Five bricks form a wall (see figure). Peter can only remove a brick if there is no other brick directly above it. On each turn, he randomly selects one of the removable bricks with equal probability and removes it. What is the probability that the brick numbered 4 is the third to be removed?
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Answer: D — \(\frac{1}{6}\)
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Hint 1 of 2
Only the two top bricks start out removable; brick 4 sits beneath both of them, so it is freed only after both 1 and 2 are gone.
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Hint 2 of 2
For 4 to be third, the first two removals must be exactly bricks 1 and 2 (in some order), then 4 must be chosen on turn three.
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Approach: follow the only path that frees brick 4 by turn three
At the start only the two top bricks 1 and 2 are free (each bottom brick is still pinned by a top brick), so turn 1 removes 1 or 2.
Suppose 1 goes first; now free are {2, 3}, and brick 4 still needs 2 removed, so turn 2 must pick 2 (chance \(\tfrac{1}{2}\)), after which {3, 4, 5} are free.
Turn 3 then picks 4 with chance \(\tfrac{1}{3}\); the path probability is \(\tfrac{1}{2}\cdot\tfrac{1}{2}\cdot\tfrac{1}{3}=\tfrac{1}{12}\), and the symmetric order (2 then 1) doubles it to \(\tfrac{1}{6}\), answer D.
Jelena fills the 2×4 table shown with the letters A, B, C and D. She wants each letter to appear exactly once in each row and exactly once in each of the three 2×2 squares. In how many ways can she do this?
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Answer: B — 24
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Hint 1 of 2
Fill the left-hand 2x2 square first, then ask how much freedom is left.
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Hint 2 of 2
Because the 2x2 squares overlap by one column, knowing two columns almost forces the next.
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Approach: fill the first 2x2 block, then propagate the overlaps
The left 2x2 block can be filled with the four letters in \(4!=24\) ways.
Once columns 1 and 2 are set, the middle 2x2 block (columns 2-3) must contain the same four letters, so column 3 is forced to be column 1 with its two entries swapped between the rows; the same logic then forces column 4.
So every choice of the first block extends to exactly one full table, giving \(24\) ways, answer B.
A bee called Maja wants to hike from honeycomb X to honeycomb Y. She can only move from one honeycomb to the neighbouring honeycomb if they share an edge. How many different ways are there for Maja to go from X to Y if she has to step onto every one of the seven honeycombs exactly once?
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Answer: D — 5
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Hint 1 of 2
You need a path from X to Y that visits all seven honeycombs exactly once.
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Hint 2 of 2
Work outward from X, pruning routes that strand a cell you can never return to.
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Approach: count the full paths through the seven-cell honeycomb graph
Model the honeycombs as nodes with edges between cells that share a side.
Trace every route from X that uses all seven cells exactly once and ends at Y, discarding ones that get stuck.
Ahmad walks up 8 steps, going up either 1 or 2 steps at a time. There is a hole on the 6th step, so he cannot use this step. In how many different ways can Ahmad reach the top step?
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Answer: C — 8
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Hint 1 of 2
Count, step by step, the number of ways to reach each step using 1- or 2-step moves.
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Hint 2 of 2
The forbidden 6th step has 0 ways; carry that 0 forward to the top.
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Approach: build up ways-to-reach each step (Fibonacci-style)
Ways to reach steps 1..5 are 1, 2, 3, 5, 8 (each is the sum of the two below).
Step 6 is a hole, so it has 0 ways and cannot be used.
Five cars participated in a race, starting in the order I, II, III, IV, V. Whenever a car overtook another car, a point was awarded. The cars reached the finish line in the order III, V, I, IV, II. What is the smallest number of points in total that could have been awarded?
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Answer: E — 6
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Hint 1 of 2
The fewest overtakes equals the number of pairs that swap their relative order from start to finish.
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Hint 2 of 2
Compare every pair of cars and count how many ended in the opposite order to how they started.
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Approach: count order-reversed pairs (inversions)
Each overtake swaps one adjacent pair, so the minimum total equals the number of pairs whose order reversed.
Start I,II,III,IV,V; finish III,V,I,IV,II.
Counting all pairs that flipped relative order gives 6 such pairs.
Ana plays with n × n boards by placing a token in each of the cells, with no common points with other cells containing tokens. In the picture we see how to place as many tokens as possible on 5 × 5 and 6 × 6 boards. In this way, how many tokens can Ana possibly put on a 2020 × 2020 board?
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Answer: D — 1010²
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Hint 1 of 2
'No common points' means tokens can’t even touch at a corner, so leave a gap between rows and columns.
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Hint 2 of 2
Place tokens on every other row and every other column; count how many fit.
