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Junior Mathematical Olympiad

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Solutions
Junior Mathematical Olympiad 2022Section A Solutions

Section A

1
A1. What is the value of 2 − 23 + 34 − 45 + 56 ?

37
Solution 60

1 30 − 40 + 45 − 48 + 50
2 − 23 + 43 − 45 + 5
6 = 60 = 37
60 .

A2. Seven consecutive odd numbers add up to 105. What is the largest of these numbers?

Solution 21
Let the fourth odd number be 𝑥. Then (𝑥−6)+(𝑥−4)+(𝑥−2)+(𝑥)+(𝑥+2)+(𝑥+4)+(𝑥+6) = 105.
This simplifies to 7𝑥 = 105, which gives 𝑥 = 15. The largest of the seven consecutive odd
numbers is 15 + 6 = 21.

A3. In a class, 55% of students scored at least 55% on a test. 65% of students scored at
most 65% on the same test. What percentage of students scored between 55% and
65% (inclusive) on the test?

Solution 20%
As 55% of students scored at least 55% on the test, 45% of the students scored less than 55%.
Also, 65% of students scored 65% or less on the test, so the percentage of students who scored
between 55% and 65% (inclusive) is 65% − 45% = 20%.

A4. What is the sum of the marked angles in this diagram?

Solution 1980°
First we consider the sum of the angles around each of the seven vertices of the three triangles,
which is 7 × 360° = 2520°. The sum of the marked angles is the previous sum minus the sum
of the interior angles of the three triangles, which is 2520° − 3 × 180° = 2520° − 540° = 1980°.

© 2022 UK Mathematics Trust www.ukmt.org.uk 2

Junior Mathematical Olympiad 2022Section A Solutions

A5. Consider the six-digit multiples of three with at least one of each of the digits 0,
1 and 2, and no other digits. What is the difference between the largest and the
smallest of these numbers?

Solution 122 208

Multiples of three have a digit sum which is divisible by three. The six-digit numbers we
are considering must have the digits 0, 1 and 2 occurring at least once. These three digits
sum to 3, which means the three remaining digits must also sum to a multiple of three. To
create the largest possible number, we want to maximise the digits used from left to right. To
create the smallest possible number, we want to minimise the digits used from left to right but
the first digit must not be 0 or else the number does not have six digits. For the largest such
six-digit number we can use another three digit 2s and for the smallest we can use another
three digits 0s. The largest possible number formed of the relevant digits is 222 210 and the
smallest possible number formed of the relevant digits is 100 002. The difference we want is
222 210 − 100 002 = 122 208.

A6. Two positive numbers 𝑎 and 𝑏, with 𝑎 > 𝑏, are such that twice their sum is equal to
three times their difference. What is the ratio 𝑎 : 𝑏?

Solution 5:1
The information given can be written as 2(𝑎 + 𝑏) = 3(𝑎 − 𝑏), which upon expansion of the
brackets gives 2𝑎 + 2𝑏 = 3𝑎 − 3𝑏. After rearrangement we have that 5𝑏 = 𝑎 and so the ratio
𝑎 : 𝑏 is 5 : 1.

A7. The diagram on the right shows a 4 by 4 square placed on top

of a 5 by 5 square, so that they have one vertex in common as
shown. One diagonal of each square is also drawn. What is the
area of the shaded region that is inside the 4 by 4 square and
between the two diagonals?

Solution 3.5
Referring to the diagram on the right, 𝐽 𝐿 is a diagonal of the square J P M
𝐽𝐾 𝐿 𝑀 therefore ∠𝑃𝐽 𝐴 = 45°. The given configuration of the two
squares means that ∠ 𝐴𝑃𝐽 = 90° therefore ∠𝐽 𝐴𝑃 = 180°−45°−90° = A
45°. Hence triangle 𝐴𝑃𝐽 is isosceles and 𝐴𝑃 = 𝐽𝑃. As 𝐽𝑃 is
the difference between the side lengths of the two squares, then
𝐴𝑃 = 𝐽𝑃 = 5 − 4 = 1. Similarly, we can deduce that 𝐶𝑄 = 1. The Q
B C
shaded area is equal to the area of triangle 𝑃𝐵𝑄 minus the area of K L
triangle 𝐴𝐵𝐶. As 𝐴𝐵 = 𝐵𝐶 = 4 − 1, this is 12 × 4 × 4 − 12 × 3 × 3 = 3.5
square units.

