Logic & Brain Teasers

Correct answer: .

There are only 12 possible birth months, so if a room holds no more than 12 kids, it is possible -- just barely -- for every single one of them to have been born in a different month, leaving no guaranteed match. The moment a 13th kid enters, though, there are more kids than months, so at least one month must hold two or more of them; this is the pigeonhole principle (n+1 items sorted into n categories always leaves one category with at least two items). The option naming 12 kids fails because 12 kids can still be spread one-per-month with zero overlap -- it's the largest number that does NOT guarantee a match, not the smallest that does. The option naming 24 vastly overshoots what's actually required; doubling the months isn't needed to force one overlap. The option naming 6 is far too small -- with only half the months represented, every kid could easily land in a different one.

Source: Pigeonhole principle (n+1 objects, n categories) -- standard combinatorics theorem; verified via university course materials (HKUST Department of Mathematics; IRMO 'The Pigeonhole Principle').

Correct answer: .

The gap between each pair of numbers isn't constant -- it grows by exactly one every time: from 1 to 2 is a gap of 1, from 2 to 4 is a gap of 2, from 4 to 7 is a gap of 3, from 7 to 11 is a gap of 4, and from 11 to 16 is a gap of 5. Following that same rule, the next gap must be 6, and 16 + 6 = 22. The option giving 21 assumes the gap stays at 5 again instead of growing to 6 -- a natural mistake if you only glance at the last gap instead of tracking how the gaps themselves are climbing. The option giving 20 assumes the gap shrinks back down to 4, which runs in the wrong direction entirely. The option giving 23 overshoots by assuming the gap jumps to 7, skipping past the steady one-at-a-time growth the rest of the sequence actually follows.

Source: Original arithmetic-progression sequence constructed for this question; correctness verified by direct calculation of the increasing common-difference pattern (1,2,3,4,5,6).

Correct answer: .

The hen has to go first because it's the only one of the three that is dangerous on both sides -- left with the fox it gets eaten, and left with the corn it eats the corn -- so it can never safely be the one left behind. Taking the fox across first leaves the hen alone with the corn on the near bank, which the hen immediately eats; taking the corn across first leaves the hen alone with the fox on the near bank, which the fox immediately eats. Only starting with the hen leaves a fox-and-corn pair behind, and since the fox has no interest in the corn, that pair is perfectly safe to leave unattended while the ranger paddles back for the next item (later swapping the hen back out for whichever item still needs to cross). The option claiming any first move works ignores that two of the three choices immediately end the puzzle in a loss before a second trip is even possible.

Source: Wikipedia, 'Wolf, goat and cabbage problem' (cross-checked against MathsIsFun.com, 'Farmer Crosses River Puzzle -- Solution').

Correct answer: .

The $27 the three friends actually paid already splits cleanly into the $25 the restaurant kept and the $2 the runner pocketed, so $27 correctly equals $25 + $2 -- there's no shortfall in that number at all. The trick is comparing the wrong two quantities: adding the friends' $27 to the runner's $2 double-counts the $2 that's already sitting inside the $27, which is what manufactures the illusion of a dollar vanishing. The option blaming the restaurant invents a debt that isn't there -- the restaurant received exactly the $25 it was owed. The option claiming the runner pocketed $3 changes the numbers given in the problem rather than explaining them. The option claiming the original bill was wrong also rewrites the setup instead of finding the actual (very ordinary) arithmetic error, which is purely in how the totals get added together afterward.

Source: Wikipedia, 'Missing dollar riddle' (cross-checked against Wolfram MathWorld, 'Missing Dollar Paradox').

Correct answer: .