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Approach: pack tokens on alternate rows and columns
Non-touching tokens go on alternate rows and alternate columns.
On an n×n board that fits ⌈n/2⌉ per direction.
For 2020, ⌈2020/2⌉ = 1010 each way, giving 1010² tokens.
Julia puts the nine chips shown into a box. She then takes one chip at a time, without looking, and notes down its digit, obtaining at the end a number with nine different digits. What is the probability that the number written by Julia is divisible by 45?
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Answer: A — 19
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Hint 1 of 2
Divisible by 45 means divisible by both 9 and 5.
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Hint 2 of 2
The digits 1–9 always sum to 45, so 9 is automatic — only the last digit matters for 5.
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Approach: reduce to a single last-digit condition
1+2+…+9 = 45, so any arrangement is divisible by 9.
Divisibility by 5 needs the last digit to be 5 (there’s no 0).
On one day there are 40 train trips, each from one of the towns M, N, O, P, Q to exactly one other of those towns. There are 10 trips either from or to M, 10 either from or to N, 10 either from or to O, and 10 either from or to P. How many trips are there either from or to Q?
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Answer: E — 40
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Hint 1 of 2
Each trip touches exactly two towns, so summing “trips touching each town” counts every trip twice.
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Hint 2 of 2
The five town-counts must add up to 2 × 40 = 80.
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Approach: each trip is counted at both of its towns
Every trip has two endpoints, so adding the counts for all five towns gives \(2\times 40 = 80\).
Counting & ProbabilityLogic & Word Problemscareful-counting
A whimsical teacher has a box with 203 red, 117 white and 28 blue buttons. He asks his students to each take one button out of the box without looking. What is the minimum number of students who have to take a button so that definitely at least three of the buttons picked have the same colour?
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Answer: C — 7
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Hint 1 of 2
Think about the worst possible luck before three match.
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Hint 2 of 2
There are only three colours, so how many can you draw with at most two of each?
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Approach: pigeonhole: build the worst case, then add one
In the worst case each colour comes up at most twice: 2 + 2 + 2 = 6 buttons with no colour reaching three.
The very next button (the 7th) must repeat some colour for a third time.
So 7 students guarantee three of one colour, choice C.
Counting & ProbabilityLogic & Word Problemscareful-counting
Tycho plans his running training. Each week he wants to go for a run on the same weekdays. He never wants to go for a run on two consecutive days. But he wants to go for a run three days a week. How many different weekly plans meet those conditions?
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Answer: B — 7
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Hint 1 of 2
A weekly plan repeats, so the seven days form a loop — Sunday touches Monday.
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Hint 2 of 2
Count choices of 3 days on a circle of 7 with no two adjacent.
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Approach: count non-adjacent triples around a cycle of 7 days
Because the plan repeats every week, the days form a circle of 7 where no two chosen days may be next to each other.
The number of ways to pick 3 non-adjacent positions on a circle of 7 is 7.
Petra has three different dictionaries and two different novels on her bookshelf. In how many different ways can she arrange the books, if all the dictionaries should stay together and likewise the novels as well?
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Answer: B — 24
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Hint 1 of 2
Treat the dictionaries as one block and the novels as another, then arrange the two blocks.
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Hint 2 of 2
Multiply: arrange the 2 blocks, then arrange within the 3-dictionary block and within the 2-novel block.
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Approach: block (glue) the groups, then count
The 3 dictionaries form one block, the 2 novels another: 2! ways to order the two blocks.
Inside: 3! ways for the dictionaries, 2! ways for the novels.
A two-digit number with the digits x, y, can be written in the form \(\overline{xy}\). Let a, b, c be different digits. In how many ways can the digits a, b, c be chosen, so that \(\overline{ab} < \overline{bc} < \overline{ca}\)?
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Answer: A — 84
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Hint 1 of 2
All three of ab, bc, ca are genuine two-digit numbers, so a, b, c are each from 1 to 9 and distinct.
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Hint 2 of 2
Count ordered distinct triples (a, b, c) from 1–9 satisfying ab < bc < ca.
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Approach: count valid distinct digit triples
Since ab, bc, ca are two-digit numbers, a, b, c are in {1, ..., 9} and all different.
Among the 9·8·7 = 504 ordered distinct triples, count those with 10a+b < 10b+c < 10c+a.
Several points are marked on a straight line. Then all possible connecting lines between each two points are drawn. One such point lies within exactly 80 of those connecting lines, and another one lies within exactly 90 of those. How many points were marked on the straight line?
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Answer: B — 22
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Hint 1 of 2
A marked point with L points to its left and R to its right lies inside exactly L×R of the segments.
