Download 2015 Mathematics Contests – The Australian Scene Part 1

MATHEMATICS CONTESTS
THE AUSTRALIAN SCENE 2015
PART 1: MATHEMATICS CHALLENGE FOR YOUNG AUSTRALIANS
KL McAvaney
A u s t r a l i a n M at h e mat i c a l O ly m p i a d C omm i t t e e
A
d e pa r t m e n t o f t h e
A u s t r a l i a n M at h e mat i c s T r u s t
Published by
AMT Publishing
Australian Mathematics Trust
University of Canberra Locked Bag 1
Canberra GPO ACT 2601
Australia
Tel: 61 2 6201 5137
www.amt.edu.au
AMTT Limited
ACN 083 950 341
Copyright ©2015 by the Australian Mathematics Trust
National Library of Australia Card Number
and ISSN 1323-6490
2
Mathematics Contests The Australian Scene 2015
SUPPORT FOR THE AUSTRALIAN MATHEMATICAL
OLYMPIAD COMMITTEE TRAINING PROGRAM
The Australian Mathematical Olympiad Committee Training Program is an activity of the Australian Mathematical
Olympiad Committee, a department of the Australian Mathematics Trust.
Trustee
The University of Canberra
Sponsors
The Mathematics/Informatics Olympiads are supported by the Australian Government Department of Education
and Training through the Mathematics and Science Participation Program.
The Australian Mathematical Olympiad Committee (AMOC) also acknowledges the significant financial support
it has received from the Australian Government towards the training of our Olympiad candidates and the
participation of our team at the International Mathematical Olympiad (IMO).
The views expressed here are those of the authors and do not necessarily represent the views of the
government.
Special thanks
With special thanks to the Australian Mathematical Society, the Australian Association of Mathematics Teachers
and all those schools, societies, families and friends who have contributed to the expense of sending the 2015
IMO team to Chiang Mai, Thailand.
Part 1: Mathematics Challenge for Young Australians 3
ACKNOWLEDGEMENTS
The Australian Mathematical Olympiad Committee (AMOC) sincerely thanks all sponsors, teachers,
mathematicians and others who have contributed in one way or another to the continued success of its
activities. The editors sincerely thank those who have assisted in the compilation of this book, in particular
the students who have provided solutions to the 2015 IMO. Thanks also to members of AMOC and Challenge
Problems Committees, Adjunct Professor Mike Clapper, staff of the Australian Mathematics Trust and others who
are acknowledged elsewhere in the book.
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Mathematics Contests The Australian Scene 2015
PREFACE
After last year, there seemed little room for improvement, but 2015 has
been even better, marked particularly by our best ever result at an IMO,
where we were placed 6th out of the 104 competing countries, finishing
ahead of all European countries (including Russia) and many other
traditional powerhouses such as Singapore, Japan and Canada. For the
first time ever, all six team members obtained Silver or better, with two
team members (Alex Gunning and Seyoon Ragavan) claiming Gold.
Alex finished fourth in the world (after his equal first place last year), and
becomes Australia’s first triple Gold medallist in any academic Olympiad.
Seyoon, who finished 19th, now has three IMOs under his belt with a year
still to go. Once again, we had three Year 12 students in the team, so there
will certainly be opportunities for new team members next year. It was also
pleasing to see, once again, an Australian-authored question on the paper.
This year, it was the prestigious Question 6, which was devised by Ivan
Guo and Ross Atkins, based on the mathematics of juggling.
In the Mathematics Ashes we tied with the British team; however, we
finished comfortably ahead of them in the IMO competition proper. Director of Training and IMO Team Leader,
Dr Angelo Di Pasquale, along with his Deputy Andrew Elvey Price and their team of former Olympians continue
to innovate and keep the training alive, fresh and, above all, of high quality. This year they adopted a specific
policy of tackling very hard questions in training, which was daunting for team members at times but seems to
have paid off. We congratulate them again on their success.
The Mathematics Challenge for Young Australians (MCYA) also continues to attract strong entries, with the
Challenge continuing to grow, helped by the gathering momentum of the new Middle Primary Division, which
began in 2014. The Enrichment stage, containing course work, allows students to broaden their knowledge
base in the areas of mathematics associated with the Olympiad programs and more advanced problem-solving
techniques. We have continued running workshops for teachers to develop confidence in managing these
programs for their more able students—this seems to be paying off with strong numbers in both the Challenge
and Enrichment stages. The final stage of the MCYA program is the Australian Intermediate Mathematics
Olympiad (AIMO). It is a delight to record that over the last two years the number of entries to AIMO has
doubled. This has been partly due to wider promotion of the competition, but more specifically a result of
the policy of offering free entry to AMC prize winners. There were some concerns in 2014 that some of the
‘new’ contestants were under-prepared for AIMO and there were more zero scores than we would have liked.
However, this year, the number of zero scores was less than 1% and the quality of papers was much higher,
revealing some significant new talent, some of whom will be rewarded with an invitation to the December AMOC
School of Excellence.
There are many people who contribute to the success of the AMOC program. These include the Director of
Training and the ex-Olympians who train the students at camps; the AMOC State Directors; and the Challenge
Director, Dr Kevin McAvaney, and the various members of his Problems Committee, who develop such original
problems, solutions and discussions each year. The AMOC Senior Problems Committee is also a major
contributor and Norm Do is continuing with his good work. The invitational program saw some outstanding
results from Australian students, with a number of perfect scores. Details of these achievements are provided in
the appropriate section of this book. As was the case last year, we are producing Mathematics Contests—The
Australian Scene in electronic form only and making it freely available through the website. We hope this will
provide greater access to the problems and section reports. This book is also available in two sections, one
containing the MCYA reports and papers and the other containing the Olympiad training program reports and
papers.
Mike Clapper
November 2015
Part 1: Mathematics Challenge for Young Australians 5
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Mathematics Contests The Australian Scene 2015
CONTENTS
Support for the Australian Mathematical Olympiad Committee Training Program 3
Acknowledgements 4
Preface5
Mathematics Challenge for Young Australians 8
Membership of MCYA Committees 10
Challenge Problems – Middle Primary
12
Challenge Problems – Upper Primary
14
Challenge Problems – Junior
16
Challenge Problems – Intermediate
19
Challenge Solutions – Middle Primary
21
Challenge Solutions – Upper Primary
24
Challenge Solutions – Junior
28
Challenge Solutions – Intermediate
36
Challenge Statistics – Middle Primary
46
Challenge Statistics – Upper Primary
47
Challenge Statistics – Junior
48
Challenge Statistics – Intermediate
49
Australian Intermediate Mathematics Olympiad
50
Australian Intermediate Mathematics Olympiad Solutions
52
Australian Intermediate Mathematics Olympiad Statistics
60
Australian Intermediate Mathematics Olympiad Results
61
Honour Roll 65
Part 1: Mathematics Challenge for Young Australians 7
MATHEMATICS CHALLENGE FOR YOUNG AUSTRALIANS
The Mathematics Challenge for Young Australians (MCYA) started on a national scale in 1992. It was set up to
cater for the needs of the top 10 percent of secondary students in Years 7–10, especially in country schools and
schools where the number of students may be quite small. Teachers with a handful of talented students spread
over a number of classes and working in isolation can find it very difficult to cater for the needs of these students.
The MCYA provides materials and an organised structure designed to enable teachers to help talented students
reach their potential. At the same time, teachers in larger schools, where there are more of these students, are
able to use the materials to better assist the students in their care.
The aims of the Mathematics Challenge for Young Australians include:
•
encouraging and fostering
– a greater interest in and awareness of the power of mathematics
– a desire to succeed in solving interesting mathematical problems
– the discovery of the joy of solving problems in mathematics
•
identifying talented young Australians, recognising their achievements nationally and providing support that will
enable them to reach their own levels of excellence
•
providing teachers with
– interesting and accessible problems and solutions as well as detailed and motivating teaching discussion
and extension materials
– comprehensive Australia-wide statistics of students’ achievements in the Challenge.
There are three independent stages in the Mathematics Challenge for Young Australians:
•
Challenge (three weeks during the period March–June)
•
Enrichment (April–September)
•
Australian Intermediate Mathematics Olympiad (September).
Challenge
Challenge now consists of four levels. Upper Primary (Years 5–6) and Middle Primary (Years 3–4) present
students with four problems each to be attempted over three weeks, students being allowed to work on the
problems in groups of up to three participants, but each to write their solutions individually. The Junior (Years
7–8) and Intermediate (Years 9–10) levels present students with six problems to be attempted over three weeks,
students being allowed to work on the problems with a partner but each must write their solutions individually.
There were 12692 entries (1166 Middle Primary, 3416 Upper Primary, 5006 Junior, 3104 Intermediate) for the
Challenge in 2015. The 2015 problems and solutions for the Challenge, together with some statistics, appear later
in this book.
Enrichment
This is a six-month program running from April to September, which consists of six different parallel stages of
comprehensive student and teacher support notes. Each student participates in only one of these stages.
The materials for all stages are designed to be a systematic structured course over a flexible 12–14 week period
between April and September. This enables schools to timetable the program at convenient times during their
school year.
Enrichment is completely independent of the earlier Challenge; however, they have the common feature of
providing challenging mathematics problems for students, as well as accessible support materials for teachers.
Newton (years 5–6) includes polyominoes, fast arithmetic, polyhedra, pre-algebra concepts, patterns, divisibility
and specific problem-solving techniques. There were 1165 entries in 2015.
Dirichlet (years 6–7) includes mathematics concerned with tessellations, arithmetic in other bases, time/distance/
speed, patterns, recurring decimals and specific problem-solving techniques. There were 1181 entries in 2015.
Euler (years 7–8) includes primes and composites, least common multiples, highest common factors, arithmetic
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Mathematics Contests The Australian Scene 2015
sequences, figurate numbers, congruence, properties of angles and pigeonhole principle. There were 1708 entries
in 2015.
Gauss (years 8–9) includes parallels, similarity, Pythagoras’ Theorem, using spreadsheets, Diophantine
equations, counting techniques and congruence. Gauss builds on the Euler program. There were 1150 entries
in 2015.
Noether (top 10% years 9–10) includes expansion and factorisation, inequalities, sequences and series, number
bases, methods of proof, congruence, circles and tangents. There were 818 entries in 2015.
Polya (top 10% year 10) (currently under revision) Topics will include angle chasing, combinatorics, number
theory, graph theory and symmetric polynomials. There were 303 entries in 2015.
Australian Intermediate Mathematics Olympiad
This four-hour competition for students up to Year 10 offers a range of challenging and interesting questions.
It is suitable for students who have performed well in the AMC (Distinction and above), and is designed as an
endpoint for students who have completed the Gauss or Noether stage. There were 1440 entries for 2015
and 19 perfect scores.
Part 1: Mathematics Challenge for Young Australians 9
MEMBERSHIP OF MCYA COMMITTEES
Mathematics Challenge for Young Australians Committee 2015
Director
Dr K McAvaney, Victoria
Challenge
Committee
Adj Prof M Clapper, Australian Mathematics Trust, ACT
Mrs B Denney, NSW
Mr A Edwards, Queensland Studies Authority
Mr B Henry, Victoria
Ms J McIntosh, AMSI, VIC
Mrs L Mottershead, New South Wales
Ms A Nakos, Temple Christian College, SA
Prof M Newman, Australian National University, ACT
Dr I Roberts, Northern Territory
Ms T Shaw, SCEGGS, NSW
Ms K Sims, New South Wales
Dr A Storozhev, Attorney General’s Department, ACT
Mr S Thornton, South Australia
Ms G Vardaro, Wesley College, VIC
Moderators
Mr W Akhurst, New South Wales
Mr R Blackman, Victoria
Ms J Breidahl, Victoria
Mr A Canning, Queensland
Dr E Casling, Australian Capital Territory
Mr B Darcy, South Australia
Mr J Dowsey, Victoria
Ms P Graham, MacKillop College, TAS
Ms J Hartnett, Queensland
Ms N Hill, Victoria
Dr N Hoffman, Edith Cowan University, WA
Ms R Jorgenson, Australian Capital Territory
Prof H Lausch, Victoria
Mr J Lawson, St Pius X School, NSW
Ms K McAsey, Victoria
Ms T McNamara, Victoria
Mr G Meiklejohn, Department of Education, QLD
Mr M O’Connor, AMSI, VIC
Mr J Oliver, Northern Territory
Mr G Pointer, Marratville High School, SA
Dr H Sims, Victoria
Mrs M Spandler, New South Wales
Ms C Stanley, Queensland
Mr P Swain, Ivanhoe Girls’ Grammar School, VIC
Dr P Swedosh, The King David School, VIC
Mrs A Thomas, New South Wales
Ms K Trudgian, Queensland
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Mathematics Contests The Australian Scene 2015
Australian Intermediate Mathematics Olympiad Problems Committee
Dr K McAvaney, Victoria (Chair)
Adj Prof M Clapper, Australian Mathematics Trust, ACT
Mr J Dowsey, University of Melbourne, VIC
Dr M Evans, International Centre of Excellence for Education in Mathematics, VIC
Mr B Henry, Victoria
Assoc Prof H Lausch, Monash University, VIC
Enrichment
Editors
Mr G R Ball, University of Sydney, NSW
Dr M Evans, International Centre of Excellence for Education in Mathematics, VIC
Mr K Hamann, South Australia
Mr B Henry, Victoria
Dr K McAvaney, Victoria
Dr A M Storozhev, Attorney General’s Department, ACT
Emeritus Prof P Taylor, Australian Capital Territory
Dr O Yevdokimov, University of Southern Queensland
Part 1: Mathematics Challenge for Young Australians 11
CHALLENGE PROBLEMS – MIDDLE PRIMARY
2015 Challenge Problems - Middle Primary
Students may work on each of these four problems in groups of up to three, but must write their solutions individually.
MP1 e-Numbers
The digits on many electronic devices look like this:
When numbers made of these digits are rotated 180◦ (
), some of them become numbers and some don’t. For
example, when 8 is rotated it remains the same, 125 becomes 521, and 14 does not become a number. Except for 0,
no number starts with 0.
All rotations referred to below are through 180◦.
a List all digits which rotate to the same digit.
b List all digits which rotate to a different digit.
c List all numbers from 10 to 50 whose rotations are numbers.
d List all numbers from 50 to 200 that stay the same when rotated.
MP2 Quazy Quilts
Joy is making a rectangular patchwork quilt from square pieces of fabric. All the square pieces are the same size. She
starts by joining two black squares into a rectangle. Then she adds a border of grey squares.
Next, she adds a border of white squares.
a How many white squares does she need?
Joy continues to add borders of grey and borders of white squares alternately.
b How many grey squares and how many white squares does she need for the next two borders?
c Joy notices a pattern in the number of squares used for each border. Describe this pattern and explain why it
always works.
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Mathematics Contests The Australian Scene 2015
1
d There are 90 squares in the outside border of the completed quilt. How many borders have been added to Joy’s
starting black rectangle?
MP3 Egg Cartons
Zoe decided to investigate the number of different ways of placing identical eggs in clear rectangular cartons of various
sizes.
For example, there is only one way of placing six eggs in a 2 × 3 carton and only two different ways of placing five
eggs.
Reflections and rotations of arrangements are not considered different. For example, all of these arrangements are the
same:
a Here are two ways of arranging four eggs in a 2 × 3 carton.
Draw four more different ways.
b Draw six different ways of arranging two eggs in a 2 × 3 carton.
c Draw six different ways of arranging three eggs in a 2 × 3 carton.
d Draw all the different ways of arranging six eggs in a 2 × 4 carton.
MP4 Condates
On many forms, the date has to be written as DD/MM/YY. For example, 25 April 2013 is written 25/04/13. Dates
such as this that use all the digits 0, 1, 2, 3, 4, 5 will be called condates.
a Find two condates in 2015.
b Find the first condate after 2015.
c Find the last condate in any century.
d Explain why no date of the form DD/MM/YY can use all the digits 1, 2, 3, 4, 5, 6.
