2022 AMC Solutions
2022 AMC Solutions
3. The shaded triangle is half the square and the square is half the rectangle, so the shaded
triangle is one-quarter of the rectangle,
hence (B).
4. 17 − 9 = 8,
hence (B).
6. Here is where there is a circle, an oval, a triangle and a rectangle in the picture:
rectangle
triangle
circle
oval
www.amt.edu.au
34 2022 AMC Middle Primary Solutions
7. Since Eve counts 10 numbers, there are 9 skips of 2 between her numbers. Since she started
at 20, she finishes at 20 + 9 × 2 = 20 + 18 = 38,
hence (E).
8. There were 5 students who chose Saturday and 3 who chose Sunday, so 8 chose Saturday
or Sunday,
hence (B).
9. From the start time to 8 am is 45 minutes. From 8 am to the finish time is 2 minutes. So
the bike ride was 45 + 2 = 47 minutes long,
hence (C).
10. As shown in the diagram, there are 7 children between Edie and Louie:
hence (A).
11. The ten-cent coins are worth $4.90 and the twenty-cent coins are worth $5.00. Together
they are worth $9.90,
hence (B).
12. Here are five vertical slices from front to back, including the missing cubes:
2×1
3×2
4×3
13. If the students get 4 stickers each, Ms Amali gives away 23 × 4 = 92 stickers. So she will
have 8 left over,
hence (B).
www.amt.edu.au
2022 AMC Middle Primary Solutions35
hence (A).
15. Alternative 1
Testing each number in turn, we see that
(A) 7 + 1 = 8 but 11 − 1 = 10
(B) 7 + 2 = 9 and 11 − 2 = 9
(C) 7 + 3 = 10 but 11 − 3 = 8
(D) 7 + 4 = 11 but 11 − 4 = 7
(E) 7 + 5 = 12 but 11 − 5 = 6
So, of the available choices, only the number 2 can be inserted into the boxes to make the
number sentence true,
hence (B).
Alternative 2
On the number line shown, 7 + and 11 − both end up at the same point:
0 1 2 3 4 5 6 7 8 9 10 11 12
+ −
Both jumps go to 9, half-way between 7 and 11. Then , the size of each jump, is 2 units,
hence (B).
16. Considering possible ages for Sally as 1, 2, 3, . . . , her brother’s age is 3, 6, 9, . . . , going up
in 3s. The difference between their ages is 2, 4, 6, . . . , going up in 2s. For this difference
to be 10, Sally will be 5 and her brother 15,
hence (B).
17. The shaded area can be divided into 5 squares identical to the one shown on the left. As
the diagram on the right shows, each of these squares can be covered by 8 of the small
tiles:
www.amt.edu.au
36 2022 AMC Middle Primary Solutions
The number of tiles needed to cover the total cross shape is then 5 × 8 = 40,
hence (E).
18. There are two pairs of numbers which multiply to give 12, (2,6) and (3,4). There are also
two pairs which add to 11, (5,6) and (4,7).
If we use 2 × 6 to give 12, we must use 4 + 7 to give 11. This leaves 1, 3 and 5 and there
is no pair which subtracts to give 1.
So we try 3 × 4 = 12 and 5 + 6 = 11 which leaves 1, 2 and 7. Now 2 − 1 = 1 so the unused
card is 7,
hence (E).
19. The path of the beam can be drawn in, taking care to follow the diagonals of the grid
squares:
1 10 3 8 5
>
6
<
11 2 9 4 7
As the numbering shows, there are 11 points where the beam of light bounces off a mirror,
hence (D).
20. The 10 marbles removed from the can weigh 115 − 85 = 30 g, so 30 marbles will weigh
3 × 30 = 90 g. Therefore the empty can weighs 115 − 90 = 25 g,
hence (C).
21. The ribbon wraps around 8 sides of the box. Each side is 20 cm, so this requires 8 × 20 =
160 cm of ribbon. Another 80 cm is required to tie the bow so Peter needs approximately
240 cm of ribbon,
hence (C).
22. Alternative 1
This table shows the balance after each deposit:
www.amt.edu.au
2022 AMC Middle Primary Solutions37
Alternative 2
After the week 1 deposits, Eliza has $20 more in the bank than Hamish.
Two weeks later (week 3), Hamish has deposited $12 more and Eliza has deposited $8
more. The difference between bank accounts has gone down by $4.
Two weeks later, the difference between banks accounts has gone down another $4.
This continues until the difference between accounts has gone down by 5 × 4 = 20 dollars,
which is 5 × 2 = 10 weeks after the first week. This is in week 11,
hence (E).
23. We try to use the smallest set of numbers 1, 2, 3 and 4. The 1 must be opposite the 2 and
then the 3 will be next to the 2 so this is impossible.
The next smallest set is 1, 2, 3 and 5 but this gives exactly the same problem. In both
cases we cannot use 1, 2 and 3 in the set, since the first diagram shows that once we’ve
placed 1 and 2, we can’t place the 3.
? 2 4 2
1 ? 1 5
The next smallest total is given by 1, 2, 4 and 5 which add to 12. The second diagram
shows that this can be done,
hence (C).
24. The three numbers are factors of 280, and each is less than 21. So they must be chosen
from 1, 2, 4, 5, 7, 10, 14, 20.
From these numbers we need three numbers that add to 21. By choosing the largest of the
three numbers first, possibilities for the other two numbers can be found. For instance,
choosing 14 means that the other two numbers add to 7, which can only be 5 + 2. Working
this way, we find just four ways to choose three numbers that add to 21:
The only solution that has product 280 is when the three numbers are 4, 7 and 10,
hence (C).
www.amt.edu.au
38 2022 AMC Middle Primary Solutions
25. If the net had the two pierced faces joined together, they would look like this:
None of the nets (A) to (E) have these two faces joined, so the two pierced faces must be
two that become joined at a common edge like this:
Here are the five nets, with arrows showing how some of the edges are joined to make the
cube. For each net, the join between the two pierced faces has a solid arrow, and the joined
faces are shown underneath:
In these diagrams, only (D) has the right orientation of the two pierced faces,
hence (D).
26. We cannot just increase the units digit because all the numbers higher than 4 are already
used.
The same applies to increasing the tens or hundreds digits. So we must increase the
thousands digit.
The smallest change will be to increase this to a 7 and then use numbers as small as
possible in the next three positions. This gives 897 012, so we will need another 258 units,
hence (258).
27. Alternative 1
Starting with 2, these are the numbers containing the digit 3: 203, 213, 223, 230, 231, 232,
233, 234, 235, 236, 237, 238, 239, 243, 253, 263, 273, 283, 293, a total of 19 numbers.
Similarly, there are also 19 numbers starting with 4 that contain 3.
Finally, all 100 numbers from 300 to 399 contain the digit 3.
In all there are 19 + 100 + 19 = 138 numbers from 200 to 500 containing the digit 3,
hence (138).
Alternative 2
Since 500 doesn’t have a 3, we can focus on the 300 numbers from 200 to 499. We first
count those that don’t have the digit 3. Here are the choices for the three digits:
2 or 4 0, 1, 2, 4, 5, 6, 7, 8 or 9 (9 choices)
(2 choices) 0, 1, 2, 4, 5, 6, 7, 8 or 9 (9 choices)
Any of these 2 × 9 × 9 = 162 choices gives a number from 200 to 499 without the digit 3.
