Junior Mathematical Olympiad

Tuesday 11 June 2024

© 2024 UK Mathematics Trust

supported by

England & Wales: Year 8 or below

Scotland: S2 or below
Northern Ireland: Year 9 or below

Instructions
1. Do not open the paper until the invigilator tells you to do so.
2. Time allowed: 2 hours.
3. The use of rulers, set squares and compasses is allowed, but calculators and protractors are
forbidden. You are strongly encouraged to use geometrical instruments to construct large, accurate
diagrams for geometry problems.
4. Start each question on an official answer sheet on which there is a QR code.
5. If you use additional sheets of (plain or lined) paper for a question, please write the following in the
top left-hand corner of each sheet. (i) The question number. (ii) The page number for that question.
(iii) The digits following the ‘:’ from the question’s answer sheet QR code.
Please do not write your name or initials on additional sheets.
6. Write on one side of the paper only. Make sure your writing and diagrams are clear and not too
faint. (Your work will be scanned for marking.)
7. Arrange your answer sheets in question order before they are collected. If you are not submitting
work for a particular problem, please remove the associated answer sheet.
8. Your answers should be fully simplified, and exact. They may contain symbols such as 𝜋, fractions,
or square roots, if appropriate, but not decimal approximations.
9. Section A - Only answers are required.
10. Section B - You should give full written solutions, including mathematical reasons as to why your
method is correct. Just stating an answer, even a correct one, will earn you very few marks; also,
incomplete or poorly presented solutions will not receive full marks.
11. Scoring Each question in Section A is worth 1 point. Each question in Section B is worth up to
10 points. Answers in Section B can be awarded partial marks depending on on the clarity of the
participant’s mathematical presentation
12. To accommodate candidates sitting in other time zones, please do not discuss the paper on the
internet until 8am GMT on Thursday 13 June, when the solutions video will be released at
ukmt.org.uk/competitions-papers. Candidates in time zones more than 5 hours ahead of GMT must
sit the paper on Wednesday 12 June (as defined locally).

Enquiries about the Junior Mathematical Olympiad should be sent to:

challenges@ukmt.org.uk
www.ukmt.org.uk
Junior Mathematical Olympiad Tuesday 11 June 2024

Section A
Try to complete Section A within 30 minutes or so. Only answers are required.

A1. What is the sum of the first nine primes?

A2. Forty-two cubes of side-length 1 cm are stuck together to form a solid cuboid. The
perimeter of the base of the cuboid is 16 cm. What is its height, in cm?

A3. 𝑃𝑄𝑅𝑆 is a square which has been divided into four regions: two identical rectangles,
one square of area 9 cm2 and a second square of area 25 cm2 . What is the area of square
𝑃𝑄𝑅𝑆, in cm2 ?

A4. Note that 49 = 4 × 9 + 4 + 9. How many two-digit numbers are equal to the product of
their digits plus the sum of their digits?

A5. The difference between an interior angle of a regular polygon and an exterior angle of
the same polygon is 150°. How many sides does the polygon have?

A6. When a group of five friends met up, Alice shook hands with one person; Bill shook
hands with two people; Cara shook hands with three people; Dhriti shook hands with
four people. How many people did Erin shake hands with?

A7. The time is 20:24 (expressed in 24-hour time). What is the angle between the hour hand
and the minute hand on an accurate analogue clock, in degrees?

A8. We make a sequence of diagrams. The first diagram consists

of a single node. The second diagram is made from the first
diagram by drawing two edges of length 1 cm from that node,
and putting a node at the other end of each new edge. After
that, we make each new diagram from the previous diagram
by adding two new edges to each node, with these new edges
each having half the length of the edges that were added in the
previous diagram. We also attach a node to the end of each
new edge. The first four diagrams are shown. Find the total
length of all the edges in the fifth diagram, in cm.

A9. The numbers 𝑥 and 𝑦 satisfy the equations:

7 5 7
xy = x(y + 1) = y(x + 1) =
6 3 2
What is the value of (𝑥 + 1)(𝑦 + 1)?
2024 )
A10. What is the last digit of 2 (2 ?

© 2024 UK Mathematics Trust www.ukmt.org.uk

Junior Mathematical Olympiad Tuesday 11 June 2024

Section B

Your solutions to Section B will have a major effect on your result.

Concentrate firstly on one or two Section B questions and then write out full solutions (not
just brief ‘answers’), including mathematical reasons as to why your method is correct.
You will have done well if you hand in full solutions to two or more Section B questions.
Do not hand in rough work.

B1. What is the smallest positive integer that only contains the digits 0 and 1, and is divisible
by 36?

B2. Natasha and Rosie are running at constant speeds in opposite directions around a running
track. Natasha takes 70 seconds to complete each lap of the track and meets Rosie every
42 seconds.
How long does it take Rosie to complete each lap?

B3. The positive integers from 1 to 𝑛 (𝑛 ≥ 2) inclusive are to be spaced equally around the
circumference of a circle so that:
(a) no two even numbers are adjacent;
(b) no two odd numbers are adjacent;
(c) no two numbers differing by 1 are adjacent.
What is the smallest value of 𝑛 for which the above is possible?

