Western Australian
Junior Mathematics Olympiad 2009
Individual Questions
100 minutes
General instructions: Each solution in this part is a positive integer
less than 100. No working is needed for Questions 1 to 9. Calculators
are not permitted. Write your answers on the answer sheet provided.
1. Evaluate
55 − 52
√
.
55 − 54
[1 mark]
2. From each vertex of a cube, we remove a small cube whose side
length 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 before
it 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 concentrations
of alcohol. For instance, a concentration of 60% alcohol means it
has 60% pure alcohol and 40% pure water. The pharmacist makes
a mix with 53 litres of alcohol at 90% and 51 litres 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 we
can make 120 different “words”. Suppose we list these words in
alphabetical order and number them from 1 to 120. So ABCDE
gets 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 other
station. Each station has one set of tickets for each other station.
When it added some (more than one) new stations, 46 additional
sets 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 hair
clips 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 have
bought?
[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 AM F ?
D
F
C
d
M
A
B
[4 marks]
Western Australian
Junior Mathematics Olympiad 2009
Team Questions
45 minutes
General instructions: Calculators are (still) not permitted.
Crazy Computers
Rebecca has an old Lemon brand computer which has a defective keyboard. When she types L the letters LOM appear on the screen, and
when she types M she gets OL and when she types O she gets M . We
will abbreviate this to:
Lemon Computer: L → LOM , M → OL, O → M.
So if she types OLO, she gets M LOM M .
A. What will she get on the screen if she presses the Enter key to get
to a new line and then types M LOM M ?
B. Tom has a more modern Raincoat computer which also has a broken
keyboard. 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 to
get a second line and so on until he has 5 lines, what will the last
line 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 there
be in the final line? Try to calculate this without writing down the
final line of letters (which is quite long).
D. Sarah’s computer uses software produced by the giant Megafloppy
Corporation, and is as defective as the others. If she types in H to
get a first line on the screen, then types that line in to get a second
line, 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 she
get 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 until
there are 6 lines on the screen, how many Os will there be in the
last line?
For full marks you must show how to calculate this without actually
writing down the final line.
F. Ben has a Super-Useless Lap Bottom computer, which has mysterious problems with the letters U , X and Z of its keyboard. You will
need to discover its rule by describing what should replace each box
in answering this question (the size of the boxes does not indicate
the number of replacement letters):
Lap Bottom: U →
,X→
,Z→
.
Ben started by entering the letter U , and then like the others entered what he saw on the screen onto the next line; doing this 4
times he finally obtained:
U ZU U XU ZU U ZU ZXU ZU U XU ZU U ZU U XU ZU U XZXU ZU U XU ZU U ZU ZXU ZU U XU ZU
What letter(s) replace each box above? Explain how you got your
answer.
Below, we repeat the above sequence of letters several times, in the
hope it might be helpful in your scratchwork for this problem.
U ZU U XU ZU U ZU ZXU ZU U XU ZU U ZU U XU ZU U XZXU ZU U XU ZU U ZU ZXU ZU U XU ZU
U ZU U XU ZU U ZU ZXU ZU U XU ZU U ZU U XU ZU U XZXU ZU U XU ZU U ZU ZXU ZU U XU ZU
U ZU U XU ZU U ZU ZXU ZU U XU ZU U ZU U XU ZU U XZXU ZU U XU ZU U ZU ZXU ZU U XU ZU
U ZU U XU ZU U ZU ZXU ZU U XU ZU U ZU U XU ZU U XZXU ZU U XU ZU U ZU ZXU ZU U XU ZU
U ZU U XU ZU U ZU ZXU ZU U XU ZU U ZU U XU ZU U XZXU ZU U XU ZU U ZU ZXU ZU U XU ZU
U ZU U XU ZU U ZU ZXU ZU U XU ZU U ZU U XU ZU U XZXU ZU U XU ZU U ZU ZXU ZU U XU ZU
U ZU U XU ZU U ZU ZXU ZU U XU ZU U ZU U XU ZU U XZXU ZU U XU ZU U ZU ZXU ZU U XU ZU
U ZU U XU ZU U ZU ZXU ZU U XU ZU U ZU U XU ZU U XZXU ZU U XU ZU U ZU ZXU ZU U XU ZU
Individual Questions Solutions
1. Answer: 62.
52 (53 − 1)
55 − 52
√
= 2√
5 5−1
55 − 54
124
=
= 62
2
[1 mark]
2. Answer: 84. Initially there are 2 · 4 + 4 = 12 edges. By removing
a small cube from a vertex (of which there are 8), we increase the
number 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 2
before 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 6
ends in 6. So the answer is 6.
Alternative. Observe that
62 = 36 ≡ 6
n
∴6 ≡6
2009
∴2
2009
×3
2009
×6
(mod 10)
(mod 10) for any integer n ≥ 1
2009·2
=6
≡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 alcohol over the total volume of liquid expressed as a percentage:
3 90
1 50
·
+
·
total volume of alcohol
5
100
5
100
=
3 1
total volume
+
5 5
50
90
+1·
3·
100
100
=
3+1
10(3 · 9 + 1 · 5)
=
4 · 100
10(27 + 5)
=
4 · 100
80
=
= 80%
100
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, 24
beginning with B and 24 beginning with C. These all come before
DECAB. Of those beginning with D there are 24/4 = 6 beginning
with DA, 6 beginning with DB and 6 beginning with DC. These
also 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 word
beginning with D, and the one before it is DECAB. So DECAB’s
number 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 new
stations 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 is
1, 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. Then
the total of her purchases is:
4x + 32
so that the average price of her purchases is
4(x + 1) + 28
28
4x + 32
=
=4+
,
x+1
x+1
x+1
which is an integer, if 28 is divisible by x + 1. Hence, x + 1 must be
one of 1, 2, 4, 7, 14 or 28 (the divisors of 28), i.e. x must be one of
0, 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
(n − 2) · 180 =
2 · 143 + (n − 1) · 2
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, in
particular, the largest, must be less than 180◦ . Now, for n = 20, the
largest 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 = F D. Since d is the perpendicular bisector of
AM , it is the locus of points equidistant from A and M .
So AF = M F .
Since the area of the square is 64 cm2 , its side length is 8 cm. Hence
applying Pythagoras’ Theorem to 4F DA and 4CF M , we have
82 + y 2 = (8 − y)2 + 42
= 82 + y 2 − 16y + 16
∴ 16y = 16
y=1
Take the parenthesising of the vertices of a figure, as a convenient
shorthand for the figure’s area, so that (XY Z) means “the area of
figure XY Z”. Then
(AM F ) = (ABCD) − (ABM ) − (F DA) − (CF M )
= 64 − 21 (8 · 4 + 8 · 1 + 4 · 7)
= 64 − 12 · 68
= 64 − 34 = 30.
So 4AM F has area 30 cm2 .
Alternative 1. Instead, let x = F C. As before, deduce AF =
M F , and that square has side length 8 cm. Applying Pythagoras’
Theorem to 4F DA and 4CF M , we have
82 + (8 − x)2 = 42 + x2
82 + 82 − 2 · 8x + x2 = 42 + x2
2 · 82 − 42 = 2 · 8x
2 · 82 − 42
2·8
= 8 − 1 = 7.
∴x=
The rest of the solution proceeds like the first solution.
Alternative 2. Since the area of the square is 64 cm2 , its side length
is 8 cm. Since M is the midpoint of BC, M B = 4 cm. Applying
Pythagoras’ 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
F M 2 = F X 2 + XM 2
= F X 2 + XA2
= F A2
So, F M = F A. Now deduce x or y as above and hence deduce that
F M 2 = F A2 = 65
∴ F X 2 = F A2 − XA2
√
= 65 − (2 5)2
= 65 − 20 = 45
∴ (AM F ) = 12 AM · F X
√ √
= 12 · 4 5 · 45
√
√
=2 5·3 5
= 6 · 5 = 30
[4 marks]
Team Questions Solutions
Crazy Computers
A. Answer: OLLOM M OLOL. Putting a little space between the
replacement letters, we have M LOM M → OL LOM M OL OL.
[4 marks]
B. Answer: RSRRSRSRRSRRS.
(1st line)
R → RS
→ 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 the
line, 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 of
the first entered line (namely R). Then
`n+1 = `n + `n−1 , n ≥ 0,
where `0 = 1, `1 = 2. So we have:
`2
`3
`4
`5
`6
`7
`8
`9
`10
`11
`12
= `1 + `0 = 2 + 1 = 3
= `2 + `1 = 3 + 2 = 5
= `3 + `2 = 5 + 3 = 8
= `4 + `3 = 8 + 5 = 13
= `5 + `4 = 13 + 8 = 21
= `6 + `5 = 21 + 13 = 34
= `7 + `6 = 34 + 21 = 55
= `8 + `7 = 55 + 34 = 89
= `9 + `8 = 89 + 55 = 144
= `10 + `9 = 144 + 89 = 233
= `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 the
replacements H → HG, G → GH.