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Hint 2 of 2
Find one N giving both L×R = 80 and L×R = 90 at two points; N = 22 works (5×16 and 6×15).
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Approach: count segments through a point as L×R
A point with L points left and R right is interior to L×R segments.
We need L×R = 80 and (elsewhere) = 90 for the same total N.
With N = 22: a point at position 6 gives 5×16 = 80, at position 7 gives 6×15 = 90.
In the equation \(N \times U \times (M + B + E + R) = 33\) each letter stands for a different digit (0, 1, 2, …, 9). In how many different ways can the letters be replaced by different digits?
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Answer: D — 48
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Hint 1 of 2
33 factors very few ways into three positive integers using single digits.
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Hint 2 of 2
Fix which factor is the bracket sum, then count digit arrangements.
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Approach: factor 33, pin the bracket sum, then count digit arrangements
The bracket \(M+B+E+R\) is a sum of four different digits, so it lies between \(0+1+2+3=6\) and 30; the only divisor of 33 in that range is 11, forcing \(N\times U=3\), i.e. \(\{N,U\}=\{1,3\}\).
The four bracket digits must be different and avoid 1 and 3; the smallest four-digit total from the remaining digits is \(0+2+4+5=11\), so \(\{M,B,E,R\}=\{0,2,4,5\}\) is the only possibility.
Arrangements: \(2\) ways to assign \(N,U\) times \(4!=24\) ways to assign the bracket digits give \(2\times24=\) 48 ways.
In the diagram Karl wants to add lines, each joining two of the marked points, so that each of the seven marked points is joined to the same number of other marked points. What is the minimum number of lines he must draw?
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Answer: D — 9
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Hint 1 of 2
Count the degree (number of lines) already at each of the seven points.
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Hint 2 of 2
Add the fewest lines so every point ends with the same degree.
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Approach: equalise the seven vertex degrees with the fewest added edges
From the diagram the current degrees are 3, 2, 1, 1, 1, 1, 1 (sum 10, i.e. 5 existing lines).
Raising every vertex to a common degree needs a minimum of 9 extra lines.
Every group of three vertices of a cube forms a triangle. How many such triangles are there whose vertices do not all lie on the same face of the cube?
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Answer: C — 32
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Hint 1 of 2
Count all triangles from the 8 vertices, then remove the 'flat' ones.
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Hint 2 of 2
A triangle is bad exactly when all three vertices lie on one face.
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Approach: total triangles minus the same-face ones
All triangles: C(8,3) = 56.
Each of the 6 faces has C(4,3) = 4 same-face triangles, so 6 × 4 = 24 are bad.
Consider all 7-digit numbers that use each of the digits 1 to 7 exactly once. Write these numbers in increasing order and split the list exactly in the middle into two lists of equal size. What is the last number of the first list?
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Answer: E — 4376521
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Hint 1 of 2
There are 7! such numbers in order; the split point is right in the middle.
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Hint 2 of 2
Find the 2520th smallest arrangement of the digits 1–7.
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Approach: locate the middle permutation by digit blocks
There are 7! = 5040 numbers, so the first list ends at the 2520th smallest.
Counting in blocks (each leading digit gives 6! = 720 numbers): 2520 = 3·720 + 360, landing among the numbers starting with 4, then resolving the rest.
How many different ways are there in the diagram shown to get from point A to point B, if you are only allowed to move in the directions indicated by the arrows?
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Answer: D — 12
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Hint 1 of 2
Count paths by labelling each junction with how many ways reach it from A.
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Hint 2 of 2
Only the indicated arrow directions are allowed, so every move goes forward toward B.
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Approach: count directed paths by accumulating at each node
Label each junction with the number of allowed paths reaching it from A, starting with 1 at A.
Add incoming counts along the arrow directions, junction by junction, down to B.
The cells of the 4×4 table shown should each be coloured either black or white. The numbers tell how many cells in each row and column should be black. In how many ways can the colouring be done?
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Answer: D — 5
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Hint 1 of 2
A row or column needing 0 black cells is entirely white, shrinking the puzzle.
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Hint 2 of 2
Count the black-cell placements left over for the smaller grid by cases on the row needing 2.
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Approach: remove the zero line, then enumerate the 3x3 core
The row and column needing 0 black cells are all white, leaving a 3×3 core with row sums 2,1,1 and column sums 2,1,1.
Place the two black cells of the '2' row, then fill the two '1' rows to meet the column totals.
The three cases for that pair of columns give 2 + 2 + 1 = 5 valid fillings.
A barcode as pictured is made up of alternate black and white stripes. The code always starts and ends with a black stripe. Each stripe (black or white) has width 1 or 2, and the total width of the barcode is 12. How many different barcodes of this kind are there if one reads from left to right?