B4. My piggy bank contains 𝑥 pound coins and 𝑦 pennies and rattles nicely. If instead it
contained 𝑦 pound coins and 𝑥 pennies, then I would only have half as much money.
What is the smallest amount of money my piggy bank could contain?

B5. A regular octagon 𝐴𝐵𝐶𝐷𝐸 𝐹𝐺𝐻 has sides of length 1. 𝐴 𝐻

The lines 𝐴𝐶 and 𝐻𝐹 meet the line going through 𝐷 and 𝐸
at 𝑃 and 𝑄 respectively. 𝐵 𝐺
What is the length of the line 𝑃𝑄?
𝐶 𝐹

𝑃 𝐷 𝐸 𝑄

B6. Let 𝐴 be the set of 2𝑛 positive integers 1, 2, 3, . . . , 2𝑛, where 𝑛 ≥ 1.

For which values of 𝑛 can this be split into 𝑛 pairs of integers in such a way that every
pair has a sum which is a multiple of 3?
As always in Olympiad problems such as this, you also need to explain why no other
values of 𝑛 are possible.

© 2024 UK Mathematics Trust www.ukmt.org.uk

 MT
UK
MT

UK
UKMT

United Kingdom
Mathematics Trust

Junior Mathematical Olympiad

© 2024 UK Mathematics Trust

supported by

Solutions
Junior Mathematical Olympiad 2024 Section A Solutions

Section A

A1. What is the sum of the first nine primes?

Solution 100
The first 9 primes are 2, 3, 5, 7, 11, 13, 17, 19, and 23.

A2. Forty-two cubes of side-length 1 cm are stuck together to form a solid cuboid. The
perimeter of the base of the cuboid is 16 cm. What is its height, in cm?

Solution 6 cm
We need to factorise 42 as 𝐿 × 𝑊 × 𝐻, with 2𝐿 + 2𝑊 = 16. The only solutions are
𝐿 = 7, 𝑊 = 1, 𝐻 = 6 and 𝐿 = 1, 𝑊 = 7, 𝐻 = 6.

A3. 𝑃𝑄𝑅𝑆 is a square which has been divided into four regions: two identical rectangles,
one square of area 9 cm2 and a second square of area 25 cm2 . What is the area of
square 𝑃𝑄𝑅𝑆, in cm2 ?

Solution 64 cm2
There are only two ways to divide up a square into two unequal squares and two identical
rectangles:

9
25 25

In each case, the side length of 𝑃𝑄𝑅𝑆 must be the sum of the side lengths of the smaller
squares, in this case 3+5=8, for an area of 64 cm2 .

A4. Note that 49 = 4 × 9 + 4 + 9. How many two-digit numbers are equal to the product
of their digits plus the sum of their digits?

Solution 9
A two-digit number 𝐴𝐵 has the value 10𝐴 + 𝐵, so we need 10𝐴 + 𝐵 = 𝐴𝐵 + 𝐴 + 𝐵. Subtracting
𝐵 from each side, and dividing by 𝐴 (since 𝐴 must be non-zero, to be a two-digit number),
we obtain 𝐵 = 9. This condition turns out to be sufficient too (check for yourself that,
assuming 𝐵 = 9, we can be sure that the condition will definitely hold), so we get 9 solutions,
corresponding to 𝐴 being one of 1,...,9.

© 2024 UK Mathematics Trust www.ukmt.org.uk 2

Junior Mathematical Olympiad 2024 Section A Solutions

A5. The difference between an interior angle of a regular polygon and an exterior angle
of the same polygon is 150°. How many sides does the polygon have?

Solution 24
Suppose the shape has 𝑛 sides. Then the exterior angle is 360°/𝑛 and the interior angle is
180° − 360°/𝑛. This tells us that 180 − 720/𝑛 = 150, ie 720/𝑛 = 30, or 𝑛 = 24.

A6. When a group of five friends met up, Alice shook hands with one person; Bill shook
hands with two people; Cara shook hands with three people; Dhriti shook hands
with four people. How many people did Erin shake hands with?

Solution 2
D shook hands with everyone, notably including A. Therefore A didn’t shake hands with anyone
other than D. C shook hands with all but one, ie all but A, notably including B. B shook hands
with only two: C and D. Therefore C and D shook hands with E, while A and B did not, so the
answer is 2.

A7. The time is 20:24 (expressed in 24-hour time). What is the angle between the hour
hand and the minute hand on an accurate analogue clock, in degrees?

Solution 108°
In this solution, all positions of hands will be expressed as clockwise angles relative to the
12:00 position at the top of the clock. Note also that, despite the time being expressed in
24-hour time, the analogue clock is of course a 12-hour clock as usual.
At 20:00, the hour hand is at a position of 8/12 ∗ 360° = 240° and the minute hand is at 0°. At
20:24, the minute hand is at a position of 24/60 ∗ 360° = 144°, and the hour hand has moved
1/12 as far as the minute hand has, ie 12° bringing it to a position of 252°, for a difference
between them of 108°. Since this result is less than 180° it is the angle between the two hands.