Then there are 300 − 162 = 138 numbers that do have the digit 3,
hence (138).
www.amt.edu.au
2022 AMC Middle Primary Solutions39
28. To make the largest possible 3-digit number, we try starting with the first digit 9. This
uses 6 bars, so the digit sum so far is 3 more than the number of bars used.
To compensate, we need at least one other digit which uses more bars than its own value,
namely 0, 1, 2 or 3. The biggest gain is from using a 0, which uses 6 more bars than its
value. With 9 and 0, the digit sum is 3 less than the number of bars used. This allows us
to use another 9 to balance the two totals. So 990 is such a number and no larger number
will work,
hence (990).
1+3+5=9 1 + 3 + 7 = 11 1 + 5 + 7 = 13
3 + 5 + 7 = 15 3 + 5 + 9 = 17 3 + 7 + 9 = 19
.. .. ..
. . .
99 + 101 + 103 = 303 99 + 101 + 105 = 305 99 + 103 + 105 = 307
101 + 103 + 105 = 309
This table has 50 full rows, so there are 3 × 50 + 1 = 151 possible totals,
hence (151).
www.amt.edu.au
40 2022 AMC Upper Primary Solutions
4
2. There are 9 small squares of equal size and 4 are shaded, so the fraction is ,
9
hence (D).
3. (Also I1)
2220 − 2022 = 198,
hence (C).
4. The smallest of the three numbers is 0.08, and the largest is 1.03,
hence (B).
6. (Also I2)
Alternative 1
Out of the 29 squares, there are 12 squares that have 3 edges on the perimeter, 4 squares
that have 2 opposite edges on the perimeter and 4 squares that have 2 adjacent edges on
the perimeter. This accounts for all 12 × 3 + 4 × 2 + 4 × 2 = 52 edges on the shape,
hence (A).
Alternative 2
Start with the central 3 × 3 square, with perimeter P = 12. Join the remaining 20 squares
one-by-one. Each join increases P by 2: an increase of 4 from the new square but then a
decrease of 2 from the join. After joining 20 squares, P = 12 + 20 × 2 = 52,
hence (A).
7. Between the numbers on the dial the tick marks divide each section into 5 parts, each
1
measuring = 0.2. So the needle is pointing at 2.4. Since this is in thousands of rpm, it
5
represents 2.4 × 1000 = 2400 rpm,
hence (E).
8. Since one litre is 1000 mL, there are 1000 − 320 = 680 mL left,
hence (C).
www.amt.edu.au
2022 AMC Upper Primary Solutions41
•
0 1 2 3 4 5 6 7 8
hence (E).
13. 46 + 364 + 591 = 1001. The missing unit digit must be a 6 to give 11 (1 carry 1). So,
including the carry, the tens digits add to 14 which means the missing digit must again be
6, to give a sum of 20 (0 carry 2). In the hundreds column, we now have 3 + 2 = 5 so the
missing number is 5 in order to get 0 carry 1. The missing numbers are 6, 6 and 5 and the
sum of these is 17,
hence (D).
14. In the original stack, the yellow triangle is on top of the blue square and under the red
circle, so in the flipped stack, the yellow triangle will be under the blue square and on top
of the red circle. Only (A) and (C) have this. Considering the position of the three shapes,
the blue square is originally on the right side of the arrangement, so it will be on the left
side when flipped. So only (A) has all the shapes in their correct positions,
hence (A).
15. Alternative 1
1 1
15 is of 45, so the missing number must be of 270, which is 90,
3 3
hence (D).
Alternative 2
Writing each division as a fraction,
270
=
45 15
To convert the fraction on the left to the fraction on the right, a factor of 3 is cancelled in
the denominator. So we need to cancel 3 in the numerator, giving 90. Checking, both 270 45
and 90
15
equal 6,
hence (D).
www.amt.edu.au
42 2022 AMC Upper Primary Solutions
16. The smallest square has side 3 cm and the largest square has side 8 cm. Consequently the
medium-sized square has side 8 − 3 = 5 cm and perimeter 5 × 4 = 20 cm,
hence (C).
17. For each marble Mei takes, she leaves two. Then Huang takes one of these two, leaving
one marble. Since 8 marbles were left, Mei took 8 and then Huang took 8, a total of 24
marbles that were originally in the bag,
hence (D).
8 2
5 7
4 6
1 3
hence (A).
19. This year, the sum of George’s grandchildren’s ages is 27 + 23 + 16 = 66. This is 12 years
less than George’s age. After each year, the total of the grandchildren’s ages will increase
by 3, whilst George ages only 1 year. So they will gain 2 years in total each year. So in 6
years, they will gain the 12 years and the sum of their ages will equal George’s age,
hence (B).
21. The number of prize-winning girls was 15% of 80, which is 12. Including the 6 boys, there
were 12 + 6 = 18 prize winners out of 80 + 70 = 150 students. As a percentage, this is
18
× 100 = 12%,
150
hence (B).
www.amt.edu.au
2022 AMC Upper Primary Solutions43
F
22. Rotating the page by 180◦ and then reflecting the image in a vertical mirror line has the
same effect as reflecting the image in a horizontal mirror line, as demonstrated by the
letter F:
F
rotate
180◦
F reflect
reflect
F
So, for a letter to look the same after this process, it must have a horizontal line of
symmetry. There are 9 such letters:
B CD E H I KOX
hence (E).
1000
23. For each class, the 1000 students will require = 40 teachers. There are 6 classes each
25
day for the students so we need 40 × 6 = 240 teacher classes. Each teacher teaches only 5
classes each day, so we need 240 ÷ 5 = 48 teachers,
hence (B).
24. There will be 6 starting with p, then 6 starting with q and 6 starting with r. So the 19th
will be the first one starting with s. This is spqr,
hence (A).
www.amt.edu.au
44 2022 AMC Upper Primary Solutions
Alternative 2
7
Completing 7 + 15 + x + 2 = 40 gives x = 16, as shown.
Completing 8 + 16 + y + 9 = 40 gives y = 7, as shown.
Each line adds to 40, so summing all 5 gives 5 × 40 = 200. This 8 15 a b
counts every number twice, so the 10 numbers add to 100.
16 c
The 7 known numbers add to 64 so a + b + c = 36. Since
a + b = 17, c = 36 − 17 = 19. The last two numbers can now 7
be found: a = 5 and b = 12. 2 9
The largest number used is c = 19,
hence (E).
27. The cake has 5 sides to be decorated and each will have a 9 × 9 array of chocolate drops.
This gives a total of 5 × 81 = 405,
hence (405).
This table has 50 full rows, so there are 3 × 50 + 1 = 151 possible totals,
hence (151).
29. Let the Albury sprinters be A, B, C, D and E. We need to choose two of these and we
have the combinations (AB, AC, AD, AE) (BC, BD, BE) (CD, CE) and (DE) giving 10
combinations. In a similar way, we find that there are 5 + 4 + 3 + 2 + 1 = 15 combinations
for the Wodonga pair. Each of the 10 Albury combinations can be combined with each of
the 15 Wodonga combinations, so we have 10 × 15 = 150 ways to choose the team,
hence (150).
www.amt.edu.au
2022 AMC Upper Primary Solutions45
30. Alternative 1
On the net, we can identify faces and visualise how they fold together:
top
12
Try 2 as the height. From the front and top faces, the length of the prism would be
12 − 2 = 10. Similarly, from the right and bottom faces, the width of the prism would be
10 − 2 = 8. Then the total width of the net would be 2 × 8 + 2 × 10 = 36, but it should
be 32.