© 2022 UK Mathematics Trust www.ukmt.org.uk 3

Junior Mathematical Olympiad 2022Section A Solutions

A8. The sum of the numbers 1 to 123 is 7626. One number is omitted so that the sum is
now a multiple of 4. How many different numbers could be omitted?

Solution 31
7626 is two more than a multiple of four, which means for the reduced sum to be a multiple of
four then the number omitted must also be two more than a multiple of four. The sequence of
relevant numbers that could be removed is 2, 6, 10, . . . , 122. If we add two to each of these
and then divide by four we get the sequence 1, 2, 3, . . . , 31. This shows that we could omit 31
different numbers.

A9. Dividing 52 by 12 gives 4 remainder 4. What is the sum of all the numbers for which
dividing by 12 gives a whole number answer which is the same as the remainder?

Solution 858
Using the example given we can write 52 = 12 × 4 + 4 = 13 × 4, and so in general we are looking
for numbers of the form 12𝑚 + 𝑚 = 13𝑚 where 𝑚 is any possible remainder when dividing by
twelve. When dividing by twelve the possible remainders are 0, 1, 2, ..., 11. The sum we want is
13 × 0 + 13 × 1 + 13 × 2 + ... + 13 × 11. This can be rewritten as 13(0 + 1 + 2 + ...11) = 13 × 66
and so the desired sum is 858.

A10. Farmer Alice has an alpaca, a cat, a dog, a gnu and a pig. She also has five fields in a
row. She wants to put one animal in each field, but doesn’t want to put two animals
in adjacent fields if their names contain the same letter. In how many different ways
can she place her animals?

Solution 4
The pig shares a letter with three other animals (alpaca, dog and gnu) and so it must have only
one neighbour. This means that the pig must go at either end of the row and must be adjacent to
the cat. The dog and gnu cannot be neighbours as they share the letter g, which means that the
alpaca must be placed between them. The dog or the gnu can be placed next to the cat. This
means that there are two choices for where the pig is placed, and for each of these choices there
are two choices for which animal is placed next to the cat, and so in total there are 2 × 2 = 4
different arrangements.

© 2022 UK Mathematics Trust www.ukmt.org.uk 4

Junior Mathematical Olympiad 2022Section B Solutions

Section B
B1. The sum of two numbers is 90.
40% of the first number is 15 more than 30% of the second number.
Find the two numbers.

Solution
Let the first number be 𝑎. Then the second number is 90 − 𝑎. Therefore 0.4𝑎 = 0.3(90 − 𝑎) + 15.
Expanding the bracket gives 0.4𝑎 = 27 − 0.3𝑎 + 15. Multiplying the equation by 10 gives
4𝑎 = 270 − 3𝑎 + 150.
After rearrangement we have 7𝑎 = 420. Therefore 𝑎 = 60 and the two numbers are 60 and 30.

© 2022 UK Mathematics Trust www.ukmt.org.uk 5

Junior Mathematical Olympiad 2022Section B Solutions

B2. In a certain quadrilateral, the four angles are

each two-digit numbers. These four numbers ‘ab’◦ ‘ac’◦ a b
can be placed in the 2 by 2 grid shown, with c d
one digit in each cell. ‘cd’◦ ‘bd’◦

Find all the possibilities for the set of four

angles.