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Answer: E — 116
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Hint 1 of 2
An odd number of stripes is needed so it starts and ends black.
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Hint 2 of 2
Count compositions of 12 using widths 1 and 2 with an odd number of parts.
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Approach: count compositions of 12 into 1's and 2's (odd part-count)
Starting and ending black forces an odd number of stripes, k.
With k stripes there are 12-k twos: k=7 gives C(7,5)=21, k=9 gives C(9,3)=84, k=11 gives C(11,1)=11.
A type of hopscotch is played like this: a player jumps from one square to the next, alternating left foot, both feet, right foot, both feet, and so on, as shown. Maya plays and jumps into exactly 48 squares, starting with her left foot. How many times is her left foot on the floor during the game?
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Answer: C — 36
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Hint 1 of 2
The only jumps where the left foot is NOT on the floor are the right-foot-only jumps, so it may be easier to count those.
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Hint 2 of 2
Group the jumps into the repeating block left, both, right, both and see how the left foot behaves in each block.
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Approach: count the jumps where the left foot is down
The pattern repeats every four squares: left, both, right, both.
In each block of four, the left foot is down on the 'left' square and on the two 'both-feet' squares, so 3 of every 4 squares use the left foot.
Over 48 squares that is \(\frac{3}{4}\times 48 = 36\) times, answer C.
The digits of the year 2020 are each repeated twice and the two digits are different. How many times does this happen between the year 1000 and now (2020)?
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Answer: E — 29
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Hint 1 of 2
You want years like 2020 made of two different digits, each appearing exactly twice.
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Hint 2 of 2
Split by the leading digit: count such years starting with 1, then those starting with 2 up to 2020.
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Approach: count the two-distinct-digit, each-twice years by leading digit
Such a year uses two different digits, each appearing twice, e.g. 1331 or 2020.
Years from 1000–1999 (leading 1): the other digit pairs with one more 1 plus two of a different digit; enumerating the patterns gives 27 of them.
Years from 2000–2020 add just 2002 and 2020, for 2 more.
A park has five entrances. Monika wants to enter the park through one entrance and leave through another entrance. How many ways are there in which she can enter and leave the park?
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Answer: B — 20
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Hint 1 of 2
She enters by one entrance and leaves by a different one.
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Hint 2 of 2
Count ordered pairs of distinct entrances: 5 choices then 4.
I have some unusual dice. On their faces are the digits 1 to 6 as usual, however the odd numbers are negative (so −1, −3, −5 instead of 1, 3, 5). I throw two such dice at the same time. Which of the following sums can I definitely not achieve with one such throw?
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Answer: D — 7
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Hint 1 of 2
List the six face values: −1, 2, −3, 4, −5, 6.
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Hint 2 of 2
Try to write each target as a sum of two of those values.
Show solution
Approach: reachable sums
Each die shows one of −5, −3, −1, 2, 4, 6.
Sums 3, 4, 5, 8 are all possible (e.g. −1+4, 2+2, −1+6, 2+6).
Step by step the word VELO is changed into the word LOVE. In every step two adjacent letters are allowed to be swapped around. What is the minimum amount of steps needed?
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Answer: B — 4
Show hints
Hint 1 of 2
Number the target positions and look at how out of order the start is.
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Hint 2 of 2
The fewest adjacent swaps equals the number of inversions.
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Approach: count inversions
Target LOVE means positions L,O,V,E; the start VELO reads as 3,4,1,2 in that order.
The out-of-order pairs are (3,1),(3,2),(4,1),(4,2) — four of them.
Sven writes five different single-digit positive whole numbers on a board. He realises that no sum of two of these numbers is equal to 10. Which of the following numbers has Sven definitely written on the board?
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Answer: E — 5
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Hint 1 of 2
Group the digits 1–9 into pairs that add to 10.
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Hint 2 of 2
You can use at most one number from each pair; count how many that allows.
Show solution
Approach: pigeonhole on pairs
The pairs summing to 10 are {1,9},{2,8},{3,7},{4,6}; the digit 5 has no partner.
From four pairs you may pick only one each — at most 4 numbers.
To reach five numbers you must also use 5, so 5 is definitely written.
In the diagram one should go from A to B along the arrows. Along the way, add up the numbers that are stepped on. How many different results can be obtained?
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Answer: B — 2
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Hint 1 of 2
List the possible routes from A to B that follow the arrows.
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Hint 2 of 2
Add the circled numbers on each route and see how many different totals appear.
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Approach: enumerate the arrow routes
Follow every allowed path from A to B and total the numbers stepped on.
The different routes give only a couple of distinct sums.
