50 challenging logic and lateral-thinking problems

Articles /50-challenging-equation-logic-lateral-thinking-problems

90 min

This collection presents fifty challenging problems and combines three approaches: modelling with an equation, reasoning directly from the data, and changing perspective through lateral thinking. Not every problem calls for computation: choosing an appropriate method is part of solving it.

How to use the collection. Read precisely, note what is actually stated, and try a strategy before opening the solution. Here, lateral thinking does not mean guessing a hidden trick: it means deliberately revisiting assumptions and separating what the text states from what we have added ourselves.

Three ways to solve

Equation: translate relationships into symbols and check the result in context. Reasoning: use invariants, counting, proportions, cases, or working backwards. Lateral thinking: look for an unnecessary assumption that is blocking the simplest reading. The label inside each solution identifies the main method, not the only possible one.

The 50 problems

The square border
A square plaza has a square flower bed at its centre. Corresponding sides are parallel, and the path between the two squares is uniformly 3 m wide. The path alone has area 204 m². What are the side lengths of the two squares?
Show the solution to problem 1
Main method: Equation
Let x be the outer side. The inner side is x − 6, because the path uses 3 m at both ends. The border area gives x² − (x − 6)² = 204. Expanding: 12x − 36 = 204, hence x = 20. The inner side is 20 − 6 = 14 m. Check: 20² − 14² = 400 − 196 = 204.
Answer. Outer side 20 m; inner side 14 m
Another route
Cutting the border along the diagonals gives four congruent trapezoids, each with area 204/4 = 51 m² and height 3 m. Their average base length is therefore 51/3 = 17 m. The bases differ by 6 m, so they lie 3 m below and above that average: 14 m and 20 m.
The escalator steps
An escalator moves upward at constant speed. Ada walks up at 1 step per second relative to the escalator and counts 30 steps before reaching the top. On the same escalator under the same conditions, Bruno walks at 2 steps per second and counts 45. How many visible steps would the stationary escalator have, and how fast is it moving? Assume constant speeds and no skipped steps.
Show the solution to problem 2
Main method: Equation
Let v be the escalator speed in steps per second and N the number of visible steps when stopped. Ada takes 30 s, so N = 30(1 + v). Bruno takes 45/2 = 22.5 s, so N = 22.5(2 + v). Equating them gives 30 + 30v = 45 + 22.5v, hence v = 2. Therefore N = 30 · 3 = 90. Bruno also covers 22.5 · 4 = 90 steps relative to the building.
Answer. 90 steps; escalator speed 2 steps per second
The two counter experiments
Two bags have the same initial composition: each contains r red counters and b blue counters, all identical except for colour. Six red counters are added to the first bag, making the probability of drawing red 2/3. From the second bag, instead, four blue counters are removed, making that probability 3/4. How many counters of each colour did each bag initially contain?
Show the solution to problem 3
Main method: Equation
Let r and b be the initial numbers. The first experiment gives (r + 6)/(r + b + 6) = 2/3, or r + 6 = 2b. The second gives r/(r + b − 4) = 3/4, or r = 3b − 12. Substitution yields 3b − 12 + 6 = 2b, so b = 6 and r = 6. Check: after the addition, 12 of 18 are red; after the removal, 6 of 8 are red.
Answer. 6 red and 6 blue counters
The three digital archives
Three archives A, B and C occupy 405 GB altogether. A and B together occupy 30 GB more than twice C. After 15 GB is moved from A to B, B becomes 3/4 the size of A. What were the three original sizes?
Show the solution to problem 4
Main method: Equation
From A + B + C = 405 and A + B = 2C + 30, we get 3C + 30 = 405; thus C = 125 and A + B = 280. After the transfer, B + 15 = 3(A − 15)/4, or 3A − 4B = 105. Using B = 280 − A gives 7A = 1225, so A = 175 and B = 105. Check: after the move, A is 160 GB and B is 120 GB, exactly three quarters of A.
Answer. A = 175 GB, B = 105 GB, C = 125 GB
The drone's echo
In a simulation, a drone flies straight towards a wall at 20 m/s. At an unknown distance it emits a sound pulse, and the echo returns to the drone exactly 1 s later. The speed of sound is 340 m/s. How far from the wall was the drone when it emitted the pulse? Ignore wind and reflection time.
Show the solution to problem 5
Main method: Equation
Let d be the initial distance. In 1 s the sound travels 340 m in total. Its outward trip to the wall is d. By the time the echo reaches the drone, the drone has moved 20 m closer, so the return trip is d − 20. Therefore d + (d − 20) = 340, giving 2d = 360 and d = 180 m. As a check, the sound reaches the wall in 180/340 s and meets the approaching drone within the one-second total.
Answer. 180 m
Another route
During the elapsed second the drone advances 20 m. The 340 m sound path is therefore twice the initial distance minus those 20 m. Add 20 to 340 and halve the result to obtain 180 m.
The two virtual candles
In a simulation, two candles have the same initial length and burn at constant rates. The first would be used up in 4 hours and the second in 6 hours. If they are lit at the same instant, when will the second candle's remaining length be twice the first one's?
Show the solution to problem 6
Main method: Equation
If the initial length is L, after t hours the remaining lengths are L(1 − t/4) and L(1 − t/6). The condition is 1 − t/6 = 2(1 − t/4). Thus 1 − t/6 = 2 − t/2, so t/3 = 1 and t = 3 hours. At that time, L/4 of the first and L/2 of the second remain, a ratio of 2.
Answer. After 3 hours
Another route
After 3 hours, the first has used three quarters of its length and retains one quarter; the second has used half and retains half. One half is twice one quarter.
The wheels on the same route
Two wheels roll without slipping over the same distance. The large wheel's circumference is 4 m greater than the small wheel's. The large wheel makes 15 fewer turns, and together the wheels make 75 turns. Find both circumferences and the distance travelled.
Show the solution to problem 7
Main method: Equation
Let n be the large wheel's turns; the small one makes n + 15. From n + n + 15 = 75, n = 30, so the small wheel makes 45 turns. If c is its circumference, equal distances give 30(c + 4) = 45c. Thus 120 = 15c and c = 8 m; the large circumference is 12 m. The distance is 45 · 8 = 360 m, also 30 · 12.
