The AMC is one of the largest competitions
in the world with around 490 000 entries
yearly or approximately one third of all
Australian secondary school students.
PROBLEM SOLVING VIA THE AMC
This book consists of a development of
techniques for solving approximately 150
problems which have been set in the
Australian Mathematics Competition for
the Westpac Awards (AMC). These problems
have been selected from the topics:
Geometry, Motion, Diophantine Equations
and Counting Techniques. Students find
these areas difficult and in the case of the
latter two, the topics are not often covered
in standard school texts.
PROBLEM SOLVING
VIA THE AMC
W ATKINS
W ATKINS
Warren Atkins graduated with a BA
from Sydney University and an MA from
Macquarie University. Warren worked as a
Mathematics teacher in NSW high schools,
then as a Senior Lecturer in the Faculty
of Education at the University of Canberra.
He is now retired but still holds the position
of Chairman of the Problems Committee
for the AMC and is Editor of Mathematics
Competition, the journal of the World
Federation of National Mathematics
Competitions.
T h e A u s T r A l i A n M AT h e M AT i c s T r u s T
e
n r i c h M e n T
s
e r i e s
IS BN 978-1-876420-07-9
5
A MT P u b l i s h i n g
Published by
A MT P u b l i s h i n g
Australian Mathematics Trust
University of Canberra ACT 2601
AUSTRALIA
Copyright ©1992 AMT Publishing
1st Reprint 1996
2nd Reprint 1999
3rd Reprint 2003
4th Reprint 2007
Telephone: +61 2 6201 5137
www.amt.edu.au
AMTT Limited ACN 083 950 341
National Library of Australia Card Number and ISSN
Australian Mathematics Trust Enrichment Series ISSN 1326-0170
Problem Solving Via the AMC
ISBN 978-1-876420-07-9
CO N T ENTS
• F O RE W O R D 3
• P RE F A C E
5
1.
DI O PHANTIN E EQ U ATIO N S S ol u tio ns
9
21
2.
COU NTIN G TEC HNIQ U ES S o l utio ns
31
64
3.
W H EN TH ING S M O VE
S o l utio ns
89
112
4.
GE O M ETR Y S o l utio ns
129
179
RE F E RE N C ES 21 3
ProblemProblem
Solving Solving
via the AMC
via the AMC
Foreward
Foreward
half
overa half
million
a million
1992 entries,
1992 entries,
the Australian
the Australian
Mathematics
Mathematics
W ithCompetition
WoverithCompetition
for thefor
Westpac
the Westpac
Awards
Awards
is one is
of one
the of
world’s
the world’s
great great
mathematics
mathematics
competitions.
competitions.
There There
are many
are reasons
many reasons
for this.
forCertainly
this. Certainly
a significant
a significant
factor factor
in its in
extraordinary
its extraordinary
successsuccess
has been
has its
been
annual
its annual
collection
collection
of motivating
of motivating
and exciting
and exciting
questions
questions
. . . an. annual
. . an annual
infusion
infusion
of
of
rich ideas
rich into
ideasthousands
into thousands
of schools
of schools
in Australia,
in Australia,
New Zealand
New Zealand
and the
and the
South South
West Pacific.
West Pacific.
“These“These
questions
questions
are intelligent,
are intelligent,
well put
well
and
puthave
and have
the property
the property
of being
of challenging
being challenging
and interesting
and interesting
. . . the .questions
. . the questions
come come
as close
astoclose
making
to making
the point
theas
point
can as
be can
done.”
be done.”
said the
said
eminent
the eminent
American
American
mathematician,
mathematician,
Paul Halmos.
Paul Halmos.
So with
Sotwo
with
important
two important
basic ingredients,
basic ingredients,
a rich source
a rich source
of beautiful
of beautiful
ques- questions and
tionsthe
and
world’s
the world’s
largestlargest
database
database
of achievement
of achievement
in mathematics,
in mathematics,
the author
the author
has taken
has up
taken
theup
challenge
the challenge
to counter
to counter
the claim
the that
claimperhaps
that perhaps
certaincertain
topics topics
shouldshould
not benot
included
be included
in the in
school
the school
curriculum
curriculum
as sixteen
as sixteen
year old
year
students’
old students’
achievement
achievement
is no better
is no better
than that
thanofthat
thirteen
of thirteen
year old
year old
students.
students.
The author
The author
has thehas
conviction
the conviction
that this
that
lack
this
oflack
improvement
of improvement
in
in
achievement
achievement
is mainly
is mainly
due to due
the to
lack
theoflack
suchofexperiences
such experiences
in the in
present
the present
school school
curriculum.
curriculum.
As a result
As a result
he hashe
produced
has produced
a booka which
book which
will provide
will provide
what Iwhat
consider
I consider
will bewill
a great
be a resource
great resource
for thefor
enrichment
the enrichment
of mathematics
of mathematics
learning
learning
in
in
secondary
secondary
schoolsschools
everywhere.
everywhere.
I congratulate
I congratulate
the author,
the author,
WarrenWarren
Atkins,Atkins,
for a splendid
for a splendid
book and
bookI wish
and Ihim
wishevery
him success
every success
with its
with
publication.
its publication.
Peter J.
Peter
OHalloran,
J. OHalloran,
Executive
Executive
Director,
Director,
Australian
Australian
Mathematics
Mathematics
Competition,
Competition,
September
September
1992. 1992.
* * ** * *
3
3
Problem Solving via the AMC
Preface
book consists of the discussion and the development of solutions
T hisof approximately
150 problems set in the Australian Mathematics
Competition for the years 1978 to 1991. These problems have been
chosen from the topics of Diophantine Equations, Counting Techniques,
Motion and Rates of Change, and Geometry.
The Australian Mathematics Competition is an annual national competition for students in years 7 to 12 from high schools and secondary
colleges. The Competition aims to stimulate the mathematical thinking
of students by the provision of a collection of stimulating and challenging
mathematical problems which can be used to enhance the teaching and
learning of mathematics. Most of the questions are appropriate for the
good average student and not just the top students, although these are
also well catered for. The Competition, whose origin was due to the foresight, energy and administrative skills of my colleague Peter O’Halloran
of the University of Canberra, began in 1978 and has expanded such that
in 1992 it is one of the largest in the world with over 500 000 student
entries, which represents approximately one third of all Australian secondary students. The competition has three divisions, Junior for years
7 and 8, Intermediate for years 9 and 10 and Senior for years 11 and 12.
Each paper consists of 30 multiple choice questions, in three sections of
10 questions, arranged in increasing order of difficulty. Some questions
are common to two of the age division papers and some are common to
all three.
This book and the selection of topics for the chapters was a direct result
of some investigation done with Gilah Leder from Monash University.
It followed on her analysis of responses to questions in the Competition
which were common to more than one of the age division papers. This
analysis showed a range of problems where students achieved no better
in the later years than students in the earlier years. While there is
the occasional problem which is very difficult, it is my view that the
lack of improvement in most of these skill areas is not due to the
inappropriateness of the skill for the particular age or grade level, but
rather that the appropriate problem solving skills are not emphasised
5
Problem Solving via the AMC
and developed sufficiently in the typical school curriculum. Hence
the selection of the chapters, Geometry and When Things Move (i.e.
speeds and rates of change), as they were perceived to be difficult by
most students, Diophantine Equations and Counting Techniques as these
topics were also seen to be difficult and, in addition, are not often covered
in standard school texts.
Acknowledgements
I would like to thank the Australian Mathematics Competition for
permission to use the questions from the Competition. In particular
I would like to thank Peter Taylor, now Executive Director of the
Australian Mathematics Trust. I would also like to thank the many
members of the Problems Committee who contributed such a wide
variety of interesting questions over a period of time.
I must give special thanks to Kevin Friel, Peter O’Halloran, Robin
Thornely and John Carty, fellow members of the Problems Committee,
and Graham Pollard, Tanya Ford and Sally Bakker of the Australian
Mathematics Competition, for their help and suggestions in improving
the manuscript. For suggestions and corrections for the second printing
I must thank Richard Bollard, of the Australian Mathematics Trust, and
George Harvey.
For the cover design and illustrations for this edition I must thank Fiona
Wilson of the Australian Mathematics Trust.
Finally I thank my wife Naida for tolerating my many hours at the
computer while I was typesetting the book.
