B
ALANCING
S
CALES
Problem
In order for the scales to balance the block on the right side must be 10 kg.
It is possible to use the same weights to weigh, for example, 4 kg.
In fact using 2 kg, 3 kg and 5 kg weights it is possible to weigh all but one value from
1 kg to 10 kg. What is that value?
(Note: You don't have to use all three weights.)
Solution
The following weights can be found, with the desired weight indicated in brackets.
Left Side  Right Side
3
2 (1)
2
(2)
3
(3)
2 5
3 (4)
5
(5)
3 5
2 (6)
2 5
(7)
3 5
(8)
2 3 5
(10)
And so it is not possible to weigh 9 kg.
What if you used 1 kg, 3 kg and 6 kg to weigh all the weights from 1 kg to 10 kg?
What about using 1 kg, 2 kg and 7 kg?
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If x + y + z = T, what must be special about x, y and z so that all the weights from 1
kg to T kg can be weighed?
(Hint: Try x + y = T to begin with.)
Problem ID: 92 (Dec 2002)     Difficulty: 1 Star     [
]
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B
EAM
O
F
L
IGHT
Problem
If a light beam is fired through the top left corner of a 2 by 3 rectangular prism block
at a 45 degrees towards the opposite wall, it will emerge from the top right corner.
If the light bounces around a 2 by 4 rectangle, it emerges from corner D.
If a beam of light is fired through the top left corner of a 2 by 50 rectangle, which
corner will it emerge from?
Solution
Consider the two diagrams.
It can be seen from this that rectangles measuring 2 2 2, 2 2 6, 2 2 10, 2 2 14, ... , 2 2 50,
that is, even non-multiples of 4, will finish in the bottom right corner, C.
Will the light ever return to A?
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Investigate other size grids, 3 3 k, 4 4 k, et cetera.
Problem ID: 63 (Jan 2002)     Difficulty: 1 Star     [
]
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B
IGGER
D
IGIT
Problem
For each of the numbers: 41, 83, 32, the first digit is greater in value than the second
digit.
How many 2-digit numbers have this property?
Solution
If we begin to list the numbers in groups: 10; 20,21; 30,31,32; 40,41,42,43; ... ;
90,91,92,93,94,95,96,97,98 ; we can see that the total number of 2-digit numbers,
for which the first digit is greater than the second digit, will be 1 + 2 + ... + 9 = 45.
How many 3-digit numbers exist for which the first digit is greater in value than both
the second digit and the third digit?
Can you generalise for n-digit numbers?
What about 3-digit numbers for which the first digit is greater than the sum of the
second and third digits?
Problem ID: 218 (30 Mar 2005)     Difficulty: 1 Star     [
]
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B
IRDS
A
ND
B
UNNIES
Problem
A cage contains birds and rabbits. There are sixteen heads and thirty-eight feet. How
many birds are there in the cage?
Solution
If we let the number of birds be represented by b and the number of rabbits be
represented by r then we get the following two equations:
b + r = 16 (1)
2b + 4r = 38 (2)
Dividing the second equation by two gives:
b + 2r = 19 (3)
If we now subtract equation (1) from equation (3) we get r = 3, and as b + r = 16 it
follows that the number of birds, b, must be 13.
What if you were told that the number of feet was equal to twice the number of
heads?
What if there were sixteen heads and twenty-eight feet?
Problem ID: 351 (17 Apr 2009)     Difficulty: 1 Star     [
]
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B
IRTHDAY
P
ARTY
Problem
At a birthday party, one-half drank only lemonade, one-third drank only cola, fifteen
people drank neither, and nobody drinks both.
How many people were at the party?
Solution
As 1/2 + 1/3 = 3/6 + 2/6 = 5/6, we know that 1/6 drank neither.
So there must have been 6  15 = 90 people at the party.
Problem ID: 274 (21 Apr 2006)     Difficulty: 1 Star     [
]
CD T
O
T
APE
Problem
Matilda wanted to transfer her favourite CD to tape.
The CD has six songs and the length of the tracks are 7:55, 9:40, 9:15, 12:45, 8:20
and 11:30; a total playing time of 59:25. After changing the order of the songs,
Matilda was able to fit all the songs, without any breaks, on a sixty minute tape. How
did she arrange the songs?
Solution
For the songs to fit on the tape, each side must be almost completely filled, as the
total playing time will leave 35 seconds unused on a sixty minute tape. If the two
longest songs were placed on the same side, 12:45 and 11:30 take up 24:15
together, leaving only 5:45, which is not enough for any other single song. So they
must go on opposite sides.
By trial we get:
Side 1 12:45, 7:55 and 9:15 (29:55 total)
Side 2 11:30, 9:40 and 8:20 (29:30 total)
If the total playing time is less than the length of the tape, is a solution always
possible?
Problem ID: 19 (Oct 2000)     Difficulty: 1 Star     [
]
C
HILDREN
I
N
A C
IRCLE
Problem
A group of children stand holding hands in a large circle and a teacher walks around
the circle giving each child in order a number 1, 2, 3, 4, ...
If number 12 is standing opposite number 30, how many children are there in the
circle?
Solution
Let c be the number of children in the circle. Half the children, 
c
/
2
, will be in-between
child 12 and child 30.
So 12 + 
c
/
2
= 30.
Therefore 
c
/
2
= 18.
That is, there are 36 children in the circle.
Problem ID: 31 (Jan 2001)     Difficulty: 1 Star     [
]
C
HOCOLATE
O
FFER
Problem
To promote the launch of a new Cradbury's chocolate bar they are offering a, "buy
four, get one free" deal. If each chocolate bar costs thirty pence, how much would
ten chocolate bars cost?
Solution
If we get five for the price of four, ten will cost the same as eight. Hence ten
chocolate bars will cost 8  0.30 = £2.40.
How much would seventeen chocolate bars cost?
What if you wanted to buy n chocolate bars?
What if it they offered a, "buy three, get one free" deal?
Can you generalise?
Problem ID: 123 (Oct 2003)     Difficulty: 1 Star     [
]
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