Answer. Circumferences 8 m and 12 m; distance 360 m
The two square mosaics
All 313 identical tiles are used to build two square mosaics. The numbers of tiles along their sides are consecutive integers. How many tiles long is each side?
Show the solution to problem 8
Main method: Equation
Let n be the smaller side; the other is n + 1. Their areas give n² + (n + 1)² = 313, or 2n² + 2n − 312 = 0. Dividing by 2: n² + n − 156 = 0 = (n − 12)(n + 13). The positive solution is n = 12, so the other side is 13. Check: 12² + 13² = 144 + 169 = 313.
Answer. 12 tiles and 13 tiles
Another route
Half of 313 is about 156.5, so both squares must have side lengths near √156.5, between 12 and 13. Testing those consecutive integers immediately gives 144 + 169 = 313.
The digits used to number a book
A book's pages are numbered from 1 onward, with no leading zeros. Printing all the page numbers uses exactly 642 digits. What is the last page number?
Show the solution to problem 9
Main method: Reasoning
Pages 1 through 9 use 9 digits. From 10 through 99 there are 90 two-digit numbers, using another 180 digits. This leaves 642 − 9 − 180 = 453 digits. Every page from 100 onward uses 3 digits, so there are 453/3 = 151 more pages: 100 through 99 + 151 = 250. Checking backwards gives 9 + 180 + 151 · 3 = 642.
Answer. Page 250
The labels on four drawers
A document is in exactly one of drawers A, B, C and D. Their labels say: A: “The document is not in B”; B: “It is in A or C”; C: “It is not in A”; D: “It is in D.” Exactly two labels are true. Which drawer contains the document?
Show the solution to problem 10
Main method: Reasoning
Test all four locations. If the document is in A, statements A and B are true while C and D are false: exactly two. If it is in B, only C is true. If it is in C, A, B and C are true. If it is in D, A, C and D are true. Only the first case has the stated number of truths, so the clues uniquely determine A.
Answer. Drawer A
The order of the six books
Six books A, B, C, D, E and F stand in a row from left to right. C is third; A is immediately next to C; D is to the left of A; B is at an end; F is immediately next to B; E is to the right of C but not at an end. What is the only possible order?
Show the solution to problem 11
Main method: Reasoning
A could be second or fourth. If A were fourth, E would have to be fifth. If B were first, F would be second and no place left of A would remain for D; if B were sixth, F would occupy the fifth place already needed by E. Therefore A is second and D, which must be to its left, is first. B can no longer be first, so it is sixth and F fifth. The only remaining place for E is fourth. The order is D, A, C, E, F, B.
Answer. D – A – C – E – F – B
The impossible handshakes
Fifteen people attend a meeting. The organiser claims that every person shook hands with exactly 5 different people, once per pair. Can this claim be true?
Show the solution to problem 12
Main method: Reasoning
No. Counting by person gives 15 · 5 = 75 handshake endpoints. Yet every handshake involves two people and is therefore counted twice, so that total must be even. Since 75 is odd, such a configuration cannot exist. The argument uses only the stated assumptions: distinct pairs and no repeated handshake.
Answer. No; parity makes it impossible
The five-bridge route
Four areas A, B, C and D are linked by five bridges: AB, AC, AD, BC and CD. An inspector wants to cross every bridge exactly once, with no other connections available. Where must she start, where must she finish, and what route works?
Show the solution to problem 13
Main method: Reasoning
The degrees, or incident bridge counts, are A = 3, B = 2, C = 3 and D = 2. On a route using every edge once, each internal visit consumes bridges in pairs, one to enter and one to leave. The only odd-degree areas must therefore be the endpoints: A and C, in either order. The route A–B–C–D–A–C proves existence; it uses AB, BC, CD, DA and AC exactly once each.
Answer. The endpoints must be A and C; for example A–B–C–D–A–C
The month with three repeated weekdays
In a certain month there are five Mondays, five Tuesdays and five Wednesdays. How many days does the month have, and on which weekday does it begin?
Show the solution to problem 14
Main method: Reasoning
Four complete weeks contain 28 days and four occurrences of every weekday. Three weekdays can have a fifth occurrence only if there are three extra days, so the month has 31 days. Those extra weekdays are consecutive and are the first, second and third days of the month. Since they are Monday, Tuesday and Wednesday, the month must start on Monday. A 30- or 29-day month could give a fifth occurrence to only two weekdays or one.
Answer. 31 days; it begins on Monday
The missing win total
Eight teams play a round-robin tournament: every pair meets once and there are no draws. Seven teams' win totals remain visible: 6, 5, 5, 4, 3, 2 and 1. The eighth value has been erased. How many wins was it?
Show the solution to problem 15
Main method: Reasoning
With 8 teams there are 8 · 7 / 2 = 28 matches. With no draws, every match produces exactly one win, so all win totals sum to 28. The seven visible values sum to 6 + 5 + 5 + 4 + 3 + 2 + 1 = 26. The eighth team therefore had 28 − 26 = 2 wins. This lies in the possible range from 0 to 7 and is determined without the individual results.
Answer. 2 wins
The large painted cube
A cube built from 4 × 4 × 4 unit cubes is painted on all six outer faces and then separated into its 64 cubes. How many unit cubes have exactly 3, 2, 1 or 0 painted faces?
Show the solution to problem 16
Main method: Reasoning
The cubes with 3 painted faces are the 8 corners. Those with 2 lie on edges but not at corners: each edge has 4 − 2 = 2, giving 12 · 2 = 24. Cubes with one painted face lie inside a face: each face has (4 − 2)² = 4, giving 6 · 4 = 24. Unpainted cubes form the inner core, (4 − 2)³ = 8. Check: 8 + 24 + 24 + 8 = 64.
Answer. 8 with 3 faces, 24 with 2, 24 with 1, 8 with 0
The diagnostic code
A code has three distinct digits. Every clue gives exact counts: 389: one digit is correct and well placed; 361: one is correct but misplaced; 953: two are correct, both misplaced; 248: none is correct; 285: one is correct but misplaced. What is the code?
Show the solution to problem 17
Main method: Reasoning
