western australian junior mathematics olympiad 2009 · western australian junior mathematics...

Western AustralianJunior Mathematics Olympiad 2009

Individual Questions 100 minutes

General instructions: Each solution in this part is a positive integerless than 100. No working is needed for Questions 1 to 9. Calculatorsare not permitted. Write your answers on the answer sheet provided.

1. Evaluate55 − 52

√55 − 54

.

[1 mark]

2. From each vertex of a cube, we remove a small cube whose sidelength is one-quarter of the side length of the original cube.How many edges does the resulting solid have? [1 mark]

3. A certain 2-digit number x has the property that if we put a 2 beforeit and a 9 afterwards we get a 4-digit number equal to 59 times x.What is x? [2 marks]

4. What is the units digit of 22009 × 32009 × 62009? [2 marks]

5. At a pharmacy, you can get disinfectant at different concentrationsof alcohol. For instance, a concentration of 60% alcohol means ithas 60% pure alcohol and 40% pure water. The pharmacist makesa mix with 3

5litres of alcohol at 90% and 1

5litres of alcohol at 50%.

How many percent is the concentration of that mix? [2 marks]

6. If we arrange the 5 letters A, B, C, D and E in different ways wecan make 120 different “words”. Suppose we list these words inalphabetical order and number them from 1 to 120. So ABCDEgets number 1 and EDCBA gets number 120.What is the number for DECAB? [3 marks]

7. Every station on the Metropolis railway sells tickets to every otherstation. Each station has one set of tickets for each other station.When it added some (more than one) new stations, 46 additionalsets of tickets had to be printed.How many stations were there initially? [3 marks]

8. At a shop, Alice bought a hat for $32 and a certain number of hairclips at $4 each. The average price of Alice’s purchases (in dollars)is an integer.What is the maximum number of hair clips that Alice could havebought? [3 marks]

9. The interior angles of a convex polygon form an arithmetic sequence:

143◦, 145◦, 147◦, . . . .

How many sides does the polygon have? [4 marks]

10. For full marks, explain how you found your solution.A square ABCD has area 64 cm2. Let M be the midpoint of BC,let d be the perpendicular bisector of AM , and let d meet CD at F .How many cm2 is the area of the triangle AMF?

A B

CD

M

d

F

[4 marks]


Team Questions 45 minutes

General instructions: Calculators are (still) not permitted.

Crazy ComputersRebecca has an old Lemon brand computer which has a defective key-board. When she types L the letters LOM appear on the screen, andwhen she types M she gets OL and when she types O she gets M . Wewill abbreviate this to:

Lemon Computer: L→ LOM , M → OL, O →M.

So if she types OLO, she gets MLOMM .

A. What will she get on the screen if she presses the Enter key to getto a new line and then types MLOMM?

B. Tom has a more modern Raincoat computer which also has a brokenkeyboard. Typing R puts RS on the screen and S puts R.

Raincoat: R→ RS, S → R.If Tom types R to give one line of the screen, types in this line toget a second line and so on until he has 5 lines, what will the lastline be? (Be careful – the first line on the screen is RS not R.)

C. If Tom kept going till he had 12 lines, how many letters would therebe in the final line? Try to calculate this without writing down thefinal line of letters (which is quite long).

D. Sarah’s computer uses software produced by the giant MegafloppyCorporation, and is as defective as the others. If she types in H toget a first line on the screen, then types that line in to get a secondline, then types that to get a third line, she finds the third line is

HGGHGHHG.

Assuming that only the G and H keys are faulty, what would sheget if she deleted everything and then typed G?

In the next question, Rebecca again uses her Lemon computer.

E. If Rebecca starts by typing O on her Lemon and continues untilthere are 6 lines on the screen, how many Os will there be in thelast line?For full marks you must show how to calculate this without actuallywriting down the final line.

F. Ben has a Super-Useless Lap Bottom computer, which has mysteri-ous problems with the letters U , X and Z of its keyboard. You willneed to discover its rule by describing what should replace each boxin answering this question (the size of the boxes does not indicatethe number of replacement letters):

Lap Bottom: U → , X → , Z → .Ben started by entering the letter U , and then like the others en-tered what he saw on the screen onto the next line; doing this 4times he finally obtained:

UZUUXUZUUZUZXUZUUXUZUUZUUXUZUUXZXUZUUXUZUUZUZXUZUUXUZU

What letter(s) replace each box above? Explain how you got youranswer.Below, we repeat the above sequence of letters several times, in thehope it might be helpful in your scratchwork for this problem.









Individual Questions Solutions

1. Answer: 62.

55 − 52

√55 − 54

=52(53 − 1)

52√

5− 1

=124

2= 62

[1 mark]

2. Answer: 84. Initially there are 2 · 4 + 4 = 12 edges. By removinga small cube from a vertex (of which there are 8), we increase thenumber of edges by 12− 3 = 9. Hence, the resulting solid has

12 + 8 · 9 = 84 edges.

[1 mark]

3. Answer: 41. Represent the 2-digit number x by ∗#. Putting 2before it and a 9 after it, we get

2∗#9 = 2009 + ∗#0

= 2009 + 10x.

We are told that this number is 59x. Thus we have

2009 + 10x = 59x

2009 = 49x

41 = x.

[2 marks]

4. Answer: 6. Since

22009 × 32009 × 62009 = 62009·2

is just a power of 6 and 6 × 6 = 36 also ends in 6, any power of 6ends in 6. So the answer is 6.

Alternative. Observe that

62 = 36 ≡ 6 (mod 10)

∴ 6n ≡ 6 (mod 10) for any integer n ≥ 1

∴ 22009 × 32009 × 62009 = 62009·2

≡ 6 (mod 10)

Hence, the last digit of 22009 × 32009 × 62009 is 6. [2 marks]

5. Answer: 80. The new concentration is the the total volume of alco-hol over the total volume of liquid expressed as a percentage:

total volume of alcohol

total volume=

3

5· 90

100+

1

5· 50

1003

5+

1

5

=3 · 90

100+ 1 · 50

1003 + 1

=10(3 · 9 + 1 · 5)

4 · 100

=10(27 + 5)

4 · 100

=80

100= 80%

So the number of percent of the new concentration is 80. [2 marks]

6. Answer: 95. There are 120/5 = 24 words beginning with A, 24beginning with B and 24 beginning with C. These all come beforeDECAB. Of those beginning with D there are 24/4 = 6 beginningwith DA, 6 beginning with DB and 6 beginning with DC. Thesealso come before DECAB. Those beginning with DE go DEABC,DEACB, DEBAC, DEBCA and DECAB. There are 5 of these.So DECAB’s number is 3× 24 + 3× 6 + 5 = 95.

Alternative. There’s less counting if one starts from the other end.There are 24 words beginning with E. Then DECBA is the last wordbeginning with D, and the one before it is DECAB. So DECAB’snumber is 120 minus the number that follow it, i.e. 120− (24 + 1) =95. [3 marks]

7. Answer: 11. If y stations are added to x already existing stations,each new station will require (x + y − 1) sets of tickets; for y newstations this is y(x + y − 1) sets. Each old station needs y sets. So:

y(x + y − 1) + xy = 46

∴ y(2x + y − 1) = 46.

Thus y must be a positive integer which is a factor of 46, i.e. it is1, 2, 23, or 46. But y > 1, and y = 23 or y = 46 imply x < 0.∴ y = 2, x = 11. Therefore there were 11 old stations. [3 marks]

8. Answer: 27. Let x be the number of hair clips Alice bought. Thenthe total of her purchases is:

4x + 32

so that the average price of her purchases is

4x + 32

x + 1=

4(x + 1) + 28

x + 1= 4 +

28

x + 1,

which is an integer, if 28 is divisible by x + 1. Hence, x + 1 must beone of 1, 2, 4, 7, 14 or 28 (the divisors of 28), i.e. x must be one of0, 1, 3, 6, 13 or 27. The largest of these is 27. [3 marks]

9. Answer: 18. Let n be the number of sides of the polygon. Then,

(n− 2) · 180 =n

2

(2 · 143 + (n− 1) · 2

)180(n− 2) = n(143 + n− 1)

= n(142 + n)

n2 − 38n + 360 = 0

(n− 18)(n− 20) = 0.

So n = 18 or n = 20. Since the n-gon is convex, all its angles, inparticular, the largest, must be less than 180◦. Now, for n = 20, thelargest angle is

143 + 19 · 2 = 181 > 180.

So, n 6= 20. On the other hand, for n = 18, the largest angle

143 + 17 · 2 = 177 < 180,

which is ok. So n = 18. [4 marks]

10. Answer: 30. Let y = FD. Since d is the perpendicular bisector ofAM , it is the locus of points equidistant from A and M .So AF = MF .Since the area of the square is 64 cm2, its side length is 8 cm. Henceapplying Pythagoras’ Theorem to 4FDA and 4CFM , we have

82 + y2 = (8− y)2 + 42

= 82 + y2 − 16y + 16

∴ 16y = 16

y = 1

Take the parenthesising of the vertices of a figure, as a convenientshorthand for the figure’s area, so that (XY Z) means “the area offigure XY Z”. Then

(AMF ) = (ABCD)− (ABM)− (FDA)− (CFM)

= 64− 12(8 · 4 + 8 · 1 + 4 · 7)

= 64− 12· 68

= 64− 34 = 30.

So 4AMF has area 30 cm2.

Alternative 1. Instead, let x = FC. As before, deduce AF =MF , and that square has side length 8 cm. Applying Pythagoras’Theorem to 4FDA and 4CFM , we have

82 + (8− x)2 = 42 + x2

82 + 82 − 2 · 8x + x2 = 42 + x2

2 · 82 − 42 = 2 · 8x

∴ x =2 · 82 − 42

2 · 8= 8− 1 = 7.

The rest of the solution proceeds like the first solution.

Alternative 2. Since the area of the square is 64 cm2, its side lengthis 8 cm. Since M is the midpoint of BC, MB = 4 cm. ApplyingPythagoras’ Theorem to 4ABM , we have

AM =√

82 + 42

= 4√

22 + 12 = 4√

5

Let the midpoint of AM be X, i.e. XM = XA. Then

FM2 = FX2 + XM2

= FX2 + XA2

= FA2

So, FM = FA. Now deduce x or y as above and hence deduce that

FM2 = FA2 = 65

∴ FX2 = FA2 −XA2

= 65− (2√

5)2

= 65− 20 = 45

∴ (AMF ) = 12AM · FX

= 12· 4√

5 ·√

45

= 2√

5 · 3√

5

= 6 · 5 = 30

[4 marks]

Team Questions Solutions

Crazy Computers

A. Answer: OLLOMMOLOL. Putting a little space between thereplacement letters, we have MLOMM → OL LOM M OL OL.

[4 marks]

B. Answer: RSRRSRSRRSRRS.

R→ RS (1st line)

→ RSR (2nd line)

→ RSRRS (3nd line)

→ RSRRSRSR (4th line)

→ RSRRSRSRRSRRS (5th line)

[5 marks]

C. 377. The numbers of letters increases by the number of Rs in theline, which is the same as the number of letters in the previous line.Let `n be the length of the nth line and let `0 = 1 be the length ofthe first entered line (namely R). Then

`n+1 = `n + `n−1, n ≥ 0,

where `0 = 1, `1 = 2. So we have:

`2 = `1 + `0 = 2 + 1 = 3

`3 = `2 + `1 = 3 + 2 = 5

`4 = `3 + `2 = 5 + 3 = 8

`5 = `4 + `3 = 8 + 5 = 13

`6 = `5 + `4 = 13 + 8 = 21

`7 = `6 + `5 = 21 + 13 = 34

`8 = `7 + `6 = 34 + 21 = 55

`9 = `8 + `7 = 55 + 34 = 89

`10 = `9 + `8 = 89 + 55 = 144

`11 = `10 + `9 = 144 + 89 = 233

`12 = `11 + `10 = 233 + 144 = 377.

So the 12th line has 377 letters.

It helps to recognise that

`n = Fn+1,

where Fn is the nth term of the Fibonacci sequence. [8 marks]

D. Answer: GH. One can show that HGGHGHHG arises from thereplacements H → HG, G→ GH. [6 marks]

E. Answer: 11. Let `n, mn, on be the number of Ls, Ms and Os,respectively on line n, or initially input in the case when n = 0.Then `0 = m0 = 0, o0 = 1 and for n ≥ 1,

`n = `n−1 + mn−1

mn = `n−1 + on−1

on = `n−1 + mn−1 = `n.

Representing this in a table we have

n `n = `n−1 + mn−1 mn = `n−1 + on−1 on = `n−1 + mn−1

0 0 0 11 0 1 02 1 0 13 1 2 14 3 2 35 5 6 56 11 10 11

So there are 11 Os in the sixth line. [12 marks]

F. Answer: U → UZU, X → ZX, Z → UX. Looking at the finalline we see the following repetitions.

UZUUXUZUUZU ZX UZUUXUZUUZUUXUZUUX ZX UZUUXUZUUZU ZX UZUUXUZU

If U → UZU then passing from the third to fourth line, we requireZ → UX. So the remaining letters must be the result of Z, i.e. wemust have X → ZX. Confirming this

U → UZU

→ UZUUXUZU

→ UZUUXUZUUZUZXUZUUXUZU

→ UZUUXUZUUZUZXUZUUXUZUUZUUXUZUUXZXUZUUXUZUUZUZXUZUUXUZU

[10 marks]




1. Below are three different views of a child’s building block and a sin-gle view of a different block.

1 2 3 4

Which is the different block? [1 mark]

2. Some horses and some jockeys are in a stable. In all, there are 71heads and 228 legs. How many jockeys are in the stable? [1 mark]

3. Four friends go fishing and catch a total of 11 fish. Each personcaught at least one fish. The following five statements each have alabel from 1 to 16. What is the sum of the labels of all the statementswhich must be true?

1: One person caught exactly 2 fish.2: One person caught exactly 3 fish.4: At least one person caught fewer than 3 fish.8: At least one person caught more than 3 fish.