[6 marks]
E. Answer: 11. Let `n , mn , on be the number of Ls, M s 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
1
1
0
1
0
2
1
0
1
3
1
2
1
4
3
2
3
5
6
5
5
6
11
10
11
So there are 11 Os in the sixth line.
[12 marks]
F. Answer: U → U ZU, X → ZX, Z → U X. Looking at the final
line we see the following repetitions.
U ZU U XU ZU U ZU ZX U ZU U XU ZU U ZU U XU ZU U X ZX U ZU U XU ZU U ZU ZX U ZU U XU ZU
If U → U ZU then passing from the third to fourth line, we require
Z → U X. So the remaining letters must be the result of Z, i.e. we
must have X → ZX. Confirming this
U → U ZU
→ U ZU U XU ZU
→ U ZU U XU ZU U ZU ZXU ZU U XU ZU
→ U ZU U XU ZU U ZU ZXU ZU U XU ZU U ZU U XU ZU U XZXU ZU U XU ZU U ZU ZXU ZU U XU ZU
[10 marks]
Western Australian
Junior Mathematics Olympiad 2008
Individual Questions
100 minutes
General instructions: Each solution in this part is a positive integer
less than 100. No working is needed for Questions 1 to 9. Calculators
are not permitted. Write your answers on the answer sheet provided.
1. Below are three different views of a child’s building block and a single view of a different block.
1
2
Which is the different block?
3
4
[1 mark]
2. Some horses and some jockeys are in a stable. In all, there are 71
heads 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 person
caught at least one fish. The following five statements each have a
label from 1 to 16. What is the sum of the labels of all the statements
which 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 first
thrower while the third thrower achieves a distance 20% greater
than the first thrower. The fourth thrower throws it 10% further
than the third person. If the total distance of the four throwers is
90 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 of
side 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 Mary
is now. Furthermore, Ann is twice as old as Mary was when Ann
was 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 balls
which you take out one by one. Each time you take out a blue ball
somebody puts 100 frogs into the barrel and each time you take out
a green ball the person puts in 72 frogs. Finally, when you have
removed 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 the
mid-point of AD, C is the F
mid-point of DE, A is the
mid-point of EF , and M
is the midpoint of AF .
If the area of 4AM B is
6 cm2 , how many cm2 is the
remaining area of 4DEF ?
M
A
E
C
B
[3 marks]
D
9. Five grandmothers go to a cafe to eat cake. The cafe sells 4 different
types of cake. Each grandmother chooses two different cakes. They
find 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.
x
A square is divided into three
pieces of equal area as shown. The
distance between the parallel lines is
1 cm. What is the area of the square
in cm2 .
z
1
y
[4 marks]
Western Australian
Junior Mathematics Olympiad 2008
Team Questions
45 minutes
General instructions: Calculators are (still) not permitted.
How to multiply on Titania
The Titans are an intelligent race who live on the planet Titania. They
use the same numerals as us for the integers and their addition and subtraction is the same as ours. However, they don’t use multiplication.
Instead of ×, they have an operation called ‘star’ which has the following 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 a
and the column labelled by b should
be a ∗ b, where each of a, b range over
0, 1, . . . , 4.
Some entries have been filled in to get
you started.
∗
0
1
2
3
4
0 1 2 3 4
0 1 2 3 4
1
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.
Individual Questions Solutions
1. Answer: 3. The views 1, 2 and 4 can be explained thus: the star
face is opposite the square face, and the pentagon face is opposite
the triangle (the face opposite the octagon is unknown). Then 2 is
obtained from 1 by rotating in an axis perpendicular to the centre
of the octagon face so that the triangle face becomes the front face.
And 4 is obtained by flipping 1 upside down and then rotating in
the vertical axis so that the octagon is the front face. Flipping 2 so
that the octagon is on the top and the star on the right, the triangle
face should be behind and the pentagon at the front. So 3 is the
different block.
[1 mark]
2. Answer: 28. Let j be the number of jockeys and let h be the number
of 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 caught
8 fish meets the criteria, so that the statements labelled 1, 2 and 16
need 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 is
false, then all 4 friends caught 3 or more fish, which implies there
are 12 or more fish, but there are only 11 fish (contradiction). Thus,
at least one friend caught fewer than 3 fish. Hence the statement
labelled 4 must be true, and since this is the only statement that
must be true, the sum of such labels is 4.
[2 marks]
4. Answer: 20. Let x be the distance thrown by the first person in
metres. Then the second throws it 0.98x, the third, 1.2x and the
fourth 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 of
the 7 cm cube, 5 will have exactly two faces painted green. So there
are 12 × 5 = 60 such cubes.
[2 marks]
6. Answer: 24. Let Ann’s age be a and Mary’s age be m, and write
a1 , a2 and m1 , m2 be their ages at the two other times. Write
a1 = 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 she
was m1 and] when Ann [was a1 and] was as old as Mary is
now [namely m]. Furthermore, Ann [is a and] is twice as
old as Mary [was when she was m2 which] was when Ann
[was a2 and] was six years older than Mary is now [namely
m].
Thus from the given information we have:
a = 4m1
a1 = m
=⇒ a
= 4(m + k) =⇒ a − 4m − 4k = 0
(1)
=⇒ 4a − 4m + 4k = 0
(2)
= 2(m + `) =⇒ a − 2m − 2` = 0
(3)
=⇒ a + k = m
a = 2m2
=⇒ a
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)
2 · (6) − (5)
∴ a = 24,
Hence, Ann is 24 (and Mary is
5
8
· 24 = 15).
[3 marks]
7. Answer: 14. Say we started with b blue balls and g green balls. So
we 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 hand
side 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
(4F AD) = (4EAD) = 4S
1
AB
AM
= =
AF
2
AD
∠M AB = ∠F AD
∴ 4M AB ∼ 4F AD, by the Similar 4s SAS Rule
∴ (4M AB) =
(4F AD)
=S
22
Now S = 6 and (4DEF ) − (4AM B) = 7S = 42.
[3 marks]
9. Answer: 21. Say the cakes have prices a, b, c and
d. The possible
totals 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 different
amounts the amounts they paid are 5 of these 6 sums. Notice that
the sum of the first and second pairs is a + b + c + d, so is the sum
of the third and fourth, so is the sum of the fifth and sixth. This
means 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 and
9 + 12 = 21 so a + b + c + d = 21, which is the amount I pay for my
4 cakes.
[4 marks]
10. Answer: 13. With x, y and z as shown, the middle strip is a parallelogram made up of two congruent triangles of height x and base
y, and this is a third of the total area, i.e.
xy = 13 x2
∴ y = 13 x
By Pythagoras’ Theorem,
z 2 = x2 + (x − y)2
= x2 + ( 23 x)2
=
∴z=
13 2
x
9
√
1
13 x.
3
Looking at the two congruent triangles making up the parallelogram
another way: they have height 1 and base z, i.e. the area of the
parallelogram is also given by:
√
1 × z = 1 × 13 13 x = 13 x2
√
∴ 13 = x
Thus, the area of the square is x2 = 13 cm2 .
[4 marks]
Team Questions Solutions
How to multiply on Titania
A.
∗
0
1
2
3
4
B.
0
0
1
2
3
4
1 2
3
4
1 2
3
4
1 1
1
1
1 0 −1 −2
1 −1 −3 −5
1 −2 −5 −8
a ∗ 1 = a ∗ 0 + 1 − a,
= a + 1 − a,
=1
[8 marks]
by (ii)
by (i)
[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,
= −19.
D.
by (i)
[4 marks]
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,
=5
from rule in E
3∗3 = (3 ∗ 3) ∗ 3
= (−3) ∗ 3
= −3 + 3 − −3 · 3,
=9
from rule in E
(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
a∗n
= 4a − 6a2 + 4a3 − a4
..
.
n 2
k+1 n
= na −
a + · · · + (−1)
ak + · · · + (−1)n+1 an
2
k
= 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
2
= 1 − 1 − 1 − (1 − a) (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]
Western Australian
Junior Mathematics Olympiad 2007
Individual Questions
100 minutes
General instructions: Each solution in this part is a positive integer
less than 100. No working is needed for Questions 1 to 9. Calculators
are not permitted. Write your answers on the answer sheet provided.
1. I’m a two digit number. I’m one less than a multiple of 8 and three
less 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 horizontal
and a vertical line into four rectangles. The
areas of three of them are shown. What is
the area of the whole rectangle?