From 248, digits 2, 4 and 8 are absent. In 285, the only possible digit is 5, present but not third. In 389, either 3 is first or 9 is third. The 5 is one of the two misplaced digits in 953. If 3 were present, 389 would fix it first, but then 361 would contain a correct, well-placed digit, contradicting its clue. Thus 3 is absent and 9 is third. In 953, 5 cannot be second; since it is not third either, it must be first. In 361, only 1 can now be present, moved from third to second. The code is 519, which satisfies all five clues.
Answer. 519
The two irregular timing tracks
A simulator contains two virtual tracks. If a track is erased from one end only, it disappears completely in 60 minutes, but its local erasure rate is not uniform and the two tracks may differ. The local work required by each section is independent of direction; simultaneous fronts act independently. A front can be started from either end at any time. With no clock, how can exactly 45 minutes be measured?
Show the solution to problem 18
Main method: Lateral thinking
At the start, activate both ends of the first track and one end of the second. The first track's two fronts jointly consume its 60 minutes of local “work”, so they meet after 30 minutes despite the uneven rate. Then activate the second end of the second track. It has work equivalent to 30 one-front minutes left; two fronts finish it in 15 minutes. The total is 30 + 15 = 45 minutes. The lateral step is not mistaking “non-uniform” for “cannot be halved”: as the simulator specifies, local erasure time is direction-independent and simultaneous fronts both contribute.
Answer. Start both ends of the first track and one end of the second; when the first ends, start the second track's other end
The three switches outside the room
Switches A, B and C are outside a closed room; exactly one powers an incandescent indicator lamp inside. You may operate the switches as often as you wish, then enter only once. No light is visible from outside. Once inside, you can see whether the lamp is on and safely tell, without touching it, whether it is warm or cold. How do you identify the right switch?
Show the solution to problem 19
Main method: Lateral thinking
Turn A on for several minutes, then turn it off. Turn B on and enter immediately. If the lamp is lit, B controls it. If it is dark but still warm, A controls it. If it is dark and cold, C controls it. The three outcomes are distinct and cover every possibility. The lateral step is to use two pieces of information—current light and residual heat—instead of treating the lamp as a device with only an on/off answer. The stated conditions make the thermal observation permitted and safe.
Answer. A on then off, B on: lit = B, dark and warm = A, dark and cold = C
Ten counters on five lines
Arrange ten point-like counters so that exactly five selected straight lines each pass through four counters. A counter may belong to more than one line, and no two counters may coincide. What figure provides such an arrangement?
Show the solution to problem 20
Main method: Lateral thinking
Draw a regular pentagram and place a counter at each of its five outer points and each of its five inner intersections, ten in all. The selected lines are the five lines containing the star's long strokes. Each contains two outer points and two inner intersections, hence four counters. An incidence check confirms it: 5 · 4 = 20 memberships, and each of the ten counters lies on two selected lines. The lateral insight is allowing the same counter to count on two lines, as the statement explicitly permits.
Answer. The five points and five intersections of a pentagram
Six sticks, four triangles
You have six rigid sticks of equal length. Without breaking or overlapping them, use each once to form four congruent equilateral triangles. Sticks may meet at their endpoints, and the construction need not be flat. What solid do you build?
Show the solution to problem 21
Main method: Lateral thinking
Build a regular tetrahedron. It has four vertices and six edges, so every edge uses one stick. Its four faces are congruent equilateral triangles, satisfying every condition exactly. The difficulty comes from the familiar but unstated assumption that the construction is planar. In three dimensions, each edge is shared by two faces, allowing six sticks to bound four triangular faces.
Answer. A regular tetrahedron
The floating ice
A block of pure ice floats in fresh water in a vertical-sided cylindrical vessel. The water level is marked. The ice melts completely; there are no trapped objects, evaporation, overflow or temperature-related volume changes. Does the level rise, fall or remain unchanged?
Show the solution to problem 22
Main method: Lateral thinking
It remains unchanged. While floating, the ice displaces a mass of water exactly equal to its own mass, by Archimedes' principle. When it melts, it produces that same mass of water. The meltwater volume therefore equals the volume previously displaced, not the whole visible and submerged volume of the ice block. The cylindrical walls turn equal volumes into equal levels. The lateral step is comparing mass and displaced volume rather than the ice's apparent volume.
Answer. The level remains unchanged
The band with one side
You have a rectangular strip of paper and a piece of tape. By joining only the two short ends, make a loop with one continuous side and one continuous boundary edge. How should the ends be oriented before joining?
Show the solution to problem 23
Main method: Lateral thinking
Before joining the short ends, rotate one end by half a turn, 180°, then tape it to the other. This makes a Möbius strip. To verify the single side, draw a centre line without lifting the pencil: it returns to its starting point after passing through what appeared to be both sides of the flat paper. To verify the single boundary, follow it with a finger: it traverses both former long edges before returning. The lateral shift is changing orientation before closure instead of making the usual cylinder.
Answer. Give one end a half-turn and form a Möbius strip
The six glasses in a row
Six identical glasses stand in a row. The first three are full of water and the last three are empty: full, full, full, empty, empty, empty. By handling only one glass once, with permission to pour its contents, make the pattern full, empty, full, empty, full, empty. Which glass do you take, and what do you do?
Show the solution to problem 24