16: Two people each caught more than 1 fish. [2 marks]

4. There are four throwers in the shot put final at the Olympic games.The distance thrown by the second thrower is 2% less than the firstthrower while the third thrower achieves a distance 20% greaterthan the first thrower. The fourth thrower throws it 10% furtherthan the third person. If the total distance of the four throwers is90 m how many metres did the first thrower throw the shot put?

[2 marks]

5. A cube of side 7 cm is painted green all over, then cut into cubes ofside 1 cm.How many of these small cubes have exactly 2 faces painted green?

[2 marks]

6. Ann is four times as old as Mary was when Ann was as old as Maryis now. Furthermore, Ann is twice as old as Mary was when Annwas six years older than Mary is now. How old is Ann? [3 marks]

7. A barrel contains a number of blue balls and a number of green ballswhich you take out one by one. Each time you take out a blue ballsomebody puts 100 frogs into the barrel and each time you take outa green ball the person puts in 72 frogs. Finally, when you haveremoved all the balls, you find there are 2008 frogs in the barrel.How many green balls were there in the barrel initially? [3 marks]

8. In the figure, B is themid-point of AD, C is themid-point of DE, A is themid-point of EF , and Mis the midpoint of AF .If the area of 4AMB is6 cm2, how many cm2 is theremaining area of 4DEF?

A

B C

D

EM

F

[3 marks]

9. Five grandmothers go to a cafe to eat cake. The cafe sells 4 differenttypes of cake. Each grandmother chooses two different cakes. Theyfind their bills are for $6, $9, $11, $12 and $15.The next day I go to the cafe and buy one of each type of cake.How much do I pay? [4 marks]

10. For full marks, explain how you found your solution.A square is divided into threepieces of equal area as shown. Thedistance between the parallel lines is1 cm. What is the area of the squarein cm2.

1

x

y

z

[4 marks]




How to multiply on TitaniaThe Titans are an intelligent race who live on the planet Titania. Theyuse the same numerals as us for the integers and their addition and sub-traction is the same as ours. However, they don’t use multiplication.Instead of ×, they have an operation called ‘star’ which has the follow-ing properties in common with ×:

For all integers a, b and c,• a ∗ c = c ∗ a and• (a ∗ b) ∗ c = a ∗ (b ∗ c).

However the other properties of star are completely different:

(i) For all integers a, a ∗ 0 = a, and(ii) For all integers a and b, a ∗ (b + 1) = (a ∗ b) + (1 − a), e.g.

7 ∗ 2 = (7 ∗ 1) + (−6).

A. Copy and complete the table shown.The entry in the row labelled by aand the column labelled by b shouldbe a ∗ b, where each of a, b range over0, 1, . . . , 4.Some entries have been filled in to getyou started.

∗ 0 1 2 3 4

0 0 1 2 3 41 1234

B. Show that for every integer a, a ∗ 1 = 1.

C. Find (−4) ∗ (−3).

D. Show that for any integer a, 4 ∗ a = 4− 3a.

E. Show that for all integers a and positive integers b, a∗b = a+b−ab.Hint. Try a = 5 and try to generalise your argument.

F. Show how to express a ∗ b, for any (positive, negative or zero)integers a, b, in terms of our addition and multiplication.

G. The Titans also use powers by defining a∗n to be

a ∗ a ∗ · · · ∗ a, where a appears n times,

for n a positive integer, e.g. a∗1 = a, a∗2 = a ∗ a, etc.Find (2∗3) ∗ (3∗2) and (3∗3)∗3.

H. Write the Titans’ power operation a∗n in terms of our operations,where n is a positive integer and a is any integer.Hint. To start, try a∗1, a∗2, a∗3 and look for a pattern.


1. Answer: 3. The views 1, 2 and 4 can be explained thus: the starface is opposite the square face, and the pentagon face is oppositethe triangle (the face opposite the octagon is unknown). Then 2 isobtained from 1 by rotating in an axis perpendicular to the centreof the octagon face so that the triangle face becomes the front face.And 4 is obtained by flipping 1 upside down and then rotating inthe vertical axis so that the octagon is the front face. Flipping 2 sothat the octagon is on the top and the star on the right, the triangleface should be behind and the pentagon at the front. So 3 is thedifferent block. [1 mark]

2. Answer: 28. Let j be the number of jockeys and let h be the numberof horses. Then

j + h = 71 =⇒ 2j + 2h = 142

2j + 4h = 228 =⇒ j + 2h = 114

=⇒ j = 142− 114

= 28

Thus the number of jockeys is 28. [1 mark]

3. Answer: 4. The example: 3 of the friends caught 1 fish and 1 caught8 fish meets the criteria, so that the statements labelled 1, 2 and 16need not be true.Also, the example: 3 of the friends caught 3 fish and 1 caught 2 fish,shows that the statement labelled 8 need not be true.Now, suppose for a contradiction that the statement labelled 4 isfalse, then all 4 friends caught 3 or more fish, which implies thereare 12 or more fish, but there are only 11 fish (contradiction). Thus,at least one friend caught fewer than 3 fish. Hence the statementlabelled 4 must be true, and since this is the only statement thatmust be true, the sum of such labels is 4. [2 marks]

4. Answer: 20. Let x be the distance thrown by the first person inmetres. Then the second throws it 0.98x, the third, 1.2x and thefourth 1.32x. Thus 4.5x = 90 and so x = 20. [2 marks]

5. Answer: 60. The cube has 12 edges. Along each of those edges ofthe 7 cm cube, 5 will have exactly two faces painted green. So thereare 12× 5 = 60 such cubes. [2 marks]

6. Answer: 24. Let Ann’s age be a and Mary’s age be m, and writea1, a2 and m1,m2 be their ages at the two other times. Writea1 = a+k (so that m1 = m+k), and a2 = a+` (so that m2 = m+`.Then rewriting the statements with their ages, we have:

Ann [is a and] is four times as old as Mary was [when shewas m1 and] when Ann [was a1 and] was as old as Mary isnow [namely m]. Furthermore, Ann [is a and] is twice asold as Mary [was when she was m2 which] was when Ann[was a2 and] was six years older than Mary is now [namelym].

Thus from the given information we have:

a = 4m1 =⇒ a = 4(m+ k) =⇒ a− 4m− 4k = 0 (1)

a1 = m =⇒ a+ k = m =⇒ 4a− 4m + 4k = 0 (2)

a = 2m2 =⇒ a = 2(m+ `) =⇒ a− 2m − 2` = 0 (3)

a2 = 6 +m =⇒ a+ ` = 6 +m =⇒ 2a− 2m + 2` = 12 (4)

Firstly, we eliminate k and `:

5a− 8m = 0, (1) + (2) (5)

3a− 4m = 12, (3) + (4) (6)

∴ a = 24, 2 · (6)− (5)

Hence, Ann is 24 (and Mary is 58· 24 = 15). [3 marks]

7. Answer: 14. Say we started with b blue balls and g green balls. Sowe must find integer solutions to the equation

100b+ 72g = 2008.

Dividing through by 4 and rearranging gives

18g = 502− 25b.

We now try values of b until we find one that makes the right handside divisible by 18. This happens when b = 10 since 502−10×25 =252 = 14× 18. So he had 14 green balls. [3 marks]

8. Answer: 42. Write (4XY Z) for the ‘area of 4XY Z’ and let S =(4ABC). Then 4ABC and 4DBC have a common altitude to C.

Hence

(4DBC) = (4ABC) = S

Similarly, (4ECA) = (4DCA) = 2S

(4FAD) = (4EAD) = 4S

AM

AF=

1

2=AB

AD∠MAB = ∠FAD

∴ 4MAB ∼ 4FAD, by the Similar 4s SAS Rule

∴ (4MAB) =(4FAD)

22= S

Now S = 6 and (4DEF )− (4AMB) = 7S = 42. [3 marks]

9. Answer: 21. Say the cakes have prices a, b, c and d. The possibletotals for the price of two cakes are the 6 =

(42

)sums a + b, c + d,

a + c, b + d, a + d, b + c. Since the grandmothers all paid differentamounts the amounts they paid are 5 of these 6 sums. Notice thatthe sum of the first and second pairs is a+ b+ c+ d, so is the sumof the third and fourth, so is the sum of the fifth and sixth. Thismeans that two pairs of grandmothers’ bills have the same total.The totals are:

6 + 9 = 15

6 + 11 = 17

6 + 12 = 18

6 + 15 = 21

9 + 11 = 20

9 + 12 = 21

9 + 15 = 24

11 + 12 = 23

11 + 15 = 26

12 + 15 = 27

We see that there is only one pair of equal sums: 6 + 15 = 21 and9 + 12 = 21 so a+ b+ c+ d = 21, which is the amount I pay for my4 cakes. [4 marks]

10. Answer: 13. With x, y and z as shown, the middle strip is a paral-lelogram made up of two congruent triangles of height x and basey, and this is a third of the total area, i.e.

xy = 13x2

∴ y = 13x

By Pythagoras’ Theorem,

z2 = x2 + (x− y)2

= x2 + (23x)2

= 139x2

∴ z = 13

√13x.

Looking at the two congruent triangles making up the parallelogramanother way: they have height 1 and base z, i.e. the area of theparallelogram is also given by:

1× z = 1× 13

√13x = 1

3x2

∴√

13 = x

Thus, the area of the square is x2 = 13 cm2. [4 marks]


How to multiply on Titania

A. ∗ 0 1 2 3 4

0 0 1 2 3 41 1 1 1 1 12 2 1 0 −1 −23 3 1 −1 −3 −54 4 1 −2 −5 −8 [8 marks]

B. a ∗ 1 = a ∗ 0 + 1− a, by (ii)

= a+ 1− a, by (i)

= 1 [3 marks]

C. Answer: −19.Rearranging (ii) we have

a ∗ b = a ∗ (b+ 1)− (1− a)

Hence

(−4) ∗ (−3) = (−4) ∗ (−2)− (1−−4)

= (−4) ∗ (−1)− (1−−4)− (1−−4)

= (−4) ∗ (−1)− 2(1−−4)

= −4 ∗ 0− 3(1−−4)

= −4− 15, by (i)

= −19. [4 marks]

D. 4 ∗ a = a ∗ 4

= a ∗ 3 + 1− a= a ∗ 2 + 2(1− a)

= a ∗ 1 + 3(1− a)

= a ∗ 0 + 4(1− a)

= a+ 4− 4a

= 4− 3a. [5 marks]

E.a ∗ b = a ∗ (b− 1) + (1− a)

= a ∗ (b− 2) + (1− a) + (1− a)

...

= a ∗ (b− k) + k(1− a)

= a ∗ 0 + b(1− a)

= a+ b− ab.[6 marks]

F. Observe that

a ∗ 0 = a

= a+ 0− a · 0.

i.e. for b = 0, the formula

a ∗ b = a+ b− ab

still works. Rearranging, a ∗ (b+ 1) = (a ∗ b) + (1− a), we have

a ∗ b = a ∗ (b+ 1)− (1− a).

Suppose b = −β, where β > 0. Then

a ∗ −β = a ∗ (−β + 1)− (1− a)

= a ∗ (−β + 2)− (1− a)− (1− a)

...

= a ∗ (−β + k)− k(1− a)

= a ∗ 0− β(1− a)

= a+−β − a(−β)

= a+ b− ab, again, since in this case b = −β.

Thus, since we already showed a ∗ b = a+ b− ab, for positive b,

a ∗ b = a+ b− ab, ∀a, b ∈ Z.[6 marks]

G. Answer: (2∗3) ∗ (3∗2) = 5, (3∗3)∗3 = 513.

2∗3 = (2 ∗ 2) ∗ 2

= 0 ∗ 2 = 2, from table in A

3∗2 = 3 ∗ 3 = −3, from table in A

(2∗3) ∗ (3∗2) = 2 ∗ (−3)

= 2 +−3− 2 · −3, from rule in E

= 5

3∗3 = (3 ∗ 3) ∗ 3

= (−3) ∗ 3

= −3 + 3−−3 · 3, from rule in E

= 9

(3∗3)∗3 = (9 ∗ 9) ∗ 9

= (9 + 9− 9 · 9) ∗ 9

= (−63) ∗ 9

= −63 + 9−−63 · 9= 9(−7 + 1 + 63)

= 9 · 57 = 513

[6 marks]

H. Method 1.Using the a ∗ b = a+ b− ab rule, again and again.

a∗1 = a = 1− (1− a)

a∗2 = a+ a− a · a= 2a− a2 = 1− (1− a)2

a∗3 = (a+ a− a · a) + a− (a+ a− a · a) · a= 4a− 6a2 + 4a3 − a4

...

a∗n = na−(n

2

)a2 + · · ·+ (−1)k+1

(n

k

)ak + · · ·+ (−1)n+1an

= 1− (1− a)n

Method 2.

a∗1 = a = 1− (1− a)

a ∗ b = a+ b− ab= 1− (1 + ab− a− b)= 1− (1− a)(1− b)

∴ a∗2 = a ∗ a= 1− (1− a)(1− a) = 1− (1− a)2

a∗3 = (a ∗ a) ∗ a

= 1−(

1−(1− (1− a)2

))(1− a)

= 1− (1− a)2(1− a) = 1− (1− a)3

...

a∗(k+1) = a∗k ∗ a

= 1−(

1−(1− (1− a)k

))(1− a)

= 1− (1− a)k(1− a) = 1− (1− a)k+1

∴ a∗n = 1− (1− a)n [7 marks]




1. I’m a two digit number. I’m one less than a multiple of 8 and threeless than a multiple of seven. What is the least number I could be?

[1 mark]

2. The diagram, which is not drawn to scale,shows a rectangle divided by a horizontaland a vertical line into four rectangles. Theareas of three of them are shown. What isthe area of the whole rectangle?

8

6 9

[2 marks]

3. From a group of girls and boys, fifteen girls depart, leaving twiceas many boys as girls. Then 45 boys depart, leaving five times asmany girls as boys. How many girls were there originally?

