2015 Mathematics Contests – The Australian Scene Part 1

MATHEMATICS CONTESTS THE AUSTRALIAN SCENE 2015

AustrAliAn MAtheMAticAl OlyMpiAd cOMMittee

A depArtMent Of the AustrAliAn MAtheMAtics trust

PART 1: MATHEMATICS CHALLENGE FOR YOUNG AUSTRALIANS

KL McAvaney

2 Mathematics Contests The Australian Scene 2015

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

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Copyright ©2015 by the Australian Mathematics Trust

National Library of Australia Card Number

and ISSN 1323-6490

Part 1: Mathematics Challenge for Young Australians 3

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.

http://www.canberra.edu.au/


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.


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



CONTENTS

Support for the Australian Mathematical Olympiad Committee Training Program 3

Acknowledgements 4

Preface 5

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


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


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.


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


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


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. Forexample, 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. Shestarts 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 italways works.

1

CHALLENGE PROBLEMS – MIDDLE PRIMARY


d There are 90 squares in the outside border of the completed quilt. How many borders have been added to Joy’sstarting black rectangle?

MP3 Egg Cartons

Zoe decided to investigate the number of different ways of placing identical eggs in clear rectangular cartons of varioussizes.

For example, there is only one way of placing six eggs in a 2 × 3 carton and only two different ways of placing fiveeggs.

Reflections and rotations of arrangements are not considered different. For example, all of these arrangements are thesame:

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. Datessuch 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.

2


2015 Challenge Problems - 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 theirends touching to form polygons. For example, with four rods, two of length 1 cm and two of length 3 cm, I can forma parallelogram which I refer to as (1, 3, 1, 3).

3 cm

1cm

3 cm

1cm

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 haveexactly 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 varioussizes.

For example, there is only one way of placing six eggs in a 2 × 3 carton and only two different ways of placing fiveeggs.

Reflections and rotations of arrangements are not considered different. For example, all of these arrangements are thesame:

3

CHALLENGE PROBLEMS – UPPER PRIMARY


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 55ways 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 for50 metres are:

Mack 55 secondsJack 1 minute 8 secondsZac 1 minute 16 secondsNick 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 boysinto 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 PBtime was included with all theirs, then the average PB time would be exactly 1 minute. What was the coach’s PBtime?

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 inPrimelandia 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 thelargest even payment that can be made using four coins and the largest even payment that can be made using threecoins?

d A bag contains six different coins. Alice, Bob and Carol take two coins each from the bag and keep them. Theyfind that they have all taken the same amount of money. What is the smallest amount of money that could havebeen in the bag? Explain your reasoning.

4


2015 Challenge Problems - 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 onthe 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 quadrilateralson 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 couldbe 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 sectorsnumbered 1 to 9. Jim has nine tiles numbered 1 to 9. When the wheel stops, he removes all tiles whose numbers arefactors or multiples of the spun number.

•

•

34

56

7

2

19

8

1 2 3

4 5 6

7 8 9

Jim then tries to use three of the remaining tiles to form a 3-digit number that is divisible by the number on thewheel. 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?

5

CHALLENGE PROBLEMS – JUNIOR


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 inPrimelandia 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. Theyfind that they have all taken the same amount of money. What is the smallest amount of money that could havebeen 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 betweentheir values.

d The new King of Primelandia decides to mint coins of prime values greater than 50T. Show that no matter whatcoins are made, there is only one set of five Primelandian coins which, when placed in ascending order, form asequence 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 aflat 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 shouldgive you the general idea.

• •2 km

Start/Finish Base of Mt KJ

3 km

•

Top of Mt KJ

Theo can run at 12km/h on the flat. Up the hill he reckons he can average 9km/h and he is good for 15km/h on thedownhill 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 willaffect 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 isat his back, he will speed up by the same amount. This applies to his speeds on the flat and on the slope. The windwill 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 legfaster, but not both. Assuming there is no wind, how much faster would he have to run in each case?

6


J6 Tessellating Hexagons

Will read in his favourite maths book that this hexagon tessellates the plane.

A B

C

D

E

F

To see if this was true, Will made 16 copies of the hexagon and glued them edge-to-edge onto a piece of paper, withno 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 dashedrectangle.

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 translationonly. Indicate three such blocks on your tessellation that are not identical.

7


2015 Challenge Problems - 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 firstequal to the digital sum of the number before it. For example: 7598, 29, 11, 2. Any digital sum sequence ends in asingle-digit number and this is called the final digital sum or FDS of the first number in the sequence. Thus the FDSof 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 offlipping over some of the coins. The number of coins that are flipped stays the same for each game but may changefrom game to game. A coin showing ‘heads’ is represented by H . A coin showing ‘tails’ is represented by T . So, whena coin is flipped it changes from H to T or from T to H . The same coin may be selected on different moves. In eachgame 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: T T T T TMove 1: T H H T TMove 2: H H T T TMove 3: H H T H H

a Starting with 14 coins and flipping over four coins on each move, explain how to finish with exactly 10 coins headsup 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 inthree moves.

c Starting with 26 coins and flipping over four coins on each move, what is the minimum number of moves needed tofinish 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 all154 coins heads up at the end of three moves.

I4 Jogging


8

CHALLENGE PROBLEMS – INTERMEDIATE


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 corneris folded to a point P on the top edge as shown, creating triangles PQR, PTS and SUV .

P Q

R

T

V

U

S

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 SUV .

c Find all positions of P so that the ratio of the sides of triangle SUV 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 smallcubes.

In the second and subsequent steps, the same procedure is applied to what remains of the original large cube after theprevious 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?

9


2015 Challenge Solutions - 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 numbersfrom 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 mustboth 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 200that 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 1 2 3 4

Squares 10 18 26 34

The number of squares from one border to the next appears to increase by 8. We now show this rule continues toapply.

10

CHALLENGE SOLUTIONS – MIDDLE PRIMARY


Except for the corner squares, each square on the previous border corresponds to a square in the new border. Eachof 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.

previous border

new border

previous border

new border

pre

vio

us

bord

er

new

bord

erprev

ious

bord

er

new

bord

er

C

C ′

C

C ′

C

C ′

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 1 2 3 4 5 6 7 8 9 10 11

Squares 10 18 26 34 42 50 58 66 74 82 90

So Joy’s completed quilt has 11 borders.

MP3 Egg Cartons

a

For each of these arrangements, any rotation or reflection is acceptable.

b


11


c


d Each arrangement of six eggs in a 2× 4 carton has two empty positions. By systematically locating the two emptypositions, we see that there are ten different ways to arrange six eggs in a 2 × 4 carton.