As we increase the height, the calculated values of width and length will both be smaller.
So try 3. Then the width is 10 − 3 = 7 and the height is 12 − 3 = 9. With these, the total
width of the net is 2 × 7 + 2 × 9 = 32.
Increasing the height beyond this will make the net narrower than 32 cm.
So the volume of the prism is 9 × 7 × 3 = 189 cm3 ,
hence (189).
Alternative 2
On the net, we can identify faces and visualise how they fold together, labelling each edge
with L, W and H:
W
L top L
12
L W L W
H left H front H right H back H
L W L W top
10 g ht
W bottom W
H front ri
L L
W
16 16
www.amt.edu.au
46 2022 AMC Junior Solutions
Junior Solutions
Solutions – Junior Division
1. The rhombus has 4 sides, each of length 9 cm, so the perimeter is 4 × 9 = 36 cm,
hence (E).
2. (Also S1)
4 − 7 = −3,
hence (C).
3. 20 × 22 = 440,
hence (B).
4. (Also UP11)
We look for the spinner which has twice as many red sections as white sections. The fourth
spinner has 4 red sections and 2 white sections and is the only one which fits,
hence (D).
6. Alternative 1
1
The square has area 2 × 2 = 4. The shaded triangle has area × 1 × 2 = 1. Therefore the
2
1
shaded triangle is of the area,
4
hence (B).
Alternative 2
The square can be divided into 4 copies of the shaded triangle, so the
1
shaded triangle is of the area,
4
hence (B).
8. Alternative 1
7 1 15 3 16 1 17 1 18 3
Converting to mixed numbers, =3 , =3 , =3 , = 4 , and = 3 . Of
2 2 4 4 5 5 4 4 5 5
1
these, only 4 is between 4 and 5,
4
hence (D).
www.amt.edu.au
2022 AMC Junior Solutions47
Alternative 2
Rather than converting these fractions to mixed numbers, note that most of them are less
8 16 20 17 1
than 4 = = = . The only one that is greater than 4 is = 4 which is less than
2 4 5 4 4
5,
hence (D).
9. (Also I5)
Since P QR is isosceles, ∠P QR = ∠P RQ. Also ∠P QR + ∠P RQ = 180◦ − 48◦ = 132◦ .
Consequently ∠P QR = ∠P RQ = 132 ÷ 2 = 66◦ ,
hence (B).
3 1
10. One-quarter of 7 days is 7 ÷ 4 = 1 = (2 − ) days.
4 4
So we count 2 days forward to midday Tuesday, then back 6 hours to 6 am Tuesday,
hence (A).
11. The hidden numbers are 60 ÷ 30 = 2, 60 ÷ 5 = 12 and 60 ÷ 4 = 15, and so their sum is
2 + 12 + 15 = 29,
hence (C).
12. The smallest square has side 3 cm and the largest square has side 9 cm. Consequently the
medium-sized square has side 9 − 3 = 6 cm and perimeter 6 × 4 = 24 cm,
hence (C).
3
14. The angle for juice is 90◦ which is times as much as the 60◦ angle for milk. So the
2
3
number of students who chose juice is × 80 = 120,
2
hence (C).
15. A number N is divisible by 3 exactly when S, the sum of its digits, is divisible by 3. So
we test the sum of the digits of the nine numbers:
www.amt.edu.au
48 2022 AMC Junior Solutions
2
16. I can see at most 15 dots per dice . Then 100 ÷ 15 = 6 , so 6 dice is too few.
3
For exactly 100 dots, write 100 = (6 × 15) + (1 × 10), so that I can arrange 7 dice by having
6 dice with 15 dots visible and 1 dice with 5 + 4 + 1 = 10 dots visible:
17. Here is how the logo is reflected, rotated and reflected again:
hence (A).
18. The patterns ××× and ◦◦◦ are not allowed in any row or column,
so several squares can be filled to avoid these. This gives all the
◦ × × ◦ ◦/××/◦
squares shown with dark background. × ◦ × ◦ ×/◦◦/×
Due to the rule that columns and rows must have three of each × ◦ ◦ ×× ◦
symbol, the five symbols with pale background are required. ◦×◦×◦×
There are two ways to complete the four remaining squares, as ×◦×◦ ◦×
shown in gold. However, in both cases, the three outlined squares ◦ × ◦ ×× ◦
have ◦ × ◦,
hence (B).
19. Alternative 1
Suppose Ash and Sash swap x cards, Ash and Tash swap y cards, and Sash and Tash swap z
cards. Then Ash’s total number of cards gives x+y = 11, Sash’s gives x+z = 8 and Tash’s
gives y + z = 15. Adding the three equations results in 2x + 2y + 2z = 11 + 8 + 15 = 34,
so x + y + z = 17. Subtracting the first equation shows that z = 6,
hence (D).
Alternative 2
There are 8 + 11 + 15 = 34 cards. Ash gives 11 cards and receives 11 cards, so he touches
22 cards and doesn’t touch 34 − 22 = 12 cards. These 12 cards must swap between Sash
and Tash, and since each swap involves 2 cards, there are 6 swaps between Sash and Tash.
Then in Sash’s 8 swaps, 8 − 6 = 2 will be with Ash. Similarly, Tash swaps 15 − 6 = 9 cards
with Ash. These values give a complete solution, confirming that the unique solution is
where Sash swaps 6 cards with Tash,
hence (D).
www.amt.edu.au
2022 AMC Junior Solutions49
20. Alternative 1 X
P Q
Let X, Y and Z be as indicated on the diagram and draw in the
line XY . By symmetry, point Z is the centre of rectangle P XY S.
1 Z
Consequently the area of SZP is the area of this rectangle and
4
1
hence the area of P QRS. There are four such unshaded triangles, S R
8 Y
1 1
so the unshaded area of P QRS is 4 × = and so the fraction that
8 2
1
is shaded is also ,
2
hence (D).
Alternative 2
P X
Consider just one quadrant of the square, with labelling as shown.
The solution to Question 6 (on page ??) shows that triangles P Y N
1 M
and P XM each has area of this smaller square and so the remaining
4
1
shaded region is of this square. This applies to each quadrant of the Y
2 N
1
larger square P QRS, so of the square is shaded,
2
hence (D).
22. A square will be shaded if the row number is a factor of the column number.
So the number of shaded squares in column 105 equals the number of factors of 105. The
factors of 105 are {1, 3, 5, 7, 15, 21, 35, 105}, so column 105 has 8 shaded squares,
hence (C).
23. If Andrew had kept walking in the rain, it would have taken him 9 minutes to get home.
Running at 1.5 times his walking speed would have taken him 9÷1.5 = 18÷3 = 6 minutes.
The total time would then be 15 + 6 = 21 minutes,
hence (C).
24. Alternative 1
1 1 1 1 1 1
− = > so minus any of the smaller fractions will also be greater than . This
2 3 6 10 2 10
gives 7 pairs.
1 1 1 1 1 1 1 2 1
If the larger fraction is , then − = < but − = > . The smaller
3 3 4 12 10 3 5 15 10
fractions will also work, giving 5 more pairs.
1 1 1
If the larger fraction is , then we have to get to − to find a pair whose difference is
4 4 7
1
greater than . The smaller fractions will also work, giving 3 more pairs.