Solution
Throughout this solution we will assume that 𝑏 ≥ 𝑐 because the problem remains unchanged if
we were to swap 𝑏 and 𝑐.
It is useful to note that a two-digit number ‘𝑥𝑦’ can be written algebraically as 10𝑥 + 𝑦 and that
the sum of the four interior angles of a quadrilateral is 360°.
Combining the previous two facts we see that 10𝑎 + 𝑏 + 10𝑎 + 𝑐 + 10𝑏 + 𝑑 + 10𝑐 + 𝑑 = 360.
This can be simplified to give:

20𝑎 + 11𝑏 + 11𝑐 + 2𝑑 = 360. (1)

If we assume that 𝑎 ≤ 7, then the maximum interior angle sum we can achieve is 20 × 7 + 11 ×
9 + 11 × 9 + 2 × 9 = 356 < 360 (where 𝑎 = 7 and 𝑏 = 𝑐 = 𝑑 = 9). As this is less than 360 we
know that 𝑎 ≥ 8, and so 𝑎 (which must be a digit) may only equal 8 or 9. We will consider
each of these two cases separately.
If 𝑎 = 8 then it must be that 𝑏 = 9, or else the four interior angles are each less than 90° and the
interior angle sum would be less than 360°. Substituting 𝑎 = 8 and 𝑏 = 9 into (1) and then
simplifying gives 11𝑐 + 2𝑑 = 101. The largest possible value of 2𝑑 is 2 × 9 = 18, and so 11𝑐
must be at least 101 − 18 = 83. This means that 𝑐 must be at least 8. If 𝑐 = 8 then 𝑑 would not
be a whole number and so not a digit. If 𝑐 = 9 then 𝑑 = 1.
If 𝑎 = 9 then 𝑏 must be equal to 8 or 9, or else the maximum interior angle sum we can achieve
is 20 × 9 + 11 × 7 + 11 × 7 + 2 × 9 = 352 < 360 (where 𝑎 = 9, 𝑏 = 𝑐 = 7 and 𝑑 = 9).
If 𝑎 = 9 and 𝑏 = 8, substituting into (1) and then simplifying gives 11𝑐 + 2𝑑 = 92. This means
that 11𝑐 must be at least 92 − 18 = 74 and so 𝑐 is at least 7. If 𝑐 = 7 then 𝑑 is not a whole
number and so not a digit. If 𝑐 = 8 then 𝑑 = 2. We know that 𝑐 cannot equal 9 because at the
start we assumed it was less than or equal to 𝑏.
If 𝑎 = 9 and 𝑏 = 9, substituting into (1) and then simplifying gives 11𝑐 + 2𝑑 = 81. This means
that 11𝑐 must be at least 81 − 18 = 63 and so 𝑐 is at least 6. If 𝑐 = 6 or 8 then 𝑑 is not a whole
number and so not a digit. If 𝑐 = 7 then 𝑑 = 2. We know that 𝑐 cannot equal 9 because then 𝑑
would be negative.
There are three sets of four angles that satisfy the problem: {89, 89, 91, 91}, {98, 98, 82, 82}
and {99, 97, 92, 72}.

© 2022 UK Mathematics Trust www.ukmt.org.uk 6

Junior Mathematical Olympiad 2022Section B Solutions

B3. You start with a regular pentagon 𝐴𝐵𝐶𝐷𝐸. Then you draw two circles: one with
centre 𝐴 and radius 𝐴𝐵, and the other with centre 𝐵 and radius 𝐵𝐴. Let the point
inside the pentagon at which these two circles intersect be 𝑋.
What is the size of ∠𝐷𝐸 𝑋?