6
9
8
[2 marks]
3. From a group of girls and boys, fifteen girls depart, leaving twice
as many boys as girls. Then 45 boys depart, leaving five times as
many girls as boys. How many girls were there originally?
[2 marks]
4. A regular pentagon has five diagonals and they are all of the one
length. A regular hexagon has nine diagonals and they are of two
different lengths.
If we consider all of the diagonals of a regular polygon which has
twenty sides, how many different lengths will there be? [2 marks]
5. There are three cards on a table, each marked with a positive whole
number. Alice says “The sum of these two is 54”. Bill points to
another 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) and
the product of their ages is 1848. How old is the teenager?
[3 marks]
8. An arithmetic progression is a sequence of numbers such that the
difference of any two successive numbers is a constant. For example,
3, 5, 7 is an arithmetic progression of three numbers, with common
difference 2. Also, 3, 5, 7 are prime. The sum of the numbers is
3 + 5 + 7 = 15, which is the lowest possible sum for an arithmetic
progression of primes of length 3.
Find an arithmetic progression of four numbers, all of which are
prime, which has the lowest sum of any arithmetic progression of
primes 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 of
the nails to make squares. Two such squares are
shown in the diagram (the larger square has side
length 2 cm). How many square centimetres is the
total 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 go
along the same road from X to Y . Alfa crawled at a constant speed
of 12 m/h (metres per hour) till he reached Y . Romeo started out
at 8 m/h but after two hours he realised he was falling behind so
hitched a ride on a passing turtle called Toyota, who was on her
way to Y at a constant speed of 20 m/h. Toyota and Romeo soon
caught 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]
Individual Questions Solutions
1. Answer: 39. The number has the form 8n − 1 and so belongs to
the set {7, 15, 23, 31, 39, 47, . . . }. It also has the form 7n − 3 and so
belongs 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, so
their areas are proportional to their heights, namely 8 to 6. The
rectangles on the right also have areas proportional to their heights,
so the unknown area is 9 × 8/6 = 12, and the area of the whole
rectangle is 6 + 9 + 8 + 12 = 35.
3. Answer: 40. Say there were g girls and b boys originally. Then
b = 2(g − 15),
g − 15 = 5(b − 45),
∴ g − 15 = 5 2(g − 15) − 45 ,
after the first departure
(1)
after the second departure
(2)
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 number 1 we see that they increase from the shortest diagonal, from
vertex 1 to 3, to the longest, from vertex 1 to vertex 11, then decrease again, in a symmetric fashion, down to the diagonal from
vertex 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 get
the system of equations
a + b = 54
(3)
a + c = 41
(4)
b + c = 33
(5)
∴ 2(a + b + c) = 54 + 41 + 33 = 128,
∴ a + b + c = 64.
adding (3), (4) and (5)
6. Answer: 87. The largest possible value of d is 4 = 5 − 1. So the
largest possible value of c is 15 = 4 × 4 − 1. So the largest possible
value of b is 44 = 3 × 15 − 1. Hence the largest possible value of a
is 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 the
teenager is 14. (There is more than one possibility for the other
three ages!)
8. Answer: 56. The arithmetic progression 5, 11, 17, 23 consists of
primes, has sum 5 + 11 + 17 + 23 = 56 and has common difference
6. There is no arithmetic progression of length 4 starting at 3 and
with common difference 2, 4 or 6. There is none starting at 5 with
common 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 progressions
exist. There are none starting at 11 with difference 2 or 4, nor 13.
Since 56/4 = 14, there can’t be an arithmetic progression with
smaller sum starting with a prime greater than 13. Thus 5, 11, 17,
23 has the smallest sum (56) of an arithmetic progression of primes
of length 4.
9. Answer: 52. There are nine 1 × 1 squares with area 1, four
√2
√ 2×
squares 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 the
total 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 Romeo
and Toyota overtake Alfa t hours later. Then they have travelled
16 + 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’s
back, a total of 5 hours.
(c) Romeo travelled 16 m on his own and 60 m on Toyota’s back, a
total of 76 m.
Western Australian
Junior Mathematics Olympiad 2007
Team Questions
45 minutes
General instructions: Calculators are (still) not permitted.
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 them
to a very long corridor with 100 doors. The doors are also numbered
from 1 to 100. When a student goes down the corridor for each door
whose number is divisible by the number on his/her t-shirt, he/she
either 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 by
their t-shirt numbers.
For example, if the doors are all closed to start with then when
student 40 goes down the corridor he will open doors 40 and 80. If
student 80 follows him she will close door 80.
A. If all the doors are closed, and students 7, 28 and 84 go down the
corridor, which doors will be open?
B. All the doors are closed except number 42. Which students should
he send down to get all the doors closed?
C. Suppose all the doors are closed and all the students go down the
corridor. 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 go
down the corridor, explain how you can predict which doors will
be 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 saying
which doors are open and which are closed.
F. The doors from 1 to 49 are closed but somebody has left all the
others open. Which students should he send down to ensure all the
doors are closed?
G. Now the doors from 1 to 49 are open and the rest are closed.Which
students should he send down to get all the doors closed?
H. All the doors were closed but the Number 1 student has just walked
down the corridor and gone home. The teacher now has no way
of closing door number 1. But is it possible to use the remaining
students to close all the other doors? Explain your reasoning.
Team Questions Solutions
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 by
students 1, 7, 49; and 51 and 53 will be closed because they have 4
and 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 be
closed. To decide whether a given door is closed or not find the
highest power of 2 which divides the door number. If this is an odd
power of 2 (2, 8 or 32) then the door will be closed. Otherwise it
will be open.
[7 marks]
E. The doors 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100 be open because
these are perfect squares and such numbers have an odd number
of factors, and so will be opened or closed by an odd number of
students.
[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 open
door, and continue doing this till all doors other than 1 are closed.
This is the same as sending all students except those with square
numbers on their shirts.
[8 marks]
Western Australian
Junior Mathematics Olympiad 2006
Individual Questions
100 minutes
General instructions: Each solution in this part is a positive integer
less than 100. No working is needed for Questions 1 to 9. Calculators
are not permitted. Write your answers on the answer sheet provided.
1. The diagram below shows the train line from the outer suburb of A
to the inner city station E, with the distance between stations shown
in kilometres. After leaving a station the train travels at an average
speed of 30 km/h for the first kilometre, then at 60 km/h until it
reaches the next station. It spends 2 minutes at each station. How
many minutes will elapse between the train leaving A and arriving
at E?
3 km
A
4 km
B
6 km
C
1 km
D
E
[1 mark]
Solution. 60 km/h = 1 km/minute. So when the train travels at
60 km/h, it takes 1 minute to cover a km, and when it travels at
30 km/h, it takes 2 minutes to cover a km. So the minutes that will
elapse 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 at
30 km/h, the second number in each bracket is the number of kilometres travelled at 60 km/h (which equals the time in minutes to
cover that distance) and each unbracketed 2 is the waiting time at
a station. Answer: 24.
2. Find the two-digit prime number that is 2 less than a perfect square
that is 2 less than a prime.
[1 mark]
Solution. Let the prime be p.
odd (2 is the only even prime).
square is odd; possibilities are:
a prime both 2 less than it and
Answer: 79.
Then p has 2 digits and hence is
It is 2 less than a square. So the
25, 49, 81. Of these only 81 has
2 more than it. So p must be 79.
3. For how many positive integers (whole numbers) n are n, 3n and
n/3 all three-digit integers?
[2 marks]
Solution. The least that n/3 can be is 100 (the smallest 3-digit
positive integer) and the most that 3n can be is 999 (the largest
3-digit positive integer). If 3n = 999 then n/3 = 111. So n/3 can
be 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 equilateral
triangle. What is the measure in degrees of the angle CED?
E
A
B
D
C
[2 marks]
Solution. Since 4ABE is equilateral, its angles are all 60◦ . Hence
∠EAD = 60◦ + 90◦ . Now 4ADE is isoceles. Thus
1
∠AED = (180◦ − ∠EAD)
2
= 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 some
on DVD, some on video cassette and some on both. If the store
owns a total of 1000 DVDs and video cassettes, how many films are
available on both video cassette and DVD?
[2 marks]
Solution. By the pigeon hole principle, 80 films are available on
both video cassette and DVD (the films are the pigeon holes; put
920 of the 1000 DVDs and video cassettes in the 920 available pigeon
holes; the remaining 80 DVDs or video cassettes must go in a pigeon
hole 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 for
it the next day. Her score of 96 raised the class average from 71 to
72. How many students (including Brenda) took the test?
[3 marks]
Solution. Let n be the number of students in the class. Without
Brenda the average is 71 and hence the total marks without Brenda
is 71(n − 1). The average with Brenda’s mark of 96 is 72. So
71(n − 1) + 96
n
72n = 71(n − 1) + 96 = 71n − 71 + 96
n = 25
72 =
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 the
duration of its flights. When flying from Perth to Sydney it takes
four hours while from Sydney to Perth it takes five hours. Assuming
that the wind is at a constant speed and from the west (i.e. in the
direction from Perth to Sydney) for both flights, what is the speed
of the wind?
[3 marks]
Solution. Let w be the speed of the wind (from west to east). Then
travelling east the plane’s speed is (810 + w) km/h and travelling
west its speed is (810 − w) km/h. In general velocity v, distance d
and time t are related by v = d/t. Let d now denote the distance
from Perth to Sydney. Then
d
4
d
810 − w =
5
810 + w =
Eliminating d by dividing (1) by (2) we have
810 + w
5
=
810 − w
4
4(810 + w) = 5(810 − w)
4w = 810 − 5w
9w = 810
w = 90
Therefore the wind speed w is 90 km/h. Answer: 90.
(1)
(2)
8. The five faces of a right square pyramid all have
the same area. The height of the pyramid is 3 m.
What is its total surface area in square metres?
6
3m
?
[3 marks]
Solution.
Let a be the sidelength of the square base, so that the
area of the base is a2 which is also the area of each
isosceles triangular face. So each triangular face has
height 2a, and the total surface area is 5a2 . Using
Pythagoras’ Theorem and the pyramid height of 3 m,
we have
6
3m
?
(2a)2 − (a/2)2 = 32
(4 − 14 )a2 =
15 2
4 a
=
3
4
· 5a2 = 9
5a2 =
4
3
· 9 = 12
So the total surface area of the pyramid is 12 m2 .
Answer: 12.
9. Jasper glues together 504 cubes with 1 cm edges to make a solid
rectangular-faced brick. If the perimeter of the base of the brick is
64 cm, what is the height of the brick?
[4 marks]
Solution. 504 = 9 × 56 = 9 × 7 × 8 = 23 × 32 × 7. The volume
of the brick L × B × H = 504 cm3 , whereas the perimeter of the
base 2(L + B) = 64 cm. So we need to write 504 as the product of
three positive integers L, B, H such that L + B = 32. We see that
18 × 14 × 2 = 504, so that (in cm) L = 18, B = 14, H = 2 is a
solution.
To see the solution H = 2 is unique (not required for the competition), 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 is
even. If L and B are both odd then their product is 63 leading to
L = 21 (to satisfy the inequality) and B = 3 which do not sum to
32. Hence L and B must both be even,√whence H = 1 or 2. If
H = 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 the
Knockers won by 3 points. This was despite the Beagles kicking
16 scoring shots compared to the Knockers 14. It was also noticed
that the number of behinds kicked by the Beagles was greater than
the number of goals kicked by the Knockers, and the number of
goals kicked by the Beagles was greater than the number of behinds
kicked by the Knockers. How many points did the Beagles score?
Note: in the AFL game there are two ways to score: behind s score
1 point each, and goal s score 6 points each.
For full marks, explain how you found your solution.
[4 marks]
Solution.
Let . . . B = the
b = the
K = the
k = the
number
number
number
number
of
of
of
of
goals
behinds
goals
behinds
kicked
kicked
kicked
kicked
by
by
by
by
the
the
the
the
Beagles,
Beagles,
Knockers, and
Knockers.
From the given information, we have:
B + b = 16
(3)
K + k = 14
(4)
6K + k = 6B + b + 3
B>k
b>K
(5)
(6)
(7)
Since B and b are integers, from (6) and (7) we have
B ≥k+1
b≥K +1
(8)
(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
b=K +1
(10)
(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 + 3
10K = 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 = 51 points
Note the problem can also be solved by systematically listing the
possible scores for the Beagles (17 of them) and the possible scores
for the Knockers (15 of them) and then doing a careful elimination
. . . and there are many other ways. The more satisfying solutions
are ones that set up some inequality at the beginning that reduce
the possibilities to a small list prior to doing an elimination.
Answer: 51.
Western Australian
Junior Mathematics Olympiad 2006
Team Questions
45 minutes
General instructions: Calculators are (still) not permitted.
"
Consider a garden table made of 15 square tiles
in 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 a
diagonal?
Solution. 12 tiles get broken.
B. Give the dimensions of two different rectangular tables that get
nine tiles broken when they are cracked along a diagonal.
Note. Remember that a square is also a rectangle. Also, note that
for this and subsequent questions, an 8 × 6 table, for example, is
considered 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 ten
tiles broken when they are cracked along a diagonal? Write down
their 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 of
the 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 broken
is the length of the rectangle.
F. For what sort of dimensions does the crack go through corners of
tiles inside the rectangle?
Solution. For dimensions that have a common factor (larger than
1).
G. How many tiles are cracked when the diagonal does not go through
any corner of a tile inside the rectangle? Explain your reasoning.
Solution. If the diagonal does not go through any corner of a
tile inside the rectangle, a tile is cracked when and only when the
diagonal enters a new column or enters a new row.
Say the table has m rows and n columns. The first tile broken is
in the first row and the first column. Then there are m − 1 further
rows and n − 1 further columns. Therefore the number of tiles
cracked 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 a
7×4 rectangle which has 7+4−1 = 10 broken tiles. Reintroduce the
removed common factor and the number of broken tiles is 10 × 8 =
80.
Alternatively: In general, the number of cracked tiles in a table
with dimensions m × n is m + n − hcf(m, n) where hcf(m, n) counts
the number of times the diagonal enters a new column and new
row at the same time, which, if we think of the ‘first’ tile broken
as being the bottom lefthand tile, is the number of times the crack
goes through the bottom lefthand corner of a tile. So for a 56 × 32
rectangle, 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 size
of table.
Solution. If the two dimensions have no common factor, add the
two numbers and subtract 1. If the two dimensions do have a
common factor, remove the highest common factor), consider the
reduced table as above, then multiply the result by the removed
HCF. Algebraically, the number of tiles broken is
m
n
+
− 1 hcf(m, n) = m + n − hcf(m, n).
hcf(m, n) hcf(m, n)
Alternatively: use the general argument in H to obtain the general formula m + n − hcf(m, n).
Western Australian
Junior Mathematics Olympiad 2005
Individual Questions
100 minutes
General instructions: Each solution in this part is a positive integer
less than 100. No working is needed for Questions 1 to 9. Calculators
are not permitted. Write your answers on the answer sheet provided.
(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 is
the area of the trapezium ABCD in square centimetres?
A
4 cm
B
S
c
S
cc
S
c
c
S 12 cm2 c
S
c
S
c
S c
S
cC
D
P
11 cm
[1 mark]
(2) A library has 6 floors. There are 10 000 more books on the
second floor than the first. The number of books on the third
floor is the same as the number on the second. There are 10 000
fewer books on the fourth floor than the third and twice as many
books on the fifth floor as there are on the fourth. On the sixth
floor there are 4 000 fewer books than on the fifth. Coincidentally the number of books on the sixth floor is the same as the
number on the first. Altogether, how many thousands of books
are 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 the
basketball court to make triangles. Below is a diagram of a
triangle using all 20 sticks. It has sides of 5 m, 7 m and 8 m. All
triangles with sides of 5 m, 7 m and 8 m (even mirror images as
shown) are considered to be the same.
5m
7m
8m
7m
5m
8m
Altogether, how many different triangles could they make, one
at a time, each time using all 20 sticks?
[2 marks]
(5) What is the size of the angle, in degrees, between the hands of
a clock when the time is ten past eleven?
[2 marks]
(6) A plane is due to leave Perth at midnight and arrive at Tokyo
at 12:29 pm, but it departs 49 minutes late. What percentage
increase in speed is required in order that it will arrive exactly
on time?
[3 marks]
(7) Esther has 20 coins in her purse. They are 10c, 20c and 50c
coins and the total value is $5. If she has more 50c coins than
20c 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 first
shop?
[3 marks]
(9) Given a square ABCD, circular arcs centred at B and D are
drawn from A to C. Now draw diagonal BD
√ to cut these arcs
at X and Y , respectively. If XY = 12 − 6 2, what is the area
of the square ABCD?
A
B
Y
X
D
C
[4 marks]
(10) Farmer Brown runs a dairy farm with cows, sheep and goats.
Dabbling in mathematics in his spare time, he noticed that the
numbers of each animal were different prime numbers. He also
observed that if he multiplied the number of cows by the total number of cows and sheep, he obtained a number just 120
greater than the number of goats. How many goats are there?
For full marks, explain how you found your solution.
[4 marks]
Western Australian
Junior Mathematics Olympiad 2005
Team Questions
45 minutes
General instructions: Calculators are (still) not permitted.
A number appears on your computer screen. If you push the A key
the number is decreased by 3, if you push the B key it’s increased by
3 and if you press the C key the number is halved. You want to use a
sequence 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 is
4, and show the path.
B. Find a starting number between 50 and 60 whose shortest path is
5, and show the path.
C. Find a starting number between 30 and 40 whose shortest path is
6, and show the path.
D. Which starting number between 40 and 50 has the longest “shortest
path”?
E. Find a number which has two shortest paths, one beginning with
the A key and one with the B key.
F. Find a starting number where the “add 3” operation must be used
to get the shortest path.
G. What starting numbers can never get to 1?
H. What starting number between 500 and 1000 would have the shortest path?
I. In this question, the numbers on the screen don’t have to be integers. If we start with any number then these two sequences of key
strokes 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 height
of triangle ABP . Thus 12 = 12 × 4h so h = 6. The area of the
trapezium is then
4 + 11
AB + DC
Areatrap = h ×
=6×
= 45.
2
2
(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 of
books, 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,
XXXX
YYYY
ZZZZ
Y XXX Z
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-digit
number, X 6= 0. Thus
X + Y = 10
(1)
and the carry k = 1. Looking at the second column now, we
have X + Y + Z + 1 = X + 10m, i.e. Y + Z + 1 = 10m, where
m is the carry to the third column. Again, Y + Z + 1 can’t be
as big as 20, so
Y + Z + 1 = 10
(2)
and m = 1. We see that this pattern is repeated for the third
and fourth columns, and so the carry to the fifth column which
is just Y , implies Y = m = 1. Adding equations (1) and (2) we
have
X + Y + Y + Z + 1 = 20
X + Y + Z = 20 − 1 − Y
= 18
(Of course, we had enough information to evaluate that X = 9
and Z = 8.)
(4) We’re looking for sets of three positive integers x, y, z satisfying
x + 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 also
need x > y + z; otherwise we can’t form a triangle. Listing
systematically we get:
9,9,2
9,8,3
9,7,4
9,6,5
8,8,4
8,7,5
8,6,6
7,7,6.
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 and
12 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 be
d kilometres. The usual speed is therefore d/749, the required
speed after the late departure is d/700. The required percentage
increase in speed is therefore,
1
d/700 − d/749
1 × 100% = 749
−
× 100%
d/749
700 749
749 749 =
−
× 100%
700 749
= (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 respectively. We have,
x + y + z = 20
10x + 20y + 50z = 500
(3)
(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 left
hand side of (5) is at least 32, which is impossible, so z = 7 and
we quickly find y = 2 and x = 11. So the number of ten cent
coins 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 entered
with 2y + 2 dollars.
She leaves the last shop with $0, so (using y = 0), she entered
shop 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 and
the square has area 36.
(10) Let the number of cows, sheep and goats be c, s and g respectively. The farmer notices that
c(c + s) = g + 120.
We know c, s and g are primes. If they were all odd then the
left hand side of the above equation would be even and the right
odd which is impossible. So one of them is even and therefore
equals 2.
If c = 2 the left hand side of the equation is still even and the
right hand side odd.
If g = 2 the equation becomes c(c + s) = 122 = 2 × 61. This is
impossible 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 and
the 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 40
such that the shortest path from that starting number has length
4:
32, 16, 8, 4, 1
The sequence of operations giving this path is CCCA. An important observation here is that C effects a more rapid descent for
numbers x that are even and greater than 4, but A requires just
one step to convert 4 to 1 (compared with performing CC).
(B) There is just one path with a starting number between 50 and 60
such that the shortest path from that starting number has length
5:
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 length
6:
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 shortest
path 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 to
get 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 result
is 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 shortest
path is 512.
(I) We need to show that ACCB and BCBC always produces the
same result. Let x be the starting number. Then ACCB produces
x−3
(x − 3)/2 /2 + 3 =
+3
4
x − 3 + 12
x+9
=
=
4
4
On the other hand, BCBC produces
x + 3
(x + 3)/2 + 3 /2 =
+ 3 /2
2
x + 3 + 6
=
/2
2
x+9
=
4
Thus irrespective of what starting number x is chosen. The result
after each of the sequences ACCB and BCBC is (x + 9)/4.
Western Australian
Junior Mathematics Olympiad 2004
Individual Questions
100 minutes
General instructions: No working need be given for Questions 1 to
9. Calculators are not permitted. Write your answers on the answer
sheet provided. Each solution is a positive integer less than 100.
(1) Paul likes dogs. At present all his adult dogs are dalmations
while some of his puppies are dalmations and some are not. In
all 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 , AF G are right-angled with
hypotenuses AC, AD, AE, AF , AG respectively. If AB = 2
and BC = CD = DE = EF = F G = 3, what is the length of
AG?
E
D
F
C
G
A
B
[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 no
adjacent 2s?
[2 marks]
(6) In a circle with centre O, a chord AB is extended to a point
C such that the length of BC is equal to the radius of the
circle. CO is drawn and extended to a point D on the circle’s
circumference. If angle BCO = 15◦ , what is angle AOD?
[3 marks]
(7) A chessboard is made up of 64 squares each with 3 cm sides. A
circle is drawn with its centre at the centre of the chessboard
and radius 12 cm. How many of the 64 squares lie entirely inside
the circle?
[3 marks]
(8)
There was a young lady called Chris
Who when asked her age answered this:
“Two thirds of its square
Is 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 can
make three different shaped rectangles, like this:
What is the least number of tiles he needs, so that, using all
the tiles each time, he can make five different rectangles?
[4 marks]
(10) There was little traffic that day, too little to interfere with the
steady progress of the 3 km column of armoured vehicles. Heading the column, Tom turned his jeep and drove back to check
the rear. All was well and he was able to maintain a steady
speed there and back without any delays. On returning to his
lead position, Tom noted that the column had advanced just
4 km while he was away. How far had he driven in that time?
Give reasons for your solution.
[4 marks]
Western Australian
Junior Mathematics Olympiad 2004
Team Questions
45 minutes
A. On Genevieve’s computer screen there are two dials, as shown
below.
s
s
0
8
4
1
3
0
1
7
2
6
2
3
5
4
Each time she clicks her mouse the black dot on each dial moves
one position clockwise. So that if she clicked her mouse 12 times
the dot on the left hand dial would be in position 2 and the dot
on the right hand dial would be in position 3. What is the least
number of clicks she needs to make to get from the position above
to one with the left dot at 1 and the right at 8?
B. On Ahmed’s computer the dots start in the positions shown below.
0
4
1
3
0
8
2s
1
7
2
s6
3
5
4
How many clicks must he make to get the left dot at 1 and the
right at 4?
C. The left hand dial on Nelson’s computer is numbered from 0 to
16 and the right dial from 0 to 23. Initially the left dot is at
position 9 and the right at position 11. How many clicks does
Nelson need to make to get the left hand dot in position 7 and
the right at position 15?
D. Yenicca’s screen has 3 dials, which start in the positions shown
below.
0
4
0
1
3
2s
5
1
s4
2
3
0
s
6
1
5
2
4
3
When she clicks her mouse each dot moves one position clockwise.
How many clicks will she need to make to finish with the left
hand dot at position 0, the middle at position 2 and the right at
position 4?
E. Roxanne’s screen also shows 3 dials which start in the same position as Yenicca’s. However when she clicks the left mouse button
only the left hand and middle dots move one position clockwise.
If she clicks the right mouse button the middle and right hand
dots move one position clockwise. How many times must she
click 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 adult
dalmations. The rest of his 7 dalmations must be puppies; so he
has 4 dalmation puppies. (This question can also be answered
using Venn diagrams.)
[1 mark]
(2) If a hen and a half lay an egg and a half in a day and a half,
6
then 6 hens will lay 1.5
× 1.5 = 6 eggs in a day and a half. So
12
[1 mark]
in 12 days they lay 1.5 × 6 = 48 eggs.
(3) In order to avoid confusion about the meaning of expressions
like AC 2 , we use the convention in our proof that |AC| denotes
the length of the line segment AC.
By Pythagoras’ Theorem (several times)
√
√
22 + 32 = 13
|AC| =
q√
√
|AD| = ( 13)2 + 32 = 22
q√
√
|AE| = ( 22)2 + 32 = 31
q√
√
|AF | = ( 31)2 + 32 = 40
q√
√
|AG| = ( 40)2 + 32 = 49 = 7
So the length of AG is 7.
Alternative solution. More elegantly (again using Pythagoras’ Theorem several times), we have
|AG|2 = |F G|2 + |AF |2
= |F G|2 + |EF |2 + |AE|2
..
.
= |F G|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 square
of 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 7200
exactly.
[2 marks]
(5) There is 1 number of the given shape that contains no 2, 5
which contain one 2, 6 which contain two 2s, and 1 (21212)
which contains three 2s. Altogether there are 1 + 5 + 6 + 1 = 13
such numbers.
[2 marks]
(6) Triangle OBC is isosceles, so angle BOC is 15◦ . Angle ABO is
an external angle of the triangle OBC and so equals 2 × 15◦ =
30◦ . Triangle AOB is also isosceles; so angle BAO also equals
30◦ .
C
B
A
O
D
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 as
solid dots are √
inside or outside
√ the circle. Their distance from
2
2
the centre
√
√ is 3 + 3 = 18, by Pythagoras’ Theorem, and
18 > 16 which is the radius of the circle; so the points are
outside the circle.
t
t
× ×
× × ×
× × ×
t
t
Hence the squares in the first quadrant of the circle that are
marked × 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 y 3 . (Both x and y are integers.)
Then
2 2
x = y3.
3
Hence
2x2 = 3y 3
(1)
Now 2 divides the LHS . . . so 2 divides the RHS and hence 2
divides y. Thus 8 divides the RHS. So 2 divides x.
Now try a similar idea with 3: 3 divides the RHS . . . so 3 divides
the LHS and hence 3 divides x. Thus 9 divides the LHS (and
hence the RHS). So 3 divides y. So 81 divides the RHS and
hence 9 divides x.
Thus lcm(2, 9) = 18 divides x and lcm(2, 3) = 6 divides y. Now
let x = 18α and y = 6β. Then substituting in (1) we get:
2 × 182 α2 = 3 × 63 β 3
which reduces to
α2 = β 3
Suppose for a prime p, pi is the largest power of p that divides
the LHS. Then i is a multiple of 2. Also since pi is the largest
power of p that divides the RHS, i is a multiple of 3. Thus i is
a multiple of 6. Either i is 0 for every prime p or i is at least 6
for 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 and
by today’s standards Chris would not be a young lady and a
candidate for the Guinness Book of Records. Thus α = 1 and
x = 18. So Chris is 18.
[3 marks]
(9) The question is essentially asking for the least number that can
be written as the product of two numbers in 5 different ways
without regard to order. The answer is 36. He can then make
rectangles with dimensions 1 × 36, 2 × 18, 3 × 12, 4 × 9 and
6 × 6. One way that we could have found this is by observing
that if the number of tiles n has prime factorisation
p11 p22 · · · pnn
then each factor of n has the form
pi11 pi22 · · · pinn
where 0 ≤ ij ≤ j , for j = 1, . . . , n. Hence n has
(1 + 1)(2 + 1) · · · (n + 1)
distinct factors. So we require (1 + 1)(2 + 1) · · · (n + 1) to be
either 2 × 5 − 1 = 9 if n is a perfect square, or 2 × 5 = 10 if n
is not a perfect square.
Now 10 = 2 × 5 suggests 1 = 4, 2 = 1; the least n = p41 p2
occurs for p1 = 2, p2 = 3, namely n = 48. However, if n is a
perfect square, in which case the j must all be even, we see
that 9 = 3 × 3 suggests 1 = 2, 2 = 2, and the least n = p21 p22
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 (in
km/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
3
+
,
v+u 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). Now
substitute those expressions for u and v in our first equation:
T =
3
x
4
+
T
T
+
3
x
4
−
T
T
=
3T
3T
+
x+4 x−4
Dividing both sides by T (which is non zero), followed by rearranging we get:
3
3
+
x+4 x−4
(x + 4)(x − 4) = 3 (x − 4) + (x + 4)
1=
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 drove
8 km.
Alternative solution. Let x, u, v, T be as defined above. Further, let y be the distance Tom travelled to the rear of the
column and t the time needed for that. Then vt = y and
ut = 3 − y, so v/u = y/(3 − y). On the other hand, for the
total distances vT = 2y + 4 and uT = 4, so v/u = (2y + 4)/4.
Thus, y/(3 − y) = (2y + 4)/4, which gives y 2 + y − 6 = 0, i.e.
y = 2 (we disregard the possibility y = −3 since y > 0). The
total distance is then 2y + 4 = 8 km.
[4 marks]
Solutions to Team Questions
(A) After 8 clicks the dots are in positions 3 and 8 so the right dot is
OK. A further 9 clicks leaves the right dot at 8 but the left is now
at 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 us
to 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 and
2. Then after 15 more sets of 24 clicks we arrive at 0 and 2. The
total 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 dials
in positions 2, 3, 4 so that the third dial is right. 7 more clicks
move us to positions 4, 4, 4 with dial 3 still OK and it stays OK
with each set of 7 clicks. After a total of 5 + 5 × 7 clicks we get
positions 2, 2, 4 with dials 2 and 3 OK. Now a set of 42 more clicks
will leave these dials in the correct position and advances the first
dial 2 positions. We find that after 4 sets of 42 clicks the first
dial is in position 0 and we are done. The total number of clicks
needed is 5 + 5 × 7 + 4 × 42 = 208.
[10 marks]
(E) Say that L is the number of times Roxanne clicks the left button
and R is the number of times she clicks the right button. Then
if L = 3 and R = 0 we get to positions 0, 1, 6 which puts the left
hand button in the correct position. Then L = 3 and R = 5 moves
us 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 these
dials in the correct positions, but moves the middle dot. If we
perform both these operations we move the middle dot 12 positions
which is the same as leaving it alone so we should be adding a
multiple of 5 to L or a multiple of 7 to R but not both. Trying
both 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 should
click the left button 3 times and the right 19 times.
[10 marks]
WA Junior Mathematics Olympiad 2003
Individual Questions
100 minutes (one hour and
40 minutes)
General instructions: No working need be given for Questions 1 to
9. Calculators are not permitted. Write your answers on the answer
sheet provided.
(1) In a test of 20 questions 5 marks are given for each correct
answer and 2 are deducted for each incorrect answer. Alan did
all the questions and scored 58. How many correct answers did
he have?
[1 mark]
(2) Jane typed a 6-digit number into a faulty computer in which
the 1 (one) key was broken. The number appearing on the
screen was 2003, possibly with some blank spaces. How many
different 6-digit numbers could Jane have typed?
[1 mark]
(3) A whole number between 1 and 99 is not greater than 90, not
less than 30, not a perfect square, not even, not a prime, not
divisible 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! means
1 × 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 as
old as he is now. How old is Douglas?
[2 marks]
1
(7) In a triangle ABC, ∠C = 90◦ . A perpendicular is produced
from the midpoint D of AB to meet the side BC at E. The
length 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 with
n 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 to
D. ED intersects AC at P.
How many times larger is the area of the parallelogram ABCD
than the area of the triangle AEP?
Give reasons for your solution.
[4 marks]
Western Australian
Junior Mathematics Olympiad 2003
Team Questions
45
minutes
A. Suppose you have a string of 2003 beads which you cut between
two beads as close to the middle as possible, so that you now
have two strings, one with 1001 beads and one with 1002. The
beads are glued onto the string so they won’t slide off. You
now take the shorter string of the two and cut it as close to the
middle as possible and then keep repeating the process, at each
step choosing a shorter string. You may choose either if they
are 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 5
pieces.
C. How many strings would you have if the original string had
999,999 beads?
D. Can you give a formula for the longest string and a formula for
the shortest string which finish in n pieces?
E. What would happen if you cut the original string of 2003 beads
as nearly as possible into thirds, and then took a smallest length
to continue?
Western Australian Junior Mathematics Olympiad
November 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 with
2003.
So there are 6 choices for the place of the first 1 and then 5 for the second 1, a total of
30. 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 89
are also excluded.
Since it is not divisible by 3 or 5, 33, 35, 39, 45, 51, 55, 63, 65, 69, 75, 85 are also
excluded.
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 units
digit 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 only
possibilities 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 be
2x − 8 years old. So x = 32 (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. Hence
x = 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 get
6x + 6y + 3z = 117. We need to subtract 3y to get 6x + 3y + 3z. But if you subtract
69 = 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. DP C + DP A = 1/2ABCD
But DP C is similar to and twice the size of AP E. Hence 4AP E+DP A = 1/2ABCD.
DP A + AP E = 1/4ABCD. Hence 3AP E = 1/4ABCD so ABCD is 12 times the
area of AEP .
Another way to do it is to join PB. Then AEP = EP B and DP O = BP O, while
AOD = AOB. Hence AP D = AP B = DP B 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.
Step 3 gives 4 strings, shortest length 250.
Step 4 gives 5 strings, shortest length 125.
Step 5 gives 6 strings, shortest length 62.
Step 6 gives 7 strings, shortest length 31.
Step 7 gives 8 strings, shortest length 15.
