APSMO OLYMPIAD

2022 : DIVISION S
WEDNESDAY 23 MARCH 2022 1
Total Time Allowed: 30 Minutes

1A. How many square centimetres are equivalent to 1 square metre? Write your
answers in the
boxes on the
back.

1B. The original price of a pair of jeans is $50. ←

Keep your
Heather purchased a pair of these jeans after a 20% discount was
applied, and had a further 10% discount applied to the already answers
discounted price. hidden by
folding
Pauline purchased a pair of jeans after a single 30% discount was backwards on
applied to the original price. this line.
How much more did Heather pay than Pauline, in dollars?

1C. Each figure in the

sequence shown is
made of identical
square tiles.
...
If the pattern is
continued, the Nth
figure will consist of
exactly 1000 square tiles.
Find N.

1D. A 4-digit “step up” number is a whole number in which the

number formed by the leftmost two digits is 1 less than the
number formed by the rightmost two digits.
For example, 1011 is a 4-digit “step up” number since 10 = 11 – 1.
How many 4-digit “step up” numbers have no repeated digits?

1E. A bookshelf holds 6 different textbooks, 5 different notebooks,

and 23 different cookbooks.
How many different pairs of books can I select from the shelf, if
the two books must be of different types?

Copyright © 2022 Australasian Problem Solving Mathematical Olympiads (APSMO) Inc. and Mathematical Olympiads for Elementary and Middle Schools. All rights reserved.
 APSMO OLYMPIAD
2022 : DIVISION S
WEDNESDAY 23 MARCH 2022 1
1A.
Student Name:

1B.
Fold here. Keep your answers hidden.

1C.

1D.

1E.

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 APSMO OLYMPIAD
2022 : DIVISION S
WEDNESDAY 23 MARCH 2022 1
Solutions and Answers
(Items in parentheses are not required)
For teacher use only. Not for Distribution.

1A: 10 000 (cm2) 1B: ($)1 1C: 333 1D: 7 1E: 283
1A. The question is: How many square centimetres are equivalent to 1 square metre?
METHOD: Draw a diagram, or model with concrete materials.

Since 1 metre is equivalent to 100 centimetres, we can use 100 Alternatively, a square with
square centimetres to construct a metre-long stick that is 1 cm 10 cm sides is also equivalent
wide. to 10 cm × 10 cm = 100 square
1 metre centimetres.

100 cm

To construct 1 square metre, we

could either use:
• 100 of the 1 cm-wide sticks, or
• 100 squares that have 10 cm
sides.
Therefore 1 square metre is
equivalent to 100 × 100 = 10 000
square centimetres.

Follow-Up: How many cubic centimetres are equivalent to 1 cubic metre? [ 1 000 000 ]

Copyright © 2022 Australasian Problem Solving Mathematical Olympiads (APSMO) Inc. and Mathematical Olympiads for Elementary and Middle Schools. All rights reserved.
 APSMO OLYMPIAD
2022 : DIVISION S
WEDNESDAY 23 MARCH 2022 1
1B. The question is: How much more did Heather pay than Pauline?
METHOD 1 Strategy: Draw a diagram.

The original price of a pair of jeans is $50. $50

When Heather went shopping, the jeans were discounted by 20%.

Since 100% of the price is $50,
20 1
20% of the price is 100 × $50 = 5 × $50 = $10. $10 $10 $10 $10 $10

Discounting by $10 means that the new price is

$50 – $10 = $40. $40 $10

Heather then received a further 10% discount off the already discounted price of $40.
10 1
10% of $40 is 100 × $40 = 10 × $40 = $4. $4 $4 $4 $4 $4 $4 $4 $4 $4 $4

Discounting by $4 means that the final price is

$40 – $4 = $36. $36 $4

Heather paid $36 for her pair of jeans.

When Pauline went shopping, the jeans were discounted by 30%.

Since 100% of the price is $50,
30 3
30% of the price is 100 × $50 = 10 × $50 = $15. $5 $5 $5 $5 $5 $5 $5 $5 $5 $5

Discounting by $15 means that the new price is

$50 – $15 = $35. $35 $5 $5 $5

Pauline paid $35 for her pair of jeans.