[2 marks]

4. A regular pentagon has five diagonals and they are all of the onelength. A regular hexagon has nine diagonals and they are of twodifferent lengths.

If we consider all of the diagonals of a regular polygon which hastwenty sides, how many different lengths will there be? [2 marks]

5. There are three cards on a table, each marked with a positive wholenumber. Alice says “The sum of these two is 54”. Bill points toanother pair of cards and says “The sum of these is 41”. Finally,Cyril points to another pair and says “The sum of these two is 33”.What is the sum of all three cards? [2 marks]

6. Given that a, b, c and d are positive integers with a < 2b, b < 3c,c < 4d and d < 5, what is the largest possible value of a? [3 marks]

7. Gwen has four children, one is a teenager (13 to 19 years old) andthe product of their ages is 1848. How old is the teenager?

[3 marks]

8. An arithmetic progression is a sequence of numbers such that thedifference of any two successive numbers is a constant. For example,3, 5, 7 is an arithmetic progression of three numbers, with commondifference 2. Also, 3, 5, 7 are prime. The sum of the numbers is3 + 5 + 7 = 15, which is the lowest possible sum for an arithmeticprogression of primes of length 3.

Find an arithmetic progression of four numbers, all of which areprime, which has the lowest sum of any arithmetic progression ofprimes of length 4. What is the sum of your four numbers?

[3 marks]

9. The picture shows a board of nails on a 1 cm grid.Jane wants to put rubber bands around some ofthe nails to make squares. Two such squares areshown in the diagram (the larger square has sidelength 2 cm). How many square centimetres is thetotal area of all the squares she can make?

[3 marks]

10. For this question you must show working to all parts.Two snails, Alfa and Romeo, both set out at the same time to goalong the same road from X to Y . Alfa crawled at a constant speedof 12 m/h (metres per hour) till he reached Y . Romeo started outat 8 m/h but after two hours he realised he was falling behind sohitched a ride on a passing turtle called Toyota, who was on herway to Y at a constant speed of 20 m/h. Toyota and Romeo sooncaught up with Alfa and two hours after doing so reached Y .(a) How many hours after leaving X did it take Romeo to catch

Alfa?(b) How many hours after starting out from X did Romeo reach Y ?(c) How many metres is it from X to Y ?

[4 marks]


1. Answer: 39. The number has the form 8n − 1 and so belongs tothe set {7, 15, 23, 31,39, 47, . . . }. It also has the form 7n− 3 and sobelongs to {4, 11, 18, 25, 32,39, 46, . . . }. So it must be 39.

2. Answer: 35. The rectangles on the left have the same width, sotheir areas are proportional to their heights, namely 8 to 6. Therectangles on the right also have areas proportional to their heights,so the unknown area is 9 × 8/6 = 12, and the area of the wholerectangle is 6 + 9 + 8 + 12 = 35.

3. Answer: 40. Say there were g girls and b boys originally. Then

b = 2(g − 15), after the first departure (1)

g − 15 = 5(b− 45), after the second departure (2)

∴ g − 15 = 5(2(g − 15)− 45

), substituting for b from (1) in (2)

∴ 5× 45 = 9(g − 15)

5× 5 = g − 15

g = 40.

4. Answer: 9. If we think about the diagonals starting at vertex num-ber 1 we see that they increase from the shortest diagonal, fromvertex 1 to 3, to the longest, from vertex 1 to vertex 11, then de-crease again, in a symmetric fashion, down to the diagonal fromvertex 1 to vertex 18. Thus there are 9 different lengths.

5. Answer: 64. If the numbers on the cards are a, b and c then we getthe system of equations

a + b = 54 (3)

a + c = 41 (4)

b + c = 33 (5)

∴ 2(a + b + c) = 54 + 41 + 33 = 128, adding (3), (4) and (5)

∴ a + b + c = 64.

6. Answer: 87. The largest possible value of d is 4 = 5 − 1. So thelargest possible value of c is 15 = 4× 4− 1. So the largest possiblevalue of b is 44 = 3 × 15− 1. Hence the largest possible value of ais 87 = 2× 44− 1.

7. Answer: 14. We have the factorisation 1848 = 2×2×2×3×7×11.The only combination of factors yielding a ‘teen’ is 2 × 7, so theteenager is 14. (There is more than one possibility for the otherthree ages!)

8. Answer: 56. The arithmetic progression 5, 11, 17, 23 consists ofprimes, has sum 5 + 11 + 17 + 23 = 56 and has common difference6. There is no arithmetic progression of length 4 starting at 3 andwith common difference 2, 4 or 6. There is none starting at 5 withcommon difference 2 or 4. If there is one starting at 7 with smaller

sum it must have common difference 2 or 4, but no such progressionsexist. There are none starting at 11 with difference 2 or 4, nor 13.Since 56/4 = 14, there can’t be an arithmetic progression withsmaller sum starting with a prime greater than 13. Thus 5, 11, 17,23 has the smallest sum (56) of an arithmetic progression of primesof length 4.

9. Answer: 52. There are nine 1 × 1 squares with area 1, four 2 × 2squares with area 4, one 3 × 3 square with area 9, four

√2 ×

√2

squares with area 2 and two√

5 ×√

5 squares with area 5. So thetotal area is 9× 1 + 4× 4 + 1× 9 + 4× 2 + 2× 5 = 52.

10. (a) Romeo covers 16 m in the first two hours. Suppose Romeoand Toyota overtake Alfa t hours later. Then they have travelled16 + 20t m. meanwhile, Alfa has travelled 12(2 + t) m. Hence

16 + 20t = 24 + 12t.

So t = 1. Hence it took a total of 3 hours to catch Alfa.(b) Romeo travelled 2 hours on his own and 3 hours on Toyota’sback, a total of 5 hours.(c) Romeo travelled 16 m on his own and 60 m on Toyota’s back, atotal of 76 m.




A teacher has a class of 100 students whom he numbers from 1 to 100.He has given each a t-shirt showing his or her number, and taken themto a very long corridor with 100 doors. The doors are also numberedfrom 1 to 100. When a student goes down the corridor for each doorwhose number is divisible by the number on his/her t-shirt, he/sheeither closes the door if it is open or opens the door if it is closed.Students don’t touch doors with numbers that are not divisible bytheir t-shirt numbers.

For example, if the doors are all closed to start with then whenstudent 40 goes down the corridor he will open doors 40 and 80. Ifstudent 80 follows him she will close door 80.

A. If all the doors are closed, and students 7, 28 and 84 go down thecorridor, which doors will be open?

B. All the doors are closed except number 42. Which students shouldhe send down to get all the doors closed?

C. Suppose all the doors are closed and all the students go down thecorridor. Which of the doors 49, 51 and 53 will be open?

D. If all the doors are closed, and students 1, 2, 4, 8, 16, 32 and 64 godown the corridor, explain how you can predict which doors willbe closed?

E. If all the doors are closed, and all the students go down the corridor,which doors will be open? Explain your answer – not just by sayingwhich doors are open and which are closed.

F. The doors from 1 to 49 are closed but somebody has left all theothers open. Which students should he send down to ensure all thedoors are closed?

G. Now the doors from 1 to 49 are open and the rest are closed.Whichstudents should he send down to get all the doors closed?

H. All the doors were closed but the Number 1 student has just walkeddown the corridor and gone home. The teacher now has no wayof closing door number 1. But is it possible to use the remainingstudents to close all the other doors? Explain your reasoning.


A. 7, 14, 21, 35, 42, 49, 63, 70, 77, 84, 91, 98. [4 marks]

B. 42 and 84. [4 marks]

C. Only 49 will be open, because 49 has 3 factors, i.e. it is moved bystudents 1, 7, 49; and 51 and 53 will be closed because they have 4and 2 factors, respectively. [6 marks]

D. Doors 2, 6, 8, 10, 14, 18, 22, 24, 26, 30, 32, 34, 38, 40, 42, 46, 50,54, 56, 58, 62, 66, 70, 72, 74, 78, 82, 86, 88, 90, 94, 96, 98 will beclosed. To decide whether a given door is closed or not find thehighest power of 2 which divides the door number. If this is an oddpower of 2 (2, 8 or 32) then the door will be closed. Otherwise itwill be open. [7 marks]

E. The doors 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100 be open becausethese are perfect squares and such numbers have an odd numberof factors, and so will be opened or closed by an odd number ofstudents. [6 marks]

F. 50, 51, 52, . . . , 99 (not 100). [6 marks]

G. 1 and 50, 51, 52, . . . , 99. [4 marks]

H. Yes, send down the student whose number is that of the first opendoor, and continue doing this till all doors other than 1 are closed.This is the same as sending all students except those with squarenumbers on their shirts. [8 marks]




1. The diagram below shows the train line from the outer suburb of Ato the inner city station E, with the distance between stations shownin kilometres. After leaving a station the train travels at an averagespeed of 30 km/h for the first kilometre, then at 60 km/h until itreaches the next station. It spends 2 minutes at each station. Howmany minutes will elapse between the train leaving A and arrivingat E?

A B C D E

3 km 4 km 6 km 1 km

[1 mark]Solution. 60 km/h = 1 km/minute. So when the train travels at60 km/h, it takes 1 minute to cover a km, and when it travels at30 km/h, it takes 2 minutes to cover a km. So the minutes that willelapse between the train leaving A and arriving at E is given by:

(2 + 2) + 2 + (2 + 3) + 2 + (2 + 5) + 2 + (2 + 0) = 24

where the first number (2) in each bracket is the time spent at30 km/h, the second number in each bracket is the number of kilo-metres travelled at 60 km/h (which equals the time in minutes tocover that distance) and each unbracketed 2 is the waiting time ata station. Answer: 24.

2. Find the two-digit prime number that is 2 less than a perfect squarethat is 2 less than a prime. [1 mark]

Solution. Let the prime be p. Then p has 2 digits and hence isodd (2 is the only even prime). It is 2 less than a square. So thesquare is odd; possibilities are: 25, 49, 81. Of these only 81 hasa prime both 2 less than it and 2 more than it. So p must be 79.Answer: 79.

3. For how many positive integers (whole numbers) n are n, 3n andn/3 all three-digit integers? [2 marks]

Solution. The least that n/3 can be is 100 (the smallest 3-digitpositive integer) and the most that 3n can be is 999 (the largest3-digit positive integer). If 3n = 999 then n/3 = 111. So n/3 canbe any integer from 100 to 111 inclusive, a total of 12 possibilities,leading to 12 possibilities for n (300, 303, . . . , 333). Answer: 12.

4. A point E lies outside a square ABCD so that ABE is an equilateraltriangle. What is the measure in degrees of the angle CED?

A B

CD

E

[2 marks]Solution. Since 4ABE is equilateral, its angles are all 60◦. Hence∠EAD = 60◦ + 90◦. Now 4ADE is isoceles. Thus

∠AED =12(180◦ − ∠EAD)

= 15◦

Similarly, ∠BEC = 15◦. So finally we have

∠CED = 60◦ − 2× 15◦ = 30◦.

Answer: 30.

5. A video store has a choice of 920 films to rent. You can rent someon DVD, some on video cassette and some on both. If the storeowns a total of 1000 DVDs and video cassettes, how many films areavailable on both video cassette and DVD? [2 marks]

Solution. By the pigeon hole principle, 80 films are available onboth video cassette and DVD (the films are the pigeon holes; put920 of the 1000 DVDs and video cassettes in the 920 available pigeonholes; the remaining 80 DVDs or video cassettes must go in a pigeonhole that already contains a DVD or video cassette). Answer: 80.

6. Brenda was sick on the day of the maths test so she had to sit forit the next day. Her score of 96 raised the class average from 71 to72. How many students (including Brenda) took the test?

[3 marks]Solution. Let n be the number of students in the class. WithoutBrenda the average is 71 and hence the total marks without Brendais 71(n− 1). The average with Brenda’s mark of 96 is 72. So

72 =71(n− 1) + 96

n72n = 71(n− 1) + 96 = 71n− 71 + 96

n = 25

So there are 25 students in the class. Answer: 25.

7. A plane flies in still air at an average speed of 810 km/h for theduration of its flights. When flying from Perth to Sydney it takesfour hours while from Sydney to Perth it takes five hours. Assumingthat the wind is at a constant speed and from the west (i.e. in thedirection from Perth to Sydney) for both flights, what is the speedof the wind? [3 marks]

Solution. Let w be the speed of the wind (from west to east). Thentravelling east the plane’s speed is (810 + w) km/h and travellingwest its speed is (810 − w) km/h. In general velocity v, distance dand time t are related by v = d/t. Let d now denote the distancefrom Perth to Sydney. Then

810 + w =d

4(1)

810− w =d

5(2)

Eliminating d by dividing (1) by (2) we have

810 + w

810− w=

54

4(810 + w) = 5(810− w)4w = 810− 5w

9w = 810w = 90

Therefore the wind speed w is 90 km/h. Answer: 90.

8. The five faces of a right square pyramid all havethe same area. The height of the pyramid is 3 m.What is its total surface area in square metres?

��

��

��

6

?

3 m

[3 marks]Solution.Let a be the sidelength of the square base, so that thearea of the base is a2 which is also the area of eachisosceles triangular face. So each triangular face hasheight 2a, and the total surface area is 5a2. UsingPythagoras’ Theorem and the pyramid height of 3 m,we have

(2a)2 − (a/2)2 = 32

(4− 14)a2 = 15

4 a2 = 34 · 5a2 = 9

5a2 = 43 · 9 = 12

So the total surface area of the pyramid is 12 m2.Answer: 12.

��

��

6

?

