inour
problem,x andy arenotrationalnumbers(astheirsquaresareirrationalnumbers).Let’s
findallrationalsolutionstotheequation.
Thegraphoftheequationisahyperbola,asshowninthepicture.
x2 1 y2
AH (cid:8)(cid:1) C D
y
(cid:16)
(cid:16)t(cid:16)
(cid:16)
(cid:16)
(cid:16) .x;y/
(cid:16)
(cid:16)
(cid:16)t
(cid:16)
(cid:16)
(cid:16) .0;1/
x
(cid:8)(cid:1) HA
Thepoint.0;1/isonthehyperbola,sowehaveonerationalsolution.If.x;y/isanother
rationalsolution(wherex 0),thentheslopeofthelinethrough.0;1/and.x;y/isalso
¤
rational.Let’scallthisslopem.Thus
y 1
m (cid:0) :
D x 0
(cid:0)
Solvingfory,weobtainy mx 1.Substitutingthisexpressionfory intotheequation
D C
ofthehyperbolayields
x2 1 .mx 1/2
C D C
m2x2 2mx 1:
D C C
Solvingforxwehave
2m
x ;
D 1 m2
(cid:0)
andhence
1 m2
y C :
D 1 m2
(cid:0)
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10 1 ElementaryProblems
Theseequations,wheremisarationalnumbernotequalto 1,areaparameterizationof
˙
all rational solutionsto the equation x2 1 y2, except for the solution.0; 1/. For
C D (cid:0)
example, if m 4=11, then .x;y/ .88=105;137=105/.The reason that .0; 1/ isn’t
D D (cid:0)
included is because it would determine a line with undefined slope. The value m 0
D
correspondstoatangentlinetothehyperbolaatthepoint.0;1/.Canyouseewhichvalues
ofmcorrespondtotheupperandlowerbranchesofthehyperbola?
Cut Down to Size
Findtwowholenumbersgreaterthan1whoseproductis999;991.
Solution
Observethat
999;991 1;000;000 9
D (cid:0)
10002 32:
D (cid:0)
Thealgebraruleforfactoringthedifferenceoftwosquarescomesinhandy:
x2 y2 .x y/.x y/:
(cid:0) D (cid:0) C
Applyingtherulewithx 1000andy 3,weobtain
D D
999;991 10002 32
D (cid:0)
.1000 3/.1000 3/
D (cid:0) C
997 1003:
D (cid:2)
Bonus: Finding thePrime Factorization
Wehavefoundtwonumbers,997and1003,whoseproductis999;991.Canthesefactors
be broken down further (i.e., do they have proper divisors?), or are they prime numbers
(seeToolkit)?Wecandivide997and1003byvariousothernumbers,suchas2,3,etc.,to
seeifthequotientsareintegers,buthowmanytrialswouldwehavetomake?
Whentestingforproperdivisorsofn,weneedonlydividenbyprimenumbers,because
ifnhasaproperdivisorthenthatdivisorhasaprimefactor.Aswetestpossibledivisors,
2, 3, 5, etc., the quotients, n=2, n=3, n=5, etc., become smaller. The “break-even point”
occursatpn,sincen=pn pn.Henceweonlyneedtosearch forprimedivisorsupto
D
pn.Iftherearenosuchdivisors,thennisaprimenumber.
As31< p997< 32,theonlyprimenumbersweneedtocheckaspossibledivisorsof
997are2,3,5,7,11,13,17,19,23,29,and31.Upondivision,wefindthatnoneofthese
primes divides 997evenly, so 997 is a prime number. To factor 1003, we workwiththe
samesetofprimes,since31< p1003< 32.Checkingthese,wehitpay-dirtwith17and
findthat1003 17 59,theproductoftwoprimes.Toverifythat59isaprime,youneed
D (cid:2)
onlycheckthat59isn’tdivisibleby2,3,5,or7,since7<p59<8.
Therefore,theprimefactorizationof999;991is17 59 997.
(cid:2) (cid:2)
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