© 2024 UK Mathematics Trust www.ukmt.org.uk 3

Junior Mathematical Olympiad 2024 Section A Solutions

A8. We make a sequence of diagrams. The first diagram

consists of a single node. The second diagram is made
from the first diagram by drawing two edges of length 1 cm
from that node, and putting a node at the other end of each
new edge. After that, we make each new diagram from the
previous diagram by adding two new edges to each node,
with these new edges each having half the length of the
edges that were added in the previous diagram. We also
attach a node to the end of each new edge. The first four
diagrams are shown. Find the total length of all the edges
in the fifth diagram, in cm.

Solution 16.25 cm
The number of nodes increases by a factor of 3 at each stage. Therefore the number of new
edges also increases by a factor of 3 relative to the number of new edges in the previous diagram.
Since each new edge is half as long as the new edges in the previous diagram, this means that
the total new length is 32 as much as the new length in the previous diagram. Adding these
terms up gives 2 + 3 + 4.5 + 6.75 = 16.25 cm.

A9. The numbers 𝑥 and 𝑦 satisfy the equations:

7 5 7
xy = x(y + 1) = y(x + 1) =
6 3 2
What is the value of (𝑥 + 1)(𝑦 + 1)?

Solution 5
There are many ways to solve this question, including explicitly solving for 𝑥 and 𝑦. But an
easy way is to multiply the second and third equations together to get 𝑥𝑦(𝑥 + 1)(𝑦 + 1) = 35/6,
and then divide that by the first equation to get (𝑥 + 1)(𝑦 + 1) = 5. Note that this division is
safe, as 𝑥𝑦 ≠ 0: we are told that it is 67 .

© 2024 UK Mathematics Trust www.ukmt.org.uk 4

Junior Mathematical Olympiad 2024 Section A Solutions

2024 )
A10. What is the last digit of 2 (2 ?

Solution 6
"Last digit" questions are quite common, and it’s tempting on a question like this to try to work
out the last digit of 22024 as a first step. Unfortunately, knowing the last digit of that expression
isn’t quite enough to answer the question. Instead, consider the last digit of 2𝑛 as we increase
𝑛, starting from 𝑛 = 1 (we’ll want to work this out with 𝑛 = 22024 ). It goes through a regular
pattern: 2, 4, 8, 6, 2, 4, 8, 6,... (can you prove this?). Since 22024 is clearly a multiple of 4, the
final answer will be the value in the sequence at positions which are a multiple of 4, ie 6. Note
that asking for the last digit of something is really asking for the remainder when we divide by
10; here, our first step is to find the remainder when we divide 22024 by 4, which is why finding
the last digit of 22024 doesn’t directly help us.

© 2024 UK Mathematics Trust www.ukmt.org.uk 5

Junior Mathematical Olympiad 2024 Section B Solutions

Section B
B1. What is the smallest positive integer that only contains the digits 0 and 1, and is
divisible by 36?

Solution
Since 36 = 4 × 9, and since 4 and 9 have no factors in common, a number is a multiple of 36 if
and only if it is a multiple of both 4 and 9. To be a multiple of 9, the digits must add up to a
multiple of 9, which means that the number of 1s must be a multiple of 9. To be a multiple of 4,
the last two digits must be a multiple of 4. Since none of 1, 10, nor 11 are multiples of 4, the
only option is to put 0s in the last two digits. Since we need to use at least 9 1s and 2 0s, we
need at least 11 digits. The solution 11111111100 is the only such 11 digit number, as we need
the two 0s to go at the end. Any other solution would have more digits, and would therefore be
larger.

B2. Natasha and Rosie are running at constant speeds in opposite directions around a
running track. Natasha takes 70 seconds to complete each lap of the track and meets
Rosie every 42 seconds.
How long does it take Rosie to complete each lap?

Solution
Let 𝑡 be the length of the track, in m, and 𝑛 and 𝑟 be Natasha’s and Rosie’s running speeds,
respectively, in m/s. From one meeting to the next, they need to have run a total distance of 𝑡
between the two of them. So we have

𝑡 = 70𝑛
𝑡 = 42(𝑛 + 𝑟)
=⇒ 42𝑟 = 28𝑛
𝑡 𝑡 𝑛 42
Rosie’s lap time is 𝑟 = 𝑛𝑟 = 70 × 28 = 105s.
An alternative method is to divide up the track into 210 equal sections. It’s tempting to assume
that it’s 210m long, but that’s making an assumption beyond what the question tells us so we
can’t call these sections metres, instead let’s call them squigs. Natasha’s running speed is 3
squigs/second. The combined running speed of the two of them is 5 squigs/second. Therefore
Rosie’s speed is 2 squigs/second, so she takes 210
2 = 105 seconds to run a lap. Note that the size
of a squig turned out not to matter – but that doesn’t mean that we could assume that without
justification.

© 2024 UK Mathematics Trust www.ukmt.org.uk 6

Junior Mathematical Olympiad 2024 Section B Solutions

B3. The positive integers from 1 to 𝑛 (𝑛 ≥ 2) inclusive are to be spaced equally around
the
circumference of a circle so that:
(a) no two even numbers are adjacent;
(b) no two odd numbers are adjacent;
(c) no two numbers differing by 1 are adjacent.
What is the smallest value of 𝑛 for which the above is possible?