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Part 1: Mathematics Challenge for Young Australians 13
CHALLENGE
PROBLEMS
PRIMARY
2015 Challenge
Problems– -UPPER
Upper Primary
Students may work on each of these four problems in groups of up to three, but must write their solutions individually.
UP1 Rod Shapes
I have a large number of thin straight steel rods of lengths 1 cm, 2 cm, 3 cm and 4 cm. I can place the rods with their
ends touching to form polygons. For example, with four rods, two of length 1 cm and two of length 3 cm, I can form
a parallelogram which I refer to as (1, 3, 1, 3).
1c
m
1c
m
3 cm
3 cm
a List all the different equilateral triangles I can make using only three rods at a time.
b Using only three rods at a time, how many different isosceles triangles can I make which are not equilateral triangles?
List them all.
c List all the different scalene triangles I can make using only three rods at a time.
d Using only four rods at a time, none of which is 4 cm, how many different quadrilaterals can I make which have
exactly one pair of parallel sides and the other pair of sides equal in length? List them all.
UP2 Egg Cartons
Zoe decided to investigate the number of different ways of placing identical eggs in clear rectangular cartons of various
sizes.
For example, there is only one way of placing six eggs in a 2 × 3 carton and only two different ways of placing five
eggs.
Reflections and rotations of arrangements are not considered different. For example, all of these arrangements are the
same:
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Mathematics Contests The Australian Scene 2015
3
a Draw four different ways of arranging two eggs in a 2 × 3 carton.
b Draw six different ways of arranging three eggs in a 2 × 3 carton.
c Draw all the different ways of arranging six eggs in a 2 × 4 carton.
d Zoe discovers that there are 55 ways of arranging three eggs in a 2 × 6 carton. Explain why there must also be 55
ways of arranging nine eggs in a 2 × 6 carton.
UP3 Seahorse Swimmers
The coach at the Seahorse Swimming Club has four boys in his Under 10 squad. Their personal best (PB) times for
50 metres are:
Mack
Jack
Zac
Nick
55 seconds
1 minute 8 seconds
1 minute 16 seconds
1 minute 3 seconds
a The coach splits the swimmers into two pairs for a practice relay race. He wants the race to be as close as possible.
Based on the PB times, who should be in each pair?
b Chloe’s PB time is 1 minute 10 seconds and Sally’s is 53 seconds. The coach arranges the two girls and four boys
into two teams of three with total PB times as close as possible.
i Why can’t the total PB times be the same for the two teams?
ii Who are in each team?
c The swimmers ask the coach what his PB time was when he was 10 years old. The coach replied that if his PB
time was included with all theirs, then the average PB time would be exactly 1 minute. What was the coach’s PB
time?
UP4 Primelandia Money
In Primelandia, the unit of currency is the Tao (T). The value of each Primelandian coin is a prime number of Taos.
So the coin with the smallest value is worth 2T. There are coins of every prime value less than 50. All payments in
Primelandia are an exact number of Taos.
a List all combinations of two coins whose sum is 50T.
b What is the smallest payment (without change) which requires at least three coins?
c Suppose you have one of each Primelandian coin with value less than 50T. What is the difference between the
largest even payment that can be made using four coins and the largest even payment that can be made using three
coins?
d A bag contains six different coins. Alice, Bob and Carol take two coins each from the bag and keep them. They
find that they have all taken the same amount of money. What is the smallest amount of money that could have
been in the bag? Explain your reasoning.
4
Part 1: Mathematics Challenge for Young Australians 15
CHALLENGE
PROBLEMS
2015 Challenge
Problems–- JUNIOR
Junior
Students may work on each of these six problems with a partner but each must write their solutions individually.
J1 Quirky Quadrilaterals
Robin and Toni were drawing quadrilaterals on square dot paper with their vertices on the dots but no other dots on
the edges. They decided to use I to represent the number of dots in the interior. Here are two examples:
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
I=1
I=6
a Toni said she could draw a rhombus with I = 2 and a kite that is not a rhombus with I = 2. Draw such quadrilaterals
on square dot paper.
b Toni drew squares with I = 4 and I = 9. Draw such squares on square dot paper.
c Robin drew a square with I = 12. Draw such a square on square dot paper.
d Toni exclaimed with excitement, ‘I can draw rhombuses with any number of interior points’. Explain how this could
be done.
J2 Indim Integers
Jim is a contestant on a TV game show called Indim. The compere spins a wheel that is divided into nine sectors
numbered 1 to 9. Jim has nine tiles numbered 1 to 9. When the wheel stops, he removes all tiles whose numbers are
factors or multiples of the spun number.
•
3
4
5
1
2
1
2
3
4
5
6
7
8
9
6
9
•
7
8
Jim then tries to use three of the remaining tiles to form a 3-digit number that is divisible by the number on the
wheel. If he can make such a number, he shouts ‘Indim’ and wins a prize.
a Show that if 1, 2, or 5 is spun, then it is impossible for Jim to win.
b How many winning numbers can Jim make if 4 is spun?
c List all the winning numbers if 6 is spun.
d What is the largest possible winning number overall?
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Mathematics Contests The Australian Scene 2015
5
J3 Primelandia Money
In Primelandia, the unit of currency is the Tao (T). The value of each Primelandian coin is a prime number of Taos.
So the coin with the smallest value is worth 2T. There are coins of every prime value less than 50. All payments in
Primelandia are an exact number of Taos.
a What is the smallest payment (without change) which requires at least three coins?
b A bag contains six different coins. Alice, Bob and Carol take two coins each from the bag and keep them. They
find that they have all taken the same amount of money. What is the smallest amount of money that could have
been in the bag? Explain your reasoning.
c Find five Primelandian coins which, when placed in ascending order, form a sequence with equal gaps of 6T between
their values.
d The new King of Primelandia decides to mint coins of prime values greater than 50T. Show that no matter what
coins are made, there is only one set of five Primelandian coins which, when placed in ascending order, form a
sequence with equal gaps of 6T between their values.
J4 Condates
On many forms, the date has to be written as DD/MM/YYYY. For example, 25 April 1736 is written 25/04/1736.
Dates such as this that use eight consecutive digits (not necessarily in order) will be called condates.
a What is the first condate after the year 2015?
b Why must there be a 0 in every condate in the years 2000 to 2999?
c Why must every condate in the years 2000 to 2999 have 0 as the first digit of the month?
d How many condates are there in the years 2000 to 2999?
J5 Jogging
Theo is training for the annual Mt Killaman Joggin 10 km Torture Track. This ‘fun’ run involves running east on a
flat track for 2 km to the base of Mt Killaman Joggin, then straight up its rather steep side a further 3 km to the top.
This is the halfway point where you turn around and run back the way you came. This simplified mudmap should
give you the general idea.
Top of Mt KJ
•
3k
•
Start/Finish
2 km
m
•
Base of Mt KJ
Theo can run at 12 km/h on the flat. Up the hill he reckons he can average 9 km/h and he is good for 15 km/h on the
downhill part of the course.
a Calculate Theo’s expected time to complete the course, assuming his estimates for his running speeds are correct.
The day before the race, Theo hears that windy conditions are expected. He knows from experience that this will
affect his speed out on the course. When the wind blows into his face, it will slow him down by 1 km/h, but when it is
at his back, he will speed up by the same amount. This applies to his speeds on the flat and on the slope. The wind
will either blow directly from the east or directly from the west.
b From which direction should he hope the wind blows if he wishes to minimise his time? Justify your answer.
c Theo would like to finish the race in 50 minutes. He decides to run either the uphill leg faster or the downhill leg
faster, but not both. Assuming there is no wind, how much faster would he have to run in each case?
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Part 1: Mathematics Challenge for Young Australians 17
J6 Tessellating Hexagons
Will read in his favourite maths book that this hexagon tessellates the plane.
D
E
C
F
A
B
To see if this was true, Will made 16 copies of the hexagon and glued them edge-to-edge onto a piece of paper, with
no overlaps and no gaps.
a Cut out 16 hexagons from the worksheet and glue them as Will might have done so they entirely cover the dashed
rectangle.
b Show that your block of 16 hexagons can be translated indefinitely to tessellate the entire plane.
c Find all points on the hexagon above that are points of symmetry of your tessellation.
d In your tessellation, select four hexagons that form one connected block which will tessellate the plane by translation
only. Indicate three such blocks on your tessellation that are not identical.
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Mathematics Contests The Australian Scene 2015
7
CHALLENGE
PROBLEMS
2015 Challenge
Problems– -INTERMEDIATE
Intermediate
Students may work on each of these six problems with a partner but each must write their solutions individually.
I1 Indim Integers
See Junior Problem 2.
I2 Digital Sums
The digital sum of an integer is the sum of all its digits. For example, the digital sum of 259 is 2 + 5 + 9 = 16.
a Find the largest even and largest odd 3-digit multiples of 7 which have a digital sum that is also a multiple of 7.
A digital sum sequence is a sequence of numbers that starts with any integer and has each number after the first
equal to the digital sum of the number before it. For example: 7598, 29, 11, 2. Any digital sum sequence ends in a
single-digit number and this is called the final digital sum or FDS of the first number in the sequence. Thus the FDS
of 7598 is 2.
b Find the three largest 3-digit multiples of 7 which have an FDS of 7.
c Find and justify a rule that produces all 3-digit multiples of 7 which have an FDS of 7.
d Find and justify a rule that produces all 3-digit multiples of 8 which have an FDS of 8.
I3 Coin Flips
I have several identical coins placed on a table. I play a game consisting of one or more moves. Each move consists of
flipping over some of the coins. The number of coins that are flipped stays the same for each game but may change
from game to game. A coin showing ‘heads’ is represented by H. A coin showing ‘tails’ is represented by T . So, when
a coin is flipped it changes from H to T or from T to H. The same coin may be selected on different moves. In each
game we start with all coins showing tails.
For example, in one game we might start with five coins and flip two at a time like this:
Start:
Move 1:
Move 2:
Move 3:
T
T
H
H
T
H
H
H
T
H
T
T
T
T
T
H
T
T
T
H
a Starting with 14 coins and flipping over four coins on each move, explain how to finish with exactly 10 coins heads
up in three moves.
b Starting with 154 coins and flipping over 52 coins on each move, explain how to finish with all coins heads up in
three moves.
c Starting with 26 coins and flipping over four coins on each move, what is the minimum number of moves needed to
finish with all coins heads up?
d Starting with 154 coins and flipping over a fixed odd number of coins on each move, explain why I cannot have all
154 coins heads up at the end of three moves.
I4 Jogging
See Junior Problem 5.
8
Part 1: Mathematics Challenge for Young Australians 19
I5 Folding Fractions
A square piece of paper has side length 1 and is shaded on the front and white on the back. The bottom-right corner
is folded to a point P on the top edge as shown, creating triangles P QR, P T S and SU V .
T
P
Q
R
S
U
V
a If P is the midpoint of the top edge of the square, find the length of QR.
b If P is the midpoint of the top edge of the square, find the side lengths of triangle SU V .
c Find all positions of P so that the ratio of the sides of triangle SU V is 7:24:25.
I6 Crumbling Cubes
A large cube is made of 1 × 1 × 1 small cubes. Small cubes are removed in a sequence of steps.
The first step consists of marking all small cubes that have at least 2 visible faces and then removing only those small
cubes.
In the second and subsequent steps, the same procedure is applied to what remains of the original large cube after the
previous step.
a Starting with a 10 × 10 × 10 cube, how many small cubes are removed at the first step?
b A different cube loses 200 small cubes at the first step. What are the dimensions of this cube?
c Beginning with a 9 × 9 × 9 cube, what is the surface area of the object that remains after the first step?
d Beginning with a 9 × 9 × 9 cube, how many small cubes are left after the third step?
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Mathematics Contests The Australian Scene 2015
9
CHALLENGE
SOLUTIONS
PRIMARY
2015 Challenge
Solutions–- MIDDLE
Middle Primary
MP1 e-Numbers
a The digits which rotate to the same digit are: 0, 1, 2, 5, 8.
b The digits which rotate to a different digit are 6 and 9.
c Only the digits in Parts a and b can be used to form numbers whose rotations are numbers. So the only numbers
from 10 to 50 whose rotations are numbers are: 11, 12, 15, 16, 18, 19, 21, 22, 25, 26, 28, 29.
d From Parts a and b, if a number and its rotation are the same, then the first and last digits of the number must
both be 1, 2, 5, 8, or the first digit is 6 and the last 9 or the first 9 and the last 6. So the numbers from 50 to 200
that stay the same when rotated are: 55, 69, 88, 96, 101, 111, 121, 151, 181.
MP2 Quazy Quilts
a The quilt after adding one grey and one white border:
The number of white squares is 18.
b The quilt after adding another grey and another white border:
There are 26 extra grey squares and 34 extra white squares.
c The table shows the number of squares in each border that we have counted so far.
Border
Squares
1
10
2
18
3
26
4
34
The number of squares from one border to the next appears to increase by 8. We now show this rule continues to
apply.
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Part 1: Mathematics Challenge for Young Australians 21
Except for the corner squares, each square on the previous border corresponds to a square in the new border. Each
of the corner squares C on the previous border corresponds to a corner square C in the new border. In addition,
there are two new squares next to each corner.
new border
C
C
previous border
C
previous border
new border
C
new border
previous border
previous border
C
C
new border
C
C
Thus the number of squares from one border to the next increases by 4 × 2 = 8.
d Alternative i
The first border has 2 + 8 squares.
The second border has 2 + 8 + 8 squares.
The third border has 2 + 8 + 8 + 8 squares.
And so on.
We want 2 + 8 + 8 + · · · = 90.
Hence the number of 8s needed is 11.
So Joy’s completed quilt has 11 borders.
Alternative ii
From Part c we have the following table.
Border
Squares
1
10
2
18
3
26
4
34
5
42
6
50
7
58
8
66
So Joy’s completed quilt has 11 borders.
MP3 Egg Cartons
a
For each of these arrangements, any rotation or reflection is acceptable.
b
For each of these arrangements, any rotation or reflection is acceptable.
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Mathematics Contests The Australian Scene 2015
11
9
74
10
82
11
90
c
For each of these arrangements, any rotation or reflection is acceptable.
d Each arrangement of six eggs in a 2 × 4 carton has two empty positions. By systematically locating the two empty
positions, we see that there are ten different ways to arrange six eggs in a 2 × 4 carton.
For each of these arrangements, any rotation or reflection is acceptable.
MP4 Condates
a Any condate in the year 2015 must be written as DD/MM/15. The remaining digits are 0, 2, 3, 4. So the month
must be 02, 03, or 04.
If the month is 02, then the day is 34 or 43, which are not allowed. If the month is 03, then the day must be 24. If
the month is 04, then the day must be 23. Thus the only condates in 2015 are 24/03/15 and 23/04/15.
b There are no condates in the years 2016, 2017, 2018, 2019 since their last digits are greater than 5.
In 2020, the remaining digits are 1, 3, 4, 5. No two of these form a month.
In 2021, the remaining digits are 0, 3, 4, 5. So the month must start with 0. Then the day must contain two of 3,
4, 5, which is impossible.
The year 2022 duplicates 2.
In 2023, the only digits remaining are 0, 1, 4, 5. The day cannot use both 4 and 5, so the month is 04 or 05. The
earliest of these is 04, in which case the day must be 15. So the first condate after 2015 is 15/04/23.
c The latest possible year ends with 54. The latest month in that year is 12. So the latest day is 30. Thus the last
condate in any century is 30/12/54.
d If DD/MM/YY contains all the digits 1, 2, 3, 4, 5, 6, then no digit can be repeated and no digit is 0. Hence the
month must be 12. So the day must contain two of the digits 3, 4, 5, 6. This is impossible.
12
Part 1: Mathematics Challenge for Young Australians 23
CHALLENGE
SOLUTIONS
PRIMARY
2015 Challenge
Solutions–- UPPER
Upper Primary
UP1 Rod Shapes
a There are four equilateral triangles: (1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4).
b To form an isosceles triangle that is not equilateral we need two rods of equal length and a third rod with a different
length. Thus we have only the following choices:
(1, 2, 2), (1, 3, 3), (1, 4, 4), (2, 1, 1), (2, 3, 3), (2, 4, 4), (3, 1, 1), (3, 2, 2), (3, 4, 4), (4, 1, 1), (4, 2, 2), (4, 3, 3).