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 monthmust 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. Ifthe 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. Theearliest 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 lastcondate 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 themonth must be 12. So the day must contain two of the digits 3, 4, 5, 6. This is impossible.

12


2015 Challenge Solutions - 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 differentlength. 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 parallelsides 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.


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c By systematically locating the two empty positions, we see that there are ten different ways to arrange six eggs ina 2 × 4 carton.

13

CHALLENGE SOLUTIONS – UPPER PRIMARY



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 fillthe empty places with 9 brown eggs. Then each placement of 3 white eggs corresponds to a placement of 9 browneggs. 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 55 sJack 1min 8 s = 68 sZac 1min 16 s = 76 sNick 1min 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 thatexclude Mack.

Pair with M MJ MZ MNTotal PB 123 131 118Other pair ZN JN JZTotal PB 139 131 144

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 totalPB 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 68with 63. Each pair then totals 131 as required. Thus Mack is paired with Zac and Jack is paired with Nick.

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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 = 385seconds. Since 385 is odd and cannot be divided into two equal amounts, the two teams cannot have the sametotal 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 havethe same total PB times. The next closest would be 1 second apart, in which case the times would be 192 s and193 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 orderof their units digits are: 70, 53, 63, 55, 76, 68. The only three that give 3 in the units digit of their total are70, 55, 68, and these do in fact total 193 s. Thus one team has Chloe, Mack, and Jack with a total PB time of70 + 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 correspondingteams that exclude Mack.

Team with M Total PB Other team Total PB

MJZ 199 NCS 186MJN 186 ZCS 199

MJC 193 ZNS 192MJS 176 ZNC 209MZN 194 JCS 191

MZC 201 JNS 184MZS 184 JNC 201

MNC 188 JZS 197MNS 171 JZC 214

MCS 178 JZN 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 Partb, 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− 43 = 7, 50− 41 = 9,50− 37 = 13, 50− 31 = 19, 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 7 = 7 12 = 5 + 7 17 = 17 22 = 3 + 193 = 3 8 = 3 + 5 13 = 13 18 = 5 + 13 23 = 234 = 2 + 2 9 = 2 + 7 14 = 3 + 11 19 = 19 24 = 5 + 195 = 5 10 = 3 + 7 15 = 2 + 13 20 = 3 + 17 25 = 2 + 236 = 3 + 3 11 = 11 16 = 3 + 13 21 = 2 + 19 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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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 evenprime. Now 27 = 2 + 25 and 25 is not prime, so 27 cannot be the sum of two primes. Hence 27T is the smallestpayment 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 fourcoins is 47 + 43 + 41 + 37 = 168. If the sum of three primes is even, then one of those primes must be 2. So thelargest 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 thesum 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 sumof 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 amultiple of 6. The next multiple of 6 is 60. The following table lists all possibilities for each sum of six coins up to72T.

Sum of six Sum of pair All possible pairs

60 20 3 + 17, 7 + 1366 22 3 + 19, 5 + 1772 24 5 + 19, 7 + 17, 11 + 13

Thus the smallest amount that could have been in the bag is 72T.

16


2015 Challenge Solutions - Junior

J1 Quirky Quadrilaterals

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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 therhombus lies on the line through the column of I dots and the other diagonal lies on the line perpendicular to thefirst diagonal and passing through the middle dot.

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17

CHALLENGE SOLUTIONS – JUNIOR


If I is an even number we can surround a south-west/north-east diagonal of I dots with a rhombus whose verticesare immediately south-west, north-east, and either side of the middle of the diagonal as the following examplesshow. Thus one diagonal of the rhombus lies on the line through the I dots and the other diagonal lies on the lineperpendicular to the first diagonal and bisecting the distance between the two middle dots.

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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 tileswould 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 itcannot 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 ifand only if it ends with a 2-digit number that is divisible by 4. The only such 2-digit multiples of 4 that can bemade 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 lastdigit 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


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 multiple994 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 8or 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 butcannot be a winning number from spinning 3 since digit 9 would have been eliminated. So 987 is not a winningnumber.

The next largest number is 986. This cannot be a winning number from spinning 9, 8, or 6. Since 986 is not divisibleby 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 divisibleby 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 awinning 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 7 = 7 12 = 5 + 7 17 = 17 22 = 3 + 193 = 3 8 = 3 + 5 13 = 13 18 = 5 + 13 23 = 234 = 2 + 2 9 = 2 + 7 14 = 3 + 11 19 = 19 24 = 5 + 195 = 5 10 = 3 + 7 15 = 2 + 13 20 = 3 + 17 25 = 2 + 236 = 3 + 3 11 = 11 16 = 3 + 13 21 = 2 + 19 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 evenprime. Now 27 = 2 + 25 and 25 is not prime, so 27 cannot be the sum of two primes. Hence 27T is the smallestpayment 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 thesum 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 sumof 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 amultiple of 6. The next multiple of 6 is 60. The following table lists all possibilities for each sum of six coins up to72T.

Sum of six Sum of pair All possible pairs

60 20 3 + 17, 7 + 1366 22 3 + 19, 5 + 1772 24 5 + 19, 7 + 17, 11 + 13

Thus the smallest amount that could have been in the bag is 72T.

19


c Examining primes less than 50 for pairs that differ by 6 and starting with small primes, we quickly find the sequence5, 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 isthe only even prime, all primes in the sequence must be odd. Hence they only end in 1, 3, 5, 7, or 9. The differencebetween any two primes in the sequence is 6, 12, 18, or 24. So no two primes in the sequence can end in the samedigit. There are five primes in the sequence so all of 1, 3, 5, 7, 9 must appear as last digits. The only prime whoselast 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 isthe only even prime, all primes in the sequence must be odd. Hence they only end in 1, 3, 5, 7, or 9. The onlyprime 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 nextfour 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 firstprime 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 atleast 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 fivenumbers 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 thesequence 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 thesequence 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 thatis 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 theoriginal 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 noavailable 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 is17/06/2345.

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 yearcannot 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. Soin 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, 7without repeating a digit. There are 5 choices for D, then 4 choices for M, and so on. So the number of condatesis 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 ofcondates is 2 × 3 × 2 × 1 = 12.

Thus the number of condates in the years 2000 to 2999 is 120 + 12 = 132.