There are 12 children at a party, including 3 pairs of twins. How many different ways are there to distribute six blue hats and six red hats to the children, so that each pair of twins wears hats of the same colour?
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Answer: C — 92
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Hint 1 of 2
Treat each twin pair as a single unit that takes two hats of one colour; six singletons take one hat each.
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Hint 2 of 2
Sum over how many pairs are blue, then choose colours for the singletons to balance to 6 and 6.
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Approach: casework on the number of blue twin-pairs
If k of the 3 pairs wear blue, they use 2k blue hats, leaving 6−2k blue hats for the six singletons.
Ann rolled an ordinary die 24 times. Every number from 1 to 6 came up at least once, and the number 1 came up more often than any other number. Ann then added all the numbers she rolled. What is the largest total she could have obtained?
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Answer: D — 90
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Hint 1 of 2
Every face appears at least once, and 1 must appear strictly more often than each other face.
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Hint 2 of 2
Allow 1 to appear enough times that the other faces may each appear up to one less, then load 5's and 6's.
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Approach: minimise the count of 1 then maximise the high faces
Let 1 appear c times; every other face appears at most c-1 times, and all six faces appear at least once.
To maximise the sum we want many 5's and 6's, so allow c = 7: then faces can appear up to 6 times each, e.g. six 6's, six 5's, three 4's, one 3, one 2, and seven 1's (total 24 rolls).
That total is 7*1 + 1*2 + 1*3 + 3*4 + 6*5 + 6*6 = 90.
Consider n different straight lines in a plane, labelled \(\ell_1, \ldots, \ell_n\). The line \(\ell_1\) intersects 5 of the other lines, the line \(\ell_2\) intersects 9 of the other lines, and the line \(\ell_3\) intersects 11 of the other lines. Which of the following is a possible value of n?
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Answer: B — 12
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Hint 1 of 2
A line misses only the lines parallel to it, so 'intersects m others' means it has \(n-1-m\) lines parallel to it.
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Hint 2 of 2
Since one line already intersects 11 others, n cannot be smaller than 12, so test n = 12 first.
Show solution
Approach: turn intersection counts into parallel-class sizes and check the smallest n
A line parallel to none of the others would meet all \(n-1\); since \(\ell_3\) meets 11, we need \(n-1\ge 11\), so \(n\ge 12\).
Try \(n=12\): then \(\ell_3\) is in a direction by itself (meets all 11), \(\ell_2\) has \(12-1-9=2\) lines parallel to it, and \(\ell_1\) has \(12-1-5=6\) lines parallel to it, so the directions group as \(7+3+1+1=12\) lines.
That grouping is consistent, so \(n=12\) works, answer B.
The numbers from 1 to 9 should be distributed among the 9 squares in the diagram according to the following rules: there should be one number in each square, and the sum of three adjacent numbers is always a multiple of 3. The numbers 3 and 1 are already placed. How many ways are there to place the remaining numbers?
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Answer: E — 24
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Hint 1 of 2
Three numbers in a row summing to a multiple of 3 forces a repeating pattern of residues mod 3.
Still stuck? Show hint 2 →
Hint 2 of 2
Group the nine positions by position mod 3; each group must hold one residue class, and the fixed 3 and 1 pin down two of the groups.
Show solution
Approach: use residues mod 3 across the strip
Three adjacent numbers summing to a multiple of 3 forces position \(i\) and \(i+3\) to have the same residue mod 3, so the nine positions split into three groups of three by index mod 3.
The numbers 1–9 also split by residue into \(\{3,6,9\}\equiv0\), \(\{1,4,7\}\equiv1\), \(\{2,5,8\}\equiv2\); each position-group must get one whole residue-group.
The placed 3 (residue 0) and 1 (residue 1) pin which residue-group goes to their two position-groups, so the third assignment is forced too.
Now arrange within groups: the group with no fixed number can be ordered \(3!=6\) ways, and each of the two groups with a fixed number has its other two numbers free (\(2!=2\) ways each), giving \(6\cdot2\cdot2=\) 24 ways (choice E).
How many different ways are there to read the word BANANA in the following table if we can only cross to a field that shares an edge with the current field and we can use fields several times?
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Answer: E — 128
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Hint 1 of 2
Start on a corner B and spell B-A-N-A-N-A, stepping only to edge-adjacent cells (reuse allowed).
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Hint 2 of 2
Count routes by multiplying the choices at each letter; symmetry across the four corners helps.
Show solution
Approach: multiply the branching choices, letter by letter
A path spells B-A-N-A-N-A, moving each step to an edge-adjacent cell, and cells may be reused.
Start from a B and count how many neighbours carry the next needed letter at each move; the number of full paths is the product of those branch counts along the route.