To form a triangle, we must have each side smaller than the sum of the other two sides. This eliminates
(2, 1, 1), (3, 1, 1), (4, 1, 1), (4, 2, 2).
So there are only eight isosceles triangles that are not equilateral triangles:
(1, 2, 2), (1, 3, 3), (1, 4, 4), (2, 3, 3), (2, 4, 4), (3, 2, 2), (3, 4, 4), (4, 3, 3).
c A scalene triangle has three unequal sides. So there are four possibilities: (1, 2, 3), (1, 2, 4), (1, 3, 4), (2, 3, 4).
To form a triangle, we must have each side smaller than the sum of the other two sides. This eliminates
(1, 2, 3), (1, 2, 4), (1, 3, 4).
So there is only one scalene triangle: (2, 3, 4).
d If the pair of parallel sides have equal length, then the other pair of sides would be parallel. So the pair of parallel
sides must have unequal length.
The possible lengths for the parallel sides are (1, 2), (1, 3), (2, 3).
So we have only the following choices for the quadrilateral:
(1, 1, 2, 1), (1, 2, 2, 2), (1, 3, 2, 3), (1, 1, 3, 1), (1, 2, 3, 2), (1, 3, 3, 3), (2, 1, 3, 1), (2, 2, 3, 2), (2, 3, 3, 3).
To form a quadrilateral, we must have each side smaller than the sum of the other three sides.
This eliminates (1, 1, 3, 1). So there are just eight required quadrilaterals.
UP2 Egg Cartons
a Any four of the following arrangements.
For each of these arrangements, any rotation or reflection is acceptable.
b
For each of these arrangements, any rotation or reflection is acceptable.
c By systematically locating the two empty positions, we see that there are ten different ways to arrange six eggs in
a 2 × 4 carton.
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Mathematics Contests The Australian Scene 2015
13
For each of these arrangements, any rotation or reflection is acceptable.
d Each of Zoe’s arrangements in the 2 × 6 carton has 3 eggs and 9 empty places. Suppose the 3 eggs are white and fill
the empty places with 9 brown eggs. Then each placement of 3 white eggs corresponds to a placement of 9 brown
eggs. So the number of ways of arranging 3 eggs equals the number of ways of arranging 9 eggs.
UP3 Seahorse Swimmers
a To compare times, first convert each PB time to seconds.
Mack
Jack
Zac
Nick
55 s
1 min 8 s = 68 s
1 min 16 s = 76 s
1 min 3 s = 63 s
Alternative i
The table shows the total PB times, in seconds, for all pairs that include Mack and for the corresponding pairs that
exclude Mack.
Pair with M
Total PB
Other pair
Total PB
MJ
123
ZN
139
MZ
131
JN
131
MN
118
JZ
144
Thus the pairs that have the same total PB time are Mack with Zac and Jack with Nick.
Alternative ii
The total PB time for the four swimmers is 55 + 68 + 76 + 63 = 262 seconds. It might be possible that the total
PB time for each pair is 262/2 = 131 seconds. To get the units digit 1 in the total we must pair 55 with 76 and 68
with 63. Each pair then totals 131 as required. Thus Mack is paired with Zac and Jack is paired with Nick.
14
Part 1: Mathematics Challenge for Young Australians 25
b
i Chloe’s PB time is 70 seconds and Sally’s is 53 seconds. So the total of all six PB times is 262 + 70 + 53 = 385
seconds. Since 385 is odd and cannot be divided into two equal amounts, the two teams cannot have the same
total PB times.
ii Alternative i
The total of all six PB times is 262 + 70 + 53 = 385 seconds. As explained in Part i, two teams cannot have
the same total PB times. The next closest would be 1 second apart, in which case the times would be 192 s and
193 s. To see if this is possible, we need to select three PB times and add them to see if they total 192 or 193.
It is easier to just add their units digits to see if we get 2 or 3. The swimmers’ PB times in increasing order
of their units digits are: 70, 53, 63, 55, 76, 68. The only three that give 3 in the units digit of their total are
70, 55, 68, and these do in fact total 193 s. Thus one team has Chloe, Mack, and Jack with a total PB time of
70 + 55 + 68 = 193 s. The other team has Sally, Nick, and Zac with a total PB time of 53 + 63 + 76 = 192 s.
Alternative ii
The table shows the total PB times, in seconds, for all teams that include Mack and for the corresponding
teams that exclude Mack.
Team with M
MJZ
MJN
MJC
MJS
MZN
MZC
MZS
MNC
MNS
MCS
Total PB
199
186
193
176
194
201
184
188
171
178
Other team
NCS
ZCS
ZNS
ZNC
JCS
JNS
JNC
JZS
JZC
JZN
Total PB
186
199
192
209
191
184
201
197
214
207
Thus the two teams that have the closest total PB time are Mack, Jack, Chloe with 193 seconds and Zac, Nick,
Sally with 192 seconds.
c If the average of seven PB times is 60 seconds, then the total of their PB times is 7 × 60 = 420 seconds. From Part
b, the total PB time for the six swimmers is 385 seconds. So the coach’s PB time was 420 − 385 = 35 seconds.
UP4 Primelandia Money
a From 50, we subtract primes from 50 down to 25 and check if the difference is prime:
50 − 47 = 3,
50 − 37 = 13,
50 − 43 = 7,
50 − 31 = 19,
50 − 41 = 9,
50 − 29 = 21.
Since 9 and 21 are not prime, the only pairs of primes that total 50 are: (47,3), (43,7), (37,13), (31,19).
b We have the following amounts that can be paid with one or two coins:
2=2
3=3
4= 2+2
5=5
6= 3+3
7=7
8= 3+5
9= 2+7
10 = 3 + 7
11 = 11
12 = 5 + 7
13 = 13
14 = 3 + 11
15 = 2 + 13
16 = 3 + 13
17 = 17
18 = 5 + 13
19 = 19
20 = 3 + 17
21 = 2 + 19
22 = 3 + 19
23 = 23
24 = 5 + 19
25 = 2 + 23
26 = 3 + 23
We now show that 27T requires at least three coins.
Alternative i
From 27, we subtract primes from 27 down to 14 and check if the difference is prime:
27 − 23 = 4, 27 − 19 = 8, 27 − 17 = 10.
None of the differences is prime. Hence 27T is the smallest payment which requires at least three coins.
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Mathematics Contests The Australian Scene 2015
15
Alternative ii
The sum of two odd primes is even. So, if 27 is the sum of two primes, then one of those primes is 2, the only even
prime. Now 27 = 2 + 25 and 25 is not prime, so 27 cannot be the sum of two primes. Hence 27T is the smallest
payment which requires at least three coins.
c The four largest primes less than 50 are 47, 43, 41, and 37. So the largest even amount that can be made from four
coins is 47 + 43 + 41 + 37 = 168. If the sum of three primes is even, then one of those primes must be 2. So the
largest even amount that can be made from three coins is 47 + 43 + 2 = 92. The difference is 168 − 92 = 76.
d The bag can’t contain a 2T coin. If it did, then the sum of the pair of coins that includes 2T would be odd and the
sum of each other pair of coins would be even.
Since all coins are odd, the sum of all six coins is even. Since the three pairs of coins have the same sum, the sum
of all six coins is divisible by 3. The sum of all six coins is at least 3 + 5 + 7 + 11 + 13 + 17 = 56, which is not a
multiple of 6. The next multiple of 6 is 60. The following table lists all possibilities for each sum of six coins up to
72T.
Sum of six
60
66
72
Sum of pair
20
22
24
All possible pairs
3 + 17, 7 + 13
3 + 19, 5 + 17
5 + 19, 7 + 17, 11 + 13
Thus the smallest amount that could have been in the bag is 72T.
16
Part 1: Mathematics Challenge for Young Australians 27
CHALLENGE
SOLUTIONS
2015 Challenge
Solutions –- JUNIOR
Junior
J1 Quirky Quadrilaterals
a
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Kite
Rhombus
Another kite
b
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I=4
I=9
c
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I = 12
d If I is an odd number we can surround a column of I dots with a rhombus whose vertices are immediately above,
below and either side of the middle of the column as the following examples show. Thus one diagonal of the
rhombus lies on the line through the column of I dots and the other diagonal lies on the line perpendicular to the
first diagonal and passing through the middle dot.
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I=1
28
•
Mathematics Contests The Australian Scene 2015
I=3
17
I=5
If I is an even number we can surround a south-west/north-east diagonal of I dots with a rhombus whose vertices
are immediately south-west, north-east, and either side of the middle of the diagonal as the following examples
show. Thus one diagonal of the rhombus lies on the line through the I dots and the other diagonal lies on the line
perpendicular to the first diagonal and bisecting the distance between the two middle dots.
•
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I =2
I =4
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I=6
These rhombuses are not unique.
J2 Indim Integers
a Since 1 is a factor of every integer, no tile remains if 1 is spun.
If 2 is spun, then all remaining tiles have an odd number. Hence any 3-digit number made from the remaining tiles
would be odd and not divisible by 2. So there is no winning number if 2 is spun.
If 5 is spun, then tiles 1 and 5 are removed. No number made from the remaining tiles is a multiple of 5 since it
cannot end in 0 or 5. So there is no winning number if 5 is spun.
b The only digits remaining after spinning 4 are 3, 5, 6, 7, 9. A number made from these digits is divisible by 4 if
and only if it ends with a 2-digit number that is divisible by 4. The only such 2-digit multiples of 4 that can be
made are 36, 56, 76, and 96.
Alternative i
So there are 12 winning numbers: 536, 736, 936, 356, 756, 956, 376, 576, 976, 396, 596, 796.
Alternative ii
From each of 36, 56, 76, 96, we form a 3-digit number by choosing one of the remaining three digits as its first digit.
So the number of winning numbers is 3 × 4 = 12.
c The only digits remaining after spinning 6 are 4, 5, 7, 8, 9. A 3-digit number is divisible by 6 if and only if its last
digit is even and the sum of all its digits is divisible by 3. So the only winning numbers if 6 is spun are: 594, 954,
894, 984, 498, 948, 798, 978.
18
Part 1: Mathematics Challenge for Young Australians 29
d From Part a, there is no winning number if 1, 2, or 5 is spun.
Alternative i
If 3 is spun, then tile 9 is excluded. So any winning number from spinning 3 must be less than 900.
From Part b, the largest winning number from spinning 4 is 976.
From Part c, the largest winning number from spinning 6 is 984.
The only 3-digit multiples of 7 larger than 984 are 987 and 994. If 7 is spun, then 987 is excluded. The multiple
994 is excluded because 9 is repeated.
All 3-different-digit numbers larger than 984 start with 98. So there are no winning numbers larger than 984 if 8
or 9 is spun.
So the largest winning number overall is 984.
Alternative ii
The largest number with three different digits is 987. This cannot be a winning number from spinning 9, 8, or 7.
Since 987 is not divisible by 6 or 4, it cannot be a winning number from spinning 6 or 4. It is divisible by 3 but
cannot be a winning number from spinning 3 since digit 9 would have been eliminated. So 987 is not a winning
number.
The next largest number is 986. This cannot be a winning number from spinning 9, 8, or 6. Since 986 is not divisible
by 7, 4, or 3, it is not a winning number from spinning 7, 4, or 3. So 986 is not a winning number.
The next largest number is 985. This cannot be a winning number from spinning 9, 8, or 5. Since 985 is not divisible
by 7, 6, 4, or 3, it is not a winning number from spinning 7, 6, 4, or 3. So 985 is not a winning number.
The next largest number is 984. This is divisible by 6 and none of its digits is a factor or multiple of 6. So it is a
winning number.
Thus the largest winning number overall is 984.
J3 Primelandia Money
a We have the following amounts that can be paid with one or two coins:
2=2
3=3
4= 2+2
5=5
6= 3+3
7=7
8= 3+5
9= 2+7
10 = 3 + 7
11 = 11
12 = 5 + 7
13 = 13
14 = 3 + 11
15 = 2 + 13
16 = 3 + 13
17 = 17
18 = 5 + 13
19 = 19
20 = 3 + 17
21 = 2 + 19
22 = 3 + 19
23 = 23
24 = 5 + 19
25 = 2 + 23
26 = 3 + 23
We now show that 27T requires at least three coins.
Alternative i
From 27, we subtract primes from 27 down to 14 and check if the difference is prime:
27 − 23 = 4, 27 − 19 = 8, 27 − 17 = 10.
None of the differences is prime. Hence 27T is the smallest payment which requires at least three coins.
Alternative ii
The sum of two odd primes is even. So, if 27 is the sum of two primes, then one of those primes is 2, the only even
prime. Now 27 = 2 + 25 and 25 is not prime, so 27 cannot be the sum of two primes. Hence 27T is the smallest
payment which requires at least three coins.
b The bag can’t contain a 2T coin. If it did, then the sum of the pair of coins that includes 2T would be odd and the
sum of each other pair of coins would be even.
Since all coins are odd, the sum of all six coins is even. Since the three pairs of coins have the same sum, the sum
of all six coins is divisible by 3. The sum of all six coins is at least 3 + 5 + 7 + 11 + 13 + 17 = 56, which is not a
multiple of 6. The next multiple of 6 is 60. The following table lists all possibilities for each sum of six coins up to
72T.
Sum of six Sum of pair
All possible pairs
60
20
3 + 17, 7 + 13
66
22
3 + 19, 5 + 17
72
24
5 + 19, 7 + 17, 11 + 13
Thus the smallest amount that could have been in the bag is 72T.
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Mathematics Contests The Australian Scene 2015
19
c Examining primes less than 50 for pairs that differ by 6 and starting with small primes, we quickly find the sequence
5, 11, 17, 23, 29.
d Alternative i
Suppose we have a sequence of five prime numbers in ascending order which are 6 apart. Since 6 is even and 2 is
the only even prime, all primes in the sequence must be odd. Hence they only end in 1, 3, 5, 7, or 9. The difference
between any two primes in the sequence is 6, 12, 18, or 24. So no two primes in the sequence can end in the same
digit. There are five primes in the sequence so all of 1, 3, 5, 7, 9 must appear as last digits. The only prime whose
last digit is 5 is 5 itself. So the sequence must be 5, 11, 17, 23, 29.
Alternative ii
Suppose we have a sequence of five prime numbers in ascending order which are 6 apart. Since 6 is even and 2 is
the only even prime, all primes in the sequence must be odd. Hence they only end in 1, 3, 5, 7, or 9. The only
prime ending in 5 is 5 itself. Therefore, if the first prime in the sequence ends in 5, the sequence is 5, 11, 17, 23, 29.
If the first prime ends in 1, then the next four primes end in 7, 3, 9, 5. If the first prime ends in 3, then the next
four primes end in 9, 5, 1, 7. If the first prime ends in 7, then the next four primes end in 3, 9, 5, 1. If the first
prime ends in 9, then the next four primes end in 5, 1, 7, 3. In each of these four cases, the prime ending in 5 is at
least 1 + 6 = 7, 3 + 6 = 9, 7 + 6 = 13, and 9 + 6 = 15 respectively. This is impossible.
So the only sequence is 5, 11, 17, 23, 29.
Alternative iii
Suppose we have a sequence of five prime numbers in ascending order which are 6 apart. Replace each of the five
numbers in the sequence with its remainder when divided by 5.
If the first remainder is 0, then the sequence of remainders is 0, 1, 2, 3, 4. If the first remainder is 1, then the
sequence of remainders is 1, 2, 3, 4, 0. If the first remainder is 2, then the sequence of remainders is 2, 3, 4, 0,
1. If the first remainder is 3, then the sequence of remainders is 3, 4, 0, 1, 2. If the first remainder is 4, then the
sequence of remainders is 4, 0, 1, 2, 3.
In all cases there is a remainder of 0, which represents a multiple of 5 in the original sequence. The only prime that
is divisible by 5 is 5 itself. Since there is no positive number that is 6 less than 5, 5 must be the first number in the
original sequence. Thus the original sequence is 5, 11, 17, 23, 29.