Solution
As the pentagon 𝐴𝐵𝐶𝐷𝐸 is regular we know that 𝐴𝐸 = 𝐴𝐵 = D
𝐵𝐶. Hence 𝐵 and 𝐸 lie on the circle with centre 𝐴 and radius
𝐴𝐵, and 𝐴 and 𝐶 lie on the circle with centre 𝐵 and radius
𝐵𝐴. E C
X
𝐴𝑋 is a radius of the circle centre 𝐴, 𝐵𝑋 is a radius of the
circle centre 𝐵, 𝐴𝐵 is a side length of the regular pentagon
and so these three lengths are equal. Hence triangle 𝐴𝑋 𝐵 is
A B
equilateral and ∠𝑋 𝐴𝐵 = 60°.
Given that ∠𝐸 𝐴𝐵 is an interior angle of a regular pentagon,
it follows that ∠𝐸 𝐴𝑋 = ∠𝐸 𝐴𝐵 − ∠𝑋 𝐴𝐵 = ( 540 5 )° − 60° =
108° − 60° = 48°.
𝐴𝐸 = 𝐴𝑋 as they are both radii of the circle centre 𝐴. Hence
triangle 𝐴𝐸 𝑋 is isosceles. It follows that ∠𝑋 𝐸 𝐴 = ∠ 𝐴𝑋 𝐸 =
( 180−48
2 )° = 66°.
As ∠𝐷𝐸 𝐴 is an interior angle of a regular pentagon, it follows
that ∠𝐷𝐸 𝑋 = ∠𝐷𝐸 𝐴 − ∠𝑋 𝐸 𝐴 = 108° − 66° = 42°.

© 2022 UK Mathematics Trust www.ukmt.org.uk 7

Junior Mathematical Olympiad 2022Section B Solutions

B4. Seth creates 𝑛 standard dice by folding up 𝑛 identical copies of

the net shown. He then repeatedly puts one on top of another
until there are none left, creating a vertical tower.
For each of the four vertical walls of the tower, he finds the total
number of dots that are visible.
Given that the four totals calculated are all odd, what are the possible values for 𝑛?

Solution
Consider a tower made up of 𝑛 dice. Let the total number of dots on the vertical wall facing us
be 𝑆. Let the total number of dots on the wall opposite this, which is facing away from us, be 𝑇.
Looking at the diagram we can see that each pair of opposite faces on a completed net will have
a total of seven dots (this is true of all standard dice), and as there are 𝑛 such pairs forming
these two walls it follows that 𝑆 + 𝑇 = 7𝑛.
If 𝑛 is odd then so is 7𝑛, which in turn means that 𝑆 + 𝑇 is also odd. We want a tower where
both 𝑆 and 𝑇 are odd, and so 𝑆 + 𝑇 would in fact be even. Therefore 𝑛 cannot be odd.
If 𝑛 is even then so is 7𝑛, which in turn means that 𝑆 + 𝑇 is also even. Therefore it may be
possible to build such a tower if 𝑛 is even. In fact, we will show that it is possible to build such
a tower for any even value of 𝑛.
Consider stacking 𝑛 = 2𝑘 dice one on top of the other so that they all have the
same orientation. That is to say that each vertical wall consists of 𝑛 identical faces
all showing the same number of dots. Next rotate the top die in the tower 180° (as
you look down from above the tower). An example, with 𝑛 = 2 × 2 = 4, is shown
in the diagram on the right.
Let 𝑅 be the total number of dots visible on one of the vertical walls of this tower
and let the number of dots on each of the first 2𝑘 − 1 faces be 𝑎. The face at the
top of the wall will have 7 − 𝑎 dots. This means the total number of dots visible
for this particular wall will be 𝑅 = (2𝑘 − 1)𝑎 + (7 − 𝑎).

If 𝑎 is odd then 𝑅 will be the sum of an odd number and an even number, and so will be odd. If
𝑎 is even then 𝑅 will be the sum of even number and an odd number, and so 𝑅 will be odd.
This means that no matter which wall we consider, the total number of dots visible will be odd.
We have shown that such a tower can be built whenever 𝑛 is even.

© 2022 UK Mathematics Trust www.ukmt.org.uk 8

Junior Mathematical Olympiad 2022Section B Solutions

B5. Charlie chooses one cell from a blank 𝑛 × 𝑛 square grid and shades it. The resulting
grid has no lines of symmetry.
In terms of 𝑛, how many different cells could be shaded?