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J5 Jogging

a On the flat, Theo will run a total of 4 km at 12 km/h and this will take 4

12× 60 = 20 minutes. He will run 3 km

uphill at 9 km/h and this will take 3

9× 60 = 20 minutes. He will run 3 km downhill at 15 km/h and this will take

3

15× 60 = 12 minutes. So the total time taken for Theo to complete the course will be 20 + 20 + 12 = 52 minutes.

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 Time

Flat east 2 km 13km/h 2

13h

Uphill 3 km 10km/h 3

10h

Downhill 3 km 14km/h 3

14h

Flat west 2 km 11km/h 2

11h

So the total time to complete the course would be ( 2

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11) × 60 ≈ 51.0 minutes.

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Section Distance Speed Time

Flat east 2 km 11km/h 2

11h

Uphill 3 km 8 km/h 3

8h

Downhill 3 km 16km/h 3

16h

Flat west 2 km 13km/h 2

13h

So the total time to complete the course would be ( 2

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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.

If the wind blows from the west, then his time on the hill will be 3

10+ 3

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then his time on the hill will be 3

8+ 3

16hours. To simplify comparison of these times we divide both by 3 and

multiply both by 2. So we want to compare 1

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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 3km uphill at 9 km/h in 20 minutes. So to gain 2 minutes uphill, he will need to run 3 km in18 minutes. Therefore his speed needs to be 3

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 run3 km in 10 minutes. Therefore his speed needs to be 3

10× 60 = 18km/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 minutesfor the hill.

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 3

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10. Hence r = 10km/h. This is 1 km/h faster.

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J6 Tessellating Hexagons

a The flipped hexagons are shaded.

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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 = FA, the sum of the three interior angles A, C, D is 360◦, and thesum 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 isA − BF − E − D − CA − BF − E − D and the top boundary isD − E − FB − AC − D − E − FB − A.So these boundaries are identical. Hence the 16-block can be translated vertically to form an infinite column withoutgaps and without overlap.

The right boundary of this column isDA − F − EB − C − DA − F − EB − C − DA repeated indefinitely.The left boundary of the column isC − 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.

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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 thetessellation 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, FA, has the smallest edge EFattached at one end but not the other. So the midpoints of edges AB, BC, DE, FA are not points of symmetryfor the tessellation.

To check the midpoint of edge CD, trace a copy of the block and place it exactly over the original block. Thenplace a pin through the midpoint of edge CD, rotate the traced copy of the block through 180◦ and notice that itagain fits exactly over the original block (except for overhang). Thus the midpoint of CD is a point of symmetryfor 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 .

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d The block of hexagons in Part a is reproduced below with hexagons labelled H1, H2, H3, H4, H ′

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H4

AB

C

D

E

FH3

A B

C

D

E

F H ′

1

A

B

C

DE

F

H ′

2

A

B

C

D E

F

H ′

4

AB

C

D

E

FH ′

3

A B

C

D

E

F H ′′

1

A

B

C

DE

F

H ′′

2

A

B

C

D E

F

H ′′

4

AB

C

D

E

FH ′′

3

A B

C

D

E

F H ′′′

1

A

B

C

DE

F

H ′′′

2

A

B

C

D E

F

H ′′′

4

AB

C

D

E

FH ′′′

3

Hexagons with the same subscript translate to one another. So there are several blocks of four hexagons thattessellate by translation. Three such blocks are: {H1, H2, H3, H4}, {H1, H ′′

2 , H3, H4}, {H ′

1, H2, H3, H ′

4}.

24


2015 Challenge Solutions - Intermediate

I1 Indim Integers


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 s 7 divides n? n s 7 divides n?

993 21 no 905 14 no984 21 no 894 21 no975 21 no 885 21 no966 21 yes 876 21 no957 21 no 867 21 no950 14 no 860 14 no948 21 no 858 21 no941 14 no 851 14 no939 21 no 849 21 no932 14 no 842 14 no923 14 no 833 14 yes914 14 no

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 and833 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 s 7 divides s? n s 7 divides s?

994 22 no 910 10 no987 24 no 903 12 no980 17 no 896 23 no973 19 no 889 25 no966 21 yes 882 18 no959 23 no 875 20 no952 16 no 868 22 no945 18 no 861 15 no938 20 no 854 17 no931 13 no 847 19 no924 15 no 840 12 no917 17 no 833 14 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 and833 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, indecreasing order, the first few 3-digit numbers n with digital sum 16 or 25.

25

CHALLENGE SOLUTIONS – INTERMEDIATE


s n 7 divides n? s n 7 divides n?

16 970 no 25 997 no961 no 988 no952 yes 979 no943 no 898 no934 no 889 yes925 no 799 no916 no907 no880 no871 no862 no853 no844 no835 no826 yes

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 DSS n DSS

994 22, 4 903 12, 3987 24, 6 896 23, 5980 17, 8 889 25, 7973 19, 10, 1 882 18, 9966 21, 3 875 20, 2959 23, 5 868 22, 4952 16, 7 861 15, 6945 18, 9 854 17, 8938 20, 2 847 19, 10, 1931 13, 4 840 12, 3924 15, 6 833 14, 5917 17, 8 826 16, 7910 10, 1

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


n F n F n F n F n F

994 4 896 5 798 6 693 9 595 1987 6 889 7 791 8 686 2 588 3980 8 882 9 784 1 679 4 581 5973 1 875 2 777 3 672 6 574 7966 3 868 4 770 5 665 8 567 9959 5 861 6 763 7 658 1 560 2952 7 854 8 756 9 651 3 553 4945 9 847 1 749 2 644 5 546 6938 2 840 3 742 4 637 7 539 8931 4 833 5 735 6 630 9 532 1924 6 826 7 728 8 623 2 525 3917 8 819 9 721 1 616 4 518 5910 1 812 2 714 3 609 6 511 7903 3 805 4 707 5 602 8 504 9

700 7

497 2 399 3 294 6 196 7490 4 392 5 287 8 189 9483 6 385 7 280 1 182 2476 8 378 9 273 3 175 4469 1 371 2 266 5 168 6462 3 364 4 259 7 161 8455 5 357 6 252 9 154 1448 7 350 8 245 2 147 3441 9 343 1 238 4 140 5434 2 336 3 231 6 133 7427 4 329 5 224 8 126 9420 6 322 7 217 1 119 2413 8 315 9 210 3 112 4406 1 308 2 203 5 105 6

301 4

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 integersn = 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-digitmultiples 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+63mthat have digital sum 16 or 25. The following table lists in decreasing order all 3-digit integers with digital sum 7.

n 7 divides n? n 7 divides n? n 7 divides n?