10
www.amt.edu.au
50 2022 AMC Junior Solutions
1
If the larger fraction is or less, then taking away any of the smaller fractions gives a
5
1
result less than so there are no more pairs.
10
In all there are 7 + 5 + 3 = 15 pairs,
hence (C).
Alternative 2
Make a table with each fraction to three decimal places, shading those where the difference
is greater than 0.1. Due to increasing and decreasing trends, not all differences need to be
calculated.
second fraction
0.333 0.250 0.200 0.167 0.143 0.125 0.111
0.500 0.167
first 0.333 — 0.083 0.133
fraction 0.250 — — 0.083 0.107
0.200 — — — 0.089
These could be out due to rounding error, but this could only occur in the thousandths
place, so there is no doubt that 15 of these differences are greater than 0.1,
hence (C).
25. If we start in a corner, such as the lower left, there are three possibilities:
2 3 6 6 5 4 2 3 4
1 4 5 1 2 3 1 6 5
3 4 5 5 4 3
2 1 6 6 1 2
It is the same for each of the two centre squares giving 4 placements.
In total there are 12 + 4 = 16 placements,
hence (A).
26. In the tens place, the only way for ♥ + ♥ to result in a digit of ♥ is if ♥ is 0 with no carry
from the units place, or if ♥ is 9 with a carry of 1 from the units place.
Clearly ♥ cannot be 0, otherwise the resulting sum ♥♥♦ would be less than the numbers
being added together. So ♥ is 9 and it follows that ♦ is 4.
To ensure the carry from the units place, must be 7. So the sum is 497 + 497 = 994,
with ♦ ♥ = 497,
hence (497).
www.amt.edu.au
2022 AMC Junior Solutions51
27. Alternative 1
Suppose the numbers have digits abcd and efgh, so that the sum of the two numbers is
If there was no carry when doing the addition, then the digits of this sum would be a + e,
b + f , c + g and d + h, and then the digit sum is a + e + b + f + c + g + d + h = 36. For
example 1234 + 8765 = 9999 has no carry and the digits sum is 36.
For each carry in the standard addition algorithm, the value in one column is reduced by
10 and the column to the left is increased by 1. So the total of the sum of all digits in the
answer is reduced by 9. So with one, two or three carries, the digit sum would go down
to 27, 18 or 9. The sum can’t be reduced to 0, and examples with three carries such as
6431 + 8572 = 15003 show that the minimum possible digit sum is 9.
Then the difference between the smallest and largest digit sums is 36 − 9 = 27,
hence (27).
Alternative 2
Let the two four-digit numbers be A and B and let C = A + B. The most C can be is
1000 × (8 + 7) + 100 × (6 + 5) + 10 × (4 + 3) + (2 + 1) = 16173. Then C , the sum of the
digits of C, is at most 36, when C = 9999. This maximum can be found in many ways,
such as A = 8765, B = 1234, which has C = 9999 and C = 36.
To find the smallest possible C , we can try to pair up digits in A and B so that most of
the digits of C are 0. For instance if A = 5876 and B = 3124 then C = 9000 and C = 9.
It seems likely that this is the smallest possible.
We claim that C is a multiple of 9, which would confirm that C is at least 9. This can be
shown using the principles behind the standard test for divisibility by 9, commonly called
casting out nines.
Let A be the sum of the digits of A and B be the sum of the digits of B. Then A + B =
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36 which is a multiple of 9. Both A and A always have
the same remainder on division by 9, so that A − A is a multiple of 9. Specifically, if
A = 1000a + 100b + 10c + d and A = a + b + c + d then A − A = 9(111a + 11b + c).
Similarly B − B is a multiple of 9. Then
C = A + B = (A − A ) + (B − B ) + (A + B ) = (A − A ) + (B − B ) + 36
28. The table below shows amounts received by all members as we increase the amount received
by trainees: $2, $3, $4, . . . Given there are twice as many legends as paddlers, these
members have been arranged into teams of 2 legends for every paddler. Since these teams
make up more than half the membership, and there are 20 trainees, there are at least 7
teams. We can work out how much is given to trainees, how much is given to the others
(the teams) and how much each team receives. From this we can check that there are a
whole number of teams.
www.amt.edu.au
52 2022 AMC Junior Solutions
For example, the first row has $2 for each trainee, 6 × 2 = $12 for each paddler and
12 + 5 = $17 for each legend. The trainees receive a total of 2 × 20 = $40, leaving
1000 − 40 = $960 for the others. The others are in their teams, with each team receiving
12 + 2 × 17 = $46. However, 960 is not a multiple of 46.
Other rows follow the same steps, helped by some of the patterns in the columns. All
the divisions are fully worked out here, although in most cases this is not necessary. This
strategy finishes fairly quickly, since each team receives more than 18 times the trainee
amount, and since there are at least 7 teams, the amount going to teams is more than 126
times the trainee amount. Since 126 × 8 > 1000, the trainee amount is $7 or less.
Tr.$ Pad.$ Leg.$ Team$ All Tr.$ All teams $ Number of teams OK?
2 12 17 46 40 960 960 ÷ 46 = 20 4046
×
3 18 23 64 60 940 940 ÷ 64 = 14 4464
×
4 24 29 82 80 920 920 ÷ 82 = 11 1882
×
5 30 35 100 100 900 900 ÷ 100 = 9 YES
75
6 36 41 115 120 880 880 ÷ 115 = 7 115 ×
44
7 42 47 136 140 860 860 ÷ 136 = 6 136 ×
So the only solution has $5 per trainee, 9 teams and the club has 20 + 9 × 3 = 47 members,
hence (47).
12 × 20 + 6 × 10 + 2 × 30 = 240 + 60 + 60 = 360,
hence (360).
www.amt.edu.au
2022 AMC Junior Solutions53
x y Subtotal
1 2–49 48
−2
2 3–48 46
−1
3 4–48 45
−2
4 5–47 43
.. .. .. −1
. . .
29 30–35 6
−2
30 31–34 4
−1
31 32–34 3
−2
32 33 1
1 + 3 + 4 + 6 + 7 + 9 + · · · + 40 + 42 + 43 + 45 + 46 + 48,
or
1 + 4 + 7 + · · · + 40 + 43 + 46
+ 48 + 45 + 42 + · · · + 9 + 6 + 3
Since there are 32 rows in the table, the above sum has 16 pairs of numbers which add to
49, so the total is 49 × 16 = 784,
hence (784).
Alternative 2
Each sum like 100 = 61 + 34 + 5 can be visualised as a row of counters of three colours,
such as 61 red counters followed by 34 green counters then 5 blue counters.
61 34 5
We’ll first work out the number of these arrangements of counters where we allow equal
numbers of some colours, and different arrangements (colours) of the same numbers are
counted separately. In this case, each choice corresponds to choosing the positions of the
two arrows in the diagram above. These arrows are in different positions, and not on the
first counter, so there are 99 possible positions.
www.amt.edu.au
54 2022 AMC Junior Solutions
www.amt.edu.au
2022 AMC Intermediate Solutions55
Intermediate Solutions
Solutions – Intermediate Division
1. (Also UP3)
2220 − 2022 = 198,
hence (C).
2. (Also UP6)
Alternative 1
Out of the 29 squares, there are 12 squares that have 3 edges on the perimeter, 4 squares
that have 2 opposite edges on the perimeter and 4 squares that have 2 adjacent edges on
the perimeter. This accounts for all 12 × 3 + 4 × 2 + 4 × 2 = 52 edges on the shape,
hence (A).