J4 Condates
a Any condate in a year starting with 20 must be DD/MM/20YY. So the month must be 1M. Then there is no
available digit for M. So there is no condate in the years 2000 to 2099.
Any condate in a year starting with 21 must be DD/MM/21YY. So the month must be 0M. Hence the day is 3D.
Then there is no available digit for D. So there is no condate in the years 2100 to 2199.
There is no condate in a year starting with 22, since 2 is repeated.
Any condate in a year starting with 23 must be DD/MM/23YY. So the month is 0M or 10. If the month is 10,
then there is no available digit for the first D in the day. So the month is 0M and the condate is 1D/0M/23YY.
The earliest year is 2345 and the earliest month in that year is 06. Hence the first condate after the year 2015 is
17/06/2345.
b In any year either 0 or 1 must appear in the month. If 0 is not in the month, then the month is 12. Hence the year
cannot start with 2. So in the years 2000 to 2999, 0 must be in the month for any condate.
c From Part b, 0 must be in the month for any condate from 2000 to 2999. If the 0 is not the first digit of the month,
then the month must be 10. Hence the remaining digits are 3, 4, 5, 6, 7, no two of which can represent a day. So
in the years 2000 to 2999, 0 must be the first digit in the month for any condate.
d From Part c, a condate in the years 2000 to 2999 has the form DD/0M/2YYY. So DD is either 1D or 3D.
If the condate is 1D/0M/2YYY, then each of the letters can be replaced by any one of the digits 3, 4, 5, 6, 7
without repeating a digit. There are 5 choices for D, then 4 choices for M, and so on. So the number of condates
is 5 × 4 × 3 × 2 × 1 = 120.
If the condate is 3D/0M/2YYY, then D is 1 and the month has 31 days. The remaining available digits are 4, 5,
6, 7. So M is 5 or 7. Thus there are 2 choices for M, then 3 choices for the first Y, and so on. So the number of
condates is 2 × 3 × 2 × 1 = 12.
Thus the number of condates in the years 2000 to 2999 is 120 + 12 = 132.
20
Part 1: Mathematics Challenge for Young Australians 31
J5 Jogging
4
a On the flat, Theo will run a total of 4 km at 12 km/h and this will take 12
× 60 = 20 minutes. He will run 3 km
3
uphill at 9 km/h and this will take 9 × 60 = 20 minutes. He will run 3 km downhill at 15 km/h and this will take
3
× 60 = 12 minutes. So the total time taken for Theo to complete the course will be 20 + 20 + 12 = 52 minutes.
15
b Alternative i
If the wind blows from the west, then Theo’s speed and times for the various sections of the course will be:
Section
Distance
Speed
Flat east
Uphill
Downhill
Flat west
2 km
3 km
3 km
2 km
13 km/h
10 km/h
14 km/h
11 km/h
2
So the total time to complete the course would be ( 13
+
3
10
+
3
14
+
Time
2
13
3
10
3
14
2
11
2
11 )
h
h
h
h
× 60 ≈ 51.0 minutes.
If the wind blows from the east, then Theo’s speed and times for the various sections of the course will be:
Section
Distance
Speed
Time
Flat east
Uphill
Downhill
Flat west
2 km
3 km
3 km
2 km
11 km/h
8 km/h
16 km/h
13 km/h
2
11 h
3
8 h
3
16 h
2
h
13
2
So the total time to complete the course would be ( 11
+
3
8
+
3
16
+
2
13 )
× 60 ≈ 53.9 minutes.
So Theo’s time will be less if the wind blows from the west.
Alternative ii
Since Theo runs both ways on the flat, he gains no advantage on the flat from either wind direction.
3
3
+ 14
hours. If the wind blows from the east,
If the wind blows from the west, then his time on the hill will be 10
3
3
then his time on the hill will be 8 + 16 hours. To simplify comparison of these times we divide both by 3 and
1
1
12
multiply both by 2. So we want to compare 15 + 17 = 12
35 with 4 + 8 = 32 . Thus the first time is shorter.
So Theo’s time will be less if the wind blows from the west.
c Alternative i
From Part a, Theo has to gain 2 minutes.
From Part a, he ran 3 km uphill at 9 km/h in 20 minutes. So to gain 2 minutes uphill, he will need to run 3 km in
3
18 minutes. Therefore his speed needs to be 18
× 60 = 10 km/h, which is 1 km/h faster.
From Part a, he ran 3 km downhill at 15 km/h in 12 minutes. So to gain 2 minutes downhill, he will need to run
3
3 km in 10 minutes. Therefore his speed needs to be 10
× 60 = 18 km/h, which is 3 km/h faster.
Alternative ii
Theo wants to complete the course in 50 minutes. From Part a, the flat takes 20 minutes. This leaves 30 minutes
for the hill.
3
If Theo’s speed uphill is r km/h and downhill is 15 km/h, then the time taken for the hill in hours is r3 + 15
=
3
1
1
3
So r = 2 − 5 = 10 . Hence r = 10 km/h. This is 1 km/h faster.
If Theo’s speed downhill is r km/h and uphill is 9 km/h, then the time taken for the hill in hours is
So 3r = 12 − 13 = 16 . Hence r = 18 km/h. This is 3 km/h faster.
32
Mathematics Contests The Australian Scene 2015
21
3
9
+ 3r =
30
60
= 12 .
30
60
= 12 .
J6 Tessellating Hexagons
a The flipped hexagons are shaded.
D
E
F
D
B
A
E
F
C
B
A
F
A C
D
C
E
B
F E
A C
D
E
B
B
D
C A
F
B
E
C
D
C A
F
A
B
D
E
C
F
E
D
B
A
F
A
B
D
E
F
E
D
B
A
F
C
F
A C
D
C
E
B
F E
A C
D
E
B
B
D
C A
F
B
E
C
D
C A
F
A
C
B
F
E
A
B
D
F
E
D
b Call the block of hexagons in the solution to Part a the 16-block.
Alternative i
We see from the 16-block that AB = ED, BC = F A, the sum of the three interior angles A, C, D is 360◦ , and the
sum of the three interior angles B, E, F is also 360◦ .
Indicating a hexagon side with a dash, the bottom boundary of the 16-block is
A − BF − E − D − CA − BF − E − D and the top boundary is
D − E − F B − AC − D − E − F B − A.
So these boundaries are identical. Hence the 16-block can be translated vertically to form an infinite column without
gaps and without overlap.
The right boundary of this column is
DA − F − EB − C − DA − F − EB − C − DA repeated indefinitely.
The left boundary of the column is
C − BE − F − AD − C − BE − F − AD − C repeated indefinitely.
So these boundaries are identical. Hence the column can be translated horizontally to tessellate the plane.
Alternative ii
Make several photocopies of the 16-block. Place them side by side and top to bottom to show how they fit together.
22
Part 1: Mathematics Challenge for Young Australians 33
c There is no point of symmetry for the tessellation inside a hexagon because the hexagon has no point of symmetry.
So all points of symmetry for the tessellation must be on the edges of the hexagons. If a point of symmetry for the
tessellation is on an edge of a hexagon, then it must be at the midpoint of that edge.
From the block of hexagons in Part a, we see that each of the edges AB, BC, DE, F A, has the smallest edge EF
attached at one end but not the other. So the midpoints of edges AB, BC, DE, F A are not points of symmetry
for the tessellation.
To check the midpoint of edge CD, trace a copy of the block and place it exactly over the original block. Then
place a pin through the midpoint of edge CD, rotate the traced copy of the block through 180◦ and notice that it
again fits exactly over the original block (except for overhang). Thus the midpoint of CD is a point of symmetry
for the tessellation. Similarly for the midpoint of EF .
So the only points of symmetry for the tessellation are the midpoints of edges CD and EF .
34
Mathematics Contests The Australian Scene 2015
23
d The block of hexagons in Part a is reproduced below with hexagons labelled H1 , H2 , H3 , H4 , H1 , H2 , H3 , H4 , and
so on as shown.
D
E
F
D
F
C
B
A
H3
A
A C
D
H4
C
B
H4
E
H3
A C
D
B
F E
E
B
B
D
C A
H1
F
B
E
C A
H1
A
B
D
E
A
B
D
E
C
H2
F
E
D
B
A
F
F
E
D
B
A
F
H4
C
H3
A C
D
H4
C
B
F E
H3
A C
D
E
B
H1
F
D
C A
C A
A
C
H2
A
B
C
H2
D
H1
F
E
B
B
E
F
C
H2
D
F
F
E
B
F
E
D
F
E
D
Hexagons with the same subscript translate to one another. So there are several blocks of four hexagons that
tessellate by translation. Three such blocks are: {H1 , H2 , H3 , H4 }, {H1 , H2 , H3 , H4 }, {H1 , H2 , H3 , H4 }.
24
Part 1: Mathematics Challenge for Young Australians 35
CHALLENGE
SOLUTIONS
2015 Challenge
Solutions– -INTERMEDIATE
Intermediate
I1 Indim Integers
See Junior Problem 2.
I2 Digital Sums
a Alternative i
The table lists 3-digit numbers n in decreasing order whose digit sums s are multiples of 7.
n
993
984
975
966
957
950
948
941
939
932
923
914
s
21
21
21
21
21
14
21
14
21
14
14
14
7 divides n?
no
no
no
yes
no
no
no
no
no
no
no
no
n
905
894
885
876
867
860
858
851
849
842
833
s
14
21
21
21
21
14
21
14
21
14
14
7 divides n?
no
no
no
no
no
no
no
no
no
no
yes
So the largest even and odd n that are multiples of 7 and have a digital sum that is also a multiple of 7 are 966 and
833 respectively.
Alternative ii
The table lists 3-digit numbers n and their digital sums s. The n are multiples of 7 in decreasing order.
n
994
987
980
973
966
959
952
945
938
931
924
917
s
22
24
17
19
21
23
16
18
20
13
15
17
7 divides s?
no
no
no
no
yes
no
no
no
no
no
no
no
n
910
903
896
889
882
875
868
861
854
847
840
833
s
10
12
23
25
18
20
22
15
17
19
12
14
7 divides s?
no
no
no
no
no
no
no
no
no
no
no
yes
So the largest even and odd n that are multiples of 7 and have a digital sum that is also a multiple of 7 are 966 and
833 respectively.
b Alternative i
The largest 3-digit number with a digital sum of 7 is 700. The digital sum s of a 3-digit number is at most 3×9 = 27.
So, if the FDS of a 3-digit number n is 7 and n > 700, then the digital sum of n is 16 or 25. The table lists, in
decreasing order, the first few 3-digit numbers n with digital sum 16 or 25.
36
Mathematics Contests The Australian Scene 2015
25
s
16
n
970
961
952
943
934
925
916
907
880
871
862
853
844
835
826
7 divides n?
no
no
yes
no
no
no
no
no
no
no
no
no
no
no
yes
s
25
n
997
988
979
898
889
799
7 divides n?
no
no
no
no
yes
no
So the three largest 3-digit multiples of 7 that have FDS 7 are: 952, 889, 826.
Alternative ii
The table lists, in decreasing order, multiples of 7 and their digital sum sequences (DSS).
n
994
987
980
973
966
959
952
945
938
931
924
917
910
DSS
22, 4
24, 6
17, 8
19, 10, 1
21, 3
23, 5
16, 7
18, 9
20, 2
13, 4
15, 6
17, 8
10, 1
n
903
896
889
882
875
868
861
854
847
840
833
826
DSS
12, 3
23, 5
25, 7
18, 9
20, 2
22, 4
15, 6
17, 8
19, 10, 1
12, 3
14, 5
16, 7
So the three largest 3-digit multiples of 7 that have FDS 7 are: 952, 889, 826.
c Alternative i
The table lists, in decreasing order, the 3-digit integers n that are multiples of 7 together with their FDSs (F).
26
Part 1: Mathematics Challenge for Young Australians 37
n
994
987
980
973
966
959
952
945
938
931
924
917
910
903
F
4
6
8
1
3
5
7
9
2
4
6
8
1
3
n
896
889
882
875
868
861
854
847
840
833
826
819
812
805
F
5
7
9
2
4
6
8
1
3
5
7
9
2
4
497
490
483
476
469
462
455
448
441
434
427
420
413
406
2
4
6
8
1
3
5
7
9
2
4
6
8
1
399
392
385
378
371
364
357
350
343
336
329
322
315
308
301
3
5
7
9
2
4
6
8
1
3
5
7
9
2
4
n
798
791
784
777
770
763
756
749
742
735
728
721
714
707
700
294
287
280
273
266
259
252
245
238
231
224
217
210
203
F
6
8
1
3
5
7
9
2
4
6
8
1
3
5
7
6
8
1
3
5
7
9
2
4
6
8
1
3
5
n
693
686
679
672
665
658
651
644
637
630
623
616
609
602
F
9
2
4
6
8
1
3
5
7
9
2
4
6
8
196
189
182
175
168
161
154
147
140
133
126
119
112
105
7
9
2
4
6
8
1
3
5
7
9
2
4
6
n
595
588
581
574
567
560
553
546
539
532
525
518
511
504
F
1
3
5
7
9
2
4
6
8
1
3
5
7
9
Thus the only 3-digit multiples of 7 that have FDS 7 are: 133, 196, 259, 322, 385, 448, 511, 574, 637, 700, 763, 826,
889, 952. These have a common difference of 63. So the only 3-digit multiples of 7 that have FDS 7 are the integers
n = 7 + 63t where t = 2, 3, . . ., 15.
Alternative ii
The three numbers from Part b, 952, 889, 826, have a common difference of 63. This suggests that all 3-digit
multiples of 7 that have FDS 7 have the form 7 + 63m. These numbers are: 133, 196, 259, 322, 385, 448, 511, 574,
637, 700, 763, 826, 889, 952. Since 7 divides 7 and 63, all these numbers are multiples of 7. By direct calculation,
each has FDS 7. But are there any other 3-digit multiples of 7 that have FDS 7?
Extending the table in the first solution to Part b shows that there are no other n besides those of the form 7 + 63m
that have digital sum 16 or 25. The following table lists in decreasing order all 3-digit integers with digital sum 7.
n
700
610
601
520
511
502
430
421
412
403
7 divides n?
yes
no
no
no
yes
no
no
no
no
no
n
340
331
322
313
304
250
241
232
223
214
7 divides n?
no
no
yes
no
no
no
no
no
no
no
n
205
160
151
142
133
124
115
106
7 divides n?
no
no
no
no
yes
no
no
no
So there are no other 3-digit multiples of 7 with FDS 7 besides those of the form 7 + 63m.
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Mathematics Contests The Australian Scene 2015
27
Alternative iii
From the table in the first solution to Part b, it appears that integers with the same FDS differ by a multiple of
9. This is equivalent to saying they have the same remainder when divided by 9. We now show that this is always
true.
Any positive integer n can be written in the form
n = a + 10b + 100c + · · ·
where a, b, . . . are the digits of n. Thus
n =
=
a + (1 + 9)b + (1 + 99)c + · · ·
(a + b + c + · · ·) + 9(b + 11c + · · ·)
So n and its digital sum have the same remainder when divided by 9. Therefore all digital sums in the digital sum
sequence for n, including its FDS, have the same remainder when divided by 9. Hence, if n is a multiple of 9, then
the FDS of n is 9, and if n is not a multiple of 9, then the FDS of n is the remainder when n is divided by 9.