SEND and MORE are both four-digit numbers, so the very largest their sum could ever reach is well under 20,000 (9999 + 9999 = 19998); that means the five-digit result MONEY can only ever have gained a single extra leading digit from the carry, and that leading digit has to be 1 -- there is no way for the addition to produce a leading 2 or higher. The option claiming M is 0 confuses 'never repeats' with the actual constraint at play, which is about the size of the sum, not repetition. The option claiming M is 9 mixes up 'biggest digit in the alphabet of digits' with what the carry arithmetic actually forces, which is the smallest possible new digit, not the largest. The option about M being 'the second letter' isn't a mathematical argument at all -- position in the written word has nothing to do with which digit a letter represents.

Source: Cryptarithm devised by Henry Dudeney, 1924 (SEND + MORE = MONEY); unique solution and carry-digit reasoning cross-verified via Programming Praxis, 'SEND + MORE = MONEY, Part 1' and multiple independent solvers.

Correct answer: .

The winning plan is: the two fastest hikers (1 and 2 minutes) cross together (2 minutes), the fastest returns with the lamp (1 minute), the two slowest hikers (5 and 10 minutes) cross together (10 minutes), the other fast hiker who was waiting on the far side brings the lamp back (2 minutes), and finally the two fastest cross together again (2 minutes) -- a total of 2+1+10+2+2 = 17 minutes. The key trick is pairing the two slow hikers together on their one and only crossing, so their large times are 'spent' just once instead of twice. The option where the fastest hiker escorts everyone individually wastes time because the fastest hiker's return trips keep adding up, reaching 19 minutes. The option crossing the slow pair first still works out to more total minutes once the necessary lamp-return trips are added on top. The option sending the two slow hikers across separately is the classic losing strategy: it forces one of their long crossing times to be paid for twice instead of once, since a slow hiker ends up escorting the lamp back as well.

Source: Wikipedia, 'Bridge and torch problem'; 17-minute optimal solution cross-checked against Nigel Coldwell's puzzle-answer archive (puzzles.nigelcoldwell.co.uk) and GeeksforGeeks, 'Puzzle: Torch and Bridge.'

Correct answer: .

Asking the self-referential question -- what the guard would say if asked about the left road -- forces both guards to give the same reliable answer: the truthful guard reports their honest 'yes' or 'no' straightforwardly, while the lying guard, asked to predict their own lie, lies about what their lie would be and ends up naming the truth anyway. Either way, a 'yes' means the left road is genuinely the way out, and a 'no' means it isn't. The question about which guard is truthful fails because every guard -- truthful or lying -- always answers 'yes' to it: the truthful one truthfully says yes, and the liar lies about being a liar, also saying yes, so the answer carries no information at all. The compound question mixing two facts into one yes/no answer can't be untangled afterward, since a single 'yes' or 'no' can't tell you which of the two facts inside it was true. Simply asking what the other guard would say and then walking down that road, without flipping the answer, is the classic trap: the two guards' opposite natures mean whichever road the other guard is reported to favor is actually the wrong one, so that road must be reversed, not followed directly, for the trick to work.

Source: Raymond Smullyan's Knights and Knaves framework (What Is the Name of This Book?, 1978); self-referential-question solution cross-verified via Wikipedia, 'Knights and Knaves' and PuzzCulture, 'The Riddle of the Two Guards.'

Correct answer: .

Splitting into groups of 3, 3, and 2 is the only first move that always leaves a puzzle small enough to finish in exactly one more weighing. If the two groups of 3 balance, the counterfeit must be one of the 2 coins set aside, and weighing those two directly against each other in the second weighing instantly reveals the heavier, fake one. If the two groups of 3 don't balance, the counterfeit is somewhere in the heavier group of 3, and weighing any 2 of those 3 against each other either shows which one is heavier or, if they balance, reveals that the untouched third coin is the fake. Weighing 4 against 4 fails because one side always tips (the fake is heavier), leaving 4 suspect coins and only one weighing left -- not enough to isolate a single coin out of 4. Weighing 1 against 1 repeatedly wastes weighings on individual coins instead of narrowing down groups, and won't reliably finish in only 2 tries. Weighing 2 against 6 leaves a group of 6 suspects if that side is heavier, which is far too many to resolve with a single remaining weighing.