Solution
𝑛 = 2 is clearly impossible as 1 and 2 would be next to each other. Therefore the digit 3 must
appear. Since the neighbours of 3 must be even, and can’t be 2 or 4, the smallest they can be is
6 and 8. This shows that we need 𝑛 ≥ 8. All that remains is to check that 𝑛 = 8 is possible.
The sequence, going around the circle, could be 1, 6, 3, 8, 5, 2, 7, 4, before coming back to the
original 1, which satisfies all the conditions.

B4. My piggy bank contains 𝑥 pound coins and 𝑦 pennies and rattles nicely. If instead it
contained 𝑦 pound coins and 𝑥 pennies, then I would only have half as much money.
What is the smallest amount of money my piggy bank could contain?

Solution
The wording here is a bit curious, with the reference to rattling. While it is easy to dismiss it as
whimsy, the important conclusion from the rattling is that at least one of 𝑥 and 𝑦 is non-zero.
The next sentence tells us that

100𝑥 + 𝑦 = 2(100𝑦 + 𝑥)

ie 98𝑥 = 199𝑦. The solution 𝑥 = 𝑦 = 0 is contradicted by the rattling, so we need to find the
smallest positive solution. Since the RHS is a multiple of 199, so must the LHS be. Since 98
has no prime factors in common with 199 (199 is, in fact, prime) we know that 𝑥 must be a
multiple of 199. So the smallest solution is 𝑥 = 199, 𝑦 = 98, for a total amount of money of
£199.98.

© 2024 UK Mathematics Trust www.ukmt.org.uk 7

Junior Mathematical Olympiad 2024 Section B Solutions

B5. A regular octagon 𝐴𝐵𝐶𝐷𝐸 𝐹𝐺𝐻 has sides of length 1. 𝐴 𝐻

The lines 𝐴𝐶 and 𝐻𝐹 meet the line going through 𝐷
and 𝐸 at 𝑃 and 𝑄 respectively. 𝐵 𝐺
What is the length of the line 𝑃𝑄?
𝐶 𝐹

𝑃 𝐷 𝐸 𝑄

Solution
Extend line 𝐵𝐶 until it meets line 𝑃𝑄 at point 𝑋.
A H

B G

C F

PX D E Q
The exterior angles of a regular octagon are 45° so the interior angles are 135°. So ∠ 𝐴𝐵𝐶 = 135°.
Triangle 𝐴𝐵𝐶 is isosceles, as two of its sides are sides of the regular octagon. Therefore we
have ∠𝐵𝐴𝐶 = ∠𝐵𝐶 𝐴. Since angles in a triangle add to 180°, these two angles add to 45° and
so we have ∠𝐵𝐴𝐶 = ∠𝐵𝐶 𝐴 = 22.5°. Since ∠𝐵𝐶 𝐴 = ∠𝑃𝐶 𝑋 (vertically opposite), we have
∠𝑃𝐶 𝑋 = 22.5°.
∠𝑋𝐶𝐷 = 45° (exterior angle of the octagon), so ∠𝑃𝐶𝐷 = ∠𝑃𝐶 𝑋 + ∠𝑋𝐶𝐷 = 67.5°. ∠𝐶𝐷𝑃 =
45° (exterior angle) and angles in triangle 𝑃𝐶𝐷 add to 180° so ∠𝐶𝑃𝐷 = 67.5°. Thus triangle
𝐶𝑃𝐷 is in fact isosceles, and so 𝑃𝐷 = 𝐶𝐷 = 1.
The same argument holds on the other side, giving 𝑃𝑄 = 3.

© 2024 UK Mathematics Trust www.ukmt.org.uk 8

Junior Mathematical Olympiad 2024 Section B Solutions

B6. Let 𝐴 be the set of 2𝑛 positive integers 1, 2, 3, . . . , 2𝑛, where 𝑛 ≥ 1.

For which values of 𝑛 can this be split into 𝑛 pairs of integers in such a way that
every pair has a sum which is a multiple of 3?
As always in Olympiad problems such as this, you also need to explain why no other
values of 𝑛 are possible.

Solution
Suppose that 𝑛 is 1 more than a multiple of three. Then 2𝑛 is one less than a multiple of three,
and we have an easy solution: pair the first with the last, the second with the second last, etc.
So in this case we can definitely achieve the required splitting.
Suppose 𝑛 is a multiple of 3, so 𝑛 = 3𝑘. Then the last four integers are 6𝑘 − 3, 6𝑘 − 2, 6𝑘 − 1, 6𝑘.
We can pair these up as (6𝑘 − 3, 6𝑘) and (6𝑘 − 2, 6𝑘 − 1). This leaves a smaller problem, with
𝑛′ = 𝑛 − 2 taking the role of 𝑛 in the original problem. Since 𝑛′ is one more than a multiple of
three, these remaining integers can also be split in the required way.
Now suppose that 𝑛 is 1 less than a multiple of 3. Consider the sum 1 + 2 + · · · + 2𝑛.
Since 𝑛 + 1 is a multiple of 3, we can split the set {1, 2, . . . , 2(𝑛 + 1)} into pairs whose
sum is a multiple of 3, as shown above. So 1 + 2 + · · · + 2(𝑛 + 1) is a multiple of 3. But
1 + 2 + · · · + 2(𝑛 + 1) = (1 + 2 + · · · + 2𝑛) + 4𝑛 + 3, and 4𝑛 + 3 is not divisible by 3 as 𝑛 is not
divisible by 3. So 1 + 2 + · · · + 2𝑛 is not divisible by 3, and it is impossible to achieve the
required split.
Alternatively, starting from the same sum 1 + 2 + · · · + 2𝑛, we can pair the terms up from the
outside in: 1 + 2𝑛, 2 + (2𝑛 − 1), etc, each of which sum to 2𝑛 + 1. There are 𝑛 such pairs, so the
total sum is 𝑛(2𝑛 + 1). Since neither 𝑛 nor 2𝑛 + 1 is a multiple of 3 in this case, the product is
also not a multiple of 3 (note that this argument relies on 3 being prime).
So in summary, any value of 𝑛 which is either a multiple of 3, or 1 more than a multiple of 3,
can be split in the required way, and no others.