So, the FDS of n is 7 if and only if n has the form n = 7 + 9r. Such an n is a multiple of 7 if and only if 7 divides
r. So n is a multiple of 7 and has FDS 7 if and only if n has the form n = 7 + 63t. Hence the only 3-digit multiples
of 7 that have FDS 7 are the integers n = 7 + 63t where t = 2, 3, . . ., 15. These are: 133, 196, 259, 322, 385, 448,
511, 574, 637, 700, 763, 826, 889, 952.
d Alternative i
The table lists, in decreasing order, the 3-digit integers n that are multiples of 8 together with their FDSs (F).
n
992
984
976
968
960
952
944
936
928
920
912
904
F
2
3
4
5
6
7
8
9
1
2
3
4
496
488
480
472
464
456
448
440
432
424
416
408
400
1
2
3
4
5
6
7
8
9
1
2
3
4
n
896
888
880
872
864
856
848
840
832
824
816
808
800
392
384
376
368
360
352
344
336
328
320
312
304
F
5
6
7
8
9
1
2
3
4
5
6
7
8
5
6
7
8
9
1
2
3
4
5
6
7
n
792
784
776
768
760
752
744
736
728
720
712
704
F
9
1
2
3
4
5
6
7
8
9
1
2
296
288
280
272
264
256
248
240
232
224
216
208
200
8
9
1
2
3
4
5
6
7
8
9
1
2
n
696
688
680
672
664
656
648
640
632
624
616
608
600
192
184
176
168
160
152
144
136
128
120
112
104
F
3
4
5
6
7
8
9
1
2
3
4
5
6
3
4
5
6
7
8
9
1
2
3
4
5
n
592
584
576
568
560
552
544
536
528
520
512
504
F
7
8
9
1
2
3
4
5
6
7
8
9
Thus the only 3-digit multiples of 8 that have FDS 8 are: 152, 224, 296, 368, 440, 512, 584, 656, 728, 800, 872,
944. These have a common difference of 72. So the only 3-digit multiples of 8 that have FDS 8 are the integers
n = 8 + 72t where t = 2, 3, . . ., 13.
Alternative ii
From the third solution to Part c, the FDS of a positive integer n is 8 if and only if n has the form n = 8 + 9r.
Such an n is a multiple of 8 if and only if 8 divides r. So n is a multiple of 8 and has FDS 8 if and only if n has the
form n = 8 + 72t. Hence the only 3-digit multiples of 8 that have FDS 8 are the integers n = 8 + 72t where t = 2,
3, . . ., 13. These are: 152, 224, 296, 368, 440, 512, 584, 656, 728, 800, 872, 944.
28
Part 1: Mathematics Challenge for Young Australians 39
I3 Coin Flips
a Here are three moves that leave exactly 10 coins heads up.
Start:
Move 1:
Move 2:
Move 3:
T T T T T T T T T T T
HHHHT T T T T T T
HHHHHHHHT T T
HHHHHHHT HHH
T
T
T
T
T
T
T
T
T
T
T
T
b Since there are 52 coins flipped in each move, the total number of coin flips is 3 × 52 = 156. We start with 154 T s
and want to finish with 154 Hs. We can achieve this by flipping 153 coins exactly once and flipping one coin three
times over the three moves. Here are three moves that leave all coins heads up.
154
Start:
T . . .T
52
102
Move 1: H . . . H T . . . T
51
51
=
51
H . . .H H T . . .T T . . .T
51
51
51
Move 2: H . . . H T H . . . H T . . . T
51
51
51
Move 3: H . . . H H H . . . H H . . . H
c Since there are 4 coins flipped in each move, the total number of coins flipped in 6 moves is at most 24. So it takes
at least 7 moves to finish with all coins heads up. We now show that this can actually be done in 7 moves.
After three moves, each flipping 4 Ts to 4 Hs, we get exactly 12 coins heads up. Then applying the three moves in
Part a to the 14 coins that are tails up, gives us exactly 12 + 10 = 22 coins heads up. One more move, flipping 4
Ts to 4 Hs, results in all coins heads up. Thus seven moves suffice to get all coins heads up.
So 7 is the minimum number of moves to get all coins heads up.
d Alternative i
Suppose m coins are flipped on each move where m is odd. Two moves give the following results.
154
Start:
T . . .T
m
154−m
Move 1: H . . . H T . . . T
m−n
n
m−n
154−2m+n
Move 2: H . . . H T . . . T H . . . H T . . . T
=
2m−2n
154−2m+2n
H . . .H
T . . .T
where 0 ≤ n ≤ m and
m − n ≤ 154 − m
To have all 154 coins with heads up after Move 3, we must have m = 154 − 2m + 2n. Thus 3m = 154 + 2n. But
3m is odd and 154 + 2n is even. So we cannot have 154 coins heads up after just three moves.
40
Mathematics Contests The Australian Scene 2015
29
Alternative ii
Three moves give the following results. Note that in a move, an even number of coins may mean 0 coins.
154
Start:
T . . .T
odd
odd
Move 1: H . . . H T . . . T
even
odd
even
odd
Move 2: H . . . H T . . . T H . . . H T . . . T or
odd
even
odd
even
H . . . H T . . . T H . . . H T . . . T giving
even
even
H . . .H T . . .T
even
even
odd
odd
Move 3: H . . . H T . . . T H . . . H T . . . T or
odd
odd
even
even
H . . . H T . . . T H . . . H T . . . T giving
odd
odd
H . . .H T . . .T
Since 154 is even, we cannot have 154 coins heads up after just three moves.
I4 Jogging
See Junior Problem 5.
I5 Folding Fractions
a Let QR = x. Then P R = 1 − x.
T
1
2
1
2
P
Q
x
1−x
R
S
1−x
U
V
In P QR Pythagoras’ theorem gives
Hence 2x = 34 and x = 38 .
1
4
+ x2 = (1 − x)2 = x2 − 2x + 1.
b Alternative i
From Part a, we have:
1
2
T
1
2
P
a
Q
b
5
8
3
8
a
R
b
S
U
b
a
V
30
Part 1: Mathematics Challenge for Young Australians 41
Since angles RQP , RP S, P T S, SU V are right angles, triangles RQP , P T S, V U S have the same angles as shown.
Hence they are similar and we have: V U :U S:SV = P T :T S:SP = RQ:QP :P R = 3:4:5.
Hence P S = 35 P T =
Then SV = 54 U S =
5
1
5
3 × 2 = 6 . So U S = U P − SP
5
5
× 16 = 24
and V U = 34 U S = 34
4
5
1
6 = 6.
1
= 18 .
6
=1−
×
Alternative ii
As in Alternative i, V U :U S:SV = P T :T S:SP = RQ:QP :P R = 3:4:5. So ST = 43 P T =
From the side of the square, we have: 1 = U V + V S + ST = 53 V S + V S +
5
So V S = 13 × 58 = 24
.
From V U S, we have: V U = 35 SV =
3
5
×
5
24
=
1
8
and U S = 45 SV =
4
5
×
2
3
5
24
= 85 V S + 23 .
= 16 .
Alternative iii
Draw RW perpendicular to T S and let T S = t and V U = r. From Part a, we have:
T
1
2
1
2
P
3
8
Q
3
8
5
8
W
t−
R
3
8
S
U
r
V r
Now apply Pythagoras’ theorem to the following triangles.
In RW S, RS 2 = RW 2 + W S 2 = 1 + (t − 38 )2 .
In RP S, RS 2 = RP 2 + P S 2 = ( 58 )2 + P S 2 .
In P T S, P S 2 = P T 2 + T S 2 = ( 12 )2 + t2 .
Hence
1 + (t − 38 )2
9
1 + t2 + 64
− 34 t
3
t
4
t
( 58 )2 + ( 12 )2 + t2
25
+ 14 + t2
64
9
1 − 14 + 64
− 25
64
3
16
3
1
−
=
−
4
64
4
4 =
4
1
2
3 × 2 = 3
=
=
=
=
=
5
So P S 2 = ( 12 )2 + ( 23 )2 = 14 + 49 = 25
36 and P S = 6 .
5
1
Hence U S = U P − SP = 1 − 6 = 6 .
Since 1 = T S + SV + r = 23 + SV + r, SV = 31 − r.
In V U S, Pythagoras’ theorem gives V S 2 = V U 2 + U S 2 . Hence
1
9
Thus V U =
42
1
8
and SV =
1
3
−
1
8
=
( 31 − r)2
+ r 2 − 23 r
2
3r
r
=
=
=
=
r 2 + ( 16 )2
1
r 2 + 36
1
1
9 − 36 =
3
1
2 × 12 =
5
24 .
Mathematics Contests The Australian Scene 2015
31
3
36
1
8
=
4
3
1
12
1
2
×
1
2
= 23 .
c Since angles RQP , RP S, P T S, SU V are right angles, triangles RQP , P T S, V U S have the same angles as shown.
Hence they are similar and we have RQ:QP :P R = P T :T S:SP = V U :U S:SV = 7:24:25 or 24:7:25.
T
Q
P
a
b
a
R
b
S
b
a
U
V
Alternative i
If P Q/QR = 7/24, let P Q = 7k. Then QR = 24k and P R = 25k. Since P R = 1 − QR, we have 25k = 1 − 24k. So
1
49k = 1 and P Q = 7 × 49
= 17 .
If P Q/QR = 24/7, let P Q = 24k. Then QR = 7k and P R = 25k. Since P R = 1 − QR, we have 25k = 1 − 7k. So
1
32k = 1 and P Q = 24 × 32
= 34 .
Alternative ii
If P Q/QR = 7/24, then P R/QR = 25/24. Since P R = 1 − QR, we have 24 − 24QR = 25QR, QR = 24/49, and
P R = 25/49. From Pythagoras, P Q2 = P R2 − QR2 = 1 − 2QR = 1 − 48/49 = 1/49. So P Q = 17 .
If P Q/QR = 24/7, then P R/QR = 25/7. Since P R = 1 − QR, we have 7 − 7QR = 25QR, QR = 7/32, and
P R = 25/32. From Pythagoras, P Q2 = P R2 − QR2 = 1 − 2QR = 1 − 7/16 = 9/16. So P Q = 34 .
Alternative iii
Let T P = x. If V U/U S = 7/24, let V U = 7k. Then U S = 24k and V S = 25k and we have:
T
P
x
1 − 32k
Q
1 − 24k
R
S
24k
25k
U
7k
V
7k
So (1 − 24k)/25k = x/7k = (1 − 32k)/24k.
Hence 25x = 7(1 − 24k) and 24x = 7(1 − 32k).
Subtracting gives x = 7(32 − 24)k = 56k.
Substituting gives 25(56k) = 7(1 − 24k).
Hence 1 = 224k and x = 56/224 = 8/32 = 1/4.
32
Part 1: Mathematics Challenge for Young Australians 43
Alternatively, in P T S, Pythagoras’ theorem gives
(1 − 24k)2
(24k)2 − 48k
16k
1
x
=
=
=
=
=
(1 − 32k)2 + (56k)2
(32k)2 − 64k + (56k)2
(562 + 322 − 242 )k 2
(142 + 82 − 62 )k = 224k
56/224 = 8/32 = 1/4
If V U/U S = 24/7, let V U = 24k. Then U S = 7k and V S = 25k. We have:
T
P
x
1 − 49k
Q
1 − 7k
R
S
7k
25k
U
24k
V
24k
So (1 − 7k)/25k = x/24k = (1 − 49k)/7k.
Hence 25x = 24(1 − 7k) and 7x = 24(1 − 49k).
Subtracting gives 18x = 24(49 − 7)k = 24(42)k, hence x = 56k.
Substituting gives 7(56k) = 24(1 − 49k).
Hence 49k = 3(1 − 49k), 4(49k) = 3, x = 56(3)/4(49) = 6/7.
Alternatively, in P T S, Pythagoras’ theorem gives
(1 − 7k)2
(7k)2 − 14k
84k
12
3
x
=
=
=
=
=
=
(1 − 49k)2 + (56k)2
(49k)2 − 98k + (56k)2
(562 + 492 − 72 )k 2
(82 + 72 − 12 )7k = (112)7k
(28)7k
6/7
I6 Crumbling Cubes
a The small cubes have at most three faces exposed. The only small cubes that have exactly three faces exposed
are the 8 on the corners of the 10 × 10 × 10 cube. The only small cubes that have exactly two faces exposed are
the 8 on each edge of the 10 × 10 × 10 cube that are not on its corners. All other cubes have less than two faces
exposed. There are 12 edges on the 10 × 10 × 10 cube, so the number of small cubes removed at the first step is
8 + (12 × 8) = 104.
b Alternative i
At the first step the 8 small cubes at the corners are removed. The other 192 cubes come equally from the 12 edges
but not from the corners. So the number of non-corner small cubes on each edge is 192/12 = 16. Hence each edge
in the original cube had a total of 18 small cubes.
Alternative ii
A cube with n small cubes along one edge will lose 8 + 12(n − 2) small cubes at the first step. So 8 + 12(n − 2) = 200,
12(n − 2) = 192, n − 2 = 16, n = 18. Thus the original cube was 18 × 18 × 18.
44
Mathematics Contests The Australian Scene 2015
33
Alternative iii
From Part a, a 10 × 10 × 10 cube loses 104 small cubes at the first step. Increasing the cube’s dimension by 1
increases the number of lost cubes by 12, one for each edge. Since 200 − 104 = 96 and 96/12 = 8, the cube that
loses 200 small cubes at the first step is 18 × 18 × 18.
c Alternative i
At the first step, each of the 6 faces of the 9 × 9 × 9 cube are converted to a 7 × 7 single layer of small cubes. The
exposed surface area of this layer is 7 × 7 small faces plus a ring of 4 × 7 small faces. So the surface area of the
remaining object is 6 × (49 + 28) = 6 × 77 = 462.
Alternative ii
First remove the middle small cube on one edge of the 9 × 9 × 9 cube. This increases the surface area by 2 small
faces. Then remove the small cubes either side. This does not change the surface area. Continue until only the two
corner cubes remain. At this stage the surface area has increased by 2 small faces. Repeating this process on all
12 edges increases the surface area by 12 × 2 small faces. At this stage all 8 corner cubes have 6 exposed faces. So
removing the 8 corner cubes reduces the surface area by 8 × 6 small faces. The surface area of the original 9 × 9 × 9
cube was 6 × 81 = 486. Hence the surface area of the remaining object is 486 + 24 − 48 = 462.
d At the first step, the original 9 × 9 × 9 cube is reduced to a 7 × 7 × 7 cube with a 7 × 7 single layer of small cubes
placed centrally on each face. At this stage the number of small cubes removed is 8 + (12 × 7) = 92.
7×7
7×
7×
7
7
At the second step the only small cubes removed are the edge cubes in the 7 × 7 single layers. This leaves a 7 × 7 × 7
cube with a 5 × 5 single layer of small cubes placed centrally on each face. So in this step, the number of small
cubes removed is 6(4 + 4 × 5) = 144.
5×5
5×
5×
5
5
At the third step the only small cubes removed are the edge cubes in the 5 × 5 single layers and the edge cubes in
the 7 × 7 × 7 cube. So in this step, the number of small cubes removed is 6(4 + 4 × 3) + 8 + (12 × 5) = 164.
Thus the number of small cubes remaining is (9 × 9 × 9) − 92 − 144 − 164 = 329.
34
Part 1: Mathematics Challenge for Young Australians 45
CHALLENGE STATISTICS – MIDDLE PRIMARY
Mean Score/School Year/Problem
Mean
Year
Number of
Students
Problem
Overall
1
2
3
4
3
486
8.8
2.1
2.4
2.3
2.1
4
653
10.6
2.6
2.9
2.7
2.6
All Years*
1166
9.8
2.3
2.7
2.6
2.4
Please note:* This total includes students who did not provide their school year.
Score Distribution %/Problem
Challenge Problem
Score
1
e-Numbers
2
Quazy Quilts
3
Egg Cartons
4
Condates
Did not attempt
0%
1%
2%
3%
0
7%
6%
10%
10%
1
15%
12%
11%
18%
2
29%
20%
18%
20%
3
32%
29%
33%
24%
4
16%
32%
26%
25%
Mean
2.3
2.7
2.6
2.4
Discrimination
Factor
0.5
0.6
0.6
0.6
Please note:
The discrimination factor for a particular problem is calculated as follows:
(1) The students are ranked in regard to their overall scores.
(2) The mean score for the top 25% of these overall ranked students is calculated for that particular problem
including no attempts. Call this mean score the ‘mean top score’.
(3) The mean score for the bottom 25% of these overall ranked students is calculated for that particular problem
including no attempts. Call this mean score the ‘mean bottom score’.