Source: Classic balance-puzzle solution (8 coins, 1 heavier, 2 weighings) verified via cut-the-knot.org, 'A Fake Among Eight Coins,' cross-checked against Wikipedia, 'Balance puzzle.'

Correct answer: .

Your first pick had only a 1-in-3 chance of hiding the prize, and nothing about the host later opening a curtain he already knew was empty changes that original 1-in-3 probability -- the host's move doesn't add new randomness, it just reveals information he had all along. That means the other 2-in-3 chance of the prize has to be sitting somewhere among the two curtains you didn't pick, and once one of those is shown to be empty, the entire 2-in-3 chance concentrates onto the single remaining unopened curtain, so switching wins about twice as often as staying. The option calling it a 50/50 coin flip ignores that the odds were set unevenly back at your very first pick and were never re-randomized afterward. The option favoring staying because your first pick was 'locked in' has the logic backwards -- the original pick isn't specially protected, it's exactly the choice that only had a 1-in-3 chance to begin with. The option about the host hesitating is pure superstition; the host's behavior carries no mathematical information, only the fact of which curtain he was forced to open does.

Source: Britannica, 'Monty Hall problem.'

Correct answer: .

The back-most student can see everyone else's cap but has no information at all about their own, so that one guess is unavoidably a coin flip -- they sacrifice it by using their turn to announce a parity count instead (say, 'red' if they see an odd number of red caps ahead of them, 'blue' if even). Every student after them can then track how the announced parity should be changing as caps get revealed one by one moving up the line, letting each of them work out their own color with total certainty from that single clue plus the guesses spoken before their own turn. That guarantees all 9 remaining students get it right, with only the very first speaker's guess left genuinely at risk. The options guaranteeing just 5 or 8 both settle for weaker strategies that throw away information the parity method captures, so they can't match the 9 that's actually achievable. The option claiming all 10 can be saved is impossible: the very first student to guess has strictly no information about their own cap from anyone else's guess (there isn't one yet), so that student's guess can never be made certain no matter how clever the prearranged code is.

Source: Classic prisoner hat-parity riddle (100 Prisoners with hats) verified via GeeksforGeeks, 'Puzzle: 100 Prisoners with Red/Black Hats,' cross-checked against TeachUA, 'The Prisoner Hat Riddle Explained: How Parity Saves the Whole Line.'

Correct answer: .

Let the poster's price be p, so the frame costs p + $1.00, and together they cost $1.10. That means p + (p + $1.00) = $1.10, which simplifies to 2p = $0.10, so p = $0.05: the poster costs five cents and the frame costs $1.05, which does add up to $1.10 and does keep the frame exactly one dollar pricier. The tempting instant guess of ten cents fails the moment you check it: if the poster were $0.10, the frame (a dollar more) would be $1.10, and the two together would total $1.20, not $1.10 -- it silently breaks the 'together they cost $1.10' condition. The guess of fifteen cents fails the same way, and even further off: a $0.15 poster would force a $1.15 frame, totalling $1.30. The remaining wrong option is simply the frame's own price mistaken for the poster's; $1.05 is what the frame costs, not the poster, and using it as the poster's price would make the frame $2.05, wildly breaking the $1.10 total. Only five cents satisfies both conditions at once: the fixed one-dollar gap between the two prices, and the fixed $1.10 total.

Source: Cognitive Reflection Test 'bat and ball' problem, originated by Shane Frederick, 'Cognitive Reflection and Decision Making' (Journal of Economic Perspectives, 2005), and popularized in Daniel Kahneman's Thinking, Fast and Slow (2011); reasoning cross-checked against Yale Law School's hosted copy of Frederick's original paper.

Correct answer: .