Alternative 2
Start with the central 3 × 3 square, with perimeter P = 12. Join the remaining 20 squares
one-by-one. Each join increases P by 2: an increase of 4 from the new square but then a
decrease of 2 from the join. After joining 20 squares, P = 12 + 20 × 2 = 52,
hence (A).
1+4 5 1
4. = = ,
9 + 16 25 5
hence (A).
5. (Also J9)
Since P QR is isosceles, ∠P QR = ∠P RQ. Also ∠P QR + ∠P RQ = 180◦ − 48◦ = 132◦ .
Consequently ∠P QR = ∠P RQ = 132 ÷ 2 = 66◦ ,
hence (B).
7(8 + 12)
6. The rectangle has area 7 × 12 = 84 and the shaded trapezium has area = 70 so
2
70 5
the fraction shaded is = ,
84 6
hence (C).
8. (Also J13)
Australia needs to buy enough petrol for another 30 days. This is 30 × 160 million = 4800
million litres,
hence (E).
www.amt.edu.au
56 2022 AMC Intermediate Solutions
9. Alternative 1
1011 − 674 = 337,
hence (A).
Alternative 2
2022( 12 − 13 ) = 2022( 16 ) = 2022 ÷ 6 = 337,
hence (A).
10. There are three s, so we divide the term on the right by three, giving 9x3 y 6 ,
hence (D).
12. Alternative 1
5 1 2 1 3 1
Let x be the number sought. Then = 1+ 1 so = 1 and = 1 + . Then
3 1+ x 3 1+ x 2 x
1 1
= so that x = 2,
2 x
hence (C).
Alternative 2
The expression has a number of add one and reciprocal operations chained together:
1 1
x +1 x +1 5
3
hence (C).
13. Alternative 1
The angles will be of the form 2x, 3x and 4x. Then
2x + 3x + 4x = 180
9x = 180
x = 20
4x = 80
hence (D).
www.amt.edu.au
2022 AMC Intermediate Solutions57
Alternative 2
The three angles have 2, 3 and 4 parts and add to 180◦ . So each part is 180◦ ÷ 9 = 20◦
and the largest angle is 20◦ × 4 = 80◦ ,
hence (D).
15
15. Daniel cycles for 15 minutes, so travels 20 × = 5 km. Luke walks for 20 minutes, so
60
20
travels 6 × = 2 km. Combined they travel 5 + 2 = 7 km,
60
hence (D).
16. Each of the 7 adults gives gifts to all 11 other people, for a total of 7 × 11 = 77 gifts.
Each of the 5 children gives gifts to all 4 other children, for a total of 5 × 4 = 20 gifts.
Altogether, the number of gifts that will be given is 77 + 20 = 97,
hence (C).
www.amt.edu.au
58 2022 AMC Intermediate Solutions
17. There are 16 equally likely outcomes from the two spins. Second spin
The diagram shows the two-digit numbers for these outcomes. 1 2 3 4
The 10 shaded cells indicate when a multiple of 11 is obtained, 1 11 12 13 14
10 5
First spin
so the probability of this is = , 2 22 22 23 24
16 8
3 33 33 33 34
4 44 44 44 44
hence (D).
1
18. The 8 unit squares each have area 1 and the 8 overlapping triangles are each of a unit
4
1
square. So the total area is 8 − 8 × = 8 − 2 = 6,
4
hence (C).
19. Suppose Nic has been learning for N days and Rick for R days. Then (R−14) = 5(N −14) =
5N − 70 and (R − 2) = 2(N − 2) = 2N − 4. Thus
R = 5N − 56 = 2N − 2
3N = 54
N = 18 and R = 2 × 18 − 2 = 34
Then R + N = 18 + 34 = 52,
hence (D).
20. In the left diagram, suppose the screen is 16 units by 9 units. Then the unshaded part of
the screen must be 12 units by 9 units, since it has the same height and its ratio of width
to height is 4 : 3. The proportion of the screen’s area which is not blacked out by the bars
12 3 1
is = , so = .
16 4 4
9 12 12 9
16 16
Similarly, in the right diagram we may assume the screen is 16 units by 12 units and the
9 3
unshaded part is 16 units by 9 units. Then the proportion not blacked out is = , so
12 4
1
r= .
4
Therefore = r, so : r = 1 : 1,
hence (C).
www.amt.edu.au
2022 AMC Intermediate Solutions59
22. The expression turns out to have the same value for all n 1:
3 (1 × 2 × 4) + (2 × 4 × 8) + · · · + (n × 2n × 4n)
(1 × 3 × 9) + (2 × 6 × 18) + · · · + (n × 3n × 9n)
3
3 8(1 + 8 + · · · + n ) 3 8 2
= 3
= =
27(1 + 8 + · · · + n ) 27 3
hence (B).
23. Alternative 1
Let the drum hold 12x litres. Then originally there are 5x litres of water and 7x litres of
5
milk. When 9 L are spilt, this is also in the ratio 5 : 7 so 9 × = 3.75 L of water and
12
7
9× = 5.25 L of milk are spilt.
12
This leaves 5x − 3.75 L of water and 7x − 5.25 L of milk. After refilling, water goes up to
5x − 3.75 + 9 = 5x + 5.25 L. The new ratio tells us that:
www.amt.edu.au
60 2022 AMC Intermediate Solutions
24. Let l be the lowest common multiple of p and q. We know that the product of any two
integers is equal to the lowest common multiple multiplied by the highest common factor,
pq prt
so pq = lt. So l = = = pr,
t t
hence (D).
25. Label the cells as shown, so that abde = bcef = degh = ef hi = 2 and
abc = def = ghi = adg = beh = cf i = 1. Then a b c
26. Alternative 1
In the hundreds column, c + a is either 9 or 10. In the units column, b + c is either 2 or 12.
If b + c = 2 then b = c = 1 and then a = 1 in the tens column. However, this puts 1 + 1 = 2
in the hundreds column. So there is no solution with b + c = 2.
So b + c = 12 and then a + b = 11 in the tens column and c + a = 9 in the hundreds
column. Combining these, 11 + 9 − 12 = (a + b) + (a + c) − (b + c) = 2a so that 2a = 8
and a = 4. Then b = 7 and c = 5 and the three-digit number abc is 475,
hence (475).
Alternative 2
Writing the numbers with their place values,
2022 = (1000 + 100c + 10a + b) + (100a + 10b + c)
= 1000 + 110a + 11b + 101c
=⇒ 1022 = 110a + 11b + 101c
Then 110a, 11b, 99c and 1023 = 11 × 93 are multiples of 11, which can all be moved to the
left-hand side
1023 − 110a − 11b − 99c = 1 + 2c
Consequently 2c + 1 is a multiple of 11. Also 3 2c + 1 19 so that 2c + 1 = 11 and
c = 5. Then 11(93 − 10a − b − 9c) = 11 so that 10a + b = 92 − 9c = 92 − 45 = 47. Hence
a = 4 and b = 7 and the number abc is 475,
hence (475).
27. We get
6 111
171 × 66 . . . 6
= 171 × × 10 − 1
9
111 sixes
= 114 × (10111 − 1)
= 113 99 . . . 9 886.
108 nines
Thus, 1 + 8 + 1 + 8 + 3 + 6 + 9 × 108 = 9 × 111 = 999,
hence (999).
www.amt.edu.au
2022 AMC Intermediate Solutions61
x y Subtotal
1 2–49 48
−2
2 3–48 46
−1
3 4–48 45
−2
4 5–47 43
.. .. .. −1
. . .