Main method: Lateral thinking
Pick up the second glass, pour all its water into the fifth, and return the second glass to its place. The second becomes empty and the fifth full; no other glass changes. The final row is full, empty, full, empty, full, empty. The lateral insight is distinguishing the one glass that is handled from its contents, which may move into a glass that is never lifted.
Answer. Pour the second glass into the fifth, then return the second to its place
Eight pieces with three cuts
In a simulation, a uniform cylindrical block must be divided into eight congruent pieces using exactly three complete planar cuts. The block may not be moved, stacked or reassembled between cuts, but the planes may have any orientation. How should the three cuts be placed?
Show the solution to problem 25
Main method: Lateral thinking
Make one horizontal cut halfway up, producing two equal cylinders stacked in place. Then make two vertical cuts along perpendicular diametral planes. Each plane passes through both layers at once and divides every circular layer into four equal quarters. The result is 2 · 4 = 8 congruent quarter-cylinders. The lateral step is not assuming that every cut must be vertical: the third-dimensional horizontal cut doubles the four portions seen from above.
Answer. One horizontal mid-plane and two perpendicular vertical diametral planes
Kits and crates
A laboratory has a number of crates and must pack some kits. With 6 kits in each crate, 13 kits remain outside. With 8 in each crate, it completely fills the crates it uses, leaves 5 crates empty, and has 3 kits left outside. How many crates and kits are there?
Show the solution to problem 26
Main method: Equation
Let c be the number of crates. Both descriptions give the same number of kits: 6c + 13 = 8(c − 5) + 3. Thus 6c + 13 = 8c − 37, so 2c = 50 and c = 25. The number of kits is 6 · 25 + 13 = 163.
Check: packing 6 per crate packs 150 and leaves 13; packing 8 fills 20 crates, leaves 5 empty, and leaves 3 kits: 20 · 8 + 3 = 163.
Answer. There are 25 crates and 163 kits.
Another route
Changing from 6 kits in 25 crates to 8 kits in only 20 crates raises packed capacity from 150 to 160. That increase of 10 matches the decrease in leftovers from 13 to 3. In terms of the number of crates, the difference is 2c − 40 = 10, which immediately gives c = 25.
The three wrongly labelled crates
Three closed crates contain, respectively, apples only, pears only, and a mixture of apples and pears. The labels Apples, Pears, and Mixed are all wrong. You may draw one fruit from one crate without looking inside. How can you determine every crate's contents with certainty?
Show the solution to problem 27
Main method: Reasoning
Draw from the crate labelled Mixed. Since every label is wrong, this crate is not mixed, so the fruit drawn reveals all of its contents. If it is an apple, that crate contains apples only. The crate labelled Pears can contain neither pears nor apples only, which has already been assigned, so it must be mixed. The crate labelled Apples then contains pears only. If a pear is drawn, the argument is symmetric.
Check: in either case each content is assigned exactly once and no label matches the actual contents.
Answer. Draw from the crate labelled Mixed; the observed fruit then determines all three crates uniquely.
The document and two padlocks
Ada must send Bruno a document in a strong box through an untrusted courier. Ada and Bruno each own a padlock and its key, but they have never exchanged keys. The box can hold both padlocks at once. How can they deliver the document without sending any key and keep the box protected during every journey?
Show the solution to problem 28
Main method: Lateral thinking
Ada locks the box with her padlock and sends it to Bruno. Without opening it, Bruno adds his padlock and sends it back. Ada removes only her own padlock and sends the box again, still secured by Bruno's lock. Bruno finally removes his lock and opens the box.
Why it works: on every journey there is at least one padlock whose key is not travelling with the box. We assume that the courier cannot force or replace the box or locks.
Answer. They use three journeys: Ada's lock, then both locks, and finally Bruno's lock alone.
The three copper tolls
A traveller passes through three gates. At each gate he gives away half the coins he has at that moment and then 2 more coins. After the third payment, he has 18 coins left. How many coins did he start with?
Show the solution to problem 29
Main method: Equation
Let x be the starting number. After all three payments, the remainder is (((x/2 − 2)/2 − 2)/2 − 2). Thus (((x/2 − 2)/2 − 2)/2 − 2) = 18. Simplifying gives x/8 − 7/2 = 18, hence x/8 = 43/2 and x = 172.
Check: from 172 he pays 86 + 2 and keeps 84; he pays 42 + 2 and keeps 40; he pays 20 + 2 and keeps 18.
Answer. He started with 172 coins.
Another route
Work backwards. Before the last gate he had to have 2 · (18 + 2) = 40 coins; before the second, 2 · (40 + 2) = 84; before the first, 2 · (84 + 2) = 172.
The three researchers' hats
From a supply of 3 blue hats and 2 green hats, one is placed on each of three researchers standing in a line. Anna sees Bruno's and Carla's hats; Bruno sees Carla's; Carla sees none. Everyone knows the supply and hears earlier answers. Anna says she does not know her colour. Bruno then says he does not know his. Carla says she now knows her colour. What is it?
Show the solution to problem 30
Main method: Reasoning
If Bruno and Carla both wore green, Anna would see that both green hats were used and would know that hers was blue. Her uncertainty therefore rules out the green-green pair. Bruno knows this. If he saw green on Carla, he would know his own hat could not be green and would conclude that it was blue. Since Bruno is also uncertain, Carla cannot be wearing green: her hat is blue.
Check: with Carla in blue, both Anna's and Bruno's uncertainty can occur; with Carla in green, Bruno would have a certain deduction.
Answer. Carla is wearing a blue hat.
Coins in the dark
There are 100 coins on a table: exactly 20 show heads and 80 show tails. You are blindfolded and cannot distinguish the faces by touch. You may move and flip coins. How can you divide them into two groups showing exactly the same number of heads, without requiring equal group sizes?
Show the solution to problem 31
Main method: Lateral thinking