Because the patch doubles in area every single day, the very last doubling step is what carries it from half-covered to fully-covered. Since the pond is completely covered on day 48, it must have been exactly half covered one day earlier, on day 47, and then doubled that half into the full surface on day 48. Day 24 is the tempting answer if you assume growth is steady and linear -- simply halving the 48-day total -- but doubling growth is not steady at all: it stays small for a long stretch and then rushes to fill the remaining space in its very last few doublings, so the midpoint in TIME (day 24) is nowhere near the midpoint in COVERAGE. Day 46 is a near-miss that's off by one day in the wrong direction: on day 46 the patch would only be a quarter covered, since it still needs two more full doublings (46 to 47, then 47 to 48) to reach total coverage. Day 12 badly underestimates how much area a doubling patch reaches after many repeated doublings, treating the growth as far slower than doubling actually is.

Source: Classic exponential-growth 'lily pad' riddle; the doubling-time reasoning (the day before full coverage is always exactly half coverage) is standard exponential-growth mathematics as presented in introductory algebra/precalculus references (e.g. OpenStax Precalculus, exponential growth and decay chapter).

Correct answer: .

Each of the 9 kids shakes hands with the 8 other kids, which looks at first like 9 x 8 = 72 handshakes. But that count secretly counts every single handshake TWICE -- once from each participant's point of view (when Kid A shakes Kid B's hand, that one handshake gets counted once while tallying A's handshakes and again while tallying B's handshakes). Dividing 72 by 2 removes that double-count and leaves the true total: 36 handshakes. The option of 81 comes from multiplying 9 by 9 instead of 9 by 8, which mistakenly includes each kid 'shaking hands with themselves' as one of their handshakes -- something the puzzle explicitly rules out. The option of 72 is exactly the halfway step described above, the count before removing the double-counting, so it's double the real answer. The option of 45 is what you'd get by running this same calculation for a DIFFERENT group size (10 kids, since 10 x 9 / 2 = 45), which is a tempting trap for anyone who misremembers how many kids showed up to the meeting.

Source: Standard combinatorics handshake-counting result (total pairings among n people equal n(n-1)/2), the same double-counting principle behind the handshake lemma in graph theory; cross-checked against Khan Academy's combinatorics 'handshake problem' materials.

Correct answer: .

A single glance at a bulb only ever tells you 'on' or 'off' -- one piece of information -- which isn't enough by itself to tell three switches apart. The trick is to create a THIRD detectable state besides simply on and off: warmth. Leaving the first switch on for a while heats up its bulb even after that switch is turned back off, so when you finally step inside, you can tell all three bulbs apart at a glance and a touch: whichever bulb is glowing belongs to whichever switch was left on right up until you entered (the second one), whichever bulb is dark but still warm belongs to the switch that was on earlier but got turned off before entering (the first one), and whichever bulb is both dark and cool was never turned on at all (the third one). Claiming it's impossible with one visit ignores that heat gives you a usable third signal beyond just light and dark. Flipping every switch on and then off again right before entering wastes that heat clue entirely, since all three bulbs would end up equally warm and equally dark, giving no way to tell them apart. Testing one switch at a time by re-entering the room after each one directly breaks the rule that the room can only be entered a single time.

Source: Classic 'three switches, one bulb' logic puzzle; the heat-based solution is cross-verified against Tanya Khovanova's Math Blog, 'The Three Light Bulbs Puzzle,' and GeeksforGeeks, 'Puzzle: 3 Bulbs and 3 Switches.'

Correct answer: .

The box labeled 'Apples & Oranges' is guaranteed to be wrong, and since it obviously can't be empty or hold some other unlisted fruit, the only two possibilities left for it are 'apples only' or 'oranges only' -- it cannot actually be a mixed box, because that's the one label it's specifically forbidden from matching. So whatever single piece of fruit you pull from it tells you, with total certainty, that box's true, single-fruit contents. From there, elimination finishes the job: the box labeled 'Apples' can't hold apples (its label is wrong) and can't hold the single fruit type you just identified in the mixed-labeled box either, so it must be the mixed box, leaving the third box as the remaining single fruit type by process of elimination. Drawing from the box labeled 'Apples' instead doesn't work nearly as cