2022 Computational and Algorithmic Thinking — Upper Primary Questions
Part A: Questions 1–6
Each question should be answered by a single choice from A to E.
Questions are worth 3 points each.
1. Lotus Bird
Lotus birds step or jump from lily pad to lily pad. They will not step into water.
A lotus bird starts on the lily pad marked
with an X. Then it moves 10 times
according to the sequence of arrows.
D X C X↑ ← ↓ → ↑ ← ↑ → ↓ ↓
Each move is to the first lily pad in that di-
rection.
A B E
Afterthese10moves,onwhichlilypaddoesthelotusbirdfinish?
(A) A (B) B (C) C (D) D (E) E
2. Juliet’s Colours
Juliet is colouring this necklace.
G
R
Juliet has a red, a green and a blue pencil. She doesn’t like blue and will use it as little
as possible. However, she won’t colour two beads next to each other the same colour.
What is the smallest number of beads she will have to colour blue?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5
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2022 Computational and Algorithmic Thinking — Upper Primary Questions
3. Doublets
In a game show, contestants are given a list of numbers. If any two numbers in the list
are the same, they can be removed.
Contestants can increase a number in the list by the click of a button. They are required
to remove all numbers in the list.
For instance, given the list 10, 8, 6, 9 they could remove all of the numbers as follows:
10 8 6 9
1 click
10 9 6 9
remove
10 6
4 clicks
10 10
remove
In that example, 1 + 4 = 5 clicks were used to remove all numbers. But you could have
done it with fewer clicks.
What is the smallest number of clicks required to remove all numbers from the following
list?
0 15 6 4 20 13 12 16
(A) 16 (B) 17 (C) 18 (D) 19 (E) 20
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2022 Computational and Algorithmic Thinking — Upper Primary Questions
4. Thatches
Jermaine is obsessed with a new style of art. She draws vertical red lines and horizontal
black lines so that they overlap each other. Depending on what order she draws the lines,
Jermaine will get different pictures. In the two examples below, the lines are numbered
in the order in which she drew them.
8 2 5 6 5 8 1 2
4 6
1 3
3 4
7 7
Another one of Jermaine’s pictures is
A B C D
In which order could she have drawn the lines?
(A) w z C x A y B D
(B) w z C x A B y D
(C) w C z x A y B D
(D) w z C A x y B D
(E) w C z A x B y D
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2022 Computational and Algorithmic Thinking — Upper Primary Questions
5. River Crossings
Winnie has to travel from Start to Finish. She can only travel along dotted paths. She
needs to make several river crossings by ferry. The cost of each river crossing is shown
by the numbers in the grey circles.
2 2
4
Start
8
6
3 2
2
7
1
5 2
2
3 2
Finish
5
2
What is the smallest cost for her to travel from Start to Finish?
(A) 14 (B) 15 (C) 16 (D) 17 (E) 18
6. Merry-go-round
When Isabelle visits the local show she uses her new camera to take photos of the merry-
go-round, her favourite ride. The merry-go-round has several magical creatures to sit
on.
Each photo Isabelle takes shows only three of the creatures as they twirl past her.
Her photos show the following creatures, in order from left to right:
Phoenix Centaur Dragon
Phoenix Centaur Faun
Unicorn Phoenix Centaur
Centaur Faun Unicorn
Faun Unicorn Phoenix
Dragon Unicorn Centaur
What is the smallest number of creatures that could be on the merry-go-round?
(A) 8 (B) 9 (C) 10 (D) 11 (E) 12
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2022 Computational and Algorithmic Thinking — Upper Primary Questions
Part B: Questions 7–9
Each question has three parts, each of which is worth 2 points.
Each part should be answered by a number in the range 0–999.
7. Last Ones Standing
In Last Ones Standing you are given a line of digits, from which you remove two
adjacent digits several times until there are two digits left.
Examples:
4 3 2 6 1 5 −→ 4 3 1 5 −→ 1 5
432615 −→ 4615 −→ 4 5
The last two digits form a two-digit number. The aim of the game is to make this number
as large as possible. (You can do better than in the examples above.)