The book was prepared in LaTEX with the diagrams designed in
MacDraw Pro.
Warren Atkins,
Faculty of Education,
University of Canberra,
March 1996.
6
Problem Solving via the AMC
Chapter 1
Diophantine Equations
equations are named after the prominent Greek mathD iophantine
ematician Diophantus of Alexandria (c. 275) who published three
works, mainly pertaining to the areas we now call algebra and the solution(s) of equations. The following is a story about his life, probably
dating from the 5th century: his boyhood lasted 16 of his life, his beard
1
grew after 12
more, after 17 more he married, 5 years later his son was
born, the son lived to half his father’s age, and the father died 4 years
after his son. This was posed as a Diophantine equation (see below),
and is generally thought to mean that he married at age 33 and died at
age 84.
The term Diophantine equation is used to refer to any equation, usually
in several variables, which arises in a problem where the solutions are
required to be integers (positive integers in many cases), or sometimes
more general rational numbers. Such an equation is
x + y = 5.
This equation has an infinite number of solutions when the set of real
numbers is considered. As a general rule however, in problem situations
that arise there are other constraints that help us narrow down the
possibilities to a small number of cases or even to a unique solution.
Consider the above equation. If we restrict the possible values of x and
y to non-negative integers (i.e. the positive integers and zero), we have
the six solutions for (x, y):
(0, 5); (1, 4); (2, 3); (3, 2); (4, 1) and (5, 0).
Note also that the pair of values x = 0 and y = 5, for example, represents
one solution to the equation.
The type of restriction where the solutions must be positive integers
readily arises in practical problems where the variables x and y represent
numbers of objects and such like. In such cases, the procedure is usually
9
Problem Solving via the AMC
to systematically list the possible solutions of the equation and then
select the ones which satisfy any additional criteria.
For a Diophantine equation there is no formula such as, for example,
the formula for the roots of a quadratic. We can, however, develop a
method, and we will usually need to take into account other aspects of
the problem context in order to fully determine the result.
Consider D1 (1985 J 23):–
Find the smallest positive integer which, when divided by
6, gives a remainder of 1 and when divided by 11, gives a
remainder of 6.
Since the integer has a remainder of 1 when divided by 6, it is of the
form
6m + 1
for m a non-negative integer. This gives the possibilities
1, 7, 13, 19, 25, 31, 37, 43, 49, 55, 61, . . .
Similarly, the positive integers which have a remainder of 6 when divided
by 11 are
6, 17, 28, 39, 50, 61, . . .
The smallest number in common to both sets is 61.
Note: This corresponds to the solution of the Diophantine equation
i.e.
6m + 1 =
6m − 11n =
11n + 6
5
which has many solutions, with m = 10 and n = 5 corresponding to the
above solution of 61.
[1985 Percentage Correct:– Junior 26% ]
10
Problem Solving via the AMC
Now try D2 (1979 I 27, S 21):–
What is the least positive integer which has remainders of 1,
1 and 5 when divided by 3, 5, and 7 respectively?
[Solution Page 21 ]
Analysing Diophantine Equations
We will in most cases need to manipulate the given equations in some
way to determine the solution.
Consider D3 (1982 J 26, I 21):–
If x and y are positive integers and
x + y + xy = 34,
find x + y.
Alternative 1:
Looking at the left hand side of the equation x + y + xy = 34, we may
see that
x + y + xy = (x + 1)(y + 1) − 1
so that the equation may be written as
(x + 1)(y + 1)
=
35
Now, since x and y are integers, x + 1 and y + 1 are also integers, so we
are looking for two factors of 35.
There are only two, 5 and 7 (other than 35 and 1, which is not possible
in this case). This tells us that
x+1=7
y+1=5
or
x+1=5
y+1=7
11
Problem Solving via the AMC
While we cannot differentiate between the two cases and determine the
actual values for x and y, we can say from the symmetry (or by adding
the equations in either case) that
x+y
=
6+4
=
10.
Alternative 2:
If we did recognise the factorisation given above, a standard approach is
to write the equation in such a way that we can systematically list the
possibilities.
This we can do by expressing one of the variables in terms of the other.