3 m

9. Jasper glues together 504 cubes with 1 cm edges to make a solidrectangular-faced brick. If the perimeter of the base of the brick is64 cm, what is the height of the brick? [4 marks]

Solution. 504 = 9 × 56 = 9 × 7 × 8 = 23 × 32 × 7. The volumeof the brick L × B × H = 504 cm3, whereas the perimeter of thebase 2(L + B) = 64 cm. So we need to write 504 as the product ofthree positive integers L,B,H such that L + B = 32. We see that18 × 14 × 2 = 504, so that (in cm) L = 18, B = 14, H = 2 is asolution.To see the solution H = 2 is unique (not required for the competi-tion), observe first that L must satisfy 16 ≤ L < 32 (taking L ≥ B).Now L and B must both be even or both odd since there sum iseven. If L and B are both odd then their product is 63 leading toL = 21 (to satisfy the inequality) and B = 3 which do not sum to32. Hence L and B must both be even, whence H = 1 or 2. IfH = 1 then the least L + B could be is 2

√504 > 2× 22 = 44 > 32.

Thus H = 2 cm. Answer: 2.

10. In the AFL Grand Final between the Knockers and the Beagles theKnockers won by 3 points. This was despite the Beagles kicking16 scoring shots compared to the Knockers 14. It was also noticedthat the number of behinds kicked by the Beagles was greater thanthe number of goals kicked by the Knockers, and the number ofgoals kicked by the Beagles was greater than the number of behindskicked by the Knockers. How many points did the Beagles score?

Note: in the AFL game there are two ways to score: behinds score1 point each, and goals score 6 points each.

For full marks, explain how you found your solution. [4 marks]Solution.

Let . . . B = the number of goals kicked by the Beagles,b = the number of behinds kicked by the Beagles,

K = the number of goals kicked by the Knockers, andk = the number of behinds kicked by the Knockers.

From the given information, we have:

B + b = 16 (3)

K + k = 14 (4)

6K + k = 6B + b + 3 (5)

B > k (6)

b > K (7)

Since B and b are integers, from (6) and (7) we have

B ≥ k + 1 (8)

b ≥ K + 1 (9)

Adding the inequalities (8) and (9) and rearranging we get

(B + b)− (K + k) ≥ 2.

But subtracting (3) and (4) gives

(B + b)− (K + k) = 2.

So the inequalities (8) and (9) must actually be equalities:

B = k + 1 (10)

b = K + 1 (11)

Substituting (10) in (3) and rearranging we have

B = 15−K (12)

Rearranging (4) we have

k = 14−K (13)

Substituting (11), (12) and (13) in (5) we have

6K + 14−K = 6(15−K) + K + 1 + 310K = 80

K = 8 (14)

Substituting (14) in (12) and (11), we have:

B = 7 (15)

b = 9 (16)

so that the Beagles scored

6B + b = 51points

Note the problem can also be solved by systematically listing thepossible scores for the Beagles (17 of them) and the possible scoresfor the Knockers (15 of them) and then doing a careful elimination. . . and there are many other ways. The more satisfying solutionsare ones that set up some inequality at the beginning that reducethe possibilities to a small list prior to doing an elimination.Answer: 51.




Consider a garden table made of 15 square tilesin a 5× 3 arrangement.The table has a straight crack along a diagonal.Seven of the individual tiles are broken. "

""

""

""

""

"

Now consider a 6× 4 rectangle.This time eight tiles are broken.

��

��

��

��

��

��

Try some other sizes. Use the squared paper provided.

A. How many tiles get broken when an 8× 6 table is cracked along adiagonal?

Solution. 12 tiles get broken.

B. Give the dimensions of two different rectangular tables that getnine tiles broken when they are cracked along a diagonal.Note. Remember that a square is also a rectangle. Also, note thatfor this and subsequent questions, an 8 × 6 table, for example, isconsidered the same as a 6× 8 table.

Solution. The possible dimensions are: 9× 1, 7× 3, 9× 3, 9× 9.

C. How many different rectangular tables can you find that get tentiles broken when they are cracked along a diagonal? Write downtheir dimensions.

Solution. The possible dimensions are: 10 × 1, 10 × 2, 10 × 5,10× 10, 9× 2, 8× 3, 7× 4, 6× 5.

D. Try some square tables. Describe what happens.Solution. The number of broken tiles is the same as the length ofthe side of the square.

E. What happens when the shorter dimension of the table is 1?Solution. Every tile is broken, i.e. the number of tiles is brokenis the length of the rectangle.

F. For what sort of dimensions does the crack go through corners oftiles inside the rectangle?

Solution. For dimensions that have a common factor (larger than1).

G. How many tiles are cracked when the diagonal does not go throughany corner of a tile inside the rectangle? Explain your reasoning.

Solution. If the diagonal does not go through any corner of atile inside the rectangle, a tile is cracked when and only when thediagonal enters a new column or enters a new row.Say the table has m rows and n columns. The first tile broken isin the first row and the first column. Then there are m− 1 furtherrows and n − 1 further columns. Therefore the number of tilescracked is 1 + (m− 1) + (n− 1) = n + m− 1.

H. Predict the number of broken tiles in a 56× 32 rectangle.Solution. Remove the highest common factor 8. Then consider a7×4 rectangle which has 7+4−1 = 10 broken tiles. Reintroduce theremoved common factor and the number of broken tiles is 10×8 =80.Alternatively: In general, the number of cracked tiles in a tablewith dimensions m×n is m+n−hcf(m,n) where hcf(m, n) countsthe number of times the diagonal enters a new column and newrow at the same time, which, if we think of the ‘first’ tile brokenas being the bottom lefthand tile, is the number of times the crackgoes through the bottom lefthand corner of a tile. So for a 56× 32rectangle, 56+32−hcf(56, 32) = 56+32− 8 = 80 tiles are broken.

I. Explain how you can predict the number of broken tiles in any sizeof table.

Solution. If the two dimensions have no common factor, add thetwo numbers and subtract 1. If the two dimensions do have acommon factor, remove the highest common factor), consider thereduced table as above, then multiply the result by the removedHCF. Algebraically, the number of tiles broken is( m

hcf(m,n)+

n

hcf(m,n)− 1

)hcf(m,n) = m + n− hcf(m,n).

Alternatively: use the general argument in H to obtain the gen-eral formula m + n− hcf(m,n).




(1) ABCD is a trapezium with AB parallel to DC, AB = 4 cm,DC = 11 cm and the area of triangle ABP is 12 cm2. What isthe area of the trapezium ABCD in square centimetres?

4 cm

12 cm2

A B

D CP

��S

SS

SS

SS

S��c

cc

cc

cc

cc

c

C

� -11 cm

[1 mark]

(2) A library has 6 floors. There are 10 000 more books on thesecond floor than the first. The number of books on the thirdfloor is the same as the number on the second. There are 10 000fewer books on the fourth floor than the third and twice as manybooks on the fifth floor as there are on the fourth. On the sixthfloor there are 4 000 fewer books than on the fifth. Coinciden-tally the number of books on the sixth floor is the same as thenumber on the first. Altogether, how many thousands of booksare there in the library?

[1 mark]

(3) XY Z is a three-digit number. Given that

XXXX + Y Y Y Y + ZZZZ = Y XXXZ,

what is X + Y + Z?[2 marks]

(4) Mathilda and her friends have 20 1-metre sticks out on thebasketball court to make triangles. Below is a diagram of atriangle using all 20 sticks. It has sides of 5 m, 7 m and 8 m. Alltriangles with sides of 5 m, 7 m and 8 m (even mirror images asshown) are considered to be the same.

8 m 8 m

5 m 5 m7 m 7 m

Altogether, how many different triangles could they make, oneat a time, each time using all 20 sticks?

[2 marks]

(5) What is the size of the angle, in degrees, between the hands ofa clock when the time is ten past eleven?

[2 marks]

(6) A plane is due to leave Perth at midnight and arrive at Tokyoat 12:29 pm, but it departs 49 minutes late. What percentageincrease in speed is required in order that it will arrive exactlyon time?

[3 marks]

(7) Esther has 20 coins in her purse. They are 10c, 20c and 50ccoins and the total value is $5. If she has more 50c coins than20c coins, how many 10c coins has she?

[3 marks]

(8) Alice spent all her money in five shops. In each shop, she spent$1 more than half of what she had when she entered that shop.How many dollars did Alice have when she entered the firstshop?

[3 marks]

(9) Given a square ABCD, circular arcs centred at B and D aredrawn from A to C. Now draw diagonal BD to cut these arcsat X and Y , respectively. If XY = 12− 6

√2, what is the area

of the square ABCD?

A B

CD

X

Y

��

��

��

��

��

��

[4 marks]

(10) Farmer Brown runs a dairy farm with cows, sheep and goats.Dabbling in mathematics in his spare time, he noticed that thenumbers of each animal were different prime numbers. He alsoobserved that if he multiplied the number of cows by the to-tal number of cows and sheep, he obtained a number just 120greater than the number of goats. How many goats are there?

For full marks, explain how you found your solution.[4 marks]




A number appears on your computer screen. If you push the A keythe number is decreased by 3, if you push the B key it’s increased by3 and if you press the C key the number is halved. You want to use asequence of key strokes to change the number to 1.

For example, there are three ways to change 16 to 1:

• Use CCCC giving the number sequence 16, 8, 4, 2, 1.• Use AAAAA giving 16, 13, 10, 7, 4, 1.• Use CCA giving 16, 8, 4, 1.

So the shortest path from 16 to 1 requires 3 key strokes.

A. Find a starting number between 30 and 40 whose shortest path is4, and show the path.

B. Find a starting number between 50 and 60 whose shortest path is5, and show the path.

C. Find a starting number between 30 and 40 whose shortest path is6, and show the path.

D. Which starting number between 40 and 50 has the longest “shortestpath”?

E. Find a number which has two shortest paths, one beginning withthe A key and one with the B key.

F. Find a starting number where the “add 3” operation must be usedto get the shortest path.

G. What starting numbers can never get to 1?

H. What starting number between 500 and 1000 would have the short-est path?

I. In this question, the numbers on the screen don’t have to be inte-gers. If we start with any number then these two sequences of keystrokes always produce the same final number: ACCB and BCBC.Explain why.

Solutions

Solutions to Individual Questions

(1) Let h be the height of the trapezium. This is also the heightof triangle ABP . Thus 12 = 1

2× 4h so h = 6. The area of the

trapezium is then

Areatrap = h× AB + DC

2= 6× 4 + 11

2= 45.

(2) Let there be x thousand books on the first floor.On the second floor there are x + 10 thousand books.On the third floor there are x + 10 thousand books.On the fourth floor there are x thousand books.On the fifth floor there are 2x thousand books.On the sixth floor there are 2x− 4 thousand books.Thus we have 2x − 4 = x and so x = 4. The total number ofbooks, in thousands, is therefore

x + x + 10 + x + 10 + x + 2x + x = 7x + 20

= 7× 4 + 20 = 48.

(3) Writing the sum in the traditional way,

XXXXY Y Y YZ Z Z Z

Y XXXZ

we observe from the last column that X + Y + Z = Z + 10k,where k is the carry to the second column, and so X +Y = 10k.Now X + Y can’t be as big as 20, and, since XY Z is a 3-digitnumber, X 6= 0. Thus

X + Y = 10 (1)

and the carry k = 1. Looking at the second column now, wehave X + Y + Z + 1 = X + 10m, i.e. Y + Z + 1 = 10m, wherem is the carry to the third column. Again, Y + Z + 1 can’t beas big as 20, so

Y + Z + 1 = 10 (2)

and m = 1. We see that this pattern is repeated for the thirdand fourth columns, and so the carry to the fifth column whichis just Y , implies Y = m = 1. Adding equations (1) and (2) wehave

X + Y + Y + Z + 1 = 20

X + Y + Z = 20− 1− Y

= 18

(Of course, we had enough information to evaluate that X = 9and Z = 8.)

(4) We’re looking for sets of three positive integers x, y, z satisfyingx + y + z = 20 and (so that we don’t count the same set twice)x ≥ y ≥ z. With x the length of the hypotenuse, we alsoneed x > y + z; otherwise we can’t form a triangle. Listingsystematically we get:

9,9,29,8,39,7,49,6,58,8,48,7,58,6,67,7,6.

So there are 8 possible triangles.(5) The minute hand is 1/6 of a revolution past the 12, which is 60

degrees and the hour hand is 5/6 of the angle between 11 and12 which is 5/6 = 1− 1/6 of 1/12 of a revolution before the 12,i.e. 25 degrees, so the angle between the hands is 85 degrees.

(6) The time usually taken for the trip is 12×60+29 = 749 minutes.In order to arrive on times the pilot must do the trip in 749−49 = 700 minutes. Let the distance between Perth and Tokyo bed kilometres. The usual speed is therefore d/749, the requiredspeed after the late departure is d/700. The required percentageincrease in speed is therefore,

d/700− d/749

d/749× 100% = 749

( 1

700− 1

749

)× 100%

=(749

700− 749

749

)× 100%

= (1.07− 1)× 100%

= 7%.

So the required answer is 7.(7) Let the number of 10c, 20c and 50c coins be x, y and z respec-

tively. We have,

x + y + z = 20 (3)

10x + 20y + 50z = 500 (4)

and z > y. Dividing (4) by 10 and subtracting (3) gives

y + 4z = 30. (5)

Since z > y this means 5z > 30 and so z ≥ 7. If z ≥ 8 the lefthand side of (5) is at least 32, which is impossible, so z = 7 andwe quickly find y = 2 and x = 11. So the number of ten centcoins is 11.

(8) In each shop she spends 2 more then she has when she leaves.So if she leaves with y dollars then she spent y + 2 and enteredwith 2y + 2 dollars.She leaves the last shop with $0, so (using y = 0), she enteredshop 5 with $2.She left shop 4 with $2, so entered it with 2× 2+2 = 6 dollars.She left shop 3 with $6, so entered it with 2×6+2 = 14 dollars.She left shop 2 with $14, so entered with 2×14+2 = 30 dollars.She left shop 1 with $30, so entered with 2×30+2 = 62 dollars.So Alice had 62 dollars when she entered the first shop.

(9) Let the side-length of the square be s. Now BX = DY = s,and, by Pythagoras’ Theorem, BD = s

√2. Therefore

XY = BX + DY −BD = (2−√

2)s.

But we’re told XY = 12 − 6√

2 = 6(2 −√

2). Thus s = 6 andthe square has area 36.