700 yes 340 no 205 no610 no 331 no 160 no601 no 322 yes 151 no520 no 313 no 142 no511 yes 304 no 133 yes502 no 250 no 124 no430 no 241 no 115 no421 no 232 no 106 no412 no 223 no403 no 214 no

So there are no other 3-digit multiples of 7 with FDS 7 besides those of the form 7 + 63m.

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 of9. This is equivalent to saying they have the same remainder when divided by 9. We now show that this is alwaystrue.

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 sumsequence for n, including its FDS, have the same remainder when divided by 9. Hence, if n is a multiple of 9, thenthe 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 dividesr. 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 multiplesof 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 F n F n F n F n F

992 2 896 5 792 9 696 3 592 7984 3 888 6 784 1 688 4 584 8976 4 880 7 776 2 680 5 576 9968 5 872 8 768 3 672 6 568 1960 6 864 9 760 4 664 7 560 2952 7 856 1 752 5 656 8 552 3944 8 848 2 744 6 648 9 544 4936 9 840 3 736 7 640 1 536 5928 1 832 4 728 8 632 2 528 6920 2 824 5 720 9 624 3 520 7912 3 816 6 712 1 616 4 512 8904 4 808 7 704 2 608 5 504 9

800 8 600 6

496 1 392 5 296 8 192 3488 2 384 6 288 9 184 4480 3 376 7 280 1 176 5472 4 368 8 272 2 168 6464 5 360 9 264 3 160 7456 6 352 1 256 4 152 8448 7 344 2 248 5 144 9440 8 336 3 240 6 136 1432 9 328 4 232 7 128 2424 1 320 5 224 8 120 3416 2 312 6 216 9 112 4408 3 304 7 208 1 104 5400 4 200 2

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 integersn = 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 theform 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


I3 Coin Flips

a Here are three moves that leave exactly 10 coins heads up.

Start: T T T T T T T T T T T T T TMove 1: H H H H T T T T T T T T T TMove 2: H H H H H H H H T T T T T TMove 3: H H H H H H H T H H H 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 sand want to finish with 154 Hs. We can achieve this by flipping 153 coins exactly once and flipping one coin threetimes over the three moves. Here are three moves that leave all coins heads up.

Start:

154︷︸︸︷

T . . .T

Move 1:

52︷︸︸︷

H . . .H

102︷︸︸︷

T . . .T

=

51︷︸︸︷

H . . .H H

51︷︸︸︷

T . . .T

51︷︸︸︷

T . . .T

Move 2:

51︷︸︸︷

H . . .H T

51︷︸︸︷

H . . .H

51︷︸︸︷

T . . .T

Move 3:

51︷︸︸︷

H . . .H H

51︷︸︸︷

H . . .H

51︷︸︸︷

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 takesat 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 inPart a to the 14 coins that are tails up, gives us exactly 12 + 10 = 22 coins heads up. One more move, flipping 4Ts 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.

Start:

154︷︸︸︷

T . . .T

Move 1:

m︷︸︸︷

H . . .H

154−m︷︸︸︷

T . . .T

Move 2:

m−n︷︸︸︷

H . . .H

n︷︸︸︷

T . . .T

m−n︷︸︸︷

H . . .H

154−2m+n︷︸︸︷

T . . .T where 0 ≤ n ≤ m andm− n ≤ 154 − m

=

2m−2n︷︸︸︷

H . . .H

154−2m+2n︷︸︸︷

T . . .T

To have all 154 coins with heads up after Move 3, we must have m = 154 − 2m + 2n. Thus 3m = 154 + 2n. But3m is odd and 154 + 2n is even. So we cannot have 154 coins heads up after just three moves.

29


Alternative ii

Three moves give the following results. Note that in a move, an even number of coins may mean 0 coins.

Start:

154︷︸︸︷

T . . .T

Move 1:

odd︷︸︸︷

H . . .H

odd︷︸︸︷

T . . .T

Move 2:

even︷︸︸︷

H . . .H

odd︷︸︸︷

T . . .T

even︷︸︸︷

H . . .H

odd︷︸︸︷

T . . .T orodd

︷︸︸︷

H . . .H

even︷︸︸︷

T . . .T

odd︷︸︸︷

H . . .H

even︷︸︸︷

T . . .T givingeven

︷︸︸︷

H . . .H

even︷︸︸︷

T . . .T

Move 3:

even︷︸︸︷

H . . .H

even︷︸︸︷

T . . .T

odd︷︸︸︷

H . . .H

odd︷︸︸︷

T . . .T orodd

︷︸︸︷

H . . .H

odd︷︸︸︷

T . . .T

even︷︸︸︷

H . . .H

even︷︸︸︷

T . . .T givingodd

︷︸︸︷

H . . .H

odd︷︸︸︷

T . . .T

Since 154 is even, we cannot have 154 coins heads up after just three moves.

I4 Jogging


I5 Folding Fractions

a Let QR = x. Then PR = 1 − x.

P Q

R

T

V

U

S

1

2

1

2

x

1 − x

1 − x

In �PQR Pythagoras’ theorem gives 1

4+ x2 = (1 − x)2 = x2 − 2x + 1.

Hence 2x = 3

4and x = 3

8.

b Alternative i

From Part a, we have:

P Q

R

T

V

U

S

1

2

1

2

3

85

8

a

ba

b

a

b

30


Since angles RQP , RPS, PTS, SUV are right angles, triangles RQP , PTS, V US have the same angles as shown.Hence they are similar and we have: V U :US:SV = PT :TS:SP = RQ:QP :PR = 3:4:5.

Hence PS = 5

3PT = 5

3× 1

2= 5

6. So US = UP − SP = 1 − 5

6= 1

6.

Then SV = 5

4US = 5

4× 1

6= 5

24and V U = 3

4US = 3

4× 1

6= 1

8.

Alternative ii

As in Alternative i, V U :US:SV = PT :TS:SP = RQ:QP :PR = 3:4:5. So ST = 4

3PT = 4

3× 1

2= 2

3.

From the side of the square, we have: 1 = UV + V S + ST = 3

5V S + V S + 2

3= 8

5V S + 2

3.

So V S = 1

3× 5

8= 5

24.

From �V US, we have: V U = 3

5SV = 3

5× 5

24= 1

8and US = 4

5SV = 4

5× 5

24= 1

6.