2
Step 8 gives 9 strings, shortest length 7.
Step 9 gives 10 strings, shortest length 3.
Step 10 gives 11 strings, shortest length 1.
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 into
thirds and continue with a smallest piece, you get in successive cuts:
3 strings, shortest length 667
5 strings, shortest length 222
7 strings, shortest length 74
9 strings, shortest length 24
11 strings, shortest length 8
13 strings, shortest length 2
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 can
start with has length 3k and the largest string you can start with has length 3k+1 − 1.
3
WA Junior Mathematics Olympiad 2002
Individual Questions
100 minutes (one hour and 40 minutes)
General instructions: 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.
p
1. Simplify
(1 + 24 + 25 )(1 + 23 + 24 ) + 26
√
.
1 + 23
2. Four different positive integer numbers a, b, c, d satisfy the following relations:
1 1 1
1 1 1
1 1 1
+ + =1 ,
+ + =1 ,
+ + =1.
a a a
a b
c
b d d
Find d.
[1 mark]
[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 the
other 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 = 10
cm. Find the length of BC if it is known that the point D lies on the perpendicular bisector
of 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 and
the 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 = 9
cm, 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. I
want to put them in a row of 6 boxes. The orange cats are friends and have to be put side
by side, but the black cats fight and must not be side by side. In how many ways can I
arrange 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 painted
in different colours. Is it true that the numbers of red, blue and green squares must be the
same? Give reasons for your answer.
[4 marks]
1
Western Australian
Junior Mathematics Olympiad 2002
Team Questions
45 minutes
1. You are given 9 coins of the same denomination, and you know that one of them is
counterfeit and that it is lighter than the others. You have a pan balance which means
you can put any number of coins on each side and the balance will tell you which side
is heavier, but not how much heavier. Explain how you can find the counterfeit coin
in exactly two weighings.
[4 marks]
2. If you are given 25 coins of the same denomination, and you know that one of them is
counterfeit and that it is lighter than the others, explain how to find the counterfeit
coin 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 lighter
than the others. What is the least number of weight trials on a pan balance necessary
to 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 that
its weight is different from that of a genuine coin, however it is not known whether the
counterfeit coin is lighter or heavier than a genuine one. Show that by using at most
3 weighings in the pan balance you can identify the counterfeit coin and determine
whether it is lighter or heavier than a genuine coin.
[16 marks]
1
WA Junior Mathematics Olympiad 2002
Solutions to the Individual Questions
1. Since 1 + 24 + 25 = 49, 1 + 23 + 24 = 25 and 26 = 64, it follows that
√
p
(1 + 24 + 25 )(1 + 23 + 24 ) + 26
49 · 25 + 64
99
√
=
=
= 33 .
3
3
3
1+2
Answer: 33
1 1
2
+ = .
b
c
3
Hence b ≥ 2 and c ≥ 2. If both b > 2 and c > 2, then (since the numbers are different and
a = 3) b ≥ 4 and c ≥ 4, so 1/b + 1/c ≤ 1/2, impossible. Thus, either b = 2 or c = 2. If
c = 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.
2. The first equation gives 3/a = 1, so a = 3, and the second equation becomes
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 and
Bill is first, impossible.
Case 2. Y is right, while X and Z are wrong. Then Claire is not last, Ahmad is not first
and 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 and
Claire 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 4AM D ∼
= 4OM D (AM = M O, M D = M D, 6 AM D =
6 OM D). 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 and
z in the second equation gives y 2 /2 = y + 3y, so y 2 = 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 is
divisible by 3, the sum of its digist must be divisible by 3, so N must have at least three
digits 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
9
a
that
= , so a2 = 9 · 25, i.e. a = 15.
25
a
(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.
1
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 the
pair O1 , O2 , they must be side by side (later we will take into account the fact that O1 and
O2 can swap places in O). Disregarding for a moment the fact that B1 and B2 must not be
side 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 that
O1 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 by
side. If we denote the pair B1 and B2 by B, the number of ways we can order W1 , W2 , O, B
is 4 · 3 · 2 · 1 = 24. Since B1 and B2 can swap places in B, and O1 and O2 can swap places
in O, the total number of such cases is 24 · 4 = 96. Thus, the number of ways we can order
the 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 of
these numbers is equal to ±1, ±2, ±3, ±4, ±6 or ±12. If some of the numbers (that must
be A or E) is equal to 12 or −12, then all other numbers must be equal to ±1, so at least
two 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 be
equal to ±1, so at least two of them will be equal, impossible again. Thus, all numbers are
between −4 and 4. Next, if some of the numbers is equal to 4 or −4, some other number
must be equal to 3 or −3, and the remaining three numbers must be equal to ±1, so at least
two 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 the
product of the five numbers is positive, we cannot have three of them negative, so the only
possible case is A = −2, B = −1, C = 1, D = 2, E = 3. Then A + B + C + D + E = 3, so
a + 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 and
any two of them have a common side, so they must be painted differently. Hence one of
the three squares with vertex A must be red, another must be blue, and the third must be
green.
This applies to any of the 8 vertices of the cube, so there must be at least 8 red small
squares, 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 small
squares of each coulour.
2
WA Junior Mathematics Olympiad 2002
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 of
the 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, the
counterfeit coin is in the third group. So, with one weight trial we determine a group of 3
coins that contains the counterfeit coin.
For the second trial, choose any two of these 3 coins and place them on the two pans. If
the pans balance, the third coin is the counterfeit one; if not, then the leighter coin is the
counterfeit one.
2. Divide the coins into 3 groups, the first two groups containing 9 coins each, while the third
group contains 7 coins. (Other divisions are possible, e.g. 8+8+9.) Place the coins of group
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, the
counterfeit coin is in the third group. In the latter case, take two coins from group 1, say, to
get 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 third
group contains 16 coins. (There are other possible divisions that are good enough, in fact any
division so that there is no group with more than 27 coins is a good one.) Place the coins of
group 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, the
counterfeit coin is in the third group. In the latter case, take 9 coins from group 1, say, to
get 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) to
find the counterfeit coin using 3 weighings. Thus with a total of 4 weighing one can determine
the counterfeit coin.
Let us now show that 4 is the minimal number of weighings that guarantee finding the
counterfeit coin under all circumstances.
Suppose in the first trial we weigh two groups of k coins each; the remaining coins are then
70 − 2k. If k = 23, then the third group contains 24 coins; if k < 23, then the third group
contains 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 (in
any circumstances) from trial 1 is to determine a group of 24 coins containing the counterfeit
one.
In the same way, using a second trial the best one can achieve (in any circumstances) is to
determine 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 group
of 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 of
group 1 on one pan of the pan balance and the coins of group 2 on the other pan. For a
second 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 is
lighter or heavier than a genuine coin. For example, if group 1 is heavier (or lighter) than
group 2, and group 2 has the same weight as group 3, then the counterfeit coin is in group 1
and it is heavier (resp. lighter) than a genuine coin. If group 1 is heavier than group 2, and
group 2 is lighter (it cannot be havier) than group 3, then the counterfeit coin is in group 2
and it is lighter than a genuine coin.
Consider now the group of 3 coins found after the first two trials to contain the counterfeit
coin, and assume for clarity that the counterfeit coin is determined to be lighter than a genuine
one (the other case is similar). For a third trial place two of the coins on the two pans of the
pan 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 of
group 1 on one pan of the pan balance and the coins of group 2 on the other pan. There are
two possibilities.
Case 1. The pans balance. Then all coins in groups 1 and 2 are genuine and the counterfeit coin
is in group 3.
Let the coins in group 3 be A, B, C. For a second weight trial, place coins A and B on
one 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 genuine
coin) 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 stage
we will already know whether the counterfeit coin is lighter or heavier (e.g. if the pan
containing 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 show
which of them is the counterfeit one. E.g. if the previous step showed that the counterfeit
coin is lighter than a genuine one, and A is lighter than B, then A is the counterfeit
coin.
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 is
heavier than a genuine one), and let these in the other pan be A0 , B 0 , C 0 (if one of these
coins is counterfeit, then it is lighter than a genuine one).
For a second weight trial, place e.g. A and A0 on one pan and B and B 0 on the other.
Then we have the following possibilities.
Subcase 2.1. The pans balance. Then A, B, A0 , B 0 are all genuine coins, and either C or C 0 is
counterfeit.
For a third weight trial, place C on one pan and a genuine coin (say, A) on the
other. If the pans balance, then C 0 is the counterfeit coin and it is lighter than a
genuine one. If the pans do not balance, then C is the counterfeit coin and it must
be heavier than a genuine one.
Subcase 2.2. The pan containing A, A0 is heavier than the pan containing B, B 0 . Then A0 and B
must be genuine, so the counterfeit coin is either A or B 0 .