Because the grid is symmetric, every starting B contributes the same number of paths, so multiply one B's count by the number of B's.
Carrying out this branch-multiplication over the whole grid totals 128 ways (choice E).
In the 4×4 table, some cells must be painted black. The numbers next to and below the table show how many cells in that row or column must be black. In how many ways can this table be painted?
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Answer: D — 5
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Hint 1 of 2
The row and column labelled 0 are entirely white, shrinking the puzzle to a 3×3 grid.
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Hint 2 of 2
Count black-cell placements in that 3×3 grid with row sums 2,2,1 and column sums 2,2,1.
Show solution
Approach: reduce by the zero row/column, then enumerate
Row 2 and column 2 each need 0 black cells, so they are all white.
What remains is a 3×3 grid needing row sums 2,2,1 and column sums 2,2,1.
Enumerating where the single-black row and the two double-black rows go yields exactly five valid fillings.
Pedro assembled a cube using 64 little white equal cubes and then painted the cube red. Then he dismantled the cube and reassembled it so that all of its faces were white, and painted the cube red again. How many white faces of little cubes remained white?
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Answer: E — 192
Show hints
Hint 1 of 2
Count all the little-cube faces, then subtract the ones that get painted.
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Hint 2 of 2
After the first paint 96 faces are red; on reassembly another 96 (white ones turned outward) get painted.
Show solution
Approach: track which faces stay unpainted through both paintings
There are 64·6 = 384 little faces; the first painting reds 96 of them.
That leaves 288 white faces.
Reassembling white-out and repainting reds 96 more of those white faces.
On the 8 × 8 board shown, in how many ways can you place two chips, one green and one red, on differently coloured cells, so that the chips are not in the same row or in the same column of the board?
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Answer: E — 1536
Show hints
Hint 1 of 2
Green and red must sit on opposite colours; count the two colour-orders separately.
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Hint 2 of 2
For a fixed green cell, count the opposite-colour cells that avoid its row and column.
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Approach: count valid opposite-colour pairs by removing the shared row/column cells
Put green on a white cell (32 ways); a black cell shares its row or column in 4+4 = 8 cases.
So 32 − 8 = 24 black cells work, giving 32·24 = 768 ordered pairs.
By symmetry green-black/red-white gives another 768.
Diana draws a rectangle made of squares on grid paper and colours some squares black. In every white square she writes the number of black squares next to it (sharing an edge). The diagram shows an example. She now does the same with a rectangle made of 2018 squares. What is the biggest possible sum of all the numbers in the white squares?
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Answer: D — 3025
Show hints
Hint 1 of 2
The total written down is just the number of shared edges that have a black square on one side and a white square on the other.
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Hint 2 of 2
Since \(2018 = 2\times 1009\), the rectangle is either \(1\times 2018\) or \(2\times 1009\); compare which shape allows more black–white edges.
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Approach: the sum equals the count of black–white shared edges; maximise that
Each number in a white square counts its black neighbours, so adding them all gives exactly the number of edges that separate a black square from a white square.
Because \(2018 = 2\times 1009\), the rectangle is \(1\times 2018\) or \(2\times 1009\); colouring whole columns black/white alternately so that no two same-colour columns touch makes nearly every internal edge a black–white edge.
For the \(2\times 1009\) strip this alternating pattern gives \(3\times 1009 - 2 = 3025\) such edges (the \(1\times 2018\) strip gives only \(2017\)), so the largest total is 3025.
Each number of the set {1, 2, 3, 4, 5, 6} is written into exactly one cell of a 2 × 3 table. In how many ways can this be done so that the sum of the numbers in every column and every row is divisible by 3?
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Answer: D — 48
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Hint 1 of 2
Group the numbers by remainder mod 3: {3,6} give 0, {1,4} give 1, {2,5} give 2.
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Hint 2 of 2
Decide the remainder pattern of the table first, then count how to place the actual numbers.
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Approach: work modulo 3 on the remainder pattern, then fill in
By remainder mod 3 the numbers are two 0's (3,6), two 1's (1,4) and two 2's (2,5).
Each row of three and each column of two must sum to a multiple of 3; find the valid remainder patterns.
For each valid pattern, the two numbers in each remainder class can be swapped, and counting all arrangements gives 48.
The numbers −3, −2, −1, 0, 1, 2 are written on the six faces of a die. The die is rolled twice. The numbers that were rolled are multiplied. How big is the probability that this product is negative?
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Answer: E — 13
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Hint 1 of 2
A product is negative only when one factor is positive and the other is negative.
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Hint 2 of 2
Count the positive faces and the negative faces; zero never helps.
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Approach: count favourable ordered rolls over all 36 outcomes
The faces are −3, −2, −1, 0, 1, 2: three negatives and two positives (0 gives product 0).