(10) Let the number of cows, sheep and goats be c, s and g respec-tively. The farmer notices that

c(c + s) = g + 120.

We know c, s and g are primes. If they were all odd then theleft hand side of the above equation would be even and the rightodd which is impossible. So one of them is even and thereforeequals 2.If c = 2 the left hand side of the equation is still even and theright hand side odd.If g = 2 the equation becomes c(c + s) = 122 = 2× 61. This isimpossible with c odd.Therefore we must have s = 2 and get

c2 + 2c = g + 120

c2 + 2c + 1 = g + 121

(c + 1)2 = g + 112

g = (c + 1)2 − 112

= (c + 1 + 11)(c + 1− 11)

Since g is a prime the first of the factors here must equal g andthe other equal 1:

g = c + 1 + 11 (6)

1 = c + 1− 11 (7)

From (7), we obtain c = 11; whence from (6) we deduce g = 23.

Solutions to Team Questions

(A) There is just one path with a starting number between 30 and 40such that the shortest path from that starting number has length4:

32, 16, 8, 4, 1

The sequence of operations giving this path is CCCA. An impor-tant observation here is that C effects a more rapid descent fornumbers x that are even and greater than 4, but A requires justone step to convert 4 to 1 (compared with performing CC).

(B) There is just one path with a starting number between 50 and 60such that the shortest path from that starting number has length5:

56, 28, 14, 7, 4, 1

The sequence of operations giving this path is CCCCA.(C) There are three paths with a starting number between 30 and 40

such that the shortest path from that starting number has length6:

37, 34, 17, 14, 7, 4, 1 (by sequence: ACACAA) or

37, 40, 20, 10, 7, 4, 1 (by sequence: BCCAAA) or

37, 40, 20, 10, 5, 2, 1 (by sequence: BCCCAC)

Thus all such paths start at 37. Thus we can see that a shortestpath can involve the B operation.

(D) The starting number between 40 and 50 that has the longest“shortest path” is 49, with paths

49, 46, 23, 20, 10, 7, 4, 1 (ACACAAA) or

49, 46, 23, 20, 10, 5, 2, 1 (ACACCAC) or

49, 52, 26, 13, 10, 7, 4, 1 (BCCAAAA) or

49, 52, 26, 13, 10, 5, 2, 1 (BCCACAC) or

49, 52, 26, 13, 16, 8, 4, 1 (BCCBCCA)

all of length 7. The path is not required.(E) Starting numbers that have two shortest paths, one beginning with

the A key and one with the B key, that are less than 100 are 13,25, 37, 41, 49, 65, 73, 89, 97.

(F) Starting numbers where the “add 3” operation must be used toget the shortest path, that are less than 100 are 29, 53, 58, 61, 77,85.

(G) The starting numbers that can never get to 1 are multiples of 3.(The way to see this is that if a number starts as a multiple of 3,then after any of the A, B or C operations is performed, the resultis still a multiple of 3, and of course 1 is not a multiple of 3.)

(H) The starting number between 500 and 1000 that has the shortestpath is 512.

(I) We need to show that ACCB and BCBC always produces thesame result. Let x be the starting number. Then ACCB produces(

(x− 3)/2)/2 + 3 =

x− 3

4+ 3

=x− 3 + 12

4=

x + 9

4On the other hand, BCBC produces(

(x + 3)/2 + 3)/2 =

(x + 3

2+ 3

)/2

=(x + 3 + 6

2

)/2

=x + 9

4Thus irrespective of what starting number x is chosen. The resultafter each of the sequences ACCB and BCBC is (x + 9)/4.



General instructions: No working need be given for Questions 1 to9. Calculators are not permitted. Write your answers on the answersheet provided. Each solution is a positive integer less than 100.

(1) Paul likes dogs. At present all his adult dogs are dalmationswhile some of his puppies are dalmations and some are not. Inall he has 11 dogs of which 7 are dalmations and 8 are puppies.How many dalmation puppies has he?

[1 mark]

(2) If a hen and a half lay an egg and a half in a day and a half,how many eggs will 6 hens lay in 12 days?

[1 mark]

(3) Triangles ABC, ACD, ADE, AEF , AFG are right-angled withhypotenuses AC, AD, AE, AF , AG respectively. If AB = 2and BC = CD = DE = EF = FG = 3, what is the length ofAG?

A B

C

DE

F

G

[2 marks]

(4) How many perfect squares divide 7200 exactly?[2 marks]

(5) How many 5 digit numbers consisting only of 1s and 2s have noadjacent 2s?

[2 marks]

(6) In a circle with centre O, a chord AB is extended to a pointC such that the length of BC is equal to the radius of thecircle. CO is drawn and extended to a point D on the circle’scircumference. If angle BCO = 15◦, what is angle AOD?

[3 marks]

(7) A chessboard is made up of 64 squares each with 3 cm sides. Acircle is drawn with its centre at the centre of the chessboardand radius 12 cm. How many of the 64 squares lie entirely insidethe circle?

[3 marks]

(8) There was a young lady called ChrisWho when asked her age answered this:“Two thirds of its squareIs a cube, I declare.”So what was the age of this Miss?

[3 marks]

(9) Will has 12 square tiles. Using all the tiles each time, he canmake three different shaped rectangles, like this:

What is the least number of tiles he needs, so that, using allthe tiles each time, he can make five different rectangles?

[4 marks]

(10) There was little traffic that day, too little to interfere with thesteady progress of the 3 km column of armoured vehicles. Head-ing the column, Tom turned his jeep and drove back to checkthe rear. All was well and he was able to maintain a steadyspeed there and back without any delays. On returning to hislead position, Tom noted that the column had advanced just4 km while he was away. How far had he driven in that time?Give reasons for your solution. [4 marks]



A. On Genevieve’s computer screen there are two dials, as shownbelow. s s

0

1

23

4

01

2

345

6

7

8

Each time she clicks her mouse the black dot on each dial movesone position clockwise. So that if she clicked her mouse 12 timesthe dot on the left hand dial would be in position 2 and the doton the right hand dial would be in position 3. What is the leastnumber of clicks she needs to make to get from the position aboveto one with the left dot at 1 and the right at 8?

B. On Ahmed’s computer the dots start in the positions shown be-low.

s s0

1

23

4

01

2

345

6

7

8

How many clicks must he make to get the left dot at 1 and theright at 4?

C. The left hand dial on Nelson’s computer is numbered from 0 to16 and the right dial from 0 to 23. Initially the left dot is atposition 9 and the right at position 11. How many clicks doesNelson need to make to get the left hand dot in position 7 andthe right at position 15?

D. Yenicca’s screen has 3 dials, which start in the positions shownbelow.

s

0

1

23

4

s0

1

23

4

5s 0

1

2

34

5

6

When she clicks her mouse each dot moves one position clockwise.How many clicks will she need to make to finish with the lefthand dot at position 0, the middle at position 2 and the right atposition 4?

E. Roxanne’s screen also shows 3 dials which start in the same posi-tion as Yenicca’s. However when she clicks the left mouse buttononly the left hand and middle dots move one position clockwise.If she clicks the right mouse button the middle and right handdots move one position clockwise. How many times must sheclick each button to finish with the left hand dot at position 0,the middle at position 2 and the right at position 4?

Solutions

Solutions to Individual Questions

(1) Since Paul has 11 dogs and 8 puppies he must have 3 adult dogs.We are told all his adult dogs are dalmations; so he has 3 adultdalmations. The rest of his 7 dalmations must be puppies; so hehas 4 dalmation puppies. (This question can also be answeredusing Venn diagrams.) [1 mark]

(2) If a hen and a half lay an egg and a half in a day and a half,then 6 hens will lay 6

1.5× 1.5 = 6 eggs in a day and a half. So

in 12 days they lay 121.5× 6 = 48 eggs. [1 mark]

(3) In order to avoid confusion about the meaning of expressionslike AC2, we use the convention in our proof that |AC| denotesthe length of the line segment AC.

By Pythagoras’ Theorem (several times)

|AC| =√

22 + 32 =√

13

|AD| =√

(√

13)2 + 32 =√

22

|AE| =√

(√

22)2 + 32 =√

31

|AF | =√

(√

31)2 + 32 =√

40

|AG| =√

(√

40)2 + 32 =√

49 = 7

So the length of AG is 7.

Alternative solution. More elegantly (again using Pythago-ras’ Theorem several times), we have

|AG|2 = |FG|2 + |AF |2

= |FG|2 + |EF |2 + |AE|2

...

= |FG|2 + |EF |2 + |DE|2 + |CD|2 + |BC|2 + |AB|2

= 5× 32 + 22

= 49

∴ |AG| = 7

So again the length of AG is 7. [2 marks]

(4) 7200 = 2 × 3600 = 2 × 602. From this we see that the squareof any divisor of 60 is a square divisor of 7200, and vice versa.So we need only count the divisors of 60. These are 1, 2, 3, 4,5, 6, 10, 12, 15, 20 and 60. So 12 perfect squares divide 7200exactly. [2 marks]

(5) There is 1 number of the given shape that contains no 2, 5which contain one 2, 6 which contain two 2s, and 1 (21212)which contains three 2s. Altogether there are 1+5+6+1 = 13such numbers. [2 marks]

(6) Triangle OBC is isosceles, so angle BOC is 15◦. Angle ABO isan external angle of the triangle OBC and so equals 2× 15◦ =30◦. Triangle AOB is also isosceles; so angle BAO also equals30◦.

A

BC

D

O

Now angle AOD is an external angle to triangle OAC, and so∠AOD = ∠BAO + ∠BCO = 30◦ + 15◦ = 45◦. [3 marks]

(7) The main question is to decide whether the points shown assolid dots are inside or outside the circle. Their distance fromthe centre is

√32 + 32 =

√18, by Pythagoras’ Theorem, and√

18 >√

16 which is the radius of the circle; so the points areoutside the circle.

× × ×× × ×× ×

t t

t t

Hence the squares in the first quadrant of the circle that aremarked × all lie inside the circle, and so in all there are 4×8 =32 squares inside the circle. [3 marks]

(8) Let her age be x and the cube be y3. (Both x and y are integers.)Then

23x2 = y3.

Hence

2x2 = 3y3 (1)

Now 2 divides the LHS . . . so 2 divides the RHS and hence 2divides y. Thus 8 divides the RHS. So 2 divides x.Now try a similar idea with 3: 3 divides the RHS . . . so 3 dividesthe LHS and hence 3 divides x. Thus 9 divides the LHS (andhence the RHS). So 3 divides y. So 81 divides the RHS andhence 9 divides x.Thus lcm(2, 9) = 18 divides x and lcm(2, 3) = 6 divides y. Nowlet x = 18α and y = 6β. Then substituting in (1) we get:

2× 182α2 = 3× 63β3

which reduces to

α2 = β3

Suppose for a prime p, pi is the largest power of p that dividesthe LHS. Then i is a multiple of 2. Also since pi is the largestpower of p that divides the RHS, i is a multiple of 3. Thus i isa multiple of 6. Either i is 0 for every prime p or i is at least 6for some prime p. If i is at least 6 for some p then at least p3

divides α in which case α ≥ 23 = 8 and x ≥ 8 × 18 = 144 andby today’s standards Chris would not be a young lady and acandidate for the Guinness Book of Records. Thus α = 1 andx = 18. So Chris is 18. [3 marks]

(9) The question is essentially asking for the least number that canbe written as the product of two numbers in 5 different wayswithout regard to order. The answer is 36. He can then makerectangles with dimensions 1 × 36, 2 × 18, 3 × 12, 4 × 9 and6 × 6. One way that we could have found this is by observingthat if the number of tiles n has prime factorisation

pε11 pε2

2 · · · pεnn

then each factor of n has the form

pi11 pi2

2 · · · pinn

where 0 ≤ ij ≤ εj, for j = 1, . . . , n. Hence n has

(ε1 + 1)(ε2 + 1) · · · (εn + 1)

distinct factors. So we require (ε1 + 1)(ε2 + 1) · · · (εn + 1) to beeither 2× 5− 1 = 9 if n is a perfect square, or 2× 5 = 10 if nis not a perfect square.

Now 10 = 2 × 5 suggests ε1 = 4, ε2 = 1; the least n = p41p2

occurs for p1 = 2, p2 = 3, namely n = 48. However, if n is aperfect square, in which case the εj must all be even, we seethat 9 = 3 × 3 suggests ε1 = 2, ε2 = 2, and the least n = p2

1p22

again occurs for p1 = 2, p2 = 3, and is n = 36. [4 marks]

(10) Let v be the jeep’s speed and u be the column’s speed (inkm/hr), let T be the total time of Tom’s round trip (in hrs),and let x be the total distance Tom travelled (in km). Then

T =3

v + u+

3

v − u, uT = 4, vT = x

Rearranging the last two equations we obtain u = 4/T , v =x/T (noting that this is allowed since T cannot be zero). Nowsubstitute those expressions for u and v in our first equation:

T =3

x

T+

4

T

+3

x

T− 4

T

=3T

x + 4+

3T

x− 4

Dividing both sides by T (which is non zero), followed by rear-ranging we get:

1 =3

x + 4+

3

x− 4

(x + 4)(x− 4) = 3((x− 4) + (x + 4)

)x2 − 16 = 6x

x2 − 6x− 16 = 0

(x− 8)(x + 2) = 0.

So x is 8 or −2 . . . but x cannot be negative. Hence Tom drove8 km.