© 2024 UK Mathematics Trust www.ukmt.org.uk 9

 MT
UK
MT

UK
UKMT

United Kingdom
Mathematics Trust

Junior Mathematical Olympiad

Wednesday 14 June 2023
© 2023 UK Mathematics Trust

supported by

England & Wales: Year 8 or below

Scotland: S2 or below
Northern Ireland: Year 9 or below

Instructions
1. Do not open the paper until the invigilator tells you to do so.
2. Time allowed: 2 hours.
3. The use of blank or lined paper for rough working, rulers and compasses is allowed; squared
paper, calculators and protractors are forbidden.
4. Start each question on an official answer sheet on which there is a QR code.
5. If you use additional sheets of (plain or lined) paper for a question, please write the following
in the top left-hand corner of each sheet. (i) The question number. (ii) The page number for
that question. (iii) The digits following the ‘:’ from the question’s answer sheet QR code.
6. Write on one side of the paper only. Make sure your writing and diagrams are clear and
not too faint. (Your work will be scanned for marking.)
7. Arrange your answer sheets in question order before they are collected. If you are not
submitting work for a particular problem, please remove the associated answer sheet.
8. Your answers should be fully simplified and exact. They may contain symbols such as 𝜋,
fractions, or square roots, if appropriate, but not decimal approximations.
9. Only answers are required to the questions in Section A.
10. For questions in Section B, you should give full written solutions, including mathematical
reasons as to why your method is correct. Just stating an answer, even a correct one, will
earn you very few marks; also, incomplete or poorly presented solutions will not receive full
marks.

Enquiries about the Junior Mathematical Olympiad should be sent to:

challenges@ukmt.org.uk
www.ukmt.org.uk
Junior Mathematical Olympiad Wednesday 14 June 2023

Section A
Try to complete Section A within 30 minutes or so. Only answers are required.

59
A1. What is the integer nearest to ?
13
A2. What is the solution of the equation 24 ÷ (3 ÷ 2) = (24 ÷ 3) ÷ 𝑚 ?

A3. Two triangles are drawn so that they overlap as shown.

What is the sum of the marked angles?

(12 + 1)(22 + 1)(32 + 1)

A4. What is the value of 2 ? Give your answer in its simplest form.
(2 − 1)(32 − 1)(42 − 1)

A5. A number line starts at −55 and ends at 55. If we start at −55, what percentage of the
way along is the number 5.5?

A6. Tea and a cake cost £4.50. Tea and an éclair cost £4. A cake and an éclair cost £6.50.
What is the cost of tea, a cake and an éclair?

A7. A 2 × 2 × 2 cm cube has a 1 × 1 × 1 cm cube removed

from it to form the shape shown in the left-hand diagram.
One of these shapes is inverted and put together with a
second of the shapes on a flat surface, as shown in the
right-hand diagram.
What is the surface area of the new shape?

A8. Alex chooses three from the six primes 2003, 2011, 2017, 2027, 2029 and 2039.
The mean of his three primes is 2023.
What is the mean of the other three primes?

A9. The diagram shows the square 𝑃𝑄𝑅𝑆, which has area 25 cm2 , 𝑆
𝑃
and the rhombus 𝑄𝑅𝑇𝑈, which has area 20 cm2 .
𝑈 𝑇
What is the area of the shaded region?

𝑄 𝑅

twenty-three 23s
!!!!!!!!!"#!!!!!!!!$
A10. What is the remainder when 23 · · · · · · 23 is divided by 32?

© 2023 UK Mathematics Trust www.ukmt.org.uk

Junior Mathematical Olympiad Wednesday 14 June 2023

Section B

Your solutions to Section B will have a major effect on your result.

Concentrate firstly on one or two Section B questions and then write out full solutions (not
just brief ‘answers’), including mathematical reasons as to why your method is correct.
You will have done well if you hand in full solutions to two or more Section B questions.
Do not hand in rough work.

1 1 1
B1. The sum of four fractions is less than 1. Three of these fractions are , and .
2 3 10
1
The fourth fraction is , where 𝑛 is a positive integer. What values could 𝑛 take?
𝑛

B2. Laura went for a training ride on her bike. She covered the first 10% of the total distance
in 20% of the total time of the ride. What was the ratio of her average speed over the
first 10% of the distance to her average speed over the remaining 90% of the distance?