29 30–35 6
−2
30 31–34 4
−1
31 32–34 3
−2
32 33 1
1 + 3 + 4 + 6 + 7 + 9 + · · · + 40 + 42 + 43 + 45 + 46 + 48,
or
1 + 4 + 7 + · · · + 40 + 43 + 46
+ 48 + 45 + 42 + · · · + 9 + 6 + 3
Since there are 32 rows in the table, the above sum has 16 pairs of numbers which add to
49, so the total is 49 × 16 = 784,
hence (784).
Alternative 2
Each sum like 100 = 61 + 34 + 5 can be visualised as a row of counters of three colours,
such as 61 red counters followed by 34 green counters then 5 blue counters.
61 34 5
We’ll first work out the number of these arrangements of counters where we allow equal
numbers of some colours, and different arrangements (colours) of the same numbers are
counted separately. In this case, each choice corresponds to choosing the positions of the
two arrows in the diagram above. These arrows are in different positions, and not on the
first counter, so there are 99 possible positions.
www.amt.edu.au
62 2022 AMC Intermediate Solutions
29. Select numbers one at a time, at each point choosing the smallest number available that
can’t be used to form a triangle with two previously-chosen numbers. The first three chosen
are 1, 2 and 3.
The fourth choice can’t be 4, since there is a triangle with sides 2, 3 and 4, due to 2+3 > 4.
So we choose 5. This works, since if a and b are chosen from {1, 2, 3} then a + b ≤ 5, so
a, b, 5 don’t make a triangle.
The same principle tells us that the fifth choice is 8 — 3,5,6 and 3,5,7 both form triangles,
but 8 can’t form a triangle with two numbers from {1, 2, 3, 5}, since 8 is the sum of the
two largest numbers.
In general each choice from here on will be the sum of the two previous choices, giving
Fibonacci numbers.
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, . . .
The first 15 of these numbers (up to 987) gives us a choice of 15 elements with no triangles,
and it seems likely that no more than 15 is possible.
To see that no more than 15 elements are possible, suppose we have our set of chosen
numbers in order: 1 a < b < c < d < e < · · · . Then a 1, b 2, and c 3.
Since b, c and d don’t form a triangle d b+ c 2 + 3 = 5. Similarly e c+ d 3 + 5 = 8,
and so on: each number is greater or equal to a corresponding Fibonacci number.
So there can’t be a 16th element in the list — it would have to be p 1597. Consequently,
the largest number of elements that can be chosen is 15,
hence (15).
30. Alternative 1
For each column of 3 horizontal lines in the grid, the path must either occupy 1 of the lines
or 3 of the lines. Make a 7-digit code from these lines. For instance, we can read across
the diagram in the question to get the code 3313111.
Consider the initial block of ‘3’s (possibly no ‘3’s) followed by a 1. Then the path must
make an ‘S’ shape like one of the following:
etc
www.amt.edu.au
2022 AMC Intermediate Solutions63
Then the path goes one column over and for the next block of ‘3’s, it creates a ‘Z’ shape
like those above, but inverted. This continues across the grid, each time with the path
being uniquely determined by the code.
There are 27 = 128 possible 7-digit codes of ‘1’s and ‘3’s, but not all of them work. To
ensure the path ends up at Bryn (rather than lower right) there must be an even number
of ‘1’s in the code. However, this is exactly half of all 128 codes, since for every code with
an even number of ‘1’s, there is a unique corresponding code with an odd number of ‘1’s
that is found by swapping ‘1’s and ‘3’s. So the number of possible paths is 128 ÷ 2 = 64,
hence (64).
Alternative 2
For a grid of 3 rows and n columns, let An be the number of paths from bottom left to
top right and let Bn be the number of paths from bottom left to bottom right. We need
to find A8 .
By inspection, we can tabulate the first few values:
n 1 2 3 4 5 6 7 8
An 1 1 2
Bn 0 1 2
For n > 3, each path will either enter the top right vertically or horizontally. Let k 0 be
the number of horizontal segments the path ends with. Then to ensure the bottom right
is visited, there is only one possibility for the k + 1 rightmost columns, as shown.
···
∗ ∗ ∗ ∗
Then for k = 1, . . . , n − 1, the number of paths in each case will be equal to the number of
paths to the point marked ∗ which is be Bn−k−1 . In addition there is the case k = n − 1,
where there is only one possibility. Consequently An = Bn−1 + Bn−2 + · · · + B1 + 1. So for
instance A4 = B3 + B2 + B1 + 1 = 2 + 1 + 0 + 1 = 4.
For Bn , a similar analysis gives us that Bn = An−1 + An−2 + · · · + A1 , so that B4 =
A3 + A2 + A1 = 2 + 1 + 1 = 4. Using these formulas, the table can be completed:
n 1 2 3 4 5 6 7 8
An 1 1 2 4 8 16 32 64
Bn 0 1 2 4 8 16 32 64
www.amt.edu.au
64 2022 AMC Senior Solutions
Senior Solutions
Solutions – Senior Division
1. (Also J2)
4 − 7 = −3,
hence (C).
2. 135 ÷ 15 = 9,
hence (B).
3. 51 + 42 + 33 + 24 + 15 = 5 + 16 + 27 + 16 + 1 = 65,
hence (E).
1 1 11 10 1
4. − = − = ,
20 22 220 220 220
hence (E).
5. (Also I11)
The diagonals of the rectangle meet at the midpoints of its two diagonals. One of these
diagonals
hasendpoints (1, 8) and (7, 4). The midpoint is found by averaging coordinates
1+7 8+4
, = (4, 6),
2 2
hence (A).
6. Alternative 1
Let the perpendicular height of the trapezium (and hence the shaded parallelogram) be
h. Then the area of the large trapezium is 12 (3 + 5)h = 4h and the area of the shaded
1
parallelogram is 1h = h. So of the area is shaded,
4
hence (C).
Alternative 2
In this grid of five rectangles, the trapezium is shaded. Since the last
two rectangles are together half shaded, the trapezium’s area is 4 grid
rectangles.
The area coloured gold is two half grid grid rectangles, for a total of 1
1
grid rectangle. So this is of the trapezium’s area,
4
hence (C).
8. The angle sum in a quadrilateral is 360◦ . Thus 140 + 90 + 2x = 360 and so 2x = 130 and
x = 65,
hence (A).
www.amt.edu.au
2022 AMC Senior Solutions65
9. Alternative 1
The only way of obtaining 12 is 11 on the large dice and 1 on the small dice. Since these
two rolls are independent,
1 1 1
P (total 12) = P (11 on large dice) × P (1 on small dice) = × =
6 2 12
hence (D).
Alternative 2
These rolls can be visualised as this 6 × 6 table of 36 equally likely outcomes:
0 0 0 1 1 1
1 1 1 1 2 2 2
3 3 3 3 4 4 4
5 5 5 5 6 6 6
7 7 7 7 8 8 8
9 9 9 9 10 10 10
11 11 11 11 12 12 12
3 1
As shown, 3 of the 36 have total 12, so the probability is = ,
36 12
hence (D).
12. Let the radii of the circles be x, y and z, with z being the radius of the largest circle.
Then x + y = 8, x + z = 9, and y + z = 13. Adding all three equations, 2x + 2y + 2z =
8 + 9 + 13 = 30 so that x + y + z = 15. Then z = 15 − 8 = 7,
hence (C).