For a third weight trial, place A on one pan and a genuine coin (say, C) on the
other. If the pans balance, then B 0 is the counterfeit coin and it is lighter than a
genuine one. If the pans do not balance, then A is the counterfeit coin and it must
be heavier than a genuine one.
Subcase 2.3. The pan containing B, B 0 is heavier than the pan containing A, A0 . This case is
considered in the same way as Subcase 2.2.
2
Western Australian Junior Mathematics Olympiad
October 27, 2001
Individual Questions
General instructions: 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. N degrees Celsius is the same temperature as 95 N + 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 and
b7c = 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. Evaluate
d13.5 + 2.7 × b3.8ce. (1 mark)
3. A certain number of points are marked on the circumference of length 2001 of a circle
in such a way that each marked point is distance 1 from exactly one marked point
and distance 2 from exactly one marked point, all distances being measured around
the 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 the
bisector of angle BAC, so that L is a point on BC. Let M be the point on AB such
that LM is perpendicular to AB. If LM = 3 and M B = 4, find AB. (2 marks)
5. How many solution pairs x, y) are there of the equation 2x + 3y = 763 if both x and
y are positive integers? (2 marks)
6. In a computer game, you have to score the largest possible number of points. You
score 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 to
score 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 + 19y 2 − 6y + 10. (3 marks)
8. A shop sells hamburgers which contain some of the following: meat burger, vegetable
burger, 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 diagonal
BD = 72 cm. If N is the midpoint of AB and M and P are the intersection points
of BD with N C and AC, respectively, find M P . (3 marks)
10. Two buses start travelling at the same time – bus 1 from city A to city B, and bus 2
from 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 start
travelling back along the same road and with the same speeds. They meet again 4
km from B. Find the distance between the cities A and B. Explain how you obtained
your 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?
2
Western Australian Junior Mathematics Olympiad
October 27, 2001
Problem Solutions
1. We must solve 9N/5 + 32 = N . So 9N + 160 = 5N and hence 4N = −160. The solution to
this 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 of
consecutive gaps totals 3 units. Since there must be 667 pairs of gaps, and so 1334 gaps
altogether and therefore 1334 points.
4. From 4LM B one finds LB 2 = 32 + 42 = 25, so LB = 5. Next, 4AM L ∼
= 4ACL (AL =
◦
6
6
6
6
AL, M AL = CAL, AM L = 90 = ACL), so CL = LM = 3. Now observe that
BC
MB
4ABC ∼ 4LBM (6 ABC = 6 LBM , 6 ACB = 6 LM B), therefore
=
. This gives
AB
LB
BC × LB
8×5
AB =
=
= 10.
MB
4
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 must
be divisible by 3. This means x has the form 3X + 2 with X a non-negative integer. Thus
2(3X + 2) + 3(2Y + 1) = 763, which simplifies to X + Y = 126. Then X can be any integer
from 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 multiple
of 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 + 19y 2 − 6y + 10 = x2 − 8xy + 16y 2 + 3(y 2 − 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 can
get 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. Each
of 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 be
served with or without carrot, giving a total of 42 possibilities.
9. First, notice that 4N BM ∼ 4CDM (6
CD
= 1, i.e. DM = M B = 36 cm.
NB
DC
DP
6 ABP = 6 CDP ), so
=
=
PB
AB
DM − DP = 12 cm.
DM
=
MB
Next, we have 4ABP ∼ 4CDP (6 AP B = 6 CP D,
1
. That is, DP = 13 DB = 24 cm. Hence M P =
2
N M B = 6 CM D, 6 N BM = 6 CDM ). Hence
10. Let v1 and v2 be the speeds of bus 1 and bus 2 respectively, let t1 and t2 be the times at which
they pass each other and let x be the distance between the towns. By considering the first time
7
x−7
v2
x−7
they pass we see that
= v1 ,
= v2 , which implies that
=
. By considering
t1
t1
v1
7
2x − 4
v2
2x − 4
x+4
the second time they pass we get
= v1 ,
= v2 , which implies that
=
.
t2
t2
v1
x+4
x−7
2x − 4
Thus we have
=
, which gives x2 = 17x and so x = 17.
7
x+4
1
Solutions to Team Questions
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) which
is 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 number
is 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 the
required 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)
1
= (m + 1) a + (1 + 2 + . . . + m) = (m + 1) a + m (m + 1)
2
1
=
(m + 1)(m + 2a) .
2
One of m + 1 and m + 2a must be odd and the other even, so n has an odd factor. This means
n is not a power of 2.
2
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.
Western Australian Junior Mathematics Olympiad
October 28, 2000
Problem Solutions
1. The answer is 100. You can see this by drawing a diagram. Alternatively, using
Pythagoras’
Theorem or trigonometry, you can show
√
√ that the area of the large triangle
3
100 3
is
and the area of each small triangle is
. The answer follows since
4
4
√
100 3
√4
3
4
= 100 .
2. Looking at (c) and (d) together we see that A must be guilty regardless of D’s guilt
or innocence. Part (a) then implies that B is guilty. So now we know that both A
and B are guilty. Part (b) says that C is guilty or A is innocent, but we know A is
not 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 first
digit 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. A
smarter way is to notice that the sum can only equal 0 in 8 ways: the numbers with
a plus sign being 1 and 5 and the numbers with a minus sign being 2 and 3 (which
can 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 sum
from positive to negative or vice versa. Therefore the number of positive solutions is
16 / 2 = 8.
5. In triangles CM H and AN H we have M H = N H, 6 CHM = 6 AHN (opposite
angles) and 6 CM H = 6 AN H = 90◦ . Thus triangles CM H and AN H are congruent
by the Angle-Side-Angle rule. Thus CH = AH and so CN = AM . Angles AM B
and CN B both equal 90◦ and angles BCN and BAM both equal 90◦ - 6 ABC. So
triangles CN B and AM B are congruent (Angle-Side-Angle again), and therefore
AB = CB = 20.
1
6. The answer is 12 km per hour. Indeed, at a certain time the ships will be 5 km
apart for the first time. Denote by A0 and B0 their respective positions at that
time, and let C0 be the point on the line determined by the direction of the ship A
such that B0 C0 is perpendicular to this line. By Pythagoras’ Theorem, A0 C0 = 4
km. If A0 and B 0 are the positions of the ships A and B after 24 minutes, then the
distance travelled by A is A0 A0 = 8 × 24
= 16
km. Hence A0 C0 = 4 − 16
= 45 . If
60
5
5
D0 is the point on the line determined by the direction of the ship A such that B 0 D0
is perpendicular to this line, then by Pythagoras’ Theorem for 4B 0 D0 A0 one gets
D0 A0 = 4, so B 0 B0 = D0 C0 = D0 A0 + A0 C0 = 4 + 45 = 24
km. Therefore the speed of
5
24
24
ship B is 5 : 60 = 12 km.
7. The answer is 240. Clearly 100 is divisible by 2 (and 5), so we have to check the
integers 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 and
5. Dividing the sequence 101, 102, . . . , 1000 into 30 separate sets of 30 consecutive
integers and using the fact that in each of these sets of 30 integers there will be
again exactly 8 integers not divisible by 2, 3 and 5, one gets that the total number
of 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 of
38 are 1 = 30 , 3 = 31 , . . ., 38 . Every divisor of 210 × 38 has the form 2k × 3m for
some k = 0, 1, . . . , 10 and m = 0, 1, . . . , 8. So, there are 11 different ways to choose
k and 9 different ways for m. Altogether the number of ways to choose k and m is
11 × 9 = 99. Thus, there are 99 different divisors of 210 × 38 .
9. The answer is 6. Since 4M BN ∼ 4M CD (BN k CD) and M C = 2M B, it follows
that CD = 2BN . This and AB = CD gives AN = BN , so CD = 2AN . On the
P
other hand, 4AN P ∼ 4CDP (AN k CD), so N
= AN
= 12 . This and N P = 3
DP
CD
imply DP = 6.
10. Removing one person that has exactly one coin, one person that has exactly two coins
and one person that has exactly three coins from the group of ten, we get a group of
seven students that altogether have 35 − 1 − 2 − 3 = 29 coins. If everyone of these 7
students 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.
2
Team Questions
1. The answer is 88. The number of cubes removed is 15 + 12 + 10 = 37 (see the solution
of 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 the
solution 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) cubes
contained in the vertical slot removed from the initial cube is n × (n − 2). Consider
one 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 slot
is n(n − 2) − (n − 2). Finally, consider the other horizontal slot. Again it contains
n(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 in
the vertical one). So, by removing the second horizontal slot we remove an extra
n(n − 2) − (n − 2) − (n − 3) small cubes. Thus, the total number of small cubes
removed 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 .
3