Choose any 20 coins for the first group and leave the other 80 in the second. Flip all 20 coins in the first group. If the selected 20 originally contained h heads, the large group contains 20 − h heads. After flipping, the h heads in the small group become tails and its 20 − h tails become heads. Both groups therefore show 20 − h heads.
Check: the argument works for every possible selection, including the extremes h = 0 and h = 20.
Answer. Set aside any 20 coins and flip every one of them; both groups will have the same number of heads.
The greenhouse shelves
A greenhouse has 17 shelves of plant pots. Each shelf holds 4 more pots than the preceding one, and there are 850 pots altogether. How many pots are on the first and last shelves?
Show the solution to problem 32
Main method: Equation
Let x be the number of pots on the first shelf. The last has x + 16 · 4 = x + 64. An arithmetic progression sums to the number of terms times the mean of its endpoints: 17(x + x + 64)/2 = 850. Thus 17(x + 32) = 850, x + 32 = 50, and x = 18. The last shelf has 18 + 64 = 82 pots.
Check: the mean of 18 and 82 is 50, and 50 · 17 = 850.
Answer. The first shelf has 18 pots and the last has 82.
Another route
The middle shelf must contain the mean, 850 / 17 = 50 pots. There are 8 increases of 4 between the first and ninth shelves, so the first has 50 − 32 = 18; by symmetry, the last has 50 + 32 = 82.
The bridge and the lamp
Four technicians must cross a bridge at night. It can carry at most two people, and their single lamp must accompany every crossing. Their individual times are 1, 2, 7, and 11 minutes; a pair moves at the slower person's pace. What is the minimum time for everyone to reach the other side?
Show the solution to problem 33
Main method: Reasoning
The optimal sequence is: 1 and 2 cross in 2 minutes; 1 returns in 1; 7 and 11 cross in 11; 2 returns in 2; 1 and 2 cross again in 2. Total: 2 + 1 + 11 + 2 + 2 = 18 minutes.
Why it is minimal: if the 7- and 11-minute technicians cross together while no fast technician is already on the far side, after their 11-minute crossing one of them must return the lamp in at least 7 minutes, and at least one further crossing is still needed: at least 11 + 7 + 1 = 19. In the only potentially optimal subcase, one fast technician is positioned on the far side in advance; apart from the slow crossing, the required movements of the two fast technicians cost at least 1 + 3 · 2 = 7, making the total at least 11 + 7 = 18. If the slow technicians cross separately, at least three forward crossings and two returns are required: the two slow crossings cost at least 7 + 11, and the other three moves cost at least 1 minute each, for at least 7 + 11 + 3 = 21. The 18-minute construction therefore attains the lower bound.
Answer. The minimum is 18 minutes.
Four indistinguishable discs
In an abstract puzzle there are two type A discs and two type B discs, identical by sight and touch, mixed in the dark. Every disc is uniform and can be split accurately along a central score. How can they be made into two portions, each equivalent to one whole A disc and one whole B disc, without identifying the types?
Show the solution to problem 34
Main method: Lateral thinking
Split all four discs into equal halves. Put one half of each of the four discs in the first portion; put the four remaining halves in the second. The first contains two halves from the two A discs, equivalent to one whole A, and two from the B discs, equivalent to one whole B. The same is true of the second portion.
Assumption: this is a purely abstract puzzle; every disc is uniform and divides exactly in half.
Answer. Split every disc in half and put one half of each disc in each portion.
Unpriced notebooks and pens
At a stationery shop, 7 notebooks and 5 pens cost €43.40, while 4 notebooks and 9 pens cost €42.00. What is the price of each item?
Show the solution to problem 35
Main method: Equation
Let q be the price of a notebook and p the price of a pen. The system is 7q + 5p = 43.40 and 4q + 9p = 42.00. Multiplying the first equation by 4 and the second by 7 gives 28q + 20p = 173.60 and 28q + 63p = 294.00. Subtraction gives 43p = 120.40, so p = 2.80. From the first equation, 7q = 43.40 − 14.00 = 29.40, hence q = 4.20.
Check: 7 · 4.20 + 5 · 2.80 = 43.40 and 4 · 4.20 + 9 · 2.80 = 42.00.
Answer. A notebook costs €4.20 and a pen costs €2.80.
Merging six numbers
The numbers 1, 2, 3, 4, 5, and 6 are written on a board. On each move, choose two numbers a and b, erase them, and replace them with a + b + ab. After five moves one number remains. What is it, regardless of which pairs were chosen?
Show the solution to problem 36
Main method: Reasoning
For the new number c, c + 1 = a + b + ab + 1 = (a + 1)(b + 1). Therefore the product of all values obtained by adding 1 to every current number is invariant: the two factors a + 1 and b + 1 are replaced by their product. Initially this product is 2 · 3 · 4 · 5 · 6 · 7 = 5040. If the final number is r, then r + 1 = 5040, so r = 5039.
Check: the proof depends on neither the order nor the chosen pairs.
Answer. The final number is always 5,039.
The bar and seven nights
A traveller owns a bar marked into 7 segments of equal value. For seven days, he must pay an inn at the end of each day so that the innkeeper holds exactly the value of the elapsed days each evening. Before arriving, he may make only two straight cuts along the marks; the innkeeper may give change using pieces already received. How should he cut and pay?
Show the solution to problem 37
Main method: Lateral thinking
Use the two cuts to make pieces worth 1, 2, and 4 segments. The net payments are: day 1, give 1; day 2, take back 1 and give 2; day 3, give 1; day 4, take back 1 and 2 and give 4; day 5, give 1; day 6, take back 1 and give 2; day 7, give 1. The innkeeper therefore holds values 1, 2, 3, 4, 5, 6, 7 on successive evenings.
Lateral idea: payment need not mean handing over a new piece without receiving anything back; making change enables the binary representation 1 + 2 + 4.
Answer. Cut the bar into pieces worth 1, 2, and 4 segments and use change to form every value from 1 to 7.
The triangle with a missing side
A panel is shaped like a right triangle. Its hypotenuse is 13 m long, and one leg is 7 m longer than the other. How long are the two legs?
Show the solution to problem 38
Main method: Equation
Let x be the shorter leg; the other is x + 7. By the Pythagorean theorem, x² + (x + 7)² = 13². Expanding gives 2x² + 14x + 49 = 169, hence x² + 7x − 60 = 0. This factors as (x + 12)(x − 5) = 0. The algebraic roots are −12 and 5, but a length must be positive, so x = 5. The other leg is 12 m.
Check: 5² + 12² = 25 + 144 = 169 = 13², and 12 − 5 = 7.
Answer. The legs are 5 m and 12 m long.
Another route
Recognise the Pythagorean triple 5, 12, 13: its legs differ by exactly 7, so it satisfies both conditions.
The number with consecutive remainders
Find the smallest positive integer that leaves remainder 1 when divided by 2, remainder 2 when divided by 3, remainder 3 when divided by 4, remainder 4 when divided by 5, and is divisible by 7.
Show the solution to problem 39
Main method: Reasoning
The first four conditions all say that adding 1 to the number gives a multiple of 2, 3, 4, and 5. Their least common multiple is 60, so the number has the form 60k − 1. It must also be divisible by 7. Since 60 ≡ 4 modulo 7, we need 4k − 1 ≡ 0, or 4k ≡ 1. The smallest positive k that works is 2, because 4 · 2 = 8 ≡ 1. The number is therefore 60 · 2 − 1 = 119.
Check: 119 leaves remainders 1, 2, 3, and 4 respectively, and equals 7 · 17.
Answer. The number is 119.
Another route
List numbers one less than a multiple of 60: 59, 119, 179, … The first one divisible by 7 is 119.
Nine dots and the imaginary square
Nine dots lie at the nodes of a 3 × 3 grid. Without lifting your pen, draw only four consecutive straight line segments that pass through every dot. You do not have to finish where you started.
Show the solution to problem 40
Main method: Lateral thinking
Assign coordinates from (0,0) at bottom left to (2,2) at top right. Draw these segments in order: from (0,0) to (2,2); then to (−1,2); then to (2,−1); finally vertically to (2,2). The first crosses the centre; the second crosses the top row; the third crosses (0,1) and (1,0); the last crosses the two missing dots in the right column.
Lateral idea: the dots suggest a square, but no rule says that the lines must remain inside its imaginary boundary.
Answer. The four-line solution extends beyond the square suggested by the nine dots.
The polygon rich in diagonals
In a convex polygon, the number of diagonals equals five times the number of sides. How many sides does it have?
Show the solution to problem 41
Main method: Equation
From each of the n vertices, diagonals can be drawn to n − 3 vertices, excluding the vertex itself and its two neighbours. This counts every diagonal twice, so the total is n(n − 3)/2. The required equation is n(n − 3)/2 = 5n. Since a polygon has n > 0, divide by n: (n − 3)/2 = 5, so n − 3 = 10 and n = 13.
Check: a 13-gon has 13 · 10 / 2 = 65 diagonals, and 5 · 13 = 65.
Answer. The polygon has 13 sides.
Another route
The ratio of diagonals to sides is (n − 3)/2. Setting it equal to 5 immediately gives n = 13.
The chocolate bar's last break
A rectangular chocolate bar is divided into a 6 × 8 grid of squares. Two players alternate. On each move, a player chooses one rectangular piece already obtained and breaks it along one complete grid line into two rectangles. Pieces may not be stacked, and a move cannot break more than one piece. Whoever makes the final possible break wins. Who wins with perfect play?
Show the solution to problem 42
Main method: Reasoning
The first player wins. Initially there is one piece; at the end, when no further break is possible, there are 6 · 8 = 48 unit squares. Every move replaces one piece with two, increasing the total number of pieces by exactly 1. Going from 1 to 48 pieces always requires 48 − 1 = 47 moves, regardless of the chosen break lines. Since 47 is odd, the first player makes the final move.
Check: no strategy changes the game's length; while a piece contains more than one unit square, it has at least one available grid line.
Answer. The first player wins because every game lasts exactly 47 moves.
A necklace from four chains
There are four separate chain segments, each made of three closed links. The goal is one simple closed necklace of 12 links, in which every link is interlocked with exactly two neighbours. Opening and then reclosing one link counts as one operation. What is the minimum number of links that must be opened, and how?
Show the solution to problem 43
Main method: Lateral thinking
Choose one segment and open all three of its links, separating them. Arrange the other three intact segments in a circle and use each open link to join the ends of two consecutive segments; then close it. This creates a 12-link necklace while opening only 3 links.
Why it is minimal: with 0 openings, the four segments remain separate. Now suppose k links were opened, where 1 ≤ k ≤ 2. Removing those k links from the final necklace, which is a cycle, leaves at most k connected paths. Yet each of the four original segments retains at least one never-opened link, because emptying a segment requires opening all 3 links. Never-opened links from different original segments cannot have become directly interlinked; after the opened links are removed, at least 4 nonempty components must therefore remain. This is a contradiction. At least 3 openings are necessary, and the construction shows that 3 suffice. The change in perspective is to sacrifice one whole segment as connectors.
Answer. The minimum is 3: open every link of one segment and use those links to join the other three.
Six recursive data blocks
An experimental archive contains 6 ordered data blocks. The second block is 3 MB larger than the first, and from the third onward each block's size equals the sum of the preceding two. The six blocks occupy 436 MB altogether. What is the size of each block?
Show the solution to problem 44
Main method: Equation
Let x be the size of the first block. In order, the six sizes are x, x + 3, 2x + 3, 3x + 6, 5x + 9, and 8x + 15. Their sum gives 20x + 36 = 436. Thus 20x = 400 and x = 20. The block sizes are 20, 23, 43, 66, 109, and 175 MB.
Check: every term from the third is the sum of the preceding two, and 20 + 23 + 43 + 66 + 109 + 175 = 436.
Answer. The blocks measure 20, 23, 43, 66, 109, and 175 MB.
Another route
Express each block as a coefficient of the first plus a constant. Both the coefficient sequence, 1, 1, 2, 3, 5, 8, and the constant sequence, 0, 3, 3, 6, 9, 15, follow the same recurrence; their sums are 20 and 36.
Two probes and one hundred levels
A test building has 100 levels. There is an unknown threshold: a probe dropped from a level at or below it survives and can be reused; above it, the probe breaks. With two identical probes, what is the minimum number of drops needed in the worst case to determine the highest safe level exactly, allowing for the possibility that no level is safe?
Show the solution to problem 45
Main method: Reasoning
The worst-case minimum is 14 drops. Test decreasing intervals with the first probe: level 14, then 27 by adding 13, then 39 by adding 12, followed by 50, 60, 69, 77, 84, 90, 95, 99, and finally 100 if needed. If the first probe breaks on drop k, enough drops remain to test levels one by one with the second probe above the last safe level: each new interval is one floor smaller than the preceding one.
Why it is minimal: with d drops and two probes, at most d + (d − 1) + … + 1 = d(d + 1)/2 levels can be covered. Thirteen drops cover only 91 levels, whereas fourteen cover 105. Thus 14 is both sufficient and necessary.
Answer. The guaranteed minimum is 14 drops.
Two guards and one question
Two doors lead respectively to the exit and to a locked room. Two guards stand beside them: one always tells the truth and the other always lies, but you do not know which is which. Both know the doors and know which guard lies. You may ask one question to only one guard. What question guarantees that you choose the exit?
Show the solution to problem 46
Main method: Lateral thinking
Ask either guard: Which door would the other guard point to as the exit? Then choose the opposite door. The truthful guard would accurately report the liar's false answer; the liar would lie about the truthful guard's correct answer. In either case, the indicated door is wrong, so the opposite one is the exit.
Assumptions: both guards know the doors, understand the question, and always follow their stated behaviour.
Answer. Ask which door the other guard would indicate, then choose the opposite door.
The range of five tyres
A rover uses exactly four tyres at a time and carries a fifth spare. Each new tyre can travel at most 40 km before reaching its wear limit. Wear is uniform, and the tyres may be swapped along the route. What is the greatest distance the rover can travel without any tyre exceeding its limit? Also show that the distance is achievable.
Show the solution to problem 47
Main method: Equation
The five tyres provide 5 · 40 = 200 tyre-kilometres of total wear. Every kilometre travelled consumes 4 tyre-kilometres, so a distance d requires 4d ≤ 200, giving d ≤ 50 km.
The bound is achievable by dividing the route into five 10 km stages and resting a different tyre on each stage. Every tyre is then used on four stages, for 4 · 10 = 40 km. After 50 km, all five have reached but not exceeded the limit.
Answer. The maximum distance is 50 km.
Another route
On average, one of the five tyres rests while four are used on each kilometre. If rest is shared equally, each tyre works for four fifths of the journey: (4/5)d = 40, hence d = 50.
Six researchers and an unavoidable trio
Six researchers attend a meeting. For every pair, all that is known is whether they have collaborated before or have never collaborated. Prove that there must be either 3 researchers who have all collaborated pairwise, or 3 researchers no pair of whom has ever collaborated.
Show the solution to problem 48
Main method: Reasoning
Choose one researcher, Ada, and consider her 5 relationships. By the pigeonhole principle, at least 3 have the same type: Ada has either collaborated with at least three people or never collaborated with at least three.
In the first case, call those people Bruno, Carla, and Diego. If any pair among them has collaborated, that pair together with Ada forms a mutually collaborating trio. If no pair among them has collaborated, Bruno, Carla, and Diego form a mutually non-collaborating trio. In the second case, reverse the roles: if any pair among the three has not collaborated, that pair with Ada is a non-collaborating trio; otherwise all three have collaborated with one another. One of the two types of trio therefore always exists.
Answer. A fully collaborating or fully non-collaborating trio must always exist.
Another route
In graph language, colour every edge of the complete graph on 6 vertices with one of two colours. The argument proves that a monochromatic triangle always exists: this is the classical result R(3,3) = 6.
The euro that is not missing
Three colleagues pay €36 for lunch, €12 each. The correct bill is €31, so the waiter brings back €5. Each person takes €1, and the remaining €2 is left as a tip. Each has spent €11 net, for a total of €33; adding the €2 tip gives €35. Where did the missing euro go?
Show the solution to problem 49
Main method: Lateral thinking
No euro is missing: the final addition combines amounts that should not be added. The €33 net expenditure already includes the €31 bill and the €2 tip: 33 = 31 + 2. To reconstruct the original €36, add the €3 returned to the colleagues to the €33, not the tip: 33 + 3 = 36.
Lateral idea: the wording encourages adding the tip to an amount that already contains it; compare matching accounting categories instead.
Answer. Nothing is missing: €31 bill + €2 tip + €3 returned = €36.
Three doors and switching
A prize is behind one of three doors; the other two hide nothing. You choose a door. The host knows where the prize is, always opens one unchosen door known to be empty, and always offers you the chance to switch to the only other closed door. If two empty doors are available, the host chooses randomly. Is it better to switch, stay, or does it make no difference?
Show the solution to problem 50
Main method: Reasoning
Switching is better. The initial choice contains the prize with probability 1/3 and misses it with probability 2/3. If the initial choice is correct, switching loses. If it is wrong, the host must open the other empty door, so the only other closed door contains the prize and switching wins. Staying therefore wins with probability 1/3, while switching wins with probability 2/3.
Check: considering the three equally likely prize locations separately, the initial choice is wrong in two of them, and switching reaches the prize in both.
Answer. Switching is better: it wins with probability 2/3 rather than 1/3.
Another route
Imagine 100 doors: you choose one, then the host opens 98 empty doors, leaving yours and one other closed. Yours retains probability 1/100; the other concentrates the 99/100 probability of the doors not initially chosen. The three-door case follows the same principle.

