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1.A three-digit code uses only the digits 1, 2 and 3, and digits may repeat. How many such codes contain at least one 3?[5]
A8
B9
C12
D19
E27
2.A number is 4 more than a third of itself. Which of the choices is that number?[5]
A3
B6
C8
D12
E18
3.Five beads of five different colours are threaded on a circular bracelet. Two bracelets count as the same if one can be turned or flipped into the other. How many different bracelets are there?[5]
A12
B20
C24
D60
E120
4.How many three-digit whole numbers contain at least one digit 7?[5]
A243
B252
C271
D280
E648
5.A box holds 4 red, 5 blue and 6 green balls, mixed up. How many must be taken out without looking to be sure of having 3 of the same colour?[5]
A3
B6
C7
D9
E13
6.One hundred apples are packed into twelve boxes. What is the largest number n for which you can always be sure that some box holds at least n apples?[5]
A8
B9
C10
D12
E100
7.The numbers 1 to 8 are on a board. A move rubs out two of them and writes their difference, larger minus smaller. After seven moves one number is left. Can it be 1?[5]
AYes, and there are many ways
BYes, but only one way
CNo, the last number is always even
DNo, the last number is always 0
EIt depends on the order of the moves
8.A class of 12 scored an average of 8. Another class of 18 scored an average of 13. What is the average for all 30 students together?[5]
A10.5
B11
C11.5
D12
E21
9.How many squares of any size can be found on a 3 by 3 board?[5]
A9
B10
C12
D13
E14
10.How many two-digit numbers have both of their digits odd?[5]
A10
B20
C25
D45
E50
11.A jug holds 900 ml of juice. A third is poured out, then 150 ml is added. How many millilitres are in the jug?[5]
A450
B600
C750
D900
E1050
12.Squares of dots grow: 1 dot, then 4, then 9, then 16. How many dots are in the seventh square?[5]
A25
B36
C42
D49
E64
13.A drawer holds 12 red socks, 10 blue socks and 8 green socks, all mixed up. You take socks out one at a time in the dark. What is the smallest number of socks you must take to be certain of having three of the same colour?[5]
A4
B7
C9
D13
E3
14.What is the last digit of 3 multiplied by itself 2026 times, that is, of 3 to the power 2026?[5]
A3
B9
C7
D1
E6
15.What is the smallest number n such that ANY collection of n whole numbers must contain two of them whose difference is divisible by 7?[5]
A8
B7
C14
D15
E4
Kangaroo Atlas · https://kangaroo-atlas.vercel.app · seed 2 · CC BY-NC-SA 4.0 · Independent project, not affiliated with Math Kangaroo in USA, NFP.
Answer key — Math Kangaroo practice
Each wrong choice carries the thinking that produces it. When a student picks C, this is what they were doing.
1.D — 19KA-0018Complementary counting
AI counted the codes that avoid 3 entirely and gave that as my answer without subtracting.
BI counted the codes with a 3 in the first position only and forgot the other two positions.
CI counted the codes with exactly one 3 and forgot the ones with two or three of them.
EI counted every possible code and forgot to remove the ones with no 3 at all.
2.B — 6KA-0045Back-solve from the answer choices
AI tested the smallest choice first and stopped at the first one that felt about right.
CI read a third of itself as three times itself and solved the wrong equation.
DI reversed the comparison, finding a number whose third is 4 less than itself in the wrong direction.
EI picked the largest choice because dividing by three made me expect a big starting number.
3.A — 12KA-0071Overcount, then correct
BI divided the 120 arrangements by 6 rather than by the 10 movements that leave a bracelet looking the same.
CI allowed for turning the bracelet but forgot it can also be flipped over.
DI halved the 120 for flipping but forgot that turning also gives the same bracelet.
EI counted every arrangement in a line, as if the bracelet had a fixed first bead.
4.B — 252KA-0075Complementary counting
AI counted the numbers made entirely of digits other than 7 in every position, including a leading zero.
CI counted the numbers with a 7 in each position separately and forgot that some were counted twice.
DI added three lots of 90 and one extra hundred, double counting the 700s.
EI counted the numbers with no 7 at all and gave that instead of subtracting it.
5.C — 7KA-0077The pigeonhole principle
AI answered with the luckiest case, where the first three balls happen to match.
BI found the worst case of two of each colour and forgot to take one more.
DI multiplied the three colours by the three balls I need, rather than thinking about the worst case.
EI worked from the number of balls of each colour rather than from the number of colours.
6.B — 9KA-0095The extremal principle
AI divided 100 by 12 and rounded down instead of up.
CI rounded the division to a convenient number without checking that a packing with a smaller maximum exists.
DI gave the number of boxes as the answer rather than a number of apples.
EI described the case where one box holds everything, which is possible but not guaranteed.
7.C — No, the last number is always evenKA-0098Invariants and monovariants
AI found sequences ending in small numbers and assumed 1 was among the reachable ones.
BI assumed a hard-looking question must have exactly one answer.
DI found one sequence ending in 0 and assumed every sequence must end there.
EI thought the order could change the outcome, when the quantity that decides it never changes.
8.B — 11KA-0112Mixtures and weighted averages
AI averaged the two averages, ignoring that the classes are different sizes.
CI weighted the averages but used the wrong class size against each one.
DI leaned towards the larger class's average without computing the totals.
EI added the two averages together instead of combining them.
9.E — 14KA-0133Counting shapes hidden inside a figure
AI counted the nine small squares and stopped, missing every larger one.
BI counted the small squares and the whole board, forgetting the middle size entirely.
CI found the 2 by 2 squares by splitting the board into blocks, getting two instead of four.
DI slid the 2 by 2 square across and down but missed one of its four positions.
10.C — 25KA-0135Digit-constraint puzzles
AI counted the odd numbers in one row of ten and treated that as the whole answer.
BI used four odd digits instead of five, forgetting that 9 is odd.
DI counted every two-digit number that is itself odd, which only fixes the last digit.
EI let the first digit be any of ten values rather than only the five odd ones.
11.C — 750KA-0137Multi-step arithmetic word problems
AI poured out half instead of a third and then forgot to add the 150 ml back.
BI poured out the third correctly but never added the 150 ml.
DI assumed the 150 ml added back was the same as the amount poured out, so nothing changed.
EI added the 150 ml to the full jug without pouring anything out first.
12.D — 49KA-0138Growth patterns and figurate numbers
AI continued the pattern to the fifth square instead of the seventh.
BI stopped one square short, giving the sixth instead of the seventh.
CI added 6 to the sixth square's 36, treating the differences as constant.
EI continued one square too far and gave the eighth.
13.B — 7KA-0157The pigeonhole principle
AI thought that one more than the number of colours must give three of a kind, which only guarantees a PAIR.
CI got to two of each colour, then added one more for each colour instead of one more in total.
DI assumed the worst case meant emptying the largest pile of 12 reds first.
EI answered with the number of socks I want rather than the number I must take to be sure of them.
14.B — 9KA-0158Last-digit behavior of products and powers
AI used the first entry of the cycle because 2026 is even, without working out where in the cycle it lands.
CI counted the cycle starting from zero, so a remainder of 2 put me one place too far along.
DI assumed a power that large must end in 1.
EI multiplied 3 by 2 and used the last digit of that.
15.A — 8KA-0175The pigeonhole principle
BI used the number of possible remainders without adding one for the pair that must collide.
CI doubled 7, thinking I needed two full sets of remainders.
DI doubled 7 and added one, applying the pigeonhole idea to the wrong number of boxes.
EI guessed from small cases without identifying what the boxes actually are.