Alternative iii

Draw RW perpendicular to TS and let TS = t and V U = r. From Part a, we have:

P Q

R

T

V

U

S

W

1

2

1

2

3

85

8

3

8

t − 3

8

r

r

Now apply Pythagoras’ theorem to the following triangles.

In �RWS, RS2 = RW 2 + WS2 = 1 + (t − 3

8)2.

In �RPS, RS2 = RP 2 + PS2 = (5

8)2 + PS2.

In �PTS, PS2 = PT 2 + TS2 = (1

2)2 + t2.

Hence1 + (t − 3

8)2 = (5

8)2 + (1

2)2 + t2

1 + t2 + 9

64− 3

4t = 25

64+ 1

4+ t2

3

4t = 1 − 1

4+ 9

64− 25

64

= 3

4− 16

64= 3

4− 1

4= 1

2

t = 4

3× 1

2= 2

3

So PS2 = (1

2)2 + (2

3)2 = 1

4+ 4

9= 25

36and PS = 5

6.

Hence US = UP − SP = 1 − 5

6= 1

6.

Since 1 = TS + SV + r = 2

3+ SV + r, SV = 1

3− r.

In �V US, Pythagoras’ theorem gives V S2 = V U2 + US2 . Hence

(1

3− r)2 = r2 + (1

6)2

1

9+ r2 − 2

3r = r2 + 1

36

2

3r = 1

9− 1

36= 3

36= 1

12

r = 3

2× 1

12= 1

8

Thus V U = 1

8and SV = 1

3− 1

8= 5

24.

31


c Since angles RQP , RPS, PTS, SUV are right angles, triangles RQP , PTS, V US have the same angles as shown.Hence they are similar and we have RQ:QP :PR = PT :TS:SP = V U :US:SV = 7:24:25 or 24:7:25.

P Q

R

T

V

U

S

a

ba

b

a

b

Alternative i

If PQ/QR = 7/24, let PQ = 7k. Then QR = 24k and PR = 25k. Since PR = 1 −QR, we have 25k = 1− 24k. So49k = 1 and PQ = 7 × 1

49= 1

7.

If PQ/QR = 24/7, let PQ = 24k. Then QR = 7k and PR = 25k. Since PR = 1 − QR, we have 25k = 1 − 7k. So32k = 1 and PQ = 24 × 1

32= 3

4.

Alternative ii

If PQ/QR = 7/24, then PR/QR = 25/24. Since PR = 1 − QR, we have 24 − 24QR = 25QR, QR = 24/49, andPR = 25/49. From Pythagoras, PQ2 = PR2 − QR2 = 1 − 2QR = 1 − 48/49 = 1/49. So PQ = 1

7.

If PQ/QR = 24/7, then PR/QR = 25/7. Since PR = 1 − QR, we have 7 − 7QR = 25QR, QR = 7/32, andPR = 25/32. From Pythagoras, PQ2 = PR2 − QR2 = 1 − 2QR = 1 − 7/16 = 9/16. So PQ = 3

4.

Alternative iii

Let TP = x. If V U/US = 7/24, let V U = 7k. Then US = 24k and V S = 25k and we have:

P Q

R

T

V

U

S

x

1 − 24k1 − 32k

25k24k

7k

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


Alternatively, in �PTS, Pythagoras’ theorem gives

(1 − 24k)2 = (1 − 32k)2 + (56k)2

(24k)2 − 48k = (32k)2 − 64k + (56k)2

16k = (562 + 322 − 242)k2

1 = (142 + 82 − 62)k = 224k

x = 56/224 = 8/32 = 1/4

If V U/US = 24/7, let V U = 24k. Then US = 7k and V S = 25k. We have:

P Q

R

T

V

U

S

x

1 − 7k1 − 49k

25k7k

24k

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 �PTS, Pythagoras’ theorem gives

(1 − 7k)2 = (1 − 49k)2 + (56k)2

(7k)2 − 14k = (49k)2 − 98k + (56k)2

84k = (562 + 492 − 72)k2

12 = (82 + 72 − 12)7k = (112)7k

3 = (28)7k

x = 6/7

I6 Crumbling Cubes

a The small cubes have at most three faces exposed. The only small cubes that have exactly three faces exposedare the 8 on the corners of the 10 × 10 × 10 cube. The only small cubes that have exactly two faces exposed arethe 8 on each edge of the 10 × 10 × 10 cube that are not on its corners. All other cubes have less than two facesexposed. There are 12 edges on the 10 × 10 × 10 cube, so the number of small cubes removed at the first step is8 + (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 edgesbut not from the corners. So the number of non-corner small cubes on each edge is 192/12 = 16. Hence each edgein 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.

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 1increases the number of lost cubes by 12, one for each edge. Since 200 − 104 = 96 and 96/12 = 8, the cube thatloses 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. Theexposed surface area of this layer is 7 × 7 small faces plus a ring of 4 × 7 small faces. So the surface area of theremaining 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 smallfaces. Then remove the small cubes either side. This does not change the surface area. Continue until only the twocorner cubes remain. At this stage the surface area has increased by 2 small faces. Repeating this process on all12 edges increases the surface area by 12 × 2 small faces. At this stage all 8 corner cubes have 6 exposed faces. Soremoving the 8 corner cubes reduces the surface area by 8× 6 small faces. The surface area of the original 9× 9× 9cube 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 cubesplaced centrally on each face. At this stage the number of small cubes removed is 8 + (12 × 7) = 92.

7 × 7

7×77× 7

At the second step the only small cubes removed are the edge cubes in the 7×7 single layers. This leaves a 7×7×7cube with a 5 × 5 single layer of small cubes placed centrally on each face. So in this step, the number of smallcubes removed is 6(4 + 4 × 5) = 144.

5 × 5

5×55× 5

At the third step the only small cubes removed are the edge cubes in the 5 × 5 single layers and the edge cubes inthe 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


CHALLENGE STATISTICS – MIDDLE PRIMARY

Score Distribution %/Problem

Score

Challenge Problem

1e-Numbers

2Quazy Quilts

3Egg Cartons

4Condates

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

Mean Score/School Year/Problem

Year Number of Students

Mean

OverallProblem

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.

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.


