Worksheet generator

1.C — 8KA-0012Casework

• AI counted each pair of digits once instead of counting both orders, such as 17 and 71.
• BI listed the pairs starting from 1 and 7 and forgot the number 80, where the second digit is zero.
• DI included 08 as a two-digit number, but a two-digit number cannot start with zero.
• EI counted both orders of every pair and then counted the pairs that reverse to themselves twice as well.

2.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.

3.C — 4KA-0020Paper folding and hole punching

• AI forgot that the punch goes through every layer at once, not just the top one.
• BI undid only one of the two folds before counting the holes.
• DI doubled for the first fold and then added two more rather than doubling a second time.
• EI counted a fold that was never made, doubling three times instead of twice.

4.E — No, it is impossibleKA-0028Parity arguments

• AI found a sequence that turned over most of the cups and assumed the last one could be fixed somehow.
• BI matched the number of moves to the number of cups without checking whether the target is reachable at all.
• CI assumed that with enough moves any arrangement can be reached.
• DI thought the choice of which cups to flip could change whether the target is reachable.

5.A — 8KA-0031Counting paths on a grid

• BI assumed closing one junction removes about half the routes and halved the total of 20.
• CI counted the routes that pass through the closed junction and gave that instead of the ones that avoid it.
• DI subtracted only the 2 routes that reach the closed junction, not all the routes that continue through it.
• EI counted every route on the open grid and forgot to remove the ones through the closed junction.

6.E — 9KA-0037Last-digit behavior of products and powers

• AI assumed a number this large has no reachable last digit and gave up on the cycle.
• BI found the cycle 7, 9, 3, 1 but used a remainder of 0 when the remainder is actually 2.
• CI found the cycle but counted its positions starting at zero, shifting my answer along by one.
• DI assumed the units digit of any power of 7 is always 7.

7.C — 4KA-0040The pigeonhole principle

• AI answered with the lucky case, where the first two socks happen to match.
• BI used the number of colours as my answer without adding one for the sock that must repeat.
• DI worked from the number of socks of each colour instead of the number of colours.
• EI took the whole drawer, guaranteeing a pair but far more socks than are needed.

8.A — blackKA-0043Invariants and monovariants

• BI noticed there are more white stones than black ones and guessed the majority colour would survive.
• CI tried a few orders, saw different-looking positions along the way, and assumed the ending must vary too.
• DI did not notice that every move takes two stones out and puts one back, so the count falls by exactly one each time.
• EI assumed the first move settles the outcome instead of looking for a quantity that no move can change.

9.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.

10.B — 0.2KA-0052Area by subtraction (shaded regions)

• AI estimated by eye from the four corner pieces without comparing them to the whole square.
• CI used a diameter of 5 instead of a radius of 5 when working out the circle's area.
• DI assumed the circle covers half the square because it touches all four sides.
• EI found the fraction covered by the circle and gave that instead of the fraction outside it.

11.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.

12.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.

13.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.

14.B — 2KA-0094Weighing and balance puzzles

• AI assumed one weighing could separate nine possibilities, but it has only three outcomes.
• CI split the coins into halves each time, which wastes the balance's third outcome.
• DI weighed the coins one against another in pairs rather than in groups.
• EI compared each coin with a known good one in turn, which always works but is nowhere near the fewest.

15.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.

16.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.

17.D — No, because the two colours cannot balanceKA-0101Coloring arguments

• AI checked that 100 divides by 4 and treated that as proof that a covering exists.
• BI assumed a covering must exist somewhere and that I simply had not found it yet.
• CI gave a reason that is not even true, since 100 really is a multiple of 4.
• EI thought orientation could rescue it, but every turn of a T covers the same mixture of colours.

18.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.

19.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.

20.D — 8KA-0134Paper folding and hole punching

• AI forgot that the punch goes through every layer, not just the top one.
• BI undid only one of the two folds before counting.
• CI doubled twice for the folds but added the second hole instead of doubling for it too.
• EI counted three folds instead of two, doubling one time too many.

21.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.

22.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.

23.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.

24.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.

25.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.

26.A — 2 hours 24 minutesKA-0159Combined work rates

• BI averaged the two times, but two taps together must be faster than either one alone.
• CI added the two times, which would be right for filling two tanks one after the other.
• DI halved the smaller of the two times, guessing that two taps must simply be twice as fast as one.
• EI found 2 hours 24 minutes correctly and then halved it a second time.

27.A — 120KA-0166Counting up to symmetry

• BI counted every arrangement in a row, so each circular seating got counted once for every rotation.
• CI divided 720 by 20 or some other number instead of by the 6 rotations.
• DI divided by 12, treating reflections as identical too, though the question only allows rotations.
• EI fixed two people rather than one, dividing by an extra factor.

28.A — 91/216KA-0168Complementary counting

• BI added 1/6 three times, which counts the overlapping cases more than once and would exceed 1 for seven rolls.
• CI worked out the probability of NO six and forgot to subtract it from 1.
• DI found the probability of three sixes instead of at least one.
• EI gave the probability for a single roll and ignored that there are three.

29.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.

30.A — 54KA-0177Overcount, then correct

• BI counted every line joining two vertices and forgot to remove the 12 sides.
• CI counted each diagonal from both of its endpoints and forgot to halve.
• DI used 12 x 10 without halving, double counting every diagonal.
• EI subtracted 24 rather than 12, removing each side twice.