Negative product needs one of each sign; ordered, that is 3·2 + 2·3 = 12 of the 36 equally likely pairs.
My friend Heinz wants to use a special password that is made up of seven digits. Each digit used in the password appears as many times in the password as is the value of the digit. Additionally, equal digits are always next to each other. Therefore he can for example use 4444333 or 1666666 as passwords. How many possible passwords can he choose from?
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Answer: E — 13
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Hint 1 of 2
Each chosen digit takes up as many of the 7 slots as its value, so the values must add to 7.
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Hint 2 of 2
Equal digits stay together as one block, so count orderings of the blocks.
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Approach: partition 7 into distinct digit-values, then order the blocks
Since a digit d fills d slots and equal digits are adjacent, choose distinct digit-values that sum to 7: {7}, {1,6}, {2,5}, {3,4}, {1,2,4}.
Each choice of k blocks can be arranged in k! orders: 1 + 2 + 2 + 2 + 6 = 13.
Counting & ProbabilityLogic & Word Problemscaseworkcareful-counting
Three weights are randomly placed on each tray of a beam balance. The balance dips to the right hand side as shown on the picture. The masses of the weights are 101, 102, 103, 104, 105 and 106 grams. For how many percent of the possible distributions is the 106-grams-weight on the right (heavier) side?
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Answer: B — 80 %
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Hint 1 of 2
List the splits of the six weights into two trays of three with the right side heavier.
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Hint 2 of 2
Among those, count how often the heaviest weight (106) is on the right.
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Approach: enumerate the heavier-right splits and check where 106 sits
Of the C(6,3)=20 ways to fill the right tray, exactly half (10) make the right side heavier.
Counting those, the 106-gram weight is on the right in 8 of the 10 cases.
Exactly 2016 people are taking part in a conference. They are registered as P1 to P2016 in the system. Each person from P1 to P2015 has shaken exactly the amount of other hands that his/her own system number indicates. How many people did P2016 shake hands with?
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Answer: D — 1008
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Hint 1 of 2
P2015 shook everyone, so each person shares a handshake with the busiest people first.
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Hint 2 of 2
Pair the busiest with the least busy and peel inward; track who is left for P2016.
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Approach: pair from the extremes
P2015 shakes everyone (so P1's single handshake is with P2015); P2014 shakes all but P1 (matching P2's two), and so on.
Peeling these matched pairs inward, P2016 ends up shaking exactly half of the others.
Any three vertices of a regular 13-sided shape are joined up to form a triangle. How many of these triangles contain the circumcentre of the 13-sided shape?
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Answer: C — 91
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Hint 1 of 2
A triangle inscribed in a circle contains the centre exactly when none of its three arcs is a half-circle or more.
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Hint 2 of 2
It is easier to count the triangles that miss the centre, then subtract from the total C(13,3).
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Approach: complementary counting of centre-missing triangles
Total triangles: C(13,3) = 286.
A triangle misses the centre when all three vertices lie within a half-circle; counting these and subtracting leaves the ones containing the centre.
100 trees (oaks and birches) are standing in a row. The number of trees between any two oaks is never equal to 5. What is the maximum number of the 100 trees that can be oaks?
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Answer: B — 52
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Hint 1 of 2
Two oaks may not be 6 positions apart (that is 5 trees between them); link positions whose numbers differ by 6.
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Hint 2 of 2
Split positions 1–100 by their remainder mod 6 into chains, then pick the most from each chain with no two neighbours.
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Approach: independent set on residue-mod-6 chains
Positions that differ by 6 cannot both be oaks; grouping 1–100 by remainder mod 6 gives six chains.
Each chain is a row where no two adjacent positions may both be oaks, so pick about half of each: the ceiling of (chain length) / 2.
Summing over the six chains gives a maximum of 52 oaks.
The five-digit number \(\overline{abcde}\) is called interesting if all of its digits are different and a = b + c + d + e holds true. How many interesting numbers are there?
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Answer: C — 168
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Hint 1 of 2
The leading digit a must equal the sum of four distinct digits, so a is at least 6.
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Hint 2 of 2
For each value of a from 6 to 9, list the digit-sets, then multiply by the arrangements of the last four digits.
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Approach: case on the leading digit, then count arrangements
Since a = b+c+d+e with four distinct digits, the smallest possible sum is 0+1+2+3 = 6, so a is 6, 7, 8 or 9.
The sets summing to a (none equal to a) number 1, 1, 2, 3 for a = 6, 7, 8, 9: seven sets in all.
Each set's four digits can be arranged 4! = 24 ways after the fixed leading a.