(4) The discrimination factor = mean top score – mean bottom score
4
Thus the discrimination factor ranges from 1 to –1. A problem with a discrimination factor of 0.4 or higher is
considered to be a good discriminator.
46
Mathematics Contests The Australian Scene 2015
CHALLENGE STATISTICS – UPPER PRIMARY
Mean Score/School Year/Problem
Mean
YEAR
Number of
Students
Problem
Overall
1
2
3
4
5
1363
8.6
1.8
2.8
2.4
1.7
6
1910
9.8
2.0
3.0
2.8
2.0
7
129
10.9
2.3
3.3
3.2
2.2
All Years*
3416
9.4
2.0
2.9
2.7
1.9
Please note:* This total includes students who did not provide their school year.
Score Distribution %/Problem
Challenge Problem
1
Rod Shapes
2
Egg Cartons
3
Seahorse
Swimmers
4
Primelandia
Money
Did not attempt
1%
0%
1%
3%
0
3%
3%
5%
11%
1
40%
6%
14%
29%
2
24%
20%
20%
26%
3
21%
38%
29%
20%
4
11%
33%
31%
12%
Mean
2.0
2.9
2.7
1.9
Discrimination
Factor
0.5
0.4
0.6
0.6
Score
Please note:
The discrimination factor for a particular problem is calculated as follows:
(1) The students are ranked in regard to their overall scores.
(2) The mean score for the top 25% of these overall ranked students is calculated for that particular problem
including no attempts. Call this mean score the ‘mean top score’.
(3) The mean score for the bottom 25% of these overall ranked students is calculated for that particular problem
including no attempts. Call this mean score the ‘mean bottom score’.
(4) The discrimination factor = mean top score – mean bottom score
4
Thus the discrimination factor ranges from 1 to –1. A problem with a discrimination factor of 0.4 or higher is
considered to be a good discriminator.
Part 1: Mathematics Challenge for Young Australians 47
CHALLENGE STATISTICS – JUNIOR
Mean Score/School Year/Problem
Mean
Year
Number of
Students
Problem
Overall
1
2
3
4
5
6
7
2607
11.0
2.6
2.3
1.9
2.0
2.2
1.1
8
2373
13.4
3.0
2.7
2.2
2.3
2.7
1.4
All Years*
5006
12.2
2.8
2.5
2.0
2.1
2.4
1.3
Please note:* This total includes students who did not provide their school year.
Score Distribution %/Problem
Challenge Problem
1
Quirky
Quadrilaterals
2
Indim
Integers
3
Primelandia
Money
4
Condates
5
Jogging
6
Tessellating
Hexagons
Did not
attempt
3%
3%
7%
8%
9%
23%
0
7%
6%
8%
15%
10%
28%
1
8%
17%
24%
14%
18%
20%
2
15%
23%
30%
19%
16%
13%
3
36%
26%
22%
31%
21%
10%
4
31%
26%
10%
13%
27%
5%
Mean
2.8
2.5
2.0
2.1
2.4
1.3
Discrimination
Factor
0.5
0.6
0.6
0.7
0.7
0.5
Score
Please note:
The discrimination factor for a particular problem is calculated as follows:
(1) The students are ranked in regard to their overall scores.
(2) The mean score for the top 25% of these overall ranked students is calculated for that particular problem
including no attempts. Call this mean score the ‘mean top score’.
(3) The mean score for the bottom 25% of these overall ranked students is calculated for that particular problem
including no attempts. Call this mean score the ‘mean bottom score’.
(4) The discrimination factor = mean top score – mean bottom score
4
Thus the discrimination factor ranges from 1 to –1. A problem with a discrimination factor of 0.4 or higher is
considered to be a good discriminator.
48
Mathematics Contests The Australian Scene 2015
CHALLENGE STATISTICS – INTERMEDIATE
Mean Score/School Year/Problem
Mean
Year
Number of
Students
Problem
Overall
1
2
3
4
5
6
9
2038
15.2
3.1
2.6
3.0
3.0
2.0
2.5
10
1055
16.6
3.3
2.9
3.2
3.2
2.2
2.7
All Years*
3104
15.7
3.2
2.7
3.0
3.1
2.0
2.6
Please note:* This total includes students who did not provide their school year.
Score Distribution %/Problem
Challenge Problem
1
Indim
Integers
2
Digital Sums
3
Coin Flips
4
Jogging
5
Folding
Fractions
6
Crumbling
Cubes
Did not
attempt
1%
3%
5%
3%
14%
9%
0
2%
6%
4%
4%
14%
11%
1
7%
11%
6%
9%
21%
7%
2
15%
26%
15%
12%
21%
21%
3
25%
16%
28%
22%
8%
25%
4
50%
38%
42%
49%
22%
27%
Mean
3.2
2.7
3.0
3.1
2.0
2.6
Discrimination
Factor
0.4
0.6
0.6
0.6
0.7
0.7
Score
Please note:
The discrimination factor for a particular problem is calculated as follows:
(1) The students are ranked in regard to their overall scores.
(2) The mean score for the top 25% of these overall ranked students is calculated for that particular problem
including no attempts. Call this mean score the ‘mean top score’.
(3) The mean score for the bottom 25% of these overall ranked students is calculated for that particular problem
including no attempts. Call this mean score the ‘mean bottom score’.
(4) The discrimination factor = mean top score – mean bottom score
4
Thus the discrimination factor ranges from 1 to –1. A problem with a discrimination factor of 0.4 or higher is
considered to be a good discriminator.
Part 1: Mathematics Challenge for Young Australians 49
2015
Australian Intermediate
Mathematics
Olympiad
- Questions
AUSTRALIAN
INTERMEDIATE
MATHEMATICS
OLYMPIAD
Time allowed: 4 hours.
NO calculators are to be used.
Questions 1 to 8 only require their numerical answers all of which are non-negative integers less than 1000.
Questions 9 and 10 require written solutions which may include proofs.
The bonus marks for the Investigation in Question 10 may be used to determine prize winners.
1. A number written in base a is 123a. The same number written in base b is 146b . What is the minimum value of
a + b?
[2 marks]
2. A circle is inscribed in a hexagon ABCDEF so that each side of the hexagon is tangent to the circle. Find the
perimeter of the hexagon if AB = 6, CD = 7, and EF = 8.
[2 marks]
3. A selection of 3 whatsits, 7 doovers and 1 thingy cost a total of $329. A selection of 4 whatsits, 10 doovers and 1
thingy cost a total of $441. What is the total cost, in dollars, of 1 whatsit, 1 doover and 1 thingy?
[3 marks]
a
2
1
, can also be written in the form + 2 , where n is a positive integer.
b
n n
If a + b = 1024, what is the value of a?
[3 marks]
4. A fraction, expressed in its lowest terms
5. Determine the smallest positive integer y for which there is a positive integer x satisfying the equation
213 + 210 + 2x = y2 .
[3 marks]
√
√
6. The large circle has radius 30/ π. Two circles with diameter 30/ π lie inside the large circle. Two more circles
lie inside the large circle so that the five circles touch each other as shown. Find the shaded area.
[4 marks]
7. Consider a shortest path along the edges of a 7 × 7 square grid from its bottom-left vertex to its top-right vertex.
How many such paths have no edge above the grid diagonal that joins these vertices?
[4 marks]
8. Determine the number of non-negative integers x that satisfy the equation
x
x
=
.
44
45
(Note: if r is any real number, then r denotes the largest integer less than or equal to r.)
50
Mathematics Contests The Australian Scene 2015
1
[4 marks]
9. A sequence is formed by the following rules: s1 = a, s2 = b and sn+2 = sn+1 + (−1)n sn for all n ≥ 1.
If a = 3 and b is an integer less than 1000, what is the largest value of b for which 2015 is a member of the sequence?
Justify your answer.
[5 marks]
10. X is a point inside an equilateral triangle ABC. Y is the foot of the perpendicular from X to AC, Z is the foot
of the perpendicular from X to AB, and W is the foot of the perpendicular from X to BC.
The ratio of the distances of X from the three sides of the triangle is 1 : 2 : 4 as shown in the diagram.
B
W
Z
4
2
X
1
A
Y
C
If the area of AZXY is 13 cm2 , find the area of ABC. Justify your answer.
[5 marks]
Investigation
If XY : XZ : XW = a : b : c, find the ratio of the areas of AZXY and ABC.
2
[2 bonus marks]
Part 1: Mathematics Challenge for Young Australians 51
AUSTRALIAN
INTERMEDIATE
MATHEMATICS
OLYMPIAD
2015
Australian Intermediate
Mathematics
Olympiad
- Solutions
SOLUTIONS
1. Method 1
123a = 146b ⇐⇒ a2 + 2a + 3 = b2 + 4b + 6
⇐⇒ (a + 1)2 + 2 = (b + 2)2 + 2
⇐⇒ (a + 1)2 = (b + 2)2
⇐⇒ a + 1 = b + 2 (a and b are positive)
⇐⇒ a = b + 1
Since the minimum value for b is 7, the minimum value for a + b is 8 + 7 = 15.
Method 2
Since the digits in any number are less than the base, b ≥ 7.
We also have a > b, otherwise a2 + 2a + 3 < b2 + 4b + 6.
If b = 7 and a = 8, then a2 + 2a + 3 = 83 = b2 + 4b + 6.
So the minimum value for a + b is 8 + 7 = 15.
2. Let AB, BC, CD, DE, EF , F A touch the circle at U , V , W , X, Y , Z respectively.
E
X
D
Y
W
F
C
Z
V
A
U
B
Since the two tangents from a point to a circle have equal length,
U B = BV , V C = CW , W D = DX, XE = EY , Y F = F Z, ZA = AU .
The perimeter of hexagon ABCDEF is
AU + U B + BV + V C + CW + W D + DX + XE + EY + Y F + F Z + ZA
= AU + U B + U B + CW + CW + W D + W D + EY + EY + Y F + Y F + AU
= 2(AU + U B + CW + W D + EY + Y F )
= 2(AB + CD + EF ) = 2(6 + 7 + 8) = 2(21) = 42.
3. Preamble
Let the required cost be x. Then, with obvious notation, we have:
3w + 7d + t = 329
4w + 10d + t = 441
w+d+t = x
Method 1
3 × (1) − 2 × (2): w + d + t = 3 × 329 − 2 × 441 = 987 − 882 = 105.
52
Mathematics Contests The Australian Scene 2015
3
(1)
(2)
(3)
Method 2
(2) − (1): w + 3d = 112.
(1) − (3): 2w + 6d = 329 − x = 2 × 112 = 224.
Then x = 329 − 224 = 105.
Method 3
10 × (1) − 7 × (2): w = (203 − 3t)/2
3 × (2) − 4 × (1): d = (7 + t)/2
Then w + d + t = 210/2 − 2t/2 + t = 105.
2
1
2n + 1
+
=
.
n n2
n2
Since 2n + 1 and n2 are coprime, a = 2n + 1 and b = n2 .
So 1024 = a + b = n2 + 2n + 1 = (n + 1)2 , hence n + 1 = 32.
4. We have
This gives a = 2n + 1 = 2 × 31 + 1 = 63.
5. Method 1
213 + 210 + 2x = y2
⇐⇒
⇐⇒
⇐⇒
⇐⇒
210 (23 + 1) + 2x = y2
(25 × 3)2 + 2x = y2
2x = y2 − 962
2x = (y + 96)(y − 96).
Since y is an integer, both y + 96 and y − 96 must be powers of 2.
Let y + 96 = 2m and y − 96 = 2n . Then 2m − 2n = 192 = 26 × 3.
Hence 2m−6 − 2n−6 = 3. So 2m−6 = 4 and 2n−6 = 1.
In particular, m = 8. Hence y = 28 − 96 = 256 − 96 = 160.
Method 2
We have y2 = 213 + 210 + 2x = 210(23 + 1 + 2x−10 ) = 210 (9 + 2x−10 ).
So we want the smallest value of 9 + 2x−10 that is a perfect square.
Since 9 + 2x−10 is odd and greater than 9, 9 + 2x−10 ≥ 25.
Since 9 + 214−10 = 25, y = 25 × 5 = 32 × 5 = 160.
Comment
Method 1 shows that 213 + 210 + 2x = y2 has only one solution.
6. The centres Y and Y of the two medium circles lie on a diameter of the large circle. By symmetry about this
diameter, the two smaller circles are congruent. Let X be the centre of the large circle and Z the centre of a small
circle.
Y
X
Z
Y
Let R and r be the radii of a medium and small circle respectively. Then ZY = R + r = ZY . Since XY = XY ,
triangles XY Z and XY Z are congruent. Hence XZ ⊥ XY .
4
Part 1: Mathematics Challenge for Young Australians 53
By Pythagoras, Y Z 2 = Y X 2 + XZ 2 . So (R + r)2 = R2 + (2R − r)2 .
Then R2 + 2Rr + r 2 = 5R2 − 4Rr + r 2 , which simplifies to 3r = 2R.
√ 2
So the large circle has area π(30/ √
π ) = 900,
each medium circle has area π(15/ √
π )2 = 225,
and each small circle has area π(10/ π )2 = 100.
Thus the shaded area is 900 − 2 × 225 − 2 × 100 = 250.
7. Method 1
Any path from the start vertex O to a vertex A must pass through either the vertex L left of A or the vertex U
underneath A. So the number of paths from O to A is the sum of the number of paths from O to L and the paths
from O to U .
•
L•
•A
•
U
O
•
There is only one path from O to any vertex on the bottom line of the grid.
So the number of paths from O to all other vertices can be progressively calculated from the second bottom row
upwards as indicated.
•
429
132 429
•
42
132 297
14
42
90
165
5
14
28
48
75
2
5
9
14
20
27
1
2
3
4
5
6
7
1
1
1
1
1
1
1
Thus the number of required paths is 429.
Method 2
To help understand the problem, consider some smaller grids.
Let p(n) equal the number of required paths on an n × n grid and let p(0) = 1.
54
Mathematics Contests The Australian Scene 2015
5
Starting with the bottom-left vertex, label the vertices of the diagonal 0, 1, 2, . . . , n.
•
n
i•
1•
0
•
Consider all the paths that touch the diagonal at vertex i but not at any of the vertices between vertex 0 and
vertex i. Each such path divides into two subpaths.
One subpath is from vertex 0 to vertex i and, except for the first and last edge, lies in the lower triangle of the
diagram above. Thus there are p(i − 1) of these subpaths.
The other subpath is from vertex i to vertex n and lies in the upper triangle in the diagram above. Thus there are
p(n − i) of these subpaths.
So the number of such paths is p(i − 1) × p(n − i).
Summing these products from i = 1 to i = n gives all required paths. Thus
p(n) = p(n − 1) + p(1)p(n − 2) + p(2)p(n − 3) + · · · + p(n − 2)p(1) + p(n − 1)
We have p(1) = 1, p(2) = 2, p(3) = 5. So
p(4) = p(3) + p(1)p(2) + p(3)p(1) + p(3) = 14,
p(5) = p(4) + p(1)p(3) + p(2)p(2) + p(3)p(1) + p(4) = 42,
p(6) = p(5) + p(1)p(4) + p(2)p(3) + p(3)p(2) + p(1)p(4) + p(5) = 132, and
p(7) = p(6) + p(1)p(5) + p(2)p(4) + p(3)p(3) + p(4)p(2) + p(5)p(1) + p(6) = 429.
8. Method 1
x
x
Let
=
= n.
44
45
Since x is non-negative, n is also non-negative.
If n = 0, then x is any integer from 0 to 44 − 1 = 43: a total of 44 values.
If n = 1, then x is any integer from 45 to 2 × 44 − 1 = 87: a total of 43 values.
If n = 2, then x is any integer from 2 × 45 = 90 to 3 × 44 − 1 = 131: a total of 42 values.
If n = k, then x is any integer from 45k to 44(k + 1) − 1 = 44k + 43: a total of (44k + 43) − (45k − 1) = 44 − k
values.
Thus, increasing n by 1 decreases the number of values of x by 1. Also the largest value of n is 43, in which case
x has only 1 value.