Heather paid $36 – $35 = $1 more than Pauline did, for her pair of jeans.

METHOD 2 Strategy: Represent each discount as the percentage that is paid.

20% off is the same as 100% – 20% = 80% of the price. 80% 20%
Heather received a 20% discount, and then a 10% discount off the already discounted price.
10% off is the same as 100% – 10% = 90% of the price. 90% 10%

Heather paid 80% × 90% × $50 = 0.8 × 0.9 × $50 = $36 for her pair of jeans.

30% off is the same as 100% – 30% = 70% of the price. 70% 30%
Pauline paid 70% × $50 = 0.7 × $50 = $35 for her pair of jeans.

Heather paid $36 – $35 = $1 more than Pauline.

Follow-Up: A calculator manufacturer needs to determine the LIST PRICE for the latest model, so that a
20% PROFIT can be made after they apply a 20% DISCOUNT off the LIST PRICE. It costs $100 to
construct one calculator. What should they use as their LIST PRICE? [ $150 ]

Copyright © 2022 Australasian Problem Solving Mathematical Olympiads (APSMO) Inc. and Mathematical Olympiads for Elementary and Middle Schools. All rights reserved.
 APSMO OLYMPIAD
2022 : DIVISION S
WEDNESDAY 23 MARCH 2022 1
1C. The question is: Find N, where the Nth figure will consist of
exactly 1000 square tiles.
...

METHOD 1 Strategy: Convert to a more convenient form, and work backwards.

We begin by listing the number of tiles Figure 1 2 3 4 ... N

that are used to construct each figure. No. of Tiles 4 7 10 13 ... 1000
The Nth figure uses 1000 tiles.

+1 +1 +1
Every figure requires 3 tiles more than
Figure 1 2 3 4 ... N
the previous figure.
4 7 10 13 ... 1000
This means that we can express the No. of Tiles
3+1 2×3 + 1 3×3 + 1 4×3 + 1 ... 999 + 1
pattern as growing by 3s.
+3 +3 +3

Working backwards, if we know the number of tiles in the figure, we can determine the corresponding
Figure number by:
• Subtracting 1, and
• Dividing by 3.
The figure that consists of 1000 tiles is figure number 999 ÷ 3 = 333.

METHOD 2 Strategy: Examine the construction of the figure, and use algebra.

We can deconstruct the figures along The tiles for each figure From the diagram, we
the following lines. can then be rearranged as can see that the number
follows. of tiles for Figure N is
equal to 3
3N + 1.
N

We want to find the value 3N + 1 = 1000

of N where the number
of tiles is 1000. 3N = 1000 – 1
= 999
N = 999 ÷ 3
= 333 The value of N is 333.

Follow-Up: Using the same sequence of shapes, what is the least value of N that would have a perimeter that
is greater than 2022? [ 337 ]

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 APSMO OLYMPIAD
2022 : DIVISION S
WEDNESDAY 23 MARCH 2022 1
1D. The question is: How many 4-digit “step up” numbers have no repeated digits?

METHOD 1 Strategy: Build a table, and eliminate numbers that do not satisfy the criteria.

Since there is only one "step-up" 1011 2021 3031 4041 5051 6061 7071 8081 9091
number for every 2-digit number, it is
reasonable to just list every "step-up" 1112 2122 3132 4142 5152 6162 7172 8182 9192
number. 1213 2223 3233 4243 5253 6263 7273 8283 9293
By eliminating all of the numbers 1314 2324 3334 4344 5354 6364 7374 8384 9394
that do have repeated digits, we can 1415 2425 3435 4445 5455 6465 7475 8485 9495
see that there are exactly 7 "step-up" 1516 2526 3536 4546 5556 6566 7576 8586 9596
numbers that have no repeated digits.
1617 2627 3637 4647 5657 6667 7677 8687 9697
1718 2728 3738 4748 5758 6768 7778 8788 9798
1819 2829 3839 4849 5859 6869 7879 8889 9899
1920 2930 3940 4950 5960 6970 7980 8990

METHOD 2 Strategy: Make an organised list.