50 challenging problems: equations, logic, and lateral thinking

Three ways to solve

The 50 problems

The square border

Another route

The escalator steps

The two counter experiments

The three digital archives

The drone's echo

Another route

The two virtual candles

Another route

The wheels on the same route

The two square mosaics

Another route

The digits used to number a book

The labels on four drawers

The order of the six books

The impossible handshakes

The five-bridge route

The month with three repeated weekdays

The missing win total

The large painted cube

The diagnostic code

The two irregular timing tracks

The three switches outside the room

Ten counters on five lines

Six sticks, four triangles

The floating ice

The band with one side

The six glasses in a row

Eight pieces with three cuts

Kits and crates

Another route

The three wrongly labelled crates

The document and two padlocks

The three copper tolls

Another route

The three researchers' hats

Coins in the dark

The greenhouse shelves

Another route

The bridge and the lamp

Four indistinguishable discs

Unpriced notebooks and pens

Merging six numbers

The bar and seven nights

The triangle with a missing side

Another route

The number with consecutive remainders

Another route

Nine dots and the imaginary square

The polygon rich in diagonals

Another route

The chocolate bar's last break

A necklace from four chains

Six recursive data blocks

Another route

Two probes and one hundred levels

Two guards and one question

The range of five tyres

Another route

Six researchers and an unavoidable trio

Another route

The euro that is not missing

Three doors and switching

Another route