13. The five numbers have integer mean, so their sum is a multiple of 5. The original four
numbers add to 26, and the fifth number must be between 3 and 11 but can’t be 4. So the
fifth number is 9, the sum of all five is 35 and the new mean is 35 ÷ 5 = 7,
hence (D).
www.amt.edu.au
66 2022 AMC Senior Solutions
√
√ √ 3
14. By Pythagoras’ theorem, BC = 22 − 12 = 3. By similar triangles, BD = .
2
3 3
So, again by Pythagoras’ theorem, CD = 3 − = ,
4 2
hence (B).
15.
√
3x + 3x+1 + 3x+2 = 13 3
√
3x (1 + 3 + 32 ) = 13 3
√ 1
3x = 3 = 3 2
1
x=
2
hence (B).
hence (D).
17. Alternative 1
1
AC
Using the labelling in the diagram, note that AZ =
4 C
3 1 3 3
and AX = AB. Hence the area of AXZ is × = 2
4 4 4 16
of the area of ABC. 6
The same applies to the other smaller triangles, so their
9 Z 6
combined area is of ABC.
16 2
The smaller equilateral triangle is formed by taking these
three triangles away from the larger equilateral triangle, A 6 X 2 B
7
leaving an area of . So the ratio is 7 : 16,
16
hence (B).
Alternative 2
Subdivide each side of the large triangle into 4, and draw a grid of 16
equilateral triangles to use as area units. Each of the shaded triangles
is half of a parallelogram with area 6, so has area 3.
So the unshaded triangle has area 16 − 9 = 7. Consequently this
7
triangle is the area of the large one,
16
hence (B).
www.amt.edu.au
2022 AMC Senior Solutions67
18. The full square pyramid is 3 times the height of piece P , and is similar to P . Consequently
it has 3 times the width and 3 times the depth, so is 33 = 27 times the volume. That is,
the full square pyramid has volume 27 × 5 = 135 cm3 .
Similarly, the pyramid formed by just P and Q has volume 23 = 8 times the volume of P ,
which is 8 × 5 = 40 cm3 .
Then piece R has volume 135 − 40 = 95 cm3 ,
hence (D).
Since each term only depends on the previous two, the repeat of the pair a1 = a6 = 1
and a2 = a7 = 2 means that the pattern 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, . . . will continue, repeating
every five terms. Consequently a100 = a5 = 1,
hence (A).
21. Alternative 1
The volume is the dotted cube minus the 8 cut-off pyramids.
√ √ √ √
The side length of the cube is 2 2, so its volume is (2 2)3 = 16 2. 2
2 2
Consider one of the cut-off pyramids. It has three dotted edges,
√ 1√ √
each of length 2. The area of its base is A = 2 2 = 1 and so 2 2 √
√ 2 2
1 2
its volume is Ah = . Since there are 8 identical corners, their √
3 √3 2
8 2
combined volume is .
3
www.amt.edu.au
68 2022 AMC Senior Solutions
√ √
√ 8 2 40 2
Thus the volume of the cuboctahedron is 16 2 − = ,
3 3
hence (B).
Alternative 2
√
As in the first solution, the original cube had squares of side 2 2 and the diagonals of these
√
squares are therefore length 4. We can make 4 vertical cuts to slice it into a 2 × 2 × 2 2
prism plus 4 rectangle-based pyramids, as shown.
√
The volume of the prism is 8 2.
Each of the
√ 4 pyramids
√ has a rectangular base of
area 2×2 2 = 4 2 and perpendicular height 1, as
1 √ 4 1
shown in the top view. So it has volume ×4 2 =
3 2
4√
2.
3 1
Then the volume of the cuboctahedron is
√ √
4√ 24 + 16 √ 40 2
8 2+4× 2= 2=
3 3 3
hence (B).
b+c f −e
x= and y=
a d
So x must be positive, but y may be negative. In order to minimise x + y, f should be less
than e. The smallest value of y occurs when f = 2, e = 9, and d = 1, giving y = −7. So
x + y is greater than −7.
7
Then with d, e, f as above and b = 3, c = 4, and a = 8, we have x = < 1 so that
8
x + y < −6. Whether or not this is the smallest possible value of x + y, we know that this
smallest value is between −7 and −5,
hence (B).
23. Alternative 1
As we are only concerned with ratios, let the radius of the semicircle be 1. Since the dashed
radii shown below are perpendicular to the corresponding tangent lines,√the large triangle
consists of 4 smaller right-angled isosceles triangles with sides 1, 1 and 2. Consequently
the large triangle has area 2.
1
x
x
2
www.amt.edu.au
2022 AMC Senior Solutions69
Let x be the side-length of the square. From the triangle formed by joining the corner of
x 2 2
the square to the centre, we have x2 + = 1, giving x = √ . Then the square has area
2 5
4 4
and the required ratio is 5
: 2 = 2 : 5,
5
hence (C).
Alternative 2
Choose a length unit so that the square has side 2, and hence area 4. By symmetry, the
centre of the semicircle is on the midpoint
√ of the lower side of the square. By Pythagoras’
theorem, the semicircle has radius 5. Draw √a copy of the diagram reflected in the diameter
of the √
semicircle, making√ a circle of radius 5 inscribed in a square. This square will have
side 2 5 and area (2 5)2 = 20. Then the original triangle has area 10 and the ratio of
areas is 4 : 10 = 2 : 5,
hence (C).
2 3 6 7 2 3 8 7 2 3 4 5 8 7 6 5
1 4 5 8 1 4 5 6 1 8 7 6 1 2 3 4
3 4 5 8 3 4 5 6 7 6 5 4
2 1 6 7 2 1 8 7 8 1 2 3
It is the same for each of the four centre squares giving 12 placements. In total there are
16 + 12 = 28 placements,
hence (D).
6 5
25. After 42 minutes I had gone of the distance and Sally had gone . So Sally’s speed is
11 11
5 6 4
of mine, and she will take × 77 minutes for the trip. After 44 minutes I had gone
6 5 7
3 3 4
and Wally had gone . So Wally’s speed is of mine and he will take × 77 for the trip.
7 4 3
Therefore the time difference will be
4 6 154
77 − = ≈ 10 minutes
3 5 15
hence (E).
www.amt.edu.au
70 2022 AMC Senior Solutions
12 × 20 + 6 × 10 + 2 × 30 = 240 + 60 + 60 = 360,
hence (360).
27. Let the square, cube and fourth power be s, c and f , respectively, with f = sc.
We are told s is even, so f is an even fourth power which, to be as small as possible, must
be of the form f = 24 × 34k for some k ≥ 1.
Since c is an odd cube factor of f , it must be of the form c = 33 , so s = 24 × 34k−3 . But
s is a square, so the index 4k − 3 must be non-negative and even.
The smallest solution is when k = = 2, so s = 24 × 32 = 144, c = 36 = 729, and
s + c = 144 + 729 = 873,
hence (873).