What is the largest two-digit number you could get from each of the lines of digits
below?
A. 5 7 3 9 4 6
B. 3 2 8 4 7 5
C. 5 7 9 4 8 4 4 6
8. Helen Hopping
Helen likes hopping in her garden which is tiled with hexagonal pavers. She starts on
the paver with the and with each hop lands on a new paver next to the one from which
she hopped. She does not hop over any flower beds, which are marked with dark double
lines.
Her favourite pavers are the ones that require the greatest number of hops to reach, using
the shortest route.
For example, in the diagram the shaded pavers
require four hops. There are 3 of them. She has 3
favourite pavers.
In each diagram below, how many favourite pavers does Helen have?
A.
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2022 Computational and Algorithmic Thinking — Upper Primary Questions
B.
C.
9. Team Photo
Each member of a sports team has a number on their back, ranging from 1 to 9. Zara
thought it would be fun to try to have the team line up to make the largest number
possible. However, as each team member arrives, they need to stand to the left or the
right of the line that already exists. They cannot go between two players who are already
there.
For example, players with 2, 4, 3 on their backs arrive in that order. The largest number
that Zara could make would be 423.
For each of the teams below, the team members arrive in the given order, and Zara
directs them to one end of the line or the other.
What are the 3rd , 5th and 7th digits of the largest number Zara can make from each of
the following?
A. 6, 4, 7, 9, 1, 8, 2, 5
B. 1, 6, 4, 3, 5, 7, 2, 9, 8
C. This time, Zara can choose to withhold one player when they arrive and direct
them to the left or right end of the line when she chooses.
Zara knows the order in which they will arrive so she can think ahead and decide
which player she should delay.
6, 4, 7, 9, 1, 8, 2, 5
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2022 Computational and Algorithmic Thinking — Upper Primary Solutions
Solutions
Part A: Questions 1–6
1. Lotus Bird
The arrows show the lotus bird’s moves.
D X C
A B E
The lotus bird finishes on lily pad B. Hence (B).
2. Juliet’s Colours
Solution 1
We start from each given colour in turn and attempt to alternate red and green in a
clockwise direction. We skip a bead when we would have to colour it the same as the
next one.
In the diagram the beads we colour have a little circle and a lower case r or g.
G r
r g
g
r R
g g
R r
g
g r
r g
g r
r
g G
R r
g r g
Any beads we skip are to be coloured blue.
There are 4 beads to be coloured blue. Hence (D).
(Note that we coloured clockwise. We could equally have coloured anticlockwise.
Different beads would be coloured blue, but the same number in total.)
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2022 Computational and Algorithmic Thinking — Upper Primary Solutions
Solution 2
If there is an even number of beads to be coloured between a green bead and a red bead,
we can colour red, green, . . . without needing to colour any beads blue.
If there is an odd number of beads we will need to colour one of them blue.
If there is an odd number of beads to be coloured between two green beads or two red
beads, we can colour red, green, . . . without needing to colour any beads blue.
If there is an even number of beads we will need to colour one of them blue.
So we can simply count the number of beads in the gaps between the coloured beads
and from that find the number to be coloured blue.
G
3
5 R
7
6
4 G
R
The bottom gap has 4 beads between a red and a green, so does not need a blue bead.
The other four gaps will each need a blue bead. Hence (D).
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2022 Computational and Algorithmic Thinking — Upper Primary Solutions
3. Doublets
The fewest clicks is achieved by making the smallest number equal to the second small-
est, the third smallest equal to the fourth smallest, and so on.
It is easier if we write the list of numbers in ascending order:
0 4 6 12 13 15 16 20
4 clicks
4 4 6 12 13 15 16 20
remove
6 12 13 15 16 20
6 clicks
12 12 13 15 16 20
remove
13 15 16 20
2 clicks
15 15 16 20
remove
16 20
4 clicks
20 20
remove
A total of 4 + 6 + 2 + 4 = 16 clicks are required. Hence (A).
4. Thatches
The top two stripes are B and D. The bottom two stripes are w and z.
So the order is w z ? ? ? ? B D.
This leaves us with w z C x A y B D and w z C A x y B D.
Since A is above x, the order is w z C x A y B D. Hence (A).
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2022 Computational and Algorithmic Thinking — Upper Primary Solutions
5. River Crossings
The numbers in the yellow circles represent the smallest cost to reach that point.
The path with the red centre shows how to reach that point with the smallest cost.
2 2
4 4
Start
0 8
6
0 9 3 2
2
7 14
1
5 2
2 15
3 13 2
Finish
5
2
The smallest cost to travel from Start to Finish is 15. Hence (B).
6. Merry-go-round
We look for the photos that have overlapping creatures. From these we build up se-
quences of creatures. Our aim is to get as much overlap as possible.
We will use P for Phoenix, C for Centaur, D for Dragon, U for Unicorn and F for Faun.
As the D only appears in the P C D and D U C photos, we will use that as our starting
point. This gives us P C D U C.
We could then extend the sequence from either end. Starting from the P end, we can
overlap two letters as we add each string.
P C F
C F U
F U P
U P C
P C D
D U C
This gives P C F U P C D U C, of length 9.
This is the best we can do with D in the middle.
We also need to check that we cannot do better with D at either end. That is D U C . . . or
. . . P C D. It is not hard to show that neither of these can result in a string with fewer
than 10 letters. Hence (B).
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2022 Computational and Algorithmic Thinking — Upper Primary Solutions
Part B: Questions 7–9
7. Last Ones Standing
As digits are removed two at a time, the leftmost digit in the final number must be the
1st , 3rd , 5th , . . . in the line of digits. This should be as large as possible. So the leftmost
digit should be the largest of the 1st , 3rd , 5th , . . . in the line of digits.
Similarly the second digit should be the largest of the 2nd , 4rd , 6th , . . . digits remaining.
In the solutions below the digits compared at each step are indicated by an ↑.
A. 5 7 3 9 4 6 → 5 7 3 9 4 6 → 59
↑ ↑ ↑ ↑ ↑ ↑
This is achieved by 5 7 3 9 4 6.
The largest two-digit number is 59.
B. 3 2 8 4 7 5 → 8 4 7 5 → 85
↑ ↑ ↑ ↑ ↑
This is achieved by 3 2 8 4 7 5
The largest two-digit number is 85.
C. 5 7 9 4 8 4 4 6 → 9 4 8 4 4 6 → 96
↑ ↑ ↑ ↑ ↑ ↑ ↑
This is achieved by 5 7 9 4 8 4 4 6
The largest two-digit number is 96.
8. Helen Hopping
The numbers in the pavers below show how many hops it would take to get to the paver
by the shortest route. The pavers that take most hops are shaded.
A.
2 3 4 7
1 4 5 6
2 3 4 7 7
There are 3 favourite pavers (each 7 hops away).
Hence 3.
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2022 Computational and Algorithmic Thinking — Upper Primary Solutions
B.
1 5
1 2 4 5
2 3 4 6
3 4 5 6
There are 2 favourite pavers (each 6 hops away).
Hence 2.
C.
6 5 6
5 4 4 5
6 2 3 4 5
1 2 4 6 6
There are 5 favourite pavers (each 6 hops away).
Hence 5.
9. Team Photo
A. 6, 4, 7, 9, 1, 8, 2, 5
When a player arrives, Zara will direct her to the left of the line if her number is
greater than that of the player on the left end of the line. Otherwise she will direct
her to the right end of the line.
6 → 64 → 764 → · · · → 97641825
Hence 612.
B. 1, 6, 4, 3, 5, 7, 2, 9, 8
1 → 61 → 614 → · · · → 976143528
Hence 645.
C. 6, 4, 7, 9, 1, 8, 2, 5
The player with 8 on her back arrives after the player with 9. Unless Zara delays
the player with 9, the player with 8 is directed to the right of the line. However,
if Zara delays the player with 9, she can direct the player with 8 to the left of the
line. Then, when all of the other players have arrived, Zara will direct player with
9 to the left of the line.
6 → 64 → 764 → · · · → 8764125 → 98764125
Hence 742.
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