CHALLENGE STATISTICS – UPPER PRIMARY


Score

Challenge Problem

1Rod Shapes

2Egg Cartons

3Seahorse Swimmers

4Primelandia

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



Mean

OverallProblem

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


YEAR Number of Students

Mean

OverallProblem

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:






4



CHALLENGE STATISTICS – JUNIOR


Score

Challenge Problem

1Quirky

Quadrilaterals

2Indim

Integers

3Primelandia

Money

4Condates

5Jogging

6Tessellating 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



Mean

OverallProblem

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:






4



CHALLENGE STATISTICS – INTERMEDIATE


Score

Challenge Problem

1Indim

Integers

2Digital Sums

3Coin Flips

4Jogging

5Folding

Fractions

6Crumbling

CubesDid 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



Mean

OverallProblem

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:






4



AUSTRALIAN INTERMEDIATE MATHEMATICS OLYMPIAD2015 Australian Intermediate Mathematics Olympiad - Questions

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 ofa + 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 theperimeter 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 1thingy cost a total of $441. What is the total cost, in dollars, of 1 whatsit, 1 doover and 1 thingy? [3 marks]

4. A fraction, expressed in its lowest termsa

b, can also be written in the form

2

n+

1

n2, where n is a positive integer.

If a + b = 1024, what is the value of a? [3 marks]

5. Determine the smallest positive integer y for which there is a positive integer x satisfying the equation213 + 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 circleslie 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

44

⌋

=

⌊

x

45

⌋

.

(Note: if r is any real number, then �r� denotes the largest integer less than or equal to r.) [4 marks]

1


9. A sequence is formed by the following rules: s1 = a, s2 = b and sn+2 = sn+1 + (−1)nsn 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 footof 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.

A

B

C

X

Y

Z

W

1

24

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 bonus marks]

2


AUSTRALIAN INTERMEDIATE MATHEMATICS OLYMPIAD SOLUTIONS

2015 Australian Intermediate Mathematics Olympiad - 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 , FA touch the circle at U , V , W , X, Y , Z respectively.

A B

C

D

E

F

U

V

W

X

Y

Z

Since the two tangents from a point to a circle have equal length,UB = BV , V C = CW , WD = DX, XE = EY , Y F = FZ, ZA = AU .

The perimeter of hexagon ABCDEF isAU + UB + BV + V C + CW + WD + DX + XE + EY + Y F + FZ + ZA= AU + UB + UB + CW + CW + WD + WD + EY + EY + Y F + Y F + AU= 2(AU + UB + CW + WD + 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 (1)

4w + 10d + t = 441 (2)

w + d + t = x (3)

Method 1

3 × (1) − 2 × (2): w + d + t = 3 × 329− 2 × 441 = 987 − 882 = 105.

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)/23 × (2) − 4 × (1): d = (7 + t)/2

Then w + d + t = 210/2− 2t/2 + t = 105.

4. We have2

n+

1

n2=

2n + 1

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.

This gives a = 2n + 1 = 2× 31 + 1 = 63.

5. Method 1213 + 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 thisdiameter, the two smaller circles are congruent. Let X be the centre of the large circle and Z the centre of a smallcircle.

X

Y

Y′

Z

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


By Pythagoras, Y Z2 = Y X2 + XZ2. So (R + r)2 = R2 + (2R − r)2.Then R2 + 2Rr + r2 = 5R2 − 4Rr + r2, which simplifies to 3r = 2R.

So the large circle has area π(30/√

π )2 = 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 Uunderneath A. So the number of paths from O to A is the sum of the number of paths from O to L and the pathsfrom O to U .

•

••

•

•

AL

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 rowupwards as indicated.

•

•

1 1 1 1 1 1 1

1 2 3 4 5 6 7

2 5 9 14 20 27

5 14 28 48 75

14 42 90 165

42 132 297

132 429

429

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.

5


Starting with the bottom-left vertex, label the vertices of the diagonal 0, 1, 2, . . . , n.

•

•

•

•

0

1

i

n

Consider all the paths that touch the diagonal at vertex i but not at any of the vertices between vertex 0 andvertex 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 thediagram 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 arep(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, andp(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

Let

⌊

x

44

⌋

=

⌊

x

45

⌋

= n.

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 − kvalues.

Thus, increasing n by 1 decreases the number of values of x by 1. Also the largest value of n is 43, in which casex has only 1 value.

Therefore the number of non-negative integer values of x is 44 + 43 + · · ·+ 1 = 1

2(44× 45) = 990.

Method 2

Let n be a non-negative integer such that

⌊

x

44

⌋

=

⌊

x

45

⌋

= n.

Then

⌊

x

44

⌋

= n ⇐⇒ 44n ≤ x < 44(n + 1) and

⌊

x

45

⌋

= n ⇐⇒ 45n ≤ x < 45(n + 1).

So

⌊

x

44

⌋

=

⌊

x

45

⌋

= n ⇐⇒ 45n ≤ x < 44(n + 1) ⇐⇒ 44n + n ≤ x < 44n + 44.

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 = 1

2(44 × 45) = 990.

6


Method 3

Let n be a non-negative integer such that

⌊

x

44

⌋

=

⌊

x

45

⌋

= n.

Then x = 44n + r where 0 ≤ r ≤ 43 and x = 45n + s where 0 ≤ s ≤ 44.

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 = 1

2(44× 45) = 990.

9. Working out the first few terms gives us an idea of how the given sequence develops:

n s2n−1 s2n

1 a b2 b − a 2b − a3 b 3b − a4 2b − a 5b − 2a5 3b − a 8b − 3a6 5b − 2a 13b− 5a7 8b − 3a 21b− 8a

It appears that the coefficients in the even terms form a Fibonacci sequence and, from the 5th term, every oddterm 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+1 + s2m

s2m+3 = s2m+2 − s2m+1 = s2m

s2m+4 = s2m+3 + s2m+2 = 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 = 3and b < 1000, none of the first five terms of the given sequence equal 2015. So we are looking for integer solutionsof 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 lessthan 1:2 and any point between AP and AB has the distance ratio greater than 1:2.

A

B

C

P

Q

1

2

1

4

7


Let Q be a point outside �ABC such that its distances from AC and BC are in the ratio 1:4. By an argumentsimilar 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 andCQ 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 theheight of �ABC. Letting | | denote area, we have

|ABC| = 1

2h × AB and

|ABC| = |AXB| + |BXC| + |CXA| = 1

2(2AB + 4BC + AC) = 1

2AB × 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 atrapezium of 3 triangles in the next layer, a trapezium of 5 triangles in the next layer, and so on. The boundaryof the combined figure is �ABC and X is one of the grid vertices as shown.

A

B

C

X

Y

Z

W

1

2

4

There are 49 small triangles in ABC and 6.5 small triangles in AZXY . Hence, after rescaling so that the area ofAZXY 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 BWXZ are similar. Let XY be1 unit and AY be x. Then BZ = 2x.

A

B

C

X

Yx

2x

1

24

Z

W

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, XC2 = 1 + (x +√

x2 − 3)2 = 2x2 − 2 + 2x√

x2 − 3and in �XWC, WC2 = 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


Squaring gives 4x2 + x2 − 3 − 4x√

x2 − 3 = 2x2 − 18 + 2x√

x2 − 3, which simplifies to 3x2 + 15 = 6x√

x2 − 3.

Squaring again gives 9x4 +90x2 +225 = 36x4−108x2. So 0 = 3x4−22x2−25 = (3x2−25)(x2 +1), giving x =5√3.

Hence, area AZXY =x

2+√

x2 − 3 =5

2√

3+

4√3

=13

2√

3and

area ABC =

√3

4(2x +

√x2 − 3)2 =

√3

4

( 10√3

+4√3

)2

=49√

3.

Since the area of AZXY is 13 cm2, the area of ABC is (49√3/

13

2√

3) × 13 = 98 cm2.

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 FH 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, XFG, XHI, BHJ are equilateral, and triangles XDE and BHJ are congruent.

F G

D

E

H

I

J

A

B

C

X

Y

1

24

Z

W

The areas of the various equilateral triangles are proportional to the square of their heights. Let the area of�FXG = 1. Then, denoting area by | |, we have:

|DEX| = 4, |XHI| = 16, |AEG| = 9, |DBI| = 36, |FHC| = 25.

|ABC| = |AEG|+ |FHC|+ |DBI| − |FXG| − |DEX| − |XHI| = 9 + 25 + 36 − 1 − 4 − 16 = 49.

|AZXY | = |AEG| − 1

2(|FXG|+ |DEX|) = 9 − 1

2(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.

A

B

C

X

Y

a

bc

Z

W

60◦120◦

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 | = |Y AZ| + |Y XZ|= 1

2(AY )(AZ) sin 60◦ + 1

2ab sin 120◦

=√

3

4((AY )(AZ) + ab)

=√

3

12((a + 2b)(b + 2a) + 3ab)

=√

3

12(2a2 + 2b2 + 8ab)

=√

3

6(a2 + b2 + 4ab)

Similarly, |CY XW | =√

3

6(a2 + c2 + 4ac) and |BWXZ| =

√3

6(b2 + c2 + 4bc).

Hence |ABC| =√

3

6(2a2 + 2b2 + 2c2 + 4ab + 4ac + 4bc) =

√3

3(a + b + c)2.

So |ABC|/|AZXY | = 2(a + b + c)2/(a2 + b2 + 4ab).

Letting a = k, b = 2k, c = 4k, and |AZXY | = 13cm2, we have |ABC| = 26(49k2)/(k2 + 4k2 + 8k2) = 98 cm2.

Investigation

Method 4 gives |ABC|/|AZXY | = 2(a + b + c)2/(a2 + b2 + 4ab).

Alternatively, as in Method 3,

|ABC| = |AEG|+ |FHC|+ |DBI| − |FXG| − |DEX| − |XHI|= (a + b)2 + (a + c)2 + (b + c)2 − a2 − b2 − c2 = (a + b + c)2.

Also |AZXY | = |AEG| − 1

2(|FXG|+ |DEX|)

= (a + b)2 − 1

2(a2 + b2)

= 2ab + 1

2(a2 + b2).

So |ABC|/|AZXY | = 2(a + b + c)2/(a2 + b2 + 4ab).

10


Distribution of Awards/School Year


Number of Awards

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

YearNumber 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

Overall MeanQuestion

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

AUSTRALIAN INTERMEDIATE MATHEMATICS OLYMPIAD STATISTICS


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 DISTINCTIONIvan 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

AUSTRALIAN INTERMEDIATE MATHEMATICS OLYMPIAD RESULTS



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



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



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


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 ChallengeDr K McAvaney Victoria, (Director) 9 years; 2006–2015; Member 1 year 2005–2006Mr B Henry Victoria (Director) 17 years; 1990–2006; Member 9 years 2007–2015Prof P J O’Halloran University of Canberra, ACT 5 years; 1990–1994Dr R A Bryce Australian National University, ACT 23 years; 1990–2012Adj Prof M Clapper Australian Mathematics Trust, ACT 3 years; 2013–2015Ms L Corcoran Australian Capital Territory 3 years; 1990–1992Ms B Denney New South Wales 6 years; 2010–2015Mr J Dowsey University of Melbourne, VIC 8 years; 1995–2002Mr A R Edwards Department of Education, QLD 26 years; 1990–2015Dr M Evans Scotch College, VIC 6 years; 1990–1995Assoc Prof H Lausch Monash University, VIC 24 years; 1990–2013Ms J McIntosh AMSI, VIC 14 years; 2002–2015Mrs L Mottershead New South Wales 24 years; 1992–2015Miss A Nakos Temple Christian College, SA 23 years; 1993–2015Dr M Newman Australian National University, ACT 26 years; 1990–2015Ms F Peel St Peter’s College, SA 2 years; 1999, 2000Dr I Roberts Northern Territory 3 years; 2013–2015Ms T Shaw SCEGGS, NSW 3 years; 2013–2015Ms K Sims New South Wales 17 years; 1999–2015Dr A Storozhev Attorney General’s Department, ACT 22 years; 1994–2015Prof P Taylor Australian Mathematics Trust, ACT 20 years; 1995–2014Mrs A Thomas New South Wales 18 years; 1990–2007Dr S Thornton South Australia 18 years; 1998–2015 Miss G Vardaro Wesley College, VIC 22 years: 1993–2006, 2008–2015Visiting membersProf E Barbeau University of Toronto, Canada 1991, 2004, 2008Prof G Berzsenyi Rose Hulman Institute of Technology, USA 1993, 2002Dr L Burjan Department of Education, Slovakia 1993Dr V Burjan Institute for Educational Research, Slovakia 1993Mrs A Ferguson Canada 1992Prof B Ferguson University of Waterloo, Canada 1992, 2005Dr D Fomin St Petersburg State University, Russia 1994Prof F Holland University College, Ireland 1994Dr A Liu University of Alberta, Canada 1995, 2006, 2009Prof Q Zhonghu Academy of Science, China 1995Dr A Gardiner University of Birmingham, United Kingdom 1996Prof P H Cheung Hong Kong 1997Prof R Dunkley University of Waterloo, Canada 1997Dr S Shirali India 1998Mr M Starck New Caledonia 1999Dr R Geretschlager Austria 1999, 2013Dr A Soifer United States of America 2000Prof M Falk de Losada Colombia 2000Mr H Groves United Kingdom 2001 Prof J Tabov Bulgaria 2001, 2010Prof A Andzans Latvia 2002 Prof Dr H-D Gronau University of Rostock, Germany 2003


