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S
HADED
G
RID
Problem
By only shading whole squares and given that 
is considered different to 
, how many different ways can a 2x2 grid be shaded?
Solution
Consider the following diagrams.
Therefore the number of possible ways to shade a 2x2 square is
1 + 4 + 6 + 4 + 1 = 16.
What about other sized grids?
Hint: Consider the total number of squares making up the grid and don't just think about
square grids; a square is either shaded or not shaded... ;)
What if you used three colours: white, grey and black?
Can you generalise for any size grid and any number of colours?
Problem ID: 87 (Nov 2002)     Difficulty: 1 Star     [
]
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S
HADED
H
EXAGON
Problem
Two congruent equilateral triangles, each with an area equal to 36 cm
2
, are placed on
top of each other so that they form a regular hexagonal overlap.
Find the area of the hexagon.
Solution
The triangles can be split up as follows:
It can be seen that the shaded area is 
6
/
9
of one triangle, that is 
2
/
3
of 36 = 24 cm
2
.
Must the triangles be equilateral?
What will the area of the overlap be if two squares, each with an area of 36 cm
2
, are
overlapped in the same way to form an octagon?
Is the octagon regular?
Problem ID: 150 (Feb 2004)     Difficulty: 1 Star     [
]
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S
HADED
R
ECTANGLE
Problem
What fraction of the diagram is shaded?
Solution
We can split the diagram as follows.
Hence 
12
/
16
3
/
4
of the diagram is shaded
Problem ID: 107 (Mar 2003)     Difficulty: 1 Star     [
]
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S
HADED
S
QUARE
Problem
The top left corner of a square is joined to the midpoint of the bottom edge, the
midpoint of the top edge is joined to the bottom right corner; the top right corner is
joined to the midpoint of the left edge, and the midpoint of the right edge is joined to
the bottom left corner.
What fraction of the square is shaded?
Solution
Consider the following diagram.
By symmetry it should be clear that each triangle is congruent. Hence 4/20 = 1/5 of
the square is shaded.
Problem ID: 229 (10 Jul 2005)     Difficulty: 1 Star     [
]
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S
HADING
P
ATTERN
Problem
In the 3x3 grid, 
6
/
9
is shaded and in the 4x4 grid, 
10
/
16
is shaded.
If a 20x20 grid was shaded in the same way, what fraction would be shaded?
Solution
By collecting data for the first few grids and choosing denominators carefully.
Size of Grid  Fraction Shaded
2x2
3
/
4
3x3
6
/
9
2
/
3
4
/
6
4x4
10
/
16
5
/
8
5x5
15
/
25
3
/
5
6
/
10
It can be seen that 
(n + 1)
/
2n
is shaded.
So in a 20x20 grid, 
21
/
40
is shaded.
Can you prove that 
(n + 1)
/
2n
is shaded on an grid?
(Hint: 1 + 2 + 3 + ... + n = ½n(n + 1).)
Problem ID: 102 (Feb 2003)     Difficulty: 1 Star     [
]
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S
IMPLE
F
RACTIONS
Problem
A fraction whose numerator (top number) is less than its denominator (bottom
number) is called a simple fraction.
1
9
2
9
4
9
5
9
7
9
8
9
There are six simple fractions involving ninths that cannot be cancelled down. How
many simple fractions with a denominator equal to 24 cannot be cancelled down?
Solution
We are looking for the set of integers less than 24 that are co-prime (have no factors
in common) with 24: 1, 5, 7, 11, 13, 17, 19 and 23.
So there are 8 simple fractions with a denominator of 24 that cannot be cancelled
down.
Investigate difference fractions.
(Hint: look at the numbers that are not co-prime to begin with.)
Problem ID: 73 (Apr 2002)     Difficulty: 1 Star     [
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S
KELETON
T
OWERS
Problem
Skeleton towers are generated as follows.
How many blocks would be needed to construct the 100th tower?
Solution
It is possible to split a tower in the following way.
And so the 100th tower can be made into a 100  100 square, requiring 10,000
blocks.
One way of constructing three dimensional skeleton towers is as follows.
How many blocks would be needed to construct the 100th tower now?
Problem ID: 54 (Nov 2001)     Difficulty: 1 Star     [
]
S
QUARE
A
GE
Problem
If you were 35 years old in the year 1225 it would be a very special time
mathematically, because 35
2
= 1225. That is, the square of your age at that moment
is equal to the year. This does not happen very often.
Augustus de Morgan, a famous mathematician, was one of those lucky people and in
1864 he wrote:
"At some point in my life the square of my age was the same as the year."
When was he born?
Solution
42
2
=1764
If he was 42 in 1764, he would have been born in 1764  42 = 1722. As he wrote it
in 1864, he would have been 1864  1722 = 142 years old at the time!!!
43
2
=1849
If he was 43 in 1849, he would have been born in 1849  42 = 1806, making him
1864  1806 = 58 years old when he wrote the statement.
44
2
=1936 (not happened by 1864)
So Augustus De Morgan must have been born in 1806.
It is likely that you know somebody with the same special property with their age.
When were they born and when is their special year?
Problem ID: 32 (Jan 2001)     Difficulty: 1 Star     [
]
S
TRAWBERRY
M
ILK
Problem
A strange cult, with a belief that strawberries contain the Elixir of life, have
succesfully isolated the gene responsible for producing the flavour in strawberry
plants. However, it seems that they have been able to combine this material with
bovine DNA to produce a highly infectious virus that affects cattle. Although the virus
causes no harm to the cow, it will cause the milk that it produces to be tainted with a
strawberry flavour. Once infected there is no cure and all future generations will
produce stawberry flavoured milk.
We were able to intercept an encoded message, sent to all members of the cult,
regarding a final meeting to discuss the release of the virus. We know that the
meeting has been arranged to take place in Athens, Greece. It is imperative that you
decode this cipher so that we can uncover the identity of the mysterious leader of
this cult and arrange his arrest.
LFI TIVZG OVZWVI, WLMZOW NXILMZOW, RMERGVH ZOO NVNYVIH GL GSV
URMZO NVVGRMT RM ZGSVMH, TIVVXV.
Solution
A little analysis of the punctuation should lead us to the suspicion that the message
ends with the location of the meeting: ATHENS, GREECE. We also note the
coincidence of the number of letters and the double E in GREECE. If this is the case
we are dealing with a monoalphabetic substitution code and we could then set about
replacing the known letters:
LFI TIVZG OVZWVI, WLMZOW NXILMZOW, RMERGVH ZOO
??R GREAT ?EA?ER, ??NAL? ?C?ONAL?, ?N??TE? A??
NVNYVIH GL GSV URMZO NVVGRMT RM ZGSVMH, TIVVXV.
?E??ERS T? THE ??NA? ?EET?NG ?N ATHENS, GREECE.
After this, it takes little effort to establish the missing letters:
OUR GREAT LEADER, DONALD MCRONALD, INVITES ALL MEMBERS TO THE
FINAL MEETING IN ATHENS, GREECE.
Although not based on the Caesar cipher (moving each letter by a fixed amount), the
system employed is not based a random translation table. Can you discover the rule?
If you write out a table of known letters:
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Z ? ? W V U T S R ? ? O N M L ? ? I H G F E ? ? ? ?
You should be able to understand, XLWV YIVZPVMT.
Problem ID: 10 (Aug 2000)     Difficulty: 1 Star     [
]
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