Therefore the number of non-negative integer values of x is 44 + 43 + · · · + 1 = 21 (44 × 45) = 990.
Method 2
x
x
Let n be a non-negative integer such that
=
= n.
44
45
x
x
= n ⇐⇒ 44n ≤ x < 44(n + 1) and
= n ⇐⇒ 45n ≤ x < 45(n + 1).
Then
44
45
x
x
=
= n ⇐⇒ 45n ≤ x < 44(n + 1) ⇐⇒ 44n + n ≤ x < 44n + 44.
So
44
45
This is the case if and only if n < 44, and then x can assume exactly 44 − n different values.
Therefore the number of non-negative integer values of x is
(44 − 0) + (44 − 1) + · · · + (44 − 43) = 44 + 43 + · · · + 1 = 21 (44 × 45) = 990.
6
Part 1: Mathematics Challenge for Young Australians 55
Method 3
x
x
=
= n.
44
45
Then x = 44n + r where 0 ≤ r ≤ 43 and x = 45n + s where 0 ≤ s ≤ 44.
Let n be a non-negative integer such that
So n = r − s. Therefore 0 ≤ n ≤ 43. Also r = n + s. Therefore n ≤ r ≤ 43.
Therefore the number of non-negative integer values of x is 44 + 43 + · · · + 1 = 21 (44 × 45) = 990.
9. Working out the first few terms gives us an idea of how the given sequence develops:
n
1
2
3
4
5
6
7
s2n−1
a
b−a
b
2b − a
3b − a
5b − 2a
8b − 3a
s2n
b
2b − a
3b − a
5b − 2a
8b − 3a
13b − 5a
21b − 8a
It appears that the coefficients in the even terms form a Fibonacci sequence and, from the 5th term, every odd
term is a repeat of the third term before it.
These observations are true for the entire sequence since, for m ≥ 1, we have:
s2m+2
s2m+3
s2m+4
=
=
=
s2m+1 + s2m
s2m+2 − s2m+1
s2m+3 + s2m+2
=
=
s2m
s2m+2 + s2m
So, defining F1 = 1, F2 = 2, and Fn = Fn−1 + Fn−2 for n ≥ 3, we have s2n = bFn − aFn−2 for n ≥ 3. Since a = 3
and b < 1000, none of the first five terms of the given sequence equal 2015. So we are looking for integer solutions
of bFn − 3Fn−2 = 2015 for n ≥ 3.
s6 = 3b − 3 = 2015, has no solution.
s8 = 5b − 6 = 2015, has no solution.
s10 = 8b − 9 = 2015 implies b = 253.
For n ≥ 6 we have b = 2015/Fn + 3Fn−2 /Fn . Since Fn increases, we have Fn ≥ 13 and Fn−2 /Fn < 1 for n ≥ 6.
Hence b < 2015/13 + 3 = 158. So the largest value of b is 253.
10. Method 1
We first show that X is uniquely defined for any given equilateral triangle ABC.
Let P be a point outside ABC such that its distances from AC and AB are in the ratio 1:2. By similar triangles,
any point on the line AP has the same property. Also any point between AP and AC has the distance ratio less
than 1:2 and any point between AP and AB has the distance ratio greater than 1:2.
B
2
4
P
Q
1
1
A
56
Mathematics Contests The Australian Scene 2015
C
7
Let Q be a point outside ABC such that its distances from AC and BC are in the ratio 1:4. By an argument
similar to that in the previous paragraph, only the points on CQ have the distance ratio equal to 1:4.
Thus the only point whose distances to AC, AB, and BC are in the ratio 1:2:4 is the point X at which AP and
CQ intersect.
Scaling if necessary, we may assume that the actual distances of X to the sides of ABC are 1, 2, 4. Let h be the
height of ABC. Letting | | denote area, we have
|ABC| = 21 h × AB and
|ABC| = |AXB| + |BXC| + |CXA| = 12 (2AB + 4BC + AC) = 12 AB × 7.
So h = 7.
Draw a 7-layer grid of equilateral triangles each of height 1, starting with a single triangle in the top layer, then a
trapezium of 3 triangles in the next layer, a trapezium of 5 triangles in the next layer, and so on. The boundary
of the combined figure is ABC and X is one of the grid vertices as shown.
B
W
4
Z
2
X
1
Y
A
C
There are 49 small triangles in ABC and 6.5 small triangles in AZXY . Hence, after rescaling so that the area of
AZXY is 13 cm2 , the area of ABC is 13 × 49/6.5 = 98 cm2 .
Method 2
Join AX, BX, CX. Since Y AZ = ZBW = 60◦, the quadrilaterals AZXY and BW XZ are similar. Let XY be
1 unit and AY be x. Then BZ = 2x.
B
2x
W
Z
4
2
X
1
A
x
Y
C
√
√
√
By Pythagoras: in AXY , AX = 1 + x2 and in AXZ, AZ = x2 − 3. Hence BW = 2 x2 − 3.
√
Since AB = AC, Y C = x + x2 − 3.
√
√
By Pythagoras: in XY C, XC 2 = 1 +√(x + x2 − 3)2 = 2x2 − 2 + 2x x2 − 3
and in XW C, W C 2 = 2x2 − 18 + 2x x2 − 3.
√
√
√
Since BA = BC, 2x + x2 − 3 = 2 x2 − 3 + 2x2 − 18 + 2x x2 − 3.
√
√
So 2x − x2 − 3 = 2x2 − 18 + 2x x2 − 3.
8
Part 1: Mathematics Challenge for Young Australians 57
√
√
√
Squaring gives 4x2 + x2 − 3 − 4x x2 − 3 = 2x2 − 18 + 2x x2 − 3, which simplifies to 3x2 + 15 = 6x x2 − 3.
5
Squaring again gives 9x4 + 90x2 + 225 = 36x4 − 108x2 . So 0 = 3x4 − 22x2 − 25 = (3x2 − 25)(x2 + 1), giving x = √ .
3
x √ 2
4
13
5
Hence, area AZXY = + x − 3 = √ + √ = √ and
2
3
2 3
√ 2 3
√
√
4 2
49
3
3 10
2
2
√ +√
(2x + x − 3) =
=√ .
area ABC =
4
4
3
3
3
13
49
Since the area of AZXY is 13 cm2 , the area of ABC is ( √ / √ ) × 13 = 98 cm2 .
3 2 3
Method 3
Let DI be the line through X parallel to AC with D on AB and I on BC.
Let EG be the line through X parallel to BC with E on AB and G on AC.
Let F H be the line through X parallel to AB with F on AC and H on BC.
Let J be a point on AB so that HJ is parallel to AC.
Triangles XDE, XF G, XHI, BHJ are equilateral, and triangles XDE and BHJ are congruent.
B
J
H
E
Z
W
4
2
X
I
D
1
A
F Y G
C
The areas of the various equilateral triangles are proportional to the square of their heights. Let the area of
F XG = 1. Then, denoting area by | |, we have:
|DEX| = 4, |XHI| = 16, |AEG| = 9, |DBI| = 36, |F HC| = 25.
|ABC| = |AEG| + |F HC| + |DBI| − |F XG| − |DEX| − |XHI| = 9 + 25 + 36 − 1 − 4 − 16 = 49.
|AZXY | = |AEG| − 12 (|F XG| + |DEX|) = 9 − 12 (1 + 4) = 6.5.
Since the area of AZXY is 13 cm2 , the area of ABC is 2 × 49 = 98 cm2 .
Method 4
Consider the general case where XY = a, XZ = b, and XW = c.
B
W
Z
60◦
A
58
Mathematics Contests The Australian Scene 2015
c
b
120
X
◦
a
Y
C
9
◦
◦
Projecting AY onto the
√ line through ZX gives AY sin
√60 − a cos 60 = b.
Hence AY = (a + 2b)/ 3. Similarly, AZ = (b + 2a)/ 3.
Letting | | denote area, we have
|AZXY |
=
=
=
=
=
=
Similarly, |CY XW | =
Hence |ABC| =
√
3
(2a2
6
√
3 2
6 (a
|Y AZ| + |Y XZ|
1
1
◦
◦
2 (AY )(AZ) sin 60 + 2 ab sin 120
√
3
4 ((AY )(AZ) + ab)
√
3
12 ((a + 2b)(b + 2a) +
√
3
2
2
12 (2a + 2b + 8ab)
√
3 2
(a + b2 + 4ab)
6
+ c2 + 4ac) and |BW XZ| =
+ 2b2 + 2c2 + 4ab + 4ac + 4bc) =
√
3 2
6 (b
√
3
(a
3
3ab)
+ c2 + 4bc).
+ b + c)2 .
So |ABC|/|AZXY | = 2(a + b + c)2 /(a2 + b2 + 4ab).
Letting a = k, b = 2k, c = 4k, and |AZXY | = 13 cm2 , we have |ABC| = 26(49k 2)/(k 2 + 4k 2 + 8k 2 ) = 98 cm2 .
Investigation
Method 4 gives |ABC|/|AZXY | = 2(a + b + c)2 /(a2 + b2 + 4ab).
Alternatively, as in Method 3,
|ABC| = |AEG| + |F HC| + |DBI| − |F XG| − |DEX| − |XHI|
= (a + b)2 + (a + c)2 + (b + c)2 − a2 − b2 − c2 = (a + b + c)2 .
Also |AZXY | = |AEG| − 12 (|F XG| + |DEX|)
= (a + b)2 − 12 (a2 + b2 )
= 2ab + 12 (a2 + b2 ).
So |ABC|/|AZXY | = 2(a + b + c)2 /(a2 + b2 + 4ab).
10
Part 1: Mathematics Challenge for Young Australians 59
AUSTRALIAN INTERMEDIATE MATHEMATICS OLYMPIAD
STATISTICS
Distribution of Awards/School Year
Number of Awards
Year
Number of
Students
Prize
High
Distinction
Distinction
Credit
Participation
8
341
3
17
39
86
196
9
414
8
45
61
99
201
10
462
11
52
89
139
171
Other
221
4
9
16
41
151
Total
1438
26
123
205
365
719
Number of Correct Answers Questions 1–8
Year
Number Correct/Question
1
2
3
4
5
6
7
8
8
119
231
282
164
128
79
48
50
9
144
298
347
224
187
138
82
86
10
176
341
377
297
219
208
103
104
Other
66
132
176
81
75
34
33
21
Total
505
1002
1182
766
609
459
266
261
Mean Score/Question/School Year
Mean Score
Year
60
Number of
Students
Question
Overall Mean
1–8
9
10
8
341
10.1
0.5
0.2
10.9
9
414
11.8
0.9
0.5
13.2
10
462
13.0
1.1
0.6
14.6
Other
221
8.6
0.4
0.2
9.3
All Years
1438
11.3
0.8
0.4
12.5
Mathematics Contests The Australian Scene 2015
AUSTRALIAN INTERMEDIATE MATHEMATICS OLYMPIAD
RESULTS
NAME
SCHOOL
YEAR
SCORE
PRIZE
Matthew Cheah
Penleigh and Essendon Grammar School, VIC
10
35
Puhua Cheng
Raffles Institution, Singapore
8
35
Ariel Pratama Junaidi
Anglo-Chinese School, Singapore
10
35
Evgeni Kayryakov
Childrens Academy 21st Century, Bulgaria
7
35
Wei Khor Jun
Raffles Institution, Singapore
8
35
Jack Liu
Brighton Grammar, VIC
9
35
Jerry Mao
Caulfield Grammar School, Wheelers Hill,VIC
9
35
Liao Meng
Anglo-Chinese School, Singapore
10
35
Nguyen Hoai Nam
Anglo-Chinese School, Singapore
10
35
Aloysius Ng Yangyi
Raffles Institution, Singapore
7
35
Kohsuke Sato
Christ Church Grammar School, WA
10
35
Yuelin Shen
Scotch College, WA
10
35
Chen Tan Xu
Raffles Institution, Singapore
7
35
Kit Victor Loh Wai
Raffles Institution, Singapore
8
35
Jianzhi Wang
Raffles Institution, Singapore
9
35
Zhe Xin
Raffles Institution, Singapore
9
35
Austin Zhang
Sydney Grammar School, NSW
10
35
Yu Zhiqiu
Anglo-Chinese School, Singapore
10
35
Lin Zien
Anglo-Chinese School, Singapore
9
35
Bobby Dey
James Ruse Agricultural High School, NSW
10
34
Goh Ethan
Raffles Institution, Singapore
7
34
Yulong Guo
Hwa Chong Institution, Singapore
9
34
Edwin Winata Hartanto
Anglo-Chinese School, Singapore
10
34
Hristo Papazov
Childrens Academy 21st Century, Bulgaria
10
34
Zhang Yansheng
Chung Cheng High School, Singapore
9
34
Guowen Zhang
St Joseph's College, QLD
9
34
HIGH DISTINCTION
Ivan Ganev
Childrens Academy 21st Century, Bulgaria
10
33
Theodore Leebrant
Anglo-Chinese School, Singapore
9
33
Yu Peng Ng
Hwa Chong Institution, Singapore
8
33
Cheng Shi
Hwa Chong Institution, Singapore
8
33
Kean Tan Wee
Raffles Institution, Singapore
7
33
Sharvil Kesarwani
Merewether High School, NSW
8
32
Hong Rui Benjamin Lee
Hwa Chong Institution, Singapore
10
32
Chenxu Li
Raffles Institution, Singapore
8
32
William Li
Barker College, NSW
9
32
Part 1: Mathematics Challenge for Young Australians 61
NAME
SCHOOL
YEAR
SCORE
Han Yang
Hwa Chong Institution, Singapore
9
32
Stanley Zhu
Melbourne Grammar School, VIC
9
32
Anand Bharadwaj
Trinity Grammar School, VIC
9
31
William Hu
Christ Church Grammar School, WA
9
31
Xianyi Huang
Baulkham Hills High School, NSW
10
31
Wanzhang Jing
James Ruse Agricultural High School, NSW
10
31
Yuhao Li
Hwa Chong Institution, Singapore
9
31
Steven Lim
Hurlstone Agricultural High School, NSW
9
31
John Min
Baulkham Hills High School, NSW
9
31
Elliott Murphy
Canberra Grammar School, ACT
10
31
Longxuan Sun
Hwa Chong Institution, Singapore
8
31
Boyan Wang
Hwa Chong Institution, Singapore
9
31
Sean Zammit
Barker College, NSW
10
31
Atul Barman
James Ruse Agricultural High School, NSW
10
30
Atanas Dinev
Childrens Academy 21st Century, Bulgaria
9
30
Bill Hu
James Ruse Agricultural High School, NSW
10
30
Phillip Huynh
Brisbane State High School, QLD
10
30
Wei Ci William Kin
Hwa Chong Institution, Singapore
10
30
Yang Lee Ker
Raffles Institution, Singapore
9
30
Forbes Mailler
Canberra Grammar School, ACT
9
30
Moses Mayer
Surya Institute, Indonesia
9
30
Zlatina Mileva
Childrens Academy 21st Century, Bulgaria
8
30
Kirill Saulov
Brisbane Grammar School, QLD
10
30
Yuxuan Seah
Raffles Institution, Singapore
8
30
Kieran Shivakumaarun
Sydney Boys High School, NSW
10
30
Liang Tan Xue
Raffles Institution, Singapore
10
30
Nicholas Tanvis
Anglo-Chinese School, Singapore
9
30
An Aloysius Wang
Hwa Chong Institution, Singapore
10
30
William Wang
Queensland Academy for Science,
Mathematics and Technology, QLD
10
30
Joshua Welling
Melrose High School, ACT
9
30
Seung Hoon Woo
Hwa Chong Institution, Singapore
10
30
Chen Yanbing
Methodist Girls' School, Singapore
10
30
Guangxuan Zhang
Raffles Institution, Singapore
10
30
Chi Zhang Yu
Raffles Institution, Singapore
7
30
Keer Chen
Presbyterian Ladies’ College, NSW
10
29
Linus Cooper
James Ruse Agricultural High School, NSW
9
29
Liam Coy
Sydney Grammar School, NSW
7
29
Rong Dai Xiang
Raffles Institution, Singapore
8
29
Hong Pei Goh
Hwa Chong Institution, Singapore
10
29
62
Mathematics Contests The Australian Scene 2015
NAME
SCHOOL
YEAR
SCORE
Tianjie Huang
Hwa Chong Institution, Singapore
9
29
Ricky Huang
James Ruse Agricultural High School, NSW
10
29
Tianjie Huang
Hwa Chong Institution, Singapore
9
29
Yu Jiahuan
Raffles Girls' School, Singapore
9
29
Tony Jiang
Scotch College, VIC
10
29
Charles Li
Camberwell Grammar School, Vic
9
29
Steven Liu
James Ruse Agricultural High School, NSW