28. Alternative 1
If the terms are a1 , a2 , . . . then an = 12 (an−1 + an+1 ) − 1 so that an−1 + 2an + an+1 = 2. We
can rewrite this equation in the following way an − an−1 + 2 = an+1 − an so the differences
between the two consecutive terms are every time increased by 2. Therefore,
an − an−1 = 2 + an−1 − an−2
= 4 + an−1 − an−2
= ···
= 2(n − 2) + a2 − a1 = 2n
Then
an = a1 + (a2 − a1 ) + (a3 − a2 ) + · · · + (an−1 − an−2 ) + (an − an−1 )
= 2 + 4 + 6 + · · · + 2n
= n(n + 1)
To find the largest value of an = n(n + 1) less than 1000, we note that a30 = 30 × 31 = 930,
then a31 = a30 + 2 × 31 = 992 and a32 = a31 + 2 × 32 = 1056. So 992 is the largest term
less than 1000,
hence (992).
www.amt.edu.au
2022 AMC Senior Solutions71
Alternative 2
a+c
If three successive terms are a, b, c then b = − 1 so that c = 2b − a + 2. Using this
2
rule the sequence continues:
2, 6, 12, 20, 30, 42, . . .
From this we claim that the nth term is n(n + 1).
Assuming this claim, we need√ the largest value of n(n + 1) less than 1000. Since n(n + 1) =
2 2
n + n ≈ n , we check n ≈ 1000 ≈ 31.6. Then 31 × 32 = 992 and 32 × 33 = 1056 so that
the largest term less that 1000 is 992.
We address the claim by strong induction on n. The statement to be proven is that for
n 1 the nth term is given by the formula n(n + 1). This is true for n = 1, 2. For
n 3, let terms n − 2, n − 1 and n be a, b, c as above, then by the inductive hypothesis,
a = (n − 2)(n − 1) and b = (n − 1)n. Then
c = 2b − a + 2 = 2n(n − 1) − (n − 2)(n − 1) + 2 = n2 + n = n(n + 1)
so that the nth term is n(n + 1), completing the induction.
By strong induction, the claim is confirmed and 992 is the largest term less than 1000,
hence (992).
29. Alternative 1
At the beginning, stage 0, Wayne has the unstamped original. So there is 1 sheet with 0
occurrences of the ‘COPY’ stamp.
After stage 1, he has the unstamped original plus two stamped copies. So there is 1 sheet
with 0 stamps, and 2 sheets with 1 stamp.
After stage 2, he has the unstamped original, the first two copies, and two stamped copies
of each of these. So there is 1 sheet with 0 stamps, 4 sheets with 1 stamp, and 4 sheets
with 2 stamps.
We set up a table to track the number of sheets with a given number of stamps. At each
stage, the number of sheets with k stamps is equal to the number with k stamps at the
previous stage (those which were fed through the copier and not re-stamped) plus twice
the number with k − 1 stamps at the previous stage (the two new copies of each which
gain one extra stamp). Also, there is only ever one copy with no stamps, being the original
sheet of paper. Since we are interested only in the number of copies with 2 stamps, the
important part of the table is as follows:
Number of stamps
Stage
0 1 2
0 1 0 0
1 1 2 0
2 1 4 4
3 1 6 12
4 1 8 24
5 1 10 40
6 1 12 60
7 1 14 84
8 1 16 112
www.amt.edu.au
72 2022 AMC Senior Solutions
This iterative process is the same as that for the coefficients of xk in (1 + 2x)n shown in
the last column of the table. When multiplying by (1 + 2x), the coefficient of xk+1 in
(1 + 2x)n+1 is twice the coefficient of xk plus the coefficient of xk+1 in (1 + 2x)n .
The x2 term in (1 + 2x)8 is 82 16 (2x)2 = 112x2 , with coefficient 112, and this is also the
number of pages with 2 stamps,
hence (112).
30. Alternative 1
We first calculate how many ways the five runners could finish the race, with each receiving
a different placing to the week before. This is equal to the number of permutations of
A, B, C, D, E such that each symbol is not in its original position. When this permutation
is written in cycle notation, it must be a 5-cycle such as (A, C, E,
5B,
D) or a 3-cycle and a
2-cycle such as (A, E, D)(B, C). There are 4! = 24 5-cycles and 3 × 2 = 20 permutations
that comprise a 3-cycle and a 2-cycle. Thus, there are 24 + 20 = 44 ways in which the five
runners could receive a different placing to the week before.
Within each of these 44 permutations there will be some “up” runners whose place in the
second race is higher than their place in the first race, and some “down” runners the other
way around.
There is only one possibility with 4 up runners and 1 down runner, with order BCDEA,
corresponding to the 5-cycle (A, E, D, C, B). The reverse permutation gives the only pos-
sibility with 4 down runners and 1 up runner, with order EABCD, corresponding to the
5-cycle (A, B, C, D, E).
www.amt.edu.au
2022 AMC Senior Solutions73
Of the remaining 42 permutations, half have 3 up and 2 down and the other half have 3
down and 2 up. This is because the reversed permutation effectively swaps race 1 and race
2, while turning an increase in placing to a decrease in placing and vice-versa. This creates
a one-to-one correspondence between these two sets of permutations.
Consequently there are 21 possibilities,
hence (21).
Alternative 2
Label the runners A to E, and we’ll use notation like C↑ to mean that C’s place in the
second race is higher than in the first. Then E↑ and A↓.
One other runner will move up in the ranking, giving 3 cases: B↑, C↑ or D↑.
In the first case, B↑, E↑, A↓, C↓ and D↓. The possibilities for the second race are then
shown in Table 1: B must move up to 1st and D must move down to 5th, C down to 4th,
then A and E can go in either order. This gives 2 possibilities.
A↓ B A↓ C E A↓ D D D E E E
B↑ A,E B↓ A,E C B↓ E A,E A,E D A,D A,D
C↓ A,E C↑ A,B,E A,B C↓ B A,E A,B,E B A,D A,B,D
D↓ C D↓ A,B,E A,B D↑ C C A,B,E C C A,B
E↑ D E↑ D D E↑ A B C A B C
2 4 2 1 2 4 1 2 3
Table 1 Table 2 Table 3
Table 2 summarises possibilities if C↑, E↑, A↓, B↓ and D↓. D must move down to 5th, and
only C or E can end up 1st, so we consider those separately. If C is 1st then 2nd must be
A or E. Also B must be 3rd or 4th. This gives the 4 possibilities. If E is first, then C must
be 2nd. This leaves A and B for 3rd and 4th as shown. So Table 2 includes 6 possibilities.
Table 3 summarises possibilities if D↑, E↑, A↓, B↓ and C↓. The 6 columns consider the
6 subcases based on 1st and 5th positions. Note that C must finish 4th or 5th, so the
subcases where A or B finish 5th must have C finishing 4th, giving very few possibilities.
Summing up these 6 subcases gives 1 + 2 + 4 + 1 + 2 + 3 = 13 possibilities.
In all, we have counted 2 + 6 + 13 = 21 possibilities,
hence (21).
www.amt.edu.au
74 2022 AMC Answer Key
Answer Key
Middle Upper
Question Junior Intermediate Senior
Primary Primary
1 C E E C C
2 D D C A B
3 B C B B E
4 B B D A E
5 A C D B A
6 D A B C C
7 E E E C E
8 B C D E A
9 C E B A D
10 A A A D C
11 B D C A B
12 E E C C C
13 B D E D D
14 A A C E B
15 B D D D B
16 B C C C D
17 E D A D B
18 E A B C D
19 D B D D A
20 C C D C B
21 C B C B B
22 E E C B B
23 C B C B C
24 C A C D D
25 D E A E E
26 258 56 497 475 360
27 138 405 27 999 873
28 990 151 47 784 992
29 56 150 360 15 112
30 151 189 784 64 21
www.amt.edu.au