B3. As Rachel travelled to school, she noticed that, at each bus stop, one passenger got off
and 𝑥 passengers got on, where 𝑥 ≥ 2. After five stops, the number of passengers on the
bus was 𝑥 times the number of passengers before the first stop. How many passengers
were on the bus before the first stop?

B4. The regular nonagon 𝐴𝐵𝐶𝐷𝐸 𝐹𝐺𝐻𝐼 shares two of its 𝐸

vertices with the regular hexagon 𝐴𝐽𝐾 𝐿 𝑀 𝐼. 𝐹 𝐷
Show that the points 𝐻, 𝑀 and 𝐷 lie on the same straight line.
𝐺 𝐿 𝐾 𝐶
𝑀 𝐽
𝐻 𝐵
𝐼 𝐴

B5. The eleven-digit number ‘𝐴123456789𝐵’ is divisible by exactly eight of the numbers 1,
2, 3, 4, 5, 6, 7, 8, 9. Find the values of 𝐴 and 𝐵, explaining why they must have these
values.

B6. The diagram shows five circles connected by five line segments.
Three colours are available to colour these circles.
In how many different ways is it possible to colour all five circles
so that circles which are connected by a line segment are coloured
differently?

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Solutions
Junior Mathematical Olympiad 2023 Section A Solutions

Section A

59
A1. What is the integer nearest to ?
13

Solution 5
59 7 1
As 59 ÷ 13 = 4 remainder 7, = 4 which is greater than 4 and smaller than 5.
13 13 2
59
So the integer nearest to is 5.
13

A2. What is the solution of the equation 24 ÷ (3 ÷ 2) = (24 ÷ 3) ÷ 𝑚 ?

1
Solution
2
The left-hand side of the equation is 24 ÷ (3 ÷ 2) = 24 × (2 ÷ 3) = 48 ÷ 3 = 16.
The right-hand side of the equation is (24 ÷ 3) ÷ 𝑚 = 8 ÷ 𝑚.
1
Hence 8 ÷ 𝑚 = 16. So 16𝑚 = 8 and therefore 𝑚 = .
2

A3. Two triangles are drawn so that they overlap as shown.

What is the sum of the marked angles?

Solution 720°
Note that the angles marked 𝑝°, 𝑞°, 𝑟°, 𝑠° are the four exterior angles
of the quadrilateral in the centre of the diagram. s◦
Therefore 𝑝 + 𝑞 + 𝑟 + 𝑠 = 360. Note also that the marked angles
in the diagram form four pairs of vertically opposite angles with
each pair containing one of the exterior angles of the quadrilateral. p◦ r◦
As vertically opposite angles are equal, the sum, in degrees, of the
marked angles is 2( 𝑝 + 𝑞 + 𝑟 + 𝑠) = 2 × 360 = 720. q◦

© 2023 UK Mathematics Trust www.ukmt.org.uk 2

Junior Mathematical Olympiad 2023 Section A Solutions

(12 + 1)(22 + 1)(32 + 1)

A4. What is the value of ? Give your answer in its simplest
(22 − 1)(32 − 1)(42 − 1)
form.

5
Solution
18
(12 + 1)(22 + 1)(32 + 1) 2 × 5 × 10 5×4×5 5
The value of is = = .
(22 − 1)(32 − 1)(42 − 1) 3 × 8 × 15 3 × 2 × 4 × 3 × 5 3 × 2 × 3
5
So the required value is .
18

A5. A number line starts at −55 and ends at 55. If we start at −55, what percentage of
the way along is the number 5.5?

Solution 55%
The distance from −55 to 55 is 110. The distance from −55 to 0 is 55 and from 0 to 5.5 it is
5.5. Therefore the distance from −55 to 5.5 is 55 + 5.5.
55 + 5.5 55 + 5.5
As a percentage of the total distance this is × 100 = × 10 = (5 + 0.5) × 10
110 11
= 50 + 5 = 55.

A6. Tea and a cake cost £4.50. Tea and an éclair cost £4. A cake and an éclair cost £6.50.
What is the cost of tea, a cake and an éclair?

Solution £7.50
Let the costs, in pounds, of tea, a cake and an éclair be 𝑡, 𝑐 and 𝑒 respectively.
Then 𝑡 + 𝑐 = 4.5, 𝑡 + 𝑒 = 4 and 𝑐 + 𝑒 = 6.5. Adding these three equations gives 2𝑡 + 2𝑐 + 2𝑒 = 15.
So the cost of tea, a cake and an éclair is £15 ÷ 2 = £7.50.

A7. A 2×2×2 cm cube has a 1×1×1 cm cube removed from

it to form the shape shown in the left-hand diagram.
One of these shapes is inverted and put together with a
second of the shapes on a flat surface, as shown in the
right-hand diagram.
What is the surface area of the new shape?

Solution 38 cm2
When two of the shapes are put together, the top and bottom faces both have area 7 cm2 . Also,
there are four square faces of area 4 cm2 and four rectangular faces of area 2 cm2 .
Hence the required surface area is (2 × 7 + 4 × 4 + 4 × 2) cm2 = (14 + 16 + 8) cm2 = 38 cm2 .