Alternative solution. Let x, u, v, T be as defined above. Fur-ther, let y be the distance Tom travelled to the rear of thecolumn and t the time needed for that. Then vt = y andut = 3 − y, so v/u = y/(3 − y). On the other hand, for thetotal distances vT = 2y + 4 and uT = 4, so v/u = (2y + 4)/4.Thus, y/(3 − y) = (2y + 4)/4, which gives y2 + y − 6 = 0, i.e.y = 2 (we disregard the possibility y = −3 since y > 0). Thetotal distance is then 2y + 4 = 8 km. [4 marks]


(A) After 8 clicks the dots are in positions 3 and 8 so the right dot isOK. A further 9 clicks leaves the right dot at 8 but the left is nowat 2. After 9 more the dots are at positions 1 and 8 as required.So we need 8 + 9 + 9 = 26 clicks. [5 marks]

(B) After 7 clicks the dots are at positions 4 and 4. 9 more takes usto positions 3 and 4, 9 more to 2 and 4, and another 9 to 1 and 4.So we need 7 + 9 + 9 + 9 = 34 clicks. [10 marks]

(C) 5 clicks move us to positions 7 and 2. 24 more take us to 14 and2. Then after 15 more sets of 24 clicks we arrive at 0 and 2. Thetotal number of clicks required is therefore 15 + 24 × 16 = 399.

[10 marks]

(D) We first get the two right hand dials correct. 5 clicks put the dialsin positions 2, 3, 4 so that the third dial is right. 7 more clicksmove us to positions 4, 4, 4 with dial 3 still OK and it stays OKwith each set of 7 clicks. After a total of 5 + 5 × 7 clicks we getpositions 2, 2, 4 with dials 2 and 3 OK. Now a set of 42 more clickswill leave these dials in the correct position and advances the firstdial 2 positions. We find that after 4 sets of 42 clicks the firstdial is in position 0 and we are done. The total number of clicksneeded is 5 + 5× 7 + 4× 42 = 208. [10 marks]

(E) Say that L is the number of times Roxanne clicks the left buttonand R is the number of times she clicks the right button. Thenif L = 3 and R = 0 we get to positions 0, 1, 6 which puts the lefthand button in the correct position. Then L = 3 and R = 5 movesus to 0, 0, 4 with the left and right dials in the correct positions.Increasing L by multiples of 5 and R by multiples of 7 leaves thesedials in the correct positions, but moves the middle dot. If weperform both these operations we move the middle dot 12 positionswhich is the same as leaving it alone so we should be adding amultiple of 5 to L or a multiple of 7 to R but not both. Tryingboth schemes we find we can get to 0, 2, 4 using (L, R) = (23, 5)or (L, R) = (3, 19). The second is the most efficient so she shouldclick the left button 3 times and the right 19 times. [10 marks]

WA Junior Mathematics Olympiad 2003

Individual Questions 100 minutes (one hour and40 minutes)

General instructions: No working need be given for Questions 1 to9. Calculators are not permitted. Write your answers on the answersheet provided.

(1) In a test of 20 questions 5 marks are given for each correctanswer and 2 are deducted for each incorrect answer. Alan didall the questions and scored 58. How many correct answers didhe have?

[1 mark]

(2) Jane typed a 6-digit number into a faulty computer in whichthe 1 (one) key was broken. The number appearing on thescreen was 2003, possibly with some blank spaces. How manydifferent 6-digit numbers could Jane have typed?

[1 mark]

(3) A whole number between 1 and 99 is not greater than 90, notless than 30, not a perfect square, not even, not a prime, notdivisible by 3 and its last digit is not 5. What is the number?

[2 marks]

(4) What is the units digit of 1! + 2! + 3! + ... + 2003!? [3! means1× 2× 3]

[2 marks]

(5) How many triples (x, y, z) of positive integers satisfy (xy)z =64?

[2 marks]

(6) Douglas is 23

as old as he will be 8 years before he is twice asold as he is now. How old is Douglas?

[2 marks]

1

(7) In a triangle ABC, ∠C = 90◦. A perpendicular is producedfrom the midpoint D of AB to meet the side BC at E. Thelength of AB is 20 and the length of AC is 12.

What is the area of triangle ACE?

[3 marks]

(8) If 2x+3y + z = 48 and 4x+3y +2z = 69 what is 6x+3y +3z?[4 marks]

(9) The sum of six consecutive positive odd integers starting withn is a perfect cube. Find the smallest possible n. [4 marks]

(10) ABCD is a parallelogram. E is the mid point of AB. Join E toD. ED intersects AC at P.

How many times larger is the area of the parallelogram ABCDthan the area of the triangle AEP?

Give reasons for your solution. [4 marks]


Team Questions 45minutes

A. Suppose you have a string of 2003 beads which you cut betweentwo beads as close to the middle as possible, so that you nowhave two strings, one with 1001 beads and one with 1002. Thebeads are glued onto the string so they won’t slide off. Younow take the shorter string of the two and cut it as close to themiddle as possible and then keep repeating the process, at eachstep choosing a shorter string. You may choose either if theyare equal. Stop when you have a string with only one bead.How many strings have you now got?

B. Find all possible initial string lengths which will finish with 5pieces.

C. How many strings would you have if the original string had999,999 beads?

D. Can you give a formula for the longest string and a formula forthe shortest string which finish in n pieces?

E. What would happen if you cut the original string of 2003 beadsas nearly as possible into thirds, and then took a smallest lengthto continue?

Western Australian Junior Mathematics OlympiadNovember 1, 2003

Problem Solutions

1. Suppose Alan solved n questions correctly. Then 58 = 5n− 2(20− n) = 7n− 40,

so 7n = 98 and therefore n = 14.

2. There are 6 places in which to type two 1’s, the remaining 4 places being filled with2003.

So there are 6 choices for the place of the first 1 and then 5 for the second 1, a total of30. But only half of these look different from each other so there are 15 possibilities.

3. From the first line of the question, we are looking for a number between 30 and 90.

From the second line it is odd and not 49 or 81.

Also it is not prime, so 31, 37, 41, 43, 47, 53, 57, 59, 61, 67, 71, 73, 79 83, 87 and 89are also excluded.

Since it is not divisible by 3 or 5, 33, 35, 39, 45, 51, 55, 63, 65, 69, 75, 85 are alsoexcluded.

The only remaining possibility is 77.

4. Since 5! and n! for all n > 5 are divisible by 10, the required units digit is the unitsdigit of 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33. hence the answer is 3.

5. First notice that 64 = 26 so x must be a power of 2 such that xyz = 64. The onlypossibilities are x = 2, yz = 6 or x = 4, yz = 3, or x = 8, yz = 2 or x = 64, yz = 1.

So altogether there are 9 solutions, (x, y, z) = (2, 1, 6), (2, 2, 3), (2, 3, 2), (2, 6, 1), (4, 1, 3),(4, 3, 1), (8, 1, 2), (8, 2, 1) and (64, 1, 1).

6. Suppose Douglas is x years old. Eight years before he is twice as old as x, he will be2x− 8 years old. So x = 2

3(2x− 8).

Hence 3x = 4x− 16, so x = 16.

7. Note that triangle ABC is four times as big as a right–angled 3− 4− 5 triangle, so,|BC| = 16.

Suppose |EC| = x so |BE| = 16 − x. Since BEA is isosceles, |AE| = 16 − x also.By Pythagoras’ Theorem, x2 = |EC|2 = (16− x)2 − 144, so 144 = 256− 32x. Hencex = 112/32 = 3.5 so the area of triangle ACE is 6× 3.5 = 21.

1

8. First notice that by adding the left sides and the right sides of the equations, you get6x + 6y + 3z = 117. We need to subtract 3y to get 6x + 3y + 3z. But if you subtract69 = 4x + 3y + 2z from 96 = 4x + 6y + 2z you get 3y = 27.

Hence 6x + 3y + 3z = 117− 27 = 90.

9. The sum of the six consecutive odd integers starting with n = n+(n+2)+ (n+4)+(n + 6) + (n + 8) + (n + 10) = 6n + 30 = 6(n + 5).

The smallest cube of the form 6(n + 5) occurs when n + 5 = 36, so n = 31.

10. DPC + DPA = 1/2ABCD

But DPC is similar to and twice the size of APE. Hence 4APE+DPA = 1/2ABCD.

DPA + APE = 1/4ABCD. Hence 3APE = 1/4ABCD so ABCD is 12 times thearea of AEP .

Another way to do it is to join PB. Then AEP = EPB and DPO = BPO, whileAOD = AOB. Hence APD = APB = DPB so they each form one sixth of ABCD.Therefore ABCD is 12 times AEP .

Team Problem Solutions

A. Step 1 gives 2 strings, shortest length 1001.

Step 2 gives 3 strings, shortest length 500.






2




B. The shortest string length to gve 5 pieces is 16, the lengths being 8, 4, 2, 1 and 1.

The longest string length to give 5 pieces is 31, the lengths being 16, 8, 4, 2, 1.

Any string length between 16 and 31 also gives 5 pieces.

C. The answer is the lowest power of 2 ≥ 106. Since 210 = 1024, 220 > 106 and 219 < 106

so the answer is 20.

D. 2n−1 and 2n − 1.

E. If you start with a string length of 2003, continually cut as nearly as possible intothirds and continue with a smallest piece, you get in successive cuts:

3 strings, shortest length 667






You cannot proceed any further. In general, if you start with a string of any length,you will finish with a shortest string of length 1 or 2.

If you start with a string of length x, the number of strings you get will be 2k + 1,where k is the highest power of 3 ≤ x.

To finish with n strings, n must be odd, say n = 2k + 1. The smallest string you canstart with has length 3k and the largest string you can start with has length 3k+1−1.

3


Individual Questions 100 minutes (one hour and 40 minutes)

General instructions: No working need be given for Questions 1 to 9. Calculators are notpermitted. For Questions 1 to 9, write the answer in the answer grid. Write your answer toQuestion 10 in the space provided.

1. Simplify√

(1 + 24 + 25)(1 + 23 + 24) + 26

√1 + 23

. [1 mark]

2. Four different positive integer numbers a, b, c, d satisfy the following relations:1a

+1a

+1a

= 1 ,1a

+1b

+1c

= 1 ,1b

+1d

+1d

= 1 .

Find d. [2 marks]

3. Three athletes Ahmad, Bill and Claire are preparing to take part in a high jump competition.At the same time some of the spectators are discussing their chances:

spectator X : ‘I think Ahmad will be first’,

spectator Y: ‘I am sure that Claire will not be the last’,

spectator Z : ‘Bill will not take first place’.

After the competition it turned out that only one of the spectators was right, while theother two were wrong. Where did Claire finish? [2 marks]

4. In a rectangle ABCD, O is the intersection point of the diagonals AC and BD and BD = 10cm. Find the length of BC if it is known that the point D lies on the perpendicular bisectorof the segment AO. [2 marks]

5. Three positive numbers are given such that :

(i) the first of the numbers is half the second;

(ii) the product of the first and the second number is equal to the sum of the second andthe third number;

(iii) the third number is three times as large as the second.

Find the first of the given numbers. [2 marks]

6. Find the smallest positive integer divisible by 15 whose every digit is 0 or 1. [2 marks]

7. A rectangle ABCD has sides AB = CD = 34 cm. E is a point on CD such that CE = 9cm, ED = 25 cm, and 6 AEB = 90◦. What is the length of BC? [3 marks]

8. I have 6 cats, two white, 2 black and 2 orange, with a male and female of each colour. Iwant to put them in a row of 6 boxes. The orange cats are friends and have to be put sideby side, but the black cats fight and must not be side by side. In how many ways can Iarrange the cats? [3 marks]

9. Five different integer numbers a, b, c, d, e (not necessarily positive) are such that

(4− a)(4− b)(4− c)(4− d)(4− e) = 12 .

Find the sum a + b + c + d + e. [4 marks]

10. All faces of a cube are divided into four equal squares and each small square is painted red,blue or green in such a way that any two small squares that have a common side are paintedin different colours. Is it true that the numbers of red, blue and green squares must be thesame? Give reasons for your answer. [4 marks]

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Western Australian

Junior Mathematics Olympiad 2002


1. You are given 9 coins of the same denomination, and you know that one of them iscounterfeit and that it is lighter than the others. You have a pan balance which meansyou can put any number of coins on each side and the balance will tell you which sideis heavier, but not how much heavier. Explain how you can find the counterfeit coinin exactly two weighings. [4 marks]

2. If you are given 25 coins of the same denomination, and you know that one of them iscounterfeit and that it is lighter than the others, explain how to find the counterfeitcoin by using at most 3 weighings on the pan balance. [8 marks]

3. It is known that there is one counterfeit coin in a collection of 70 and that it is lighterthan the others. What is the least number of weight trials on a pan balance necessaryto identify the counterfeit coin? Explain how you obtained your answer. [12 marks]

4. It is known that there is one counterfeit coin in a collection of 9 and it is known thatits weight is different from that of a genuine coin, however it is not known whether thecounterfeit coin is lighter or heavier than a genuine one. Show that by using at most3 weighings in the pan balance you can identify the counterfeit coin and determinewhether it is lighter or heavier than a genuine coin. [16 marks]

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Solutions to the Individual Questions

1. Since 1 + 24 + 25 = 49, 1 + 23 + 24 = 25 and 26 = 64, it follows that√(1 + 24 + 25)(1 + 23 + 24) + 26

√1 + 23

=√

49 · 25 + 643

=993

= 33 .

Answer: 33

2. The first equation gives 3/a = 1, so a = 3, and the second equation becomes1b

+1c

=23

.

Hence b ≥ 2 and c ≥ 2. If both b > 2 and c > 2, then (since the numbers are different anda = 3) b ≥ 4 and c ≥ 4, so 1/b + 1/c ≤ 1/2, impossible. Thus, either b = 2 or c = 2. Ifc = 2, then b = 6 and the last equation becomes 2/d = 5/6, impossible since d is an integer.Thus b = 2, c = 6, and then the 3rd equation gives 2/d = 1/2, so d = 4.

Answer: 4

3. There are three possible cases to consider:

Case 1. X is right, while Y and Z are wrong. Then Ahmad must be first, Claire is last andBill is first, impossible.

Case 2. Y is right, while X and Z are wrong. Then Claire is not last, Ahmad is not firstand Bill is first. Hence Claire is second and Ahmad must be last. This case is possible.

Case 3. Z is right, while X and Y are wrong. Then Bill is not first, Ahmad is not first andClaire is last, impossible.

Thus, only the second case is possible, so Claire must have taken second place.