In this case it does not matter which of the two variables in terms of the
other as the equation is symmetric in x and y, although in some cases
there may be advantages in choosing a particular one for ease of further
analysis.
In this case, the equation can be rearranged in the following way to
express x in terms of y.
x + y + xy
x + xy
x(1 + y)
=
=
=
x =
34
34 − y
34 − y
34 − y
1+y
Remembering that x and y are integral, we can now substitute integers
for y to obtain the corresponding x.
y
1
2
3
4
5
6
7
···
33 32 31 30
29 28
27
=6
=4
···
2
3
4
5
6
7
8
There are no more integral values until x becomes negative, hence
x
x+y
=
6 + 4 (or 4 + 6)
=
10
is the only solution.
If we had noted the symmetry in x and y in the original equation and
the fact that
29
< 5, we could have stopped at the entry for y = 5 in
6
12
Problem Solving via the AMC
the table, as the other values may be obtained by interchanging the x
and the y.
[1982 Percentage Correct:– Junior 31%, Intermediate 54% ]
And consider D4 (1984 J 22, I 20, S 12):–
Kathryn has 20 coins in her purse. They are 10c, 20c and 50c
coins and the total value of the coins is $5. If she has more 50c
coins than 10c coins, how many 10c coins has she?
The first step is to obtain an equation.
Let the number of 50c coins be x and the number of 20c coins be y.
The number of 10c coins is then 20 − (x + y) as there are 20 coins in all.
Since the total value of the coins is $5 = 500 c, we then obtain the
equation
50x + 20y + 10(20 − (x + y))
5x + 2y + 20 − x − y
4x + y
=
=
500
50
=
30
Here we have a typical Diophantine equation, where we are only
interested in positive integer solutions as x and y are representing
numbers of things (coins in this case).
Expressing y in terms of x, we have
y
=
30 − 4x
and remembering that there are 20 coins in all, we can now systematically
list the possibilities.
We can readily see that there are no possibilities for x = 1, 2 and 3, as a
substitution shows that the number of coins to make up $5 would exceed
20.
13
Problem Solving via the AMC
Listing the possibilities, we get
50c (x)
4
5
6
7
20c (y)
14
10
6
2
10c
2
5
8
11
†
Again, we can see there are no solutions for x > 7 as then y would be
negative.
We now know that it is possible to make up $5 with 20 coins that are
50c, 20c or 10c coins in 4 different ways.
However, in this problem, we are given that there are more 50c coins
than 10c coins. Scanning the table of possibilities we can easily see that
the line marked with † gives us the only solution of:–
four 50c, fourteen 20c and two 10c coins.
[1984 Percentage Correct:– Junior 22%, Intermediate 28%, Senior 45%]
Now try D5 (1991 J 27 I 22):–
A shop buys 40 pens of three different types at a cost of $40.
If the pens cost 25c, $1 and $5 each, and there are more $1
pens than $5 pens, how many 25c pens were bought?
[Solution Page 22 ]
And try D6 (1984 I 29):–
Articles X, Y and Z are for sale. Article X can be bought at
the rate of eight for $1. Article Y costs $1 each and article
Z costs $10 each. You buy a selection of each and find that
you have purchased exactly 100 articles at a cost of $100. How
many articles of type Y did you purchase?
[Solution Page 23 ]
Further Techniques
We can develop further techniques for handling indeterminate equations.
14
Problem Solving via the AMC
Consider D7 (1983 I 29, S 27):–
One of the solutions to the equation
19x + 83y
=
1983
in positive integers x and y is obviously given by
(x, y) = (100, 1).
It turns out that there is exactly one more pair of positive
integers (x, y) which satisfies the above equation. What is
that number pair?
While there are alternative methods, sometimes shorter, in particular
cases (one will be developed later in Alternative 2 for this question), we
can develop a general technique as outlined by Euclid. This method is
as follows:–
Alternative 1:
From the given equation
19x + 83y
we get
=
19x =
x =
=
i.e.
x =
where u must be integral
and
u =
and hence
7y + 19u =
1983
1983 − 83y
1983 − 83y
19
7 − 7y
19
104 − 4y + u
104 − 4y +
7 − 7y
19
7
This last equation is another Diophantine equation, similar to the first,
but with smaller coefficients.
15