© 2023 UK Mathematics Trust www.ukmt.org.uk 3

Junior Mathematical Olympiad 2023 Section A Solutions

A8. Alex chooses three from the six primes 2003, 2011, 2017, 2027, 2029 and 2039.
The mean of his three primes is 2023.
What is the mean of the other three primes?

Solution 2019
The sum of the six primes is 2003 + 2011 + 2017 + 2027 + 2029 + 2039 = 12 126.
The sum of the three primes which Alex chooses is 3 × 2023 = 6069.
12 126 − 6069 6057
Therefore the mean of the other three primes is = = 2019.
3 3

A9. The diagram shows the square 𝑃𝑄𝑅𝑆, which has area 𝑆
𝑃
25 cm2 , and the rhombus 𝑄𝑅𝑇𝑈, which has area 20 cm2 .
𝑈 𝑇
What is the area of the shaded region?

𝑄 𝑅

Solution 11 cm2
Let the perpendicular from 𝑈 to 𝑄𝑅 meet 𝑄𝑅 at 𝑉 and let the P S
point where 𝑈𝑇 and 𝑆𝑅 intersect be 𝑊, as shown.
Let the lengths of 𝑈𝑉 and 𝑄𝑉 be ℎ and 𝑥 respectively. U T
W
Square 𝑃𝑄𝑅𝑆 has area 25 cm2 , so its side-length is 5 cm. h
Hence rhombus 𝑄𝑅𝑇𝑈 has side-length 5 cm and area 20 cm2 .
Q x V R
20
Therefore ℎ is cm = 4 cm.
5
√
By Pythagoras’ Theorem, 𝑄𝑈 2 = 𝑄𝑉 2 + 𝑉𝑈 2 . Therefore 𝑥 is 52 − 42 cm = 3 cm.
So 𝑈𝑊 = 𝑉 𝑅 = (5 − 3) cm = 2 cm.
Hence the area of trapezium 𝑈𝑄𝑅𝑊 is 12 × (2 + 5) × 4 cm2 = 14 cm2 .
Therefore the area of the shaded region is (25 − 14) cm2 = 11 cm2 .

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Junior Mathematical Olympiad 2023 Section A Solutions

twenty-three 23s
!!!!!!!!!"#!!!!!!!!$
A10. What is the remainder when 23 · · · · · · 23 is divided by 32?

Solution 3
First note that 100 000 = 105 = 25 × 55 . So 100 000 is a multiple of 32.
twenty 23s
!!!!!!!!!"#!!!!!!!!$
Therefore the 46-digit number 23 · · · · · · 23 200 000 is a multiple of 32.

twenty-three 23s
!!!!!!!!!"#!!!!!!!!$
Hence, when 23 · · · · · · 23 is divided by 32, the remainder is equal to the remainder when
32 323 is divided by 32. As 32 323 = 32 × 1010 + 3, the required remainder is 3.

© 2023 UK Mathematics Trust www.ukmt.org.uk 5

Junior Mathematical Olympiad 2023 Section B Solutions

Section B
1 1 1
B1. The sum of four fractions is less than 1. Three of these fractions are , and .
2 3 10
1
The fourth fraction is , where 𝑛 is a positive integer. What values could 𝑛 take?
𝑛

Solution
1 1 1 1 15 + 10 + 3 1
We are given that + + + < 1. Therefore + < 1.
2 3 10 𝑛 30 𝑛
28 1 1 1
Hence + < 1. So < . Therefore 𝑛 could be any positive integer greater than 15.
30 𝑛 𝑛 15

B2. Laura went for a training ride on her bike. She covered the first 10% of the total
distance in 20% of the total time of the ride. What was the ratio of her average speed
over the first 10% of the distance to her average speed over the remaining 90% of
the distance?

Solution
Let the distance Laura cycled be 10𝑥 and let the time it took her be 5𝑡. Then she cycled a
distance 𝑥 in a time 𝑡, followed by a distance 9𝑥 in a time 4𝑡.
𝑥
So Laura’s average speed over the first 10% of the distance was and her average speed over
𝑡
9𝑥
the remaining 90% of the distance was .
4𝑡
𝑥 9𝑥 9
Hence the ratio of the speeds is : = 1 : = 4 : 9.
𝑡 4𝑡 4

B3. As Rachel travelled to school, she noticed that, at each bus stop, one passenger got
off and 𝑥 passengers got on, where 𝑥 ≥ 2. After five stops, the number of passengers
on the bus was 𝑥 times the number of passengers before the first stop. How many
passengers were on the bus before the first stop?

Solution
Let the number of passengers who were on the bus before the first stop be 𝑛.
Then, after five stops, the number of passengers on the bus was 𝑛 − 5 + 5𝑥.
So 𝑛 − 5 + 5𝑥 = 𝑛𝑥, that is, 𝑛𝑥 − 𝑛 = 5𝑥 − 5. Therefore 𝑛(𝑥 − 1) = 5(𝑥 − 1).
5(𝑥 − 1)
Hence 𝑛 = = 5, as 𝑥 ≠ 1.
𝑥−1
So there were five passengers on the bus before the first stop.

© 2023 UK Mathematics Trust www.ukmt.org.uk 6

Junior Mathematical Olympiad 2023 Section B Solutions

B4. The regular nonagon 𝐴𝐵𝐶𝐷𝐸 𝐹𝐺𝐻𝐼 shares two of its 𝐸

vertices with the regular hexagon 𝐴𝐽𝐾 𝐿 𝑀 𝐼. 𝐹 𝐷
Show that the points 𝐻, 𝑀 and 𝐷 lie on the same straight
𝐺 𝐿 𝐾 𝐶
line.
𝑀 𝐽
𝐻 𝐵
𝐼 𝐴

Solution
First note that the exterior angles of a hexagon and nonagon are 𝐸
360° 360° 𝐹 𝐷
and , that is, 60° and 40° respectively.
6 9
Hence the corresponding interior angles are 120° and 140° 𝐺 𝐿 𝐾 𝐶
respectively.
Consider the hexagon 𝐻𝐼 𝐴𝐵𝐶𝐷. The sum of its interior angles is 𝑀 𝐽
𝐻 𝐵
6 × 120° = 720°. Also ∠𝐻𝐼 𝐴 = ∠𝐼 𝐴𝐵 = ∠ 𝐴𝐵𝐶 = ∠𝐵𝐶𝐷 = 140°.
From the symmetry of the hexagon, we see that ∠𝐷𝐻𝐼 = ∠𝐶𝐷𝐻. 𝐼 𝐴
So ∠𝐷𝐻𝐼 = (720 − 4 × 140)° ÷ 2 = 80°.

Now consider triangle 𝐻𝐼 𝑀. Note that ∠𝐻𝐼 𝑀 = (140 − 120)° = 20°.

The side-lengths of the regular nonagon and regular hexagon are equal, so 𝐻𝐼 = 𝑀 𝐼.
Hence triangle 𝐻𝐼 𝑀 is isosceles and ∠𝑀 𝐻𝐼 = ∠𝐼 𝑀 𝐻 = (180 − 20)° ÷ 2 = 80°.
So ∠𝑀 𝐻𝐼 = ∠𝐷𝐻𝐼. Therefore 𝐻, 𝑀 and 𝐷 lie on the same straight line.

B5. The eleven-digit number ‘𝐴123456789𝐵’ is divisible by exactly eight of the numbers
1, 2, 3, 4, 5, 6, 7, 8, 9. Find the values of 𝐴 and 𝐵, explaining why they must have
these values.

Solution
Let the eleven-digit number ‘𝐴123456789𝐵’ be 𝑁.
First note that 𝑁 is divisible by least two of 2, 4 and 8. Therefore it is divisible by 4. So 𝐵 is
not 5. A number is divisible by 4 if, and only if, its last two digits form a number divisible by 4.
Hence 𝐵 is not 0 as 90 is not divisible by 4. So 𝑁 is not divisible by 5, but is divisible by each
of the numbers 1, 2, 3, 4, 6, 7, 8, 9. A number is divisible by 8 if, and only if, its last three
digits form a number divisible by 8. Therefore, for 𝑁 to be divisible by 8, 𝐵 is 6 as 896 is the
only number between 890 and 899 inclusive which is divisible by 8. 𝑁 is also divisible by 9,
which means that the sum of its digits is also divisible by 9. Hence 𝐴 + 51 is divisible by 9.
Therefore 𝐴 is 3 and, as has been shown, 𝐵 is 6.
(As 31 234 567 896 is divisible by both 8 and 9, it is clearly also divisible by 1, 2, 3, 4 and 6.
It is left to the reader to check that 31 234 567 896 is also divisible by 7.)

© 2023 UK Mathematics Trust www.ukmt.org.uk 7

Junior Mathematical Olympiad 2023 Section B Solutions

B6. The diagram shows five circles connected by five line segments.
Three colours are available to colour these circles.
In how many different ways is it possible to colour all five
circles so that circles which are connected by a line segment are
coloured differently?

Solution
We first look at the number of different ways of colouring the four A D
circles at the corners of the square.
If circles 𝐵 and 𝐷 are given the same colour, then they may be E
coloured in three different ways. For each of these, two colours are
available for both circle 𝐴 and circle 𝐶. Therefore, if 𝐵 and 𝐷 have
the same colour, then the four corner squares may be coloured in
3 × 2 × 2 = 12 different ways. B C

If circles 𝐵 and 𝐷 are coloured differently, then we can choose three colours for 𝐵 and, for
each of these, two different colours are available for 𝐷. So 𝐵 and 𝐷 may then be coloured in
2 × 3 = 6 different ways. Now, however, only one colour is available for circles 𝐴 and 𝐶, so if
circles 𝐵 and 𝐷 are coloured differently then there are six ways of colouring the corner circles.
Hence in total the four corner circles may be coloured in 12 + 6 = 18 different ways.
For each of these 18 possibilities, circle 𝐸 must be coloured differently from circle 𝐶 and thus
two colours are available for it. Therefore, in total, there are 2 × 18 = 36 different ways of
colouring the five circles so that connected circles are coloured differently.

© 2023 UK Mathematics Trust www.ukmt.org.uk 8