Answer: 2

4. If M is the midpoint of AO, then 4AMD ∼= 4OMD (AM = MO, MD = MD, 6 AMD =6 OMD). Hence AD = OD = BD/2 = 5 cm, and therefore BC = 5 cm.

Answer: 5

5. If x, y, z are the given numbers, we have x = y/2, xy = y + z, z = 3y. Substituting x andz in the second equation gives y2/2 = y + 3y, so y2 = 8y. Since y > 0, this implies y = 8,and therefore x = 4.

Answer: 4

6. If N is such a number then N is divisible by 5, so its last digit must be 0. Since N isdivisible by 3, the sum of its digist must be divisible by 3, so N must have at least threedigits 1. The smallest such number is N = 1110.

Answer: 1110

7. Since 6 AED = 90◦ − 6 BEC = 6 EBC, we have 4AED ∼ 4EBC. If a = BC, it follows

thata

25=

9a, so a2 = 9 · 25, i.e. a = 15.

(Alternative solution: use Pythagoras Theorem.)

Answer: 15

8. Begin by ignoring the sex of the cats.

If the 2 Os are in the first 2 boxes there are 3 ways OOWBBW, OOBWBW, OOWBWB.

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If they’re in the next 2 boxes there are 4 ways: BOOWBW, BOOWWB, BOOBWW,WOOBWB.

If they’re in middle 2 boxes there are 4 ways: WBOOWB, WBOOBW, BWOOWB, BWOOBW.

4th and 5th is the same as 2nd and 3rd: 4 ways.

5th and 6th is same as 1st and 2nd : 3 ways.

Total number = 3 + 4 + 4 + 4 + 3 = 18.

Since each colour can be M, F or F, M we have to multiply by 23 = 8, so number of ways= 8× 18 = 144.

Alternative Solution: Let the cats be W1, W2, O1, O2, B1, B2. We will denote by O thepair O1, O2, they must be side by side (later we will take into account the fact that O1 andO2 can swap places in O). Disregarding for a moment the fact that B1 and B2 must not beside by side, the number of possible ways we can order W1,W2, O,B1, B2 is 5·4·3·2·1 = 120.Since O1 and O2 can swap places in O, the number of ways we can order the cats so thatO1 and O2 are side by side is 2 · 120 = 240.

From this number we have to subtract the number of cases when B1 and B2 are side byside. If we denote the pair B1 and B2 by B, the number of ways we can order W1,W2, O,Bis 4 · 3 · 2 · 1 = 24. Since B1 and B2 can swap places in B, and O1 and O2 can swap placesin O, the total number of such cases is 24 · 4 = 96. Thus, the number of ways we can orderthe cats satisfying both requirements is 240− 96 = 144.

Answer: 144

9. Renaming the numbers if necessary, we may assume that

A = 4− a < B = 4− b < C = 4− c < D = 2− d < E = 4− e .

The numbers A,B, C, D, E are different integers satisfying A B C D E = 12 , so each ofthese numbers is equal to ±1, ±2, ±3, ±4, ±6 or ±12. If some of the numbers (that mustbe A or E) is equal to 12 or −12, then all other numbers must be equal to ±1, so at leasttwo of them will be equal, impossible. Similarly, if some of the numbers is equal to 6 or −6,some other number must be equal to 2 or −2, and the remaining three numbers must beequal to ±1, so at least two of them will be equal, impossible again. Thus, all numbers arebetween −4 and 4. Next, if some of the numbers is equal to 4 or −4, some other numbermust be equal to 3 or −3, and the remaining three numbers must be equal to ±1, so at leasttwo of them will be equal, impossible again. Thus, A,B, C, D, E are five of the numbers−3,−2,−1, 1, 2, 3. Obviously if A = −3, then E ≤ 2, and if E = 3, then A ≥ −2. Since theproduct of the five numbers is positive, we cannot have three of them negative, so the onlypossible case is A = −2, B = −1, C = 1, D = 2, E = 3. Then A + B + C + D + E = 3, soa + b + c + d + e = 20− (A + B + C + D + E) = 17.

Answer: 17

10. The answer is yes – there must be exactly 8 red, 8 blue and 8 green squares.

Let A be an arbitrary vertex of the cube. There are three small squares with vertex A andany two of them have a common side, so they must be painted differently. Hence one ofthe three squares with vertex A must be red, another must be blue, and the third must begreen.

This applies to any of the 8 vertices of the cube, so there must be at least 8 red smallsquares, at least 8 blue and at least 8 green.

However the total number of small squares is 6 · 4 = 24, so there must be exactly 8 smallsquares of each coulour.

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Solutions to the Team Questions

1. Divide the coins into three groups of 3 coins each. Place e.g. the coins of group 1 on one ofthe pans of the pan ballance and the coins of group 2 on the other pan.

If the pans do not balance, the counterfeit coin is in the lighter pan. If the pans balance, thecounterfeit coin is in the third group. So, with one weight trial we determine a group of 3coins that contains the counterfeit coin.

For the second trial, choose any two of these 3 coins and place them on the two pans. Ifthe pans balance, the third coin is the counterfeit one; if not, then the leighter coin is thecounterfeit one.

2. Divide the coins into 3 groups, the first two groups containing 9 coins each, while the thirdgroup contains 7 coins. (Other divisions are possible, e.g. 8+8+9.) Place the coins of group1 on one pan of the pan balance and the coins of group 2 on the other pan.

If the pans do not balance, the counterfeit coin is in the lighter pan. If the pans balance, thecounterfeit coin is in the third group. In the latter case, take two coins from group 1, say, toget a group of 9 coins containing the counterfeit one.

So, with one weight trial we determine a group of 9 coins that contains the counterfeit coin.

Then proceed as in Problem 1 above to find the counterfeit coin using 2 weighings. Thus,with a total of 3 weighing one can determine the counterfeit coin.

3. Divide the coins into 3 groups, the first two groups containing 27 coins each, while the thirdgroup contains 16 coins. (There are other possible divisions that are good enough, in fact anydivision so that there is no group with more than 27 coins is a good one.) Place the coins ofgroup 1 on one pan of the pan balance and the coins of group 2 on the other pan.

If the pans do not balance, the counterfeit coin is in the lighter pan. If the pans balance, thecounterfeit coin is in the third group. In the latter case, take 9 coins from group 1, say, toget a group of 27 coins containing the counterfeit one.

So, with one weight trial we determine a group of 27 coins that contains the counterfeit coin.

Then proceed as in Problem 2 above (dividing the 27 coins into 3 groups of 9 coins each) tofind the counterfeit coin using 3 weighings. Thus with a total of 4 weighing one can determinethe counterfeit coin.

Let us now show that 4 is the minimal number of weighings that guarantee finding thecounterfeit coin under all circumstances.

Suppose in the first trial we weigh two groups of k coins each; the remaining coins are then70 − 2k. If k = 23, then the third group contains 24 coins; if k < 23, then the third groupcontains more than 24 coins; if k = 24 or larger, then the third group contains 22 coins or less.Assuming that the result of our weight trial is the least favorable, the best we can achieve (inany circumstances) from trial 1 is to determine a group of 24 coins containing the counterfeitone.

In the same way, using a second trial the best one can achieve (in any circumstances) is todetermine a group of 24

3 = 8 coins containing the counterfeit one.

Similarly, the third trial (assuming least favorable results again) will at best give us a groupof 3 coins containing the counterfeit one, so we need one more trial.

4. Divide the coins into 3 groups of 3 coins each, and for a first weight trial place the coins ofgroup 1 on one pan of the pan balance and the coins of group 2 on the other pan. For asecond weight trial do the same, say, with groups 2 and 3. As a result of these two trials one

1

determines which of the three groups contains the counterfeit coin and whether this coin islighter or heavier than a genuine coin. For example, if group 1 is heavier (or lighter) thangroup 2, and group 2 has the same weight as group 3, then the counterfeit coin is in group 1and it is heavier (resp. lighter) than a genuine coin. If group 1 is heavier than group 2, andgroup 2 is lighter (it cannot be havier) than group 3, then the counterfeit coin is in group 2and it is lighter than a genuine coin.

Consider now the group of 3 coins found after the first two trials to contain the counterfeitcoin, and assume for clarity that the counterfeit coin is determined to be lighter than a genuineone (the other case is similar). For a third trial place two of the coins on the two pans of thepan balance. If the pans balance, the third coin is the counterfeit one. If one pan is lighter,the coin in it is the counterfeit one.

Alternative Solution. Divide the coins into 3 groups of 3 coins each and place the coins ofgroup 1 on one pan of the pan balance and the coins of group 2 on the other pan. There aretwo possibilities.

Case 1. The pans balance. Then all coins in groups 1 and 2 are genuine and the counterfeit coinis in group 3.Let the coins in group 3 be A,B, C. For a second weight trial, place coins A and B onone pan and two genuine coins (say from group 1) on the other. If the pans balance,then coin C is the counterfeit one, and a third weight trial (comparing C with a genuinecoin) will determine whether C is lighter or heavier than a genuine coin.If the pans do not balance, then either A or B is counterfeit. Moreover at this stagewe will already know whether the counterfeit coin is lighter or heavier (e.g. if the pancontaining A and B is lighter, then the counterfeit coin is lighter than a genuine one).For a third weight trial, place A and B on different pans of the balance, this will showwhich of them is the counterfeit one. E.g. if the previous step showed that the counterfeitcoin is lighter than a genuine one, and A is lighter than B, then A is the counterfeitcoin.

Case 2. One pan is heavier. Then the coins in group 3 are all genuine.Let the coins in the heavier pan be A,B, C (if one of these coins is counterfeit, then it isheavier than a genuine one), and let these in the other pan be A′, B′, C ′ (if one of thesecoins is counterfeit, then it is lighter than a genuine one).For a second weight trial, place e.g. A and A′ on one pan and B and B′ on the other.Then we have the following possibilities.

Subcase 2.1. The pans balance. Then A,B, A′, B′ are all genuine coins, and either C or C ′ iscounterfeit.For a third weight trial, place C on one pan and a genuine coin (say, A) on theother. If the pans balance, then C ′ is the counterfeit coin and it is lighter than agenuine one. If the pans do not balance, then C is the counterfeit coin and it mustbe heavier than a genuine one.

Subcase 2.2. The pan containing A,A′ is heavier than the pan containing B,B′. Then A′ and Bmust be genuine, so the counterfeit coin is either A or B′.For a third weight trial, place A on one pan and a genuine coin (say, C) on theother. If the pans balance, then B′ is the counterfeit coin and it is lighter than agenuine one. If the pans do not balance, then A is the counterfeit coin and it mustbe heavier than a genuine one.

Subcase 2.3. The pan containing B,B′ is heavier than the pan containing A,A′. This case isconsidered in the same way as Subcase 2.2.

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Western Australian Junior Mathematics OlympiadOctober 27, 2001

Individual Questions

General instructions: No working need be given for Questions 1 to 9. Calculators arenot permitted. For Questions 1 to 9, write the answer in the answer grid. Write youranswer to Question 10 in the space provided.

1. N degrees Celsius is the same temperature as 95N + 32 degrees Fahrenheit. What

temperatures have the same measure on both scales? (1 mark)

2. bxc means the greatest integer which is not more than x, so that b4.9c = 4 andb7c = 7, and dxe means the least integer which is not less than x, so that d4.9e =5 and d7e = 7. We call bxc the floor of x and dxe the ceiling of x. Evaluated13.5 + 2.7× b3.8ce. (1 mark)

3. A certain number of points are marked on the circumference of length 2001 of a circlein such a way that each marked point is distance 1 from exactly one marked pointand distance 2 from exactly one marked point, all distances being measured aroundthe circle. How many points are there? (1 mark)

4. Let ABC be a right-angled triangle with 6 ACB = 90 degrees, and let AL be thebisector of angle BAC, so that L is a point on BC. Let M be the point on AB suchthat LM is perpendicular to AB. If LM = 3 and MB = 4, find AB. (2 marks)

5. How many solution pairs x, y) are there of the equation 2x+ 3y = 763 if both x andy are positive integers? (2 marks)

6. In a computer game, you have to score the largest possible number of points. Youscore 7 points each time you find a jewel and 4 points each time you find a sword.There is no limit to the number of points you can score. Of course it is impossible toscore 5 or 6 points. What is the largest number of points it is impossible to score?(2 marks)

7. Find the least possible value of the expression x2 − 8xy + 19y2 − 6y + 10. (3 marks)

8. A shop sells hamburgers which contain some of the following: meat burger, vegetableburger, lettuce, tomato, carrot, mayonnaise and tomato sauce.

(a) You must have a meat burger or a vegetable burger, but cant have both.(b) You can also have any number of the other ingredients, even none, but:(c) If you have a meat burger you can also have tomato sauce, but not if you have a

1

vegetable burger.(d) If you have lettuce or tomato or both you can have mayonnaise, but not otherwise.

How many different hamburgers can be constructed according to these rules?(3 marks)

9. Let ABCD be a trapezium with AB parallel to CD, AB = 2CD and the diagonalBD = 72 cm. If N is the midpoint of AB and M and P are the intersection pointsof BD with NC and AC, respectively, find MP . (3 marks)

10. Two buses start travelling at the same time – bus 1 from city A to city B, and bus 2from city B to city A using the same road. Both buses travel with constant speeds.For the first time they meet 7km from A. After both buses reach their destinations(cities B and A respectively, possibly at different times), they immediately starttravelling back along the same road and with the same speeds. They meet again 4km from B. Find the distance between the cities A and B. Explain how you obtainedyour answer. (4 marks)

Team Questions

1. Find six consecutive positive integers whose sum is 513.

2. Find a set of at least two consecutive positive integers whose sum is 30.

3. There are three possible solutions to question 2. Can you find them all?

4. Find a set of at least two consecutive positive integers whose sum is 56.

5. Show how any odd integer can be written as the sum of at least 2 consecutive integers.

6. Some positive integers cant be written in this way. What are they?

7. Can you prove your answer to question 6?

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Problem Solutions

1. We must solve 9N/5 + 32 = N . So 9N + 160 = 5N and hence 4N = −160. The solution tothis is N = −40.

2. d13.5 + 2.7× b3.8ce = d13.5 + 2.7× 3e = d13.5 + 8.1e = d21.6e = 22.