10
29
Hilton Nguyen
Sydney Technical High School, NSW
9
29
James Nguyen
Baulkham Hills High School, NSW
10
29
Trung Nguyen
Penleigh and Essendon Grammar School, VIC
10
29
James Phillips
Canberra Grammar School, ACT
9
29
Ryan Stocks
Radford College, ACT
9
29
Hadyn Tang
Trinity Grammar School, VIC
6
29
Stanve Avrilium Widjaja
Surya Institute, Indonesia
7
29
Wang Yihe
Anglo-Chinese School, Singapore
9
29
Sun Yue
Raffles Girls' School, Singapore
9
29
Wang Beini
Raffles Girls' School, Singapore
10
28
Chwa Channe
Raffles Girls' School, Singapore
8
28
Keiran Hamley
All Saints Anglican School, QLD
10
28
Lee Shi Hao
Anglo-Chinese School, Singapore
9
28
Zhu Jiexiu
Raffles Girls' School, Singapore
9
28
Jodie Lee
Seymour College, SA
10
28
Yu Hsin Lee
Hwa Chong Institution, Singapore
9
28
Phillip Liang
James Ruse Agricultural High School, NSW
9
28
Anthony Ma
Shore School, NSW
10
28
Dzaki Muhammad
Surya Institute, Indonesia
9
28
Daniel Qin
Scotch College, VIC
10
28
Sang Ta
Randwick Boys High School, NSW
8
28
Ruiqian Tong
Presbyterian Ladies’ College, VIC
9
28
Hu Xing Yi
Methodist Girls' School, Singapore
10
28
Zhao Yiyang
Methodist Girls' School, Singapore
10
28
Claire Yung
Lyneham High School, ACT
10
28
Xuan Ang Ben
Raffles Institution, Singapore
9
27
Hantian Chen
James Ruse Agricultural High School, NSW
10
27
Jasmine Jiawei Chen
Pymble Ladies’ College, NSW
10
27
Harry Dinh
James Ruse Agricultural High School, NSW
10
27
Tan Jian Yee
Chung Cheng High School, Singapore
10
27
Daniel Jones
All Saints Anglican Senior School, QLD
10
27
Winfred Kong
Hwa Chong Institution, Singapore
10
27
Adrian Law
James Ruse Agricultural High School, NSW
10
27
Part 1: Mathematics Challenge for Young Australians 63
NAME
SCHOOL
YEAR
SCORE
Jason Leung
James Ruse Agricultural High School, NSW
7
27
Sabrina Natashya
Liandra
Surya Institute, Indonesia
9
27
Angela (Yunyun) Ran
The Mac.Robertson Girls’ High School, VIC
9
27
Yi Shen Xin
Raffles Institution, Singapore
7
27
Peter Tong
Yarra Valley Grammar, VIC
9
27
Jordan Truong
Sydney Technical High School, NSW
10
27
Andrew Virgona
Concord High School, NSW
9
27
Tommy Wei
Scotch College, VIC
9
27
Tianyi Xu
Sydney Boys High School, NSW
9
27
Wu Zhen
Anglo-Chinese School, Singapore
10
27
Gordon Zhuang
Sydney Boys High School, NSW
9
27
Zhou Zihan
Raffles Girls' School, Singapore
8
27
Amit Ben-Harim
McKinnon Secondary College, VIC
9
26
Hu Chen
The King's School, NSW
10
26
Merry Xiao Die Chu
North Sydney Girls' High School, NSW
9
26
Li Haocheng
Anglo-Chinese School, Singapore
9
26
William Hu
Rossmoyne Senior High School, WA
10
26
Laeeque Jamdar
Baulkham Hills High School, NSW
9
26
Yasiru Jayasoora
James Ruse Agricultural High School, NSW
7
26
Arun Jha
Perth Modern School, WA
10
26
Tony Li
Sydney Boys High School, NSW
10
26
Zefeng Jeff Li
Glen Waverley Secondary College, VIC
8
26
Adrian Lo
Newington College, NSW
7
26
Lionel Maizels
Norwood Secondary College, VIC
8
26
Marcus Rees
Taroona High School, TAS
8
26
Elva Ren
Presbyterian Ladies College, VIC
10
26
Aidan Smith
All Saints' College, WA
8
26
Jacob Smith
All Saints' College, WA
9
26
Keane Teo
Hwa Chong Institution, Singapore
10
26
Jeffrey Wang
Shore School, NSW
10
26
Xinlu Xu
Presbyterian Ladies’ College, Sydney, NSW
10
26
Jason (Yi) Yang
James Ruse Agricultural High School, NSW
8
26
Shukai Zhang
Hwa Chong Institution, Singapore
10
26
Yanjun Zhang
Hwa Chong Institution, Singapore
9
26
Kevin Zhu
James Ruse Agricultural High School, NSW
10
26
Jonathan Zuk
Elwood College, VIC
8
26
64
Mathematics Contests The Australian Scene 2015
HONOUR ROLL
Because of changing titles and affiliations, the most senior title achieved and later affiliations are generally used,
except for the Interim committee, where they are listed as they were at the time.
Mathematics Challenge for Young Australians
Problems Committee for Challenge
Dr K McAvaney
Mr B Henry
Prof P J O’Halloran
Dr R A Bryce
Adj Prof M Clapper
Ms L Corcoran
Ms B Denney
Mr J Dowsey
Mr A R Edwards
Dr M Evans
Assoc Prof H Lausch
Ms J McIntosh
Mrs L Mottershead
Miss A Nakos
Dr M Newman
Ms F Peel
Dr I Roberts
Ms T Shaw
Ms K Sims
Dr A Storozhev
Prof P Taylor
Mrs A Thomas
Dr S Thornton
Miss G Vardaro
Victoria, (Director)
Member
Victoria (Director)
Member
University of Canberra, ACT
Australian National University, ACT
Australian Mathematics Trust, ACT
Australian Capital Territory
New South Wales
University of Melbourne, VIC
Department of Education, QLD
Scotch College, VIC
Monash University, VIC
AMSI, VIC
New South Wales
Temple Christian College, SA
Australian National University, ACT
St Peter’s College, SA
Northern Territory
SCEGGS, NSW
New South Wales
Attorney General’s Department, ACT
Australian Mathematics Trust, ACT
New South Wales
South Australia
Wesley College, VIC
9 years; 2006–2015; 1 year 2005–2006
17 years; 1990–2006; 9 years 2007–2015
5 years; 1990–1994
23 years; 1990–2012
3 years; 2013–2015
3 years; 1990–1992
6 years; 2010–2015
8 years; 1995–2002
26 years; 1990–2015
6 years; 1990–1995
24 years; 1990–2013
14 years; 2002–2015
24 years; 1992–2015
23 years; 1993–2015
26 years; 1990–2015
2 years; 1999, 2000
3 years; 2013–2015
3 years; 2013–2015
17 years; 1999–2015
22 years; 1994–2015
20 years; 1995–2014
18 years; 1990–2007
18 years; 1998–2015
22 years: 1993–2006,
2008–2015
University of Toronto, Canada
Rose Hulman Institute of Technology, USA
Department of Education, Slovakia
Institute for Educational Research, Slovakia
Canada
University of Waterloo, Canada
St Petersburg State University, Russia
University College, Ireland
University of Alberta, Canada
Academy of Science, China
University of Birmingham, United Kingdom
Hong Kong
University of Waterloo, Canada
India
New Caledonia
Austria
United States of America
Colombia
United Kingdom
Bulgaria
Latvia
University of Rostock, Germany
1991,
1993,
1993
1993
1992
1992,
1994
1994
1995,
1995
1996
1997
1997
1998
1999
1999,
2000
2000
2001
2001,
2002
2003
Visiting members
Prof E Barbeau
Prof G Berzsenyi
Dr L Burjan
Dr V Burjan
Mrs A Ferguson
Prof B Ferguson
Dr D Fomin
Prof F Holland
Dr A Liu
Prof Q Zhonghu
Dr A Gardiner
Prof P H Cheung
Prof R Dunkley
Dr S Shirali
Mr M Starck
Dr R Geretschlager
Dr A Soifer
Prof M Falk de Losada
Mr H Groves
Prof J Tabov
Prof A Andzans
Prof Dr H-D Gronau
2004, 2008
2002
2005
2006, 2009
2013
2010
Part 1: Mathematics Challenge for Young Australians 65
Prof J Webb
Mr A Parris
Dr A McBride
Prof P Vaderlind
Prof A Jobbings
Assoc Prof D Wells
University of Cape Town, South Africa
Lynwood High School, New Zealand
University of Strathclyde, United Kingdom
Stockholm University, Sweden
United Kingdom
United States of America
Moderators for Challenge
Mr W Akhurst
Ms N Andrews
Prof E Barbeau
Mr R Blackman
Ms J Breidahl
Ms S Brink
Prof J C Burns
Mr A. Canning
Mrs F Cannon
Mr J Carty
Dr E Casling
Mr B Darcy
Ms B Denney
Mr J Dowsey
Br K Friel
Dr D Fomin
Mrs P Forster
Mr T Freiberg
Mr W Galvin
Mr M Gardner
Ms P Graham
Mr B Harridge
Ms J Hartnett
Mr G Harvey
Ms I Hill
Ms N Hill
Dr N Hoffman
Prof F Holland
Mr D Jones
Ms R Jorgenson
Assoc Prof H Lausch
Mr J Lawson
Mr R Longmuir
Ms K McAsey
Dr K McAvaney
Ms J McIntosh
Ms N McKinnon
Ms T McNamara
Mr G Meiklejohn
Mr M O’Connor
Mr J Oliver
Mr S Palmer
Dr W Palmer
Mr G Pointer
Prof H Reiter
Mr M Richardson
Mr G Samson
Mr J Sattler
Mr A Saunder
Mr W Scott
66
New South Wales
ACER, Camberwell, VIC
University of Toronto, Canada
Victoria
St Paul’s Woodleigh, VIC
Glen Iris, VIC
Australian Defence Force Academy, ACT
Queensland
New South Wales
ACT Department of Education, ACT
Australian Capital Territory
South Australia
New South Wales
Victoria
Trinity Catholic College, NSW
St Petersburg University, Russia
Penrhos College, WA
Queensland
University of Newcastle, NSW
North Virginia, USA
Tasmania
University of Melbourne, VIC
Queensland
Australian Capital Territory
South Australia
Victoria
Edith Cowan University, WA
University College, Ireland
Coff’s Harbour High School, NSW
Australian Capital Territory
Victoria
St Pius X School, NSW
China
Victoria
Victoria
AMSI, VIC
Victoria
Victoria
Queensland School Curriculum Council, QLD
AMSI, VIC
Northern Territory
New South Wales
University of Sydney, NSW
South Australia
University of North Carolina, USA
Yarraville Primary School, VIC
Nedlands Primary School, WA
Parramatta High School, NSW
Victoria
Seven Hills West Public School, NSW
Mathematics Contests The Australian Scene 2015
2003, 2011
2004
2007
2009, 2012
2014
2015
Moderators for Challenge continued
Mr R Shaw
Ms T Shaw
Dr B Sims
Dr H Sims
Ms K Sims
Prof J Smit
Mrs M Spandler
Mr G Spyker
Ms C Stanley
Dr E Strzelecki
Mr P Swain
Dr P Swedosh
Prof J Tabov
Mrs A Thomas
Ms K Trudgian
Prof J Webb
Ms J Vincent
Hale School, WA
New South Wales
University of Newcastle, NSW
Victoria
New South Wales
The Netherlands
New South Wales
Curtin University, WA
Queensland
Monash University, VIC
Ivanhoe Girls Grammar School, VIC
The King David School, VIC
Academy of Sciences, Bulgaria
New South Wales
Queensland
University of Capetown, South Africa
Melbourne Girls Grammar School, VIC
Mathematics Enrichment Development
Enrichment Committee — Development Team (1992–1995)
Mr B Henry
Prof P O’Halloran
Mr G Ball
Dr M Evans
Mr K Hamann
Assoc Prof H Lausch
Dr A Storozhev
Victoria (Chairman)
University of Canberra, ACT (Director)
University of Sydney, NSW
Scotch College, VIC
South Australia
Monash University, VIC
Australian Mathematics Trust, ACT
Polya Development Team (1992–1995)
Mr G Ball
University of Sydney, NSW (Editor)
Mr K Hamann
South Australia (Editor)
Prof J Burns
Australian Defence Force Academy, ACT
Mr J Carty
Merici College, ACT
Dr H Gastineau-Hill
University of Sydney, NSW
Mr B Henry
Victoria
Assoc Prof H Lausch
Monash University, VIC
Prof P O’Halloran
University of Canberra, ACT
Dr A Storozhev
Australian Mathematics Trust, ACT
Euler Development Team (1992–1995)
Dr M Evans
Scotch College, VIC (Editor)
Mr B Henry
Victoria (Editor)
Mr L Doolan
Melbourne Grammar School, VIC
Mr K Hamann
South Australia
Assoc Prof H Lausch
Monash University, VIC
Prof P O’Halloran
University of Canberra, ACT
Mrs A Thomas
Meriden School, NSW
Gauss Development Team (1993–1995)
Dr M Evans
Scotch College, VIC (Editor)
Mr B Henry
Victoria (Editor)
Mr W Atkins
University of Canberra, ACT
Mr G Ball
University of Sydney, NSW
Prof J Burns
Australian Defence Force Academy, ACT
Mr L Doolan
Melbourne Grammar School, VIC
Mr A Edwards
Mildura High School, VIC
Mr N Gale
Hornby High School, New Zealand
Dr N Hoffman
Edith Cowan University, WA
Part 1: Mathematics Challenge for Young Australians 67
Prof P O’Halloran
Dr W Pender
Mr R Vardas
University of Canberra, ACT
Sydney Grammar School, NSW
Dulwich Hill High School, NSW
Noether Development Team (1994–1995)
Dr M Evans
Scotch College, VIC (Editor)
Dr A Storozhev
Australian Mathematics Trust, ACT (Editor)
Mr B Henry
Victoria
Dr D Fomin
St Petersburg University, Russia
Mr G Harvey
New South Wales
Newton Development Team (2001–2002)
Mr B Henry
Victoria (Editor)
Mr J Dowsey
University of Melbourne, VIC
Mrs L Mottershead
New South Wales
Ms G Vardaro
Annesley College, SA
Ms A Nakos
Temple Christian College, SA
Mrs A Thomas
New South Wales
Dirichlet Development Team (2001–2003)
Mr B Henry
Victoria (Editor)
Mr A Edwards
Ormiston College, QLD
Ms A Nakos
Temple Christian College, SA
Mrs L Mottershead
New South Wales
Mrs K Sims
Chapman Primary School, ACT
Mrs A Thomas
New South Wales
Australian Intermediate Mathematics Olympiad Committee
Dr K McAvaney
Adj Prof M Clapper
Mr J Dowsey
Dr M Evans
Mr B Henry
Assoc Prof H Lausch
Mr R Longmuir
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Victoria (Chair)
Australian Mathematics Trust, ACT
University of Melbourne, VIC
AMSI, VIC
Victoria (Chair)
Member
Monash University, VIC
China
Mathematics Contests The Australian Scene 2015
9 years; 2007–2015
2 years; 2014–2015
17 years; 1999–2015
17 years; 1999–2015
8 years; 1999–2006
9 years; 2007–2015
17 years; 1999–2015
2 years; 1999–2000