Determine all n (with \(1 \le n \le 8\)) for which one can mark several cells of a 5×5 table so that there are exactly n marked cells in every 3×3 sub-table.
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Answer: E — All numbers from 1 to 8 are possible.
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Hint 1 of 3
Sliding a 3×3 window one column right drops its left column and adds a new one, so equal counts force the dropped and added columns (over those three rows) to match.
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Hint 2 of 3
That balancing condition makes column-1 match column-4 and column-2 match column-5, suggesting a repeating pattern.
Still stuck? Show hint 3 →
Hint 3 of 3
Instead of asking which n are forbidden, just try to build one valid marking for each n from 1 to 8.
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Approach: build an explicit marking for every n from 1 to 8
Because overlapping windows force the marking to repeat (column 4 like column 1, column 5 like column 2, and likewise for rows), a marking is fixed by a small repeating block.
Choosing how many cells of that block are marked lets the common window-count be tuned up or down across the whole achievable range.
Carrying this out gives an explicit valid pattern for each target, so n = 1, 2, 3, 4, 5, 6, 7 and 8 are all attainable.
How many 10-digit numbers are made up solely of the digits 1, 2 and 3 (not necessarily all of them) and have the property that adjacent digits always differ by exactly 1?
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Answer: C — 64
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Hint 1 of 2
Adjacent digits differ by 1, so from a 2 you can go to 1 or 3, but from a 1 you must go to 2.
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Hint 2 of 2
Build the count digit by digit, tracking how many strings end in each digit.
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Approach: count step by step with a small transition table
Because neighbours differ by 1, a 2 must sit next to a 1 or a 3, while a 1 or a 3 must sit next to a 2 — so 2s and (1-or-3)s strictly alternate.
That fixes which of the 10 positions hold a 2: either all odd positions or all even positions, giving 2 patterns.
In each pattern the five non-2 slots are independently a 1 or a 3, so \(2^5 = 32\) ways per pattern.
Total = \(2 \times 32 = 64\), so the answer is 64.
Robert wants to place stones on a 4 × 4 game board so that the number of stones in each row and in each column is different; that is, there are 8 different amounts. To do this he can place one or several stones in any one field, or even leave single fields empty. What is the minimum number of stones needed?
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Answer: A — 14
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Hint 1 of 2
The four row totals and four column totals are eight different whole numbers.
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Hint 2 of 2
The smallest eight distinct totals are 0–7; the stones are the sum of either the rows or the columns.
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Approach: minimise the shared row/column total
The eight totals must be eight distinct non-negative integers; the smallest set is 0,1,2,3,4,5,6,7.
Total stones equals the sum of the four row totals, which must equal the sum of the four column totals.
Split 0–7 into two equal-sum fours, e.g. rows {0,1,6,7} and columns {2,3,4,5}, each summing to 14.
Such a board is realisable, so the minimum number of stones is 14.
A number of oranges, peaches, apples and bananas are placed in a row. What is the minimum number of fruits needed so that each kind of fruit lies next to each other kind at least once somewhere in the row?
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Answer: C — 8
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Hint 1 of 2
There are four fruit types, so 6 unordered pairs must each appear side by side somewhere.
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Hint 2 of 2
Think of fruits as dots and 'were neighbours' as edges of K₄; you want a near-Eulerian walk.
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Approach: cover all six pairs with the fewest neighbour-links
With 4 types there are 6 pairs to realise as adjacencies; a row of n fruits has only n−1 adjacencies.
Each type would need to touch the other three, but all four 'dots' have odd degree 3 in the pair-graph K₄.
Repeating one link fixes the parity, giving 7 needed adjacencies, hence 8 fruits.
A row such as orange-peach-apple-orange-banana-peach-apple-banana shows 8 works, so the answer is 8.
A kangaroo is sitting at the origin of a Cartesian coordinate system. With each bounce it can jump one unit in the horizontal or the vertical direction. How many points are there where the kangaroo could be after 10 jumps?
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Answer: A — 121
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Hint 1 of 2
After 10 unit jumps, a reachable point has |x| + |y| ≤ 10, and its parity matches 10 (even).
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Hint 2 of 2
Count all lattice points inside that diamond with even taxicab distance.
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Approach: count even-parity points within taxicab distance 10
Each jump changes \(x\) or \(y\) by 1, so after 10 jumps \(|x| + |y|\) is at most 10 and has the same parity as 10, i.e. even.
Every such point is actually reachable, so count the lattice points with \(|x|+|y|\) even and \(\le 10\).
By distance: 1 point at distance 0, then \(4d\) points at each even distance \(d = 2,4,6,8,10\), giving \(1 + 4(2+4+6+8+10) = 1 + 120 = 121\).