A "step-up" number is formed by putting First 2-digit number 10 11 12 13 14

together two 2-digit numbers, where the
second 2-digit number exceeds the first "Step-up" number 1011 1112 1213 1314 1415
2-digit number by 1.
First 2-digit number 15 16 17 18 19
We can begin by listing some "step-up"
"Step-up" number 1516 1617 1718 1819 1920
numbers in an organised way.
If the ones value of the first 2-digit number is in the range 0 - 8, both of the 2-digit numbers will have
the same tens digit.
The first "step-up" number with no repeated digits is 1920.

The only way to create a "step-up" First 2-digit number 19 29 39 49 59

number with no repeated digits is by
selecting the first 2-digit number so that "Step-up" number 1920 2930 3940 4950 5960
its ones value is 9.
First 2-digit number 69 79 89 99
"Step-up" number 6970 7980 8990 99100
(exceeds 4 digits)

We can see that 8990 has a repeated digit.

First 2-digit number 19 29 39 49 59
Therefore, by inspection, there are 7 "Step-up" number 1920 2930 3940 4950 5960
"step-up" numbers that have no repeated
digits. First 2-digit number 69 79 89 99
"Step-up" number 6970 7980 8990 99100

Follow-Up: An enhanced “step-up” number is a 4-digit number such that the leftmost 2-digit number is any
value less than the rightmost 2-digit number. For example, 2730 is included because 27 < 30.
How many enhanced four-digit “step-up” numbers are there in total? [ 4005 ]

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 APSMO OLYMPIAD
2022 : DIVISION S
WEDNESDAY 23 MARCH 2022 1
1E. The question is: How many different pairs of books can I select from the shelf, if the two books must be of
different types?
METHOD: Draw a diagram or build a table, and solve a simpler related problem.
Suppose there were just 2 different textbooks, 3 different notebooks, and 4 different cookbooks.

We could select 1 textbook, and A pair comprising A pair comprising 1 notebook

1 notebook. 1 textbook and and 1 cookbook can occur in
1 cookbook can occur in 3 × 4 = 12 ways.
Using a representation such as a table
2 × 4 = 8 ways.
or a tree diagram, we can see that the
pairing can occur in 2 × 3 = 6 ways.
T1 T2 N1 N2 N3

T1 T2 C1 T1 C1 T2 C1 C1 N1 C1 N2 C1 N3 C1

N1 T1 N1 T2 N1 C2 T1 C2 T2 C2 C2 N1 C2 N2 C2 N3 C2

N2 T1 N2 T2 N2 C3 T1 C3 T2 C3 C3 N1 C3 N2 C3 N3 C3

N3 T1 N3 T2 N3 C4 T1 C4 T2 C4 C4 N1 C4 N2 C4 N3 C4

T1 T2 T1 T2 N1 N2 N3

N1 N2 N3 N1 N2 N3 C1 C2 C3 C4 C1 C2 C3 C4 C1 C2 C3 C4 C1 C2 C3 C4 C1 C2 C3 C4

With 2 different textbooks, 3 different notebooks, and 4 different cookbooks, there would be
(2 × 3) + (2 × 4) + (3 × 4)
= 6 + 8 + 12
= 26 different pairs of books, where the two books are of different types.

Using what we have noticed in the above pattern, for 6 different With 6 different textbooks, 5
textbooks and 5 different notebooks, we see that there would be different notebooks, and 23
6 × 5 = 30 different pairs comprising 1 textbook and 1 notebook. different cookbooks, there would be
(6 × 5) + (6 × 23) + (5 × 23)
T1 T2 T3 T4 T5 T6
= 30 + 138 + 115
N1 T1 N1 T2 N1 T3 N1 T4 N1 T5 N1 T6 N1
= 283 different pairs of books,
N2 T1 N2 T2 N2 T3 N2 T4 N2 T5 N2 T6 N2
where the two books are of
different types.
N3 T1 N3 T2 N3 T3 N3 T4 N3 T5 N3 T6 N3

N4 T1 N4 T2 N4 T3 N4 T4 N4 T5 N4 T6 N4

N5 T1 N5 T2 N5 T3 N5 T4 N5 T5 N5 T6 N5

Follow-Up: Suppose there are T different textbooks, N different notebooks, and C different cookbooks.
T + N + C = 34. Find the greatest possible number of pairs of 2 books of different types. [ 385 ]

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