Solution
If the cell shaded is reflected in a line of symmetry of the blank grid, it must be reflected to a
different cell or else that line is a line of symmetry of the resulting grid. In addition, shading a
single cell cannot introduce a new line of symmetry because reflecting in any other line will
cause at least one cell to reflect to a position outside of the grid. Therefore it must be that the
cell shaded is such that no line of symmetry of the blank grid passes through it.
We will first count the number of cells which do have a line of symmetry passing through them.
This count will differ depending on the parity of 𝑛, that is whether 𝑛 is even or odd. We can
then subtract this from the total number of cells, which is 𝑛2 in both cases.
For 𝑛 even, the only lines of symmetry of the blank 𝑛 × 𝑛 grid
that pass through cells of the grid are the two main diagonals as
pictured. Each of these lines pass through 𝑛 cells, and no cell lies
on both lines. Hence there are 2𝑛 such cells when 𝑛 is even.

n even

For 𝑛 odd, there are four lines of symmetry of the blank 𝑛 × 𝑛

grid. Two are the main diagonals and the other two are horizontal
and vertical lines which pass through the centre of the grid as
pictured. These four lines each pass through 𝑛 cells but the centre
cell lies on all of them. Hence there are 𝑛 + 𝑛 + 𝑛 + 𝑛 − 3 = 4𝑛 − 3
such cells when 𝑛 is odd.

n odd

If 𝑛 is even, then the number of cells which can be shaded in order to avoid a line of symmetry
is 𝑛2 − 2𝑛.
If 𝑛 is odd, then the number of cells which can be shaded in order to avoid a line of symmetry
is 𝑛2 − (4𝑛 − 3) = 𝑛2 − 4𝑛 + 3.

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Junior Mathematical Olympiad 2022Section B Solutions

B6. The descriptors ‘even’, ‘factors of 240’, ‘multiple of 3’, ‘odd’,

a b c
‘prime’ and ‘square’ are to be placed in some order as row and
column headings around the grid in positions 𝑎, 𝑏, 𝑐, 𝑑, 𝑒 and 𝑓 . d
e
The digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 are to be placed in the empty f
cells inside the grid so that each digit satisfies both the relevant
row and column headings.
(i) Show that it is possible to complete the grid.
(ii) In how many different ways can the grid be completed?

Solution
(i) Below is one possible way to complete the grid:
multiple
even odd
of 3
factor of 240 8 6 5
prime 2 3 7
square 4 9 1

(ii) The number 8 only has two of the given properties, and so if it is to be placed into the grid
then the descriptors even and factors of 240 must be paired together as its column and
row headings.
Whole numbers are either even or odd but not both and so the descriptors even and odd
must both be column headings or both be row headings.
No whole number can be both prime and square and so these two descriptors must both
be column headings or both be row headings.
This means that the three descriptors placed as column headings must be one of the two
sets {even, odd, multiple of 3} or {factors of 240, prime, square} and the row headings
the other.
For each set there are three choices for the descriptor which comes first, then for each
of these choices there are two choices for which descriptor comes second and finally a
single choice for which comes last. Therefore each of these two sets of descriptors can be
arranged in 3 × 2 × 1 = 6 different ways before being placed into the grid. Hence there are
2 × 6 × 6 = 72 different ways to place the descriptors into the grid.
Next we need to count in how many ways the numbers can be placed into the grid.
Firstly we consider the column (or row) labelled even. It must contain a factor of 240, a
prime and a square. Respectively these must be 8 as an even factor of 240 (it only has
these two properties), 2 as the only even prime and 4 as the only even square less than 10.
Secondly we consider the column (or row) labelled multiple of three. It must contain a
factor of 240, a prime and a square. Respectively these must be 6 as a multiple of 3 and a
factor of 240, 3 as the only multiple of 3 which is prime and 9 as the only multiple of 3
less than 10 which is square.
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Junior Mathematical Olympiad 2022Section B Solutions

Finally we consider the column (or row) labelled odd. It must contain a factor of 240, a
prime and a square. Respectively these must be 5 as an odd factor of 240, 7 as the only
remaining odd prime and 1 as the only remaining odd square.
Therefore once the descriptors have been placed into the grid, there is only one way to
place the numbers into the grid.
In conclusion, there are 72 ways to complete the grid.

© 2022 UK Mathematics Trust www.ukmt.org.uk 11