3. The gaps between adjacent points must be alternately 1 unit and 2 units, so any pair ofconsecutive gaps totals 3 units. Since there must be 667 pairs of gaps, and so 1334 gapsaltogether and therefore 1334 points.

4. From 4LMB one finds LB2 = 32 + 42 = 25, so LB = 5. Next, 4AML ∼= 4ACL (AL =AL, 6 MAL = 6 CAL, 6 AML = 90◦ = 6 ACL), so CL = LM = 3. Now observe that

4ABC ∼ 4LBM (6 ABC = 6 LBM , 6 ACB = 6 LMB), thereforeBC

AB=MB

LB. This gives

AB =BC × LBMB

=8× 5

4= 10.

5. It’s clear that y must be odd so we can write y = 2Y + 1 for some non-negative integer Y .Also 763 − 2x must be divisible by 3. Now 763 − 2x = (3 × 254) + (1 − 2x) so 1 − 2x mustbe divisible by 3. This means x has the form 3X + 2 with X a non-negative integer. Thus2(3X + 2) + 3(2Y + 1) = 763, which simplifies to X + Y = 126. Then X can be any integerfrom 0 to 126, and so there are 127 solutions.

6. The answer is 17. By trial and error we find that 17 can’t be expressed as the sum of a multipleof 7 plus a multiple of 4. However 18 = 2× 7 + 4, 19 = 7 + 3× 4, 20 = 5× 4 and 21 = 3× 7.After this we can get 22 by adding 4 onto 18, 23 by adding 4 onto 19 and so on.

7. We note that:

x2 − 8xy+ 19y2 − 6y+ 10 = x2 − 8xy+ 16y2 + 3(y2 − 2y+ 1) + 7 = (x− 4y)2 + 3(y− 1)2 + 7 .

Each of the squared terms is at least 0, so the whole expression must be at least 7, and we canget 7 if we set y = 1 and x = 4. So the answer is 7.

8. We have 3 basic types of burgers: vegetable, meat with sauce or meat without sauce. Eachof these is accompanied by one of the following 7 lettuce, tomato, mayonnaise combinations:LTM, LM, TM, LT, L, T, none of these. This gives 21 possibilities. Each of these 21 can beserved with or without carrot, giving a total of 42 possibilities.

9. First, notice that 4NBM ∼ 4CDM (6 NMB = 6 CMD, 6 NBM = 6 CDM). HenceDM

MB=

CD

NB= 1, i.e. DM = MB = 36 cm. Next, we have 4ABP ∼ 4CDP (6 APB = 6 CPD,

6 ABP = 6 CDP ), soDP

PB=

DC

AB=

12

. That is, DP = 13DB = 24 cm. Hence MP =

DM −DP = 12 cm.

10. Let v1 and v2 be the speeds of bus 1 and bus 2 respectively, let t1 and t2 be the times at whichthey pass each other and let x be the distance between the towns. By considering the first time

they pass we see that7t1

= v1 ,x− 7t1

= v2 , which implies thatv2

v1=x− 7

7. By considering

the second time they pass we getx+ 4t2

= v1 ,2x− 4t2

= v2 , which implies thatv2

v1=

2x− 4x+ 4

.

Thus we havex− 7

7=

2x− 4x+ 4

, which gives x2 = 17x and so x = 17.

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1. 513 = 83 + 84 + 85 + 86 + 87 + 88.

2. 30 = 9 + 10 + 11 = 6 + 7 + 8 + 9 = 4 + 5 + 6 + 7 + 8.

3. See 2. above.

4. 56 = 5 + 6 + 7 + 8 + 9 + 10 + 11.

5. Any odd number can be written as 2n+ 1 for some integer n. But 2n+ 1 = n+ (n+ 1) whichis the sum of two consecutive integers.

6. Powers of 2 (including 1 = 20).

7. If a number is not a power of 2 then it has an odd factor greater than 1. So, say our numberis n = ab, where a = 2k + 1 is odd. Then

n = (b− k) + (b− k + 1) + (b− k + 2) + . . .+ b+ (b+ 1) + . . .+ (b+ k − 1) + (b+ k) ,

which has the required form. So, anything that is not a power of 2 can be written in therequired way.

If n can be written in the required form there must be positive integers a and m such that

n = a+ (a+ 1) + . . .+ (a+m)

= (m+ 1) a+ (1 + 2 + . . .+m) = (m+ 1) a+12m (m+ 1)

=12

(m+ 1)(m+ 2a) .

One of m+ 1 and m+ 2a must be odd and the other even, so n has an odd factor. This meansn is not a power of 2.

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Individual Part of the Olympiad

100 minutes (one hour and 40 minutes)

No working need be given for Questions 1 to 9. Calculators are not permitted. For Questions 1 to 9, write the answer in the answer grid. Write your answer to Question 10 in the space provided.

1. How many non-overlapping equilateral triangles with 1 centimetre sides can we fit inside an equilateral triangle with 10 centimetre sides? (1 mark)

2. Four accused people face trial. It is known that: (a) If A is guilty, then B is guilty, (b) If B is guilty, then C is guilty or A is not guilty, (c) If D is not guilty, then A is guilty and C is not guilty, (d) If D is guilty, then A is guilty. How many of the accused must be guilty? (1 mark)

3. Find the number of integers between 100 and 500 such that the sum of their digits is 10. (1 mark)

4. In the expression S = a - b + c - d the symbols a, b, c, d are replaced by 1, 2, 3, 4 in any order with no repetitions allowed. There are 24 possible replacements. In how many of these will S be greater than 0? (1 mark)

5. Let ABC be an acute angled triangle. Let M be a point on BC such that AM is perpendicular to BC. Let N be a point on AB such that CN is perpendicular to AB. If H is the intersection point of AM and CN and it is given that HM = HN and BC = 20, find AB. (2 marks)

6. The visibility at sea, on a certain day, is 5 kilometres. Ships A and B (which start a long way apart) are t ravelling in opposite directions on courses which are parallel and 3 kilometres apart. The two ships are in sight of one another for 24 minutes. If ship A is travelling at 8 kilometres per hour, how fast is ship B travelling? (2 marks)

7. How many integers between 100 and 1000 are there that are not exactly divisible by 2, 3 or 5? (2 marks)

8. The number 6 is divisible by 1, 2, 3 and 6, so 6 has 4 divisors. How many divisors has 6718464 = (2^10) x (3^8)? (Remark. ''a^b'' means ''a to the power of b''.) (2 marks)

9. Let M be the point on the extension of the side BC of a parallelogram ABCD such that B is between M and C and MB = BC. If P and N are the intersection points of MD with AC and AB, respectively, and PN = 3, find DP. (2 marks)

10. Ten students have altogether 35 coins. It turns out that at least one of them has exactly one coin, at least one has exactly two coins, and at least one has exactly three coins. Explain why you can be sure that at least one student has 5 or more coins. (4 marks)

Team Part of the Olympiad

45 minutes

1. A cube, consisting of 125 cubes each with side 1 centimetre, is drilled through in three places. The holes are rectangular-shaped with cross-section 1 centimetre by 3 centimetres (see picture) and go all the way through the large cube. How many small cubes remain after the drilling? 2. Instead of a 5 x 5 x 5 cube, say you have a 7 x 7 x 7 cube, with three slots having cross-section 1 centimetre by 5 centimetre drilled through the middle. How many small cubes remain?

3. Now suppose you have an n x n x n cube, with n an odd number. Three slots, each with cross-section 1 centimetre by n-2 centimetres are drilled through the middle. Find a formula for the number of small cubes that remain.

4. Briefly explain how you obtained your answer to question 3.

Write your answers on the accompanying sheet.


Problem Solutions

1. The answer is 100. You can see this by drawing a diagram. Alternatively, usingPythagoras’ Theorem or trigonometry, you can show that the area of the large triangle

is100√

3

4and the area of each small triangle is

√3

4. The answer follows since

100√

34√3

4

= 100 .

2. Looking at (c) and (d) together we see that A must be guilty regardless of D’s guiltor innocence. Part (a) then implies that B is guilty. So now we know that both Aand B are guilty. Part (b) says that C is guilty or A is innocent, but we know A isnot innocent, so C must be guilty. Now we know A, B and C are all guilty. Part (c)says that if D were not guilty then C would be not guilty, which we know is not true.So D must be guilty as well and the answer is 4.

3. We put the numbers in blocks. We have 10 numbers with first digit being 1: 109,118, 127,..., 190, then 9 with first digit being 2: 208, 217,..., 280, then 8 with firstdigit being 3, and 4 with first digit being 4. The total number of integers is then 10+ 9 + 8 + 7 = 34.

4. We could do this by writing down all 24 possibities, but this is a bit tedious. Asmarter way is to notice that the sum can only equal 0 in 8 ways: the numbers witha plus sign being 1 and 5 and the numbers with a minus sign being 2 and 3 (whichcan happen in 4 ways) or the other way round (another 4 ways). This leaves 24 - 8= 16 non-zero sums. There must be an equal number of positive and negative sums,since swapping the values of a and b, and the values of c and d will change the sumfrom positive to negative or vice versa. Therefore the number of positive solutions is16 / 2 = 8.

5. In triangles CMH and ANH we have MH = NH, 6 CHM = 6 AHN (oppositeangles) and 6 CMH = 6 ANH = 90◦. Thus triangles CMH and ANH are congruentby the Angle-Side-Angle rule. Thus CH = AH and so CN = AM . Angles AMBand CNB both equal 90◦ and angles BCN and BAM both equal 90◦ - 6 ABC. Sotriangles CNB and AMB are congruent (Angle-Side-Angle again), and thereforeAB = CB = 20.

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6. The answer is 12 km per hour. Indeed, at a certain time the ships will be 5 kmapart for the first time. Denote by A0 and B0 their respective positions at thattime, and let C0 be the point on the line determined by the direction of the ship Asuch that B0C0 is perpendicular to this line. By Pythagoras’ Theorem, A0C0 = 4km. If A′ and B′ are the positions of the ships A and B after 24 minutes, then thedistance travelled by A is A0A

′ = 8 × 2460

= 165

km. Hence A′C0 = 4 − 165

= 45. If

D0 is the point on the line determined by the direction of the ship A such that B′D0

is perpendicular to this line, then by Pythagoras’ Theorem for 4B′D0A′ one gets

D0A′ = 4, so B′B0 = D0C0 = D0A

′ + A′C0 = 4 + 45

= 245

km. Therefore the speed ofship B is 24

5: 24

60= 12 km.

7. The answer is 240. Clearly 100 is divisible by 2 (and 5), so we have to check theintegers 101, 102, . . . , 999, 1000. Consider the first 30 of them: 101, 102, 103, . . . , 130.By a direct inspection, one can see that exactly 8 of them are not divisible by 2, 3 and5. Dividing the sequence 101, 102, . . . , 1000 into 30 separate sets of 30 consecutiveintegers and using the fact that in each of these sets of 30 integers there will beagain exactly 8 integers not divisible by 2, 3 and 5, one gets that the total numberof integers between 101 and 1000 not divisible by 2, 3 and 5 is 30× 8 = 240.

8. The answer is 99. The divisors of 210 are 1 = 20, 2 = 21, 22, . . ., 210, while these of38 are 1 = 30, 3 = 31, . . ., 38. Every divisor of 210 × 38 has the form 2k × 3m forsome k = 0, 1, . . . , 10 and m = 0, 1, . . . , 8. So, there are 11 different ways to choosek and 9 different ways for m. Altogether the number of ways to choose k and m is11× 9 = 99. Thus, there are 99 different divisors of 210 × 38.

9. The answer is 6. Since 4MBN ∼ 4MCD (BN ‖ CD) and MC = 2MB, it followsthat CD = 2BN . This and AB = CD gives AN = BN , so CD = 2AN . On theother hand, 4ANP ∼ 4CDP (AN ‖ CD), so NP

DP= AN

CD= 1

2. This and NP = 3

imply DP = 6.

10. Removing one person that has exactly one coin, one person that has exactly two coinsand one person that has exactly three coins from the group of ten, we get a group ofseven students that altogether have 35− 1− 2− 3 = 29 coins. If everyone of these 7students has 4 coins or less, then altogether they would have ≤ 7× 4 = 28 coins. So,at least one of these 7 students must have 5 coins or more.

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Team Questions

1. The answer is 88. The number of cubes removed is 15+12+10 = 37 (see the solutionof problem 3 below) so the number of cubes that remain is 125− 37 = 88.

2. The answer is 252. The number of cubes removed is 35 + 30 + 26 = 91 (see thesolution of problem 3 below) so the number of cubes that remain is 343− 91 = 252.

3. The answer is n3 − 3n2 + 9n− 7. The number of small (i.e. of size 1× 1× 1) cubescontained in the vertical slot removed from the initial cube is n× (n− 2). Considerone of the horizontal slots removed from the cube. It contains n(n− 2) small cubes,however n− 2 of them have already been counted in the removal of the vertical slot.So, the number of additional small cubes removed by removing the horizontal slotis n(n − 2) − (n − 2). Finally, consider the other horizontal slot. Again it containsn(n − 2) small cubes. However n − 2 of them are contained in the vertical slot,while another n − 3 are contained in the first horizontal slot removed (and not inthe vertical one). So, by removing the second horizontal slot we remove an extran(n − 2) − (n − 2) − (n − 3) small cubes. Thus, the total number of small cubesremoved is

n(n−2)+[n(n−2)−(n−2)]+[n(n−2)−(n−2)−(n−3)] = 3n(n−2)−3n+7 = 3n2−9n+7 .

Hence the number of small cubes remaining is

n3 − [3n2 − 9n+ 7] = n3 − 3n2 + 9n− 7 .

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western australian junior mathematics olympiad 2009 · western australian junior mathematics...

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