Prof J Webb University of Cape Town, South Africa 2003, 2011Mr A Parris Lynwood High School, New Zealand 2004Dr A McBride University of Strathclyde, United Kingdom 2007Prof P Vaderlind Stockholm University, Sweden 2009, 2012Prof A Jobbings United Kingdom 2014 Assoc Prof D Wells United States of America 2015

Moderators for ChallengeMr W Akhurst New South WalesMs N Andrews ACER, Camberwell, VICProf E Barbeau University of Toronto, CanadaMr R Blackman VictoriaMs J Breidahl St Paul’s Woodleigh, VICMs S Brink Glen Iris, VICProf J C Burns Australian Defence Force Academy, ACTMr A. Canning Queensland Mrs F Cannon New South WalesMr J Carty ACT Department of Education, ACTDr E Casling Australian Capital TerritoryMr B Darcy South AustraliaMs B Denney New South WalesMr J Dowsey VictoriaBr K Friel Trinity Catholic College, NSWDr D Fomin St Petersburg University, RussiaMrs P Forster Penrhos College, WAMr T Freiberg QueenslandMr W Galvin University of Newcastle, NSWMr M Gardner North Virginia, USAMs P Graham TasmaniaMr B Harridge University of Melbourne, VICMs J Hartnett Queensland Mr G Harvey Australian Capital TerritoryMs I Hill South AustraliaMs N Hill VictoriaDr N Hoffman Edith Cowan University, WAProf F Holland University College, IrelandMr D Jones Coff’s Harbour High School, NSWMs R Jorgenson Australian Capital TerritoryAssoc Prof H Lausch VictoriaMr J Lawson St Pius X School, NSWMr R Longmuir ChinaMs K McAsey Victoria Dr K McAvaney VictoriaMs J McIntosh AMSI, VICMs N McKinnon VictoriaMs T McNamara VictoriaMr G Meiklejohn Queensland School Curriculum Council, QLDMr M O’Connor AMSI, VICMr J Oliver Northern TerritoryMr S Palmer New South WalesDr W Palmer University of Sydney, NSWMr G Pointer South AustraliaProf H Reiter University of North Carolina, USAMr M Richardson Yarraville Primary School, VICMr G Samson Nedlands Primary School, WAMr J Sattler Parramatta High School, NSWMr A Saunder VictoriaMr W Scott Seven Hills West Public School, NSW


Moderators for Challenge continuedMr R Shaw Hale School, WAMs T Shaw New South WalesDr B Sims University of Newcastle, NSWDr H Sims VictoriaMs K Sims New South WalesProf J Smit The NetherlandsMrs M Spandler New South WalesMr G Spyker Curtin University, WAMs C Stanley Queensland Dr E Strzelecki Monash University, VICMr P Swain Ivanhoe Girls Grammar School, VICDr P Swedosh The King David School, VICProf J Tabov Academy of Sciences, BulgariaMrs A Thomas New South WalesMs K Trudgian QueenslandProf J Webb University of Capetown, South AfricaMs J Vincent Melbourne Girls Grammar School, VIC

Mathematics Enrichment Development

Enrichment Committee — Development Team (1992–1995)Mr B Henry Victoria (Chairman)Prof P O’Halloran University of Canberra, ACT (Director)Mr G Ball University of Sydney, NSWDr M Evans Scotch College, VICMr K Hamann South AustraliaAssoc Prof H Lausch Monash University, VICDr A Storozhev 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, ACTMr J Carty Merici College, ACTDr H Gastineau-Hill University of Sydney, NSWMr B Henry VictoriaAssoc Prof H Lausch Monash University, VICProf P O’Halloran University of Canberra, ACTDr A Storozhev Australian Mathematics Trust, ACTEuler Development Team (1992–1995)Dr M Evans Scotch College, VIC (Editor)Mr B Henry Victoria (Editor)Mr L Doolan Melbourne Grammar School, VICMr K Hamann South AustraliaAssoc Prof H Lausch Monash University, VICProf P O’Halloran University of Canberra, ACTMrs A Thomas Meriden School, NSWGauss Development Team (1993–1995)Dr M Evans Scotch College, VIC (Editor)Mr B Henry Victoria (Editor)Mr W Atkins University of Canberra, ACTMr G Ball University of Sydney, NSWProf J Burns Australian Defence Force Academy, ACTMr L Doolan Melbourne Grammar School, VICMr A Edwards Mildura High School, VICMr N Gale Hornby High School, New ZealandDr N Hoffman Edith Cowan University, WA


Prof P O’Halloran University of Canberra, ACTDr W Pender Sydney Grammar School, NSWMr R Vardas Dulwich Hill High School, NSWNoether Development Team (1994–1995)Dr M Evans Scotch College, VIC (Editor)Dr A Storozhev Australian Mathematics Trust, ACT (Editor)Mr B Henry VictoriaDr D Fomin St Petersburg University, RussiaMr G Harvey New South WalesNewton Development Team (2001–2002)Mr B Henry Victoria (Editor)Mr J Dowsey University of Melbourne, VICMrs L Mottershead New South WalesMs G Vardaro Annesley College, SAMs A Nakos Temple Christian College, SAMrs A Thomas New South WalesDirichlet Development Team (2001–2003)Mr B Henry Victoria (Editor)Mr A Edwards Ormiston College, QLDMs A Nakos Temple Christian College, SAMrs L Mottershead New South WalesMrs K Sims Chapman Primary School, ACTMrs A Thomas New South Wales

Australian Intermediate Mathematics Olympiad CommitteeDr K McAvaney Victoria (Chair) 9 years; 2007–2015Adj Prof M Clapper Australian Mathematics Trust, ACT 2 years; 2014–2015Mr J Dowsey University of Melbourne, VIC 17 years; 1999–2015Dr M Evans AMSI, VIC 17 years; 1999–2015Mr B Henry Victoria (Chair) 8 years; 1999–2006 Member 9 years; 2007–2015Assoc Prof H Lausch Monash University, VIC 17 years; 1999–2015Mr R Longmuir China 2 years; 1999–2000

Documents

2015 Mathematics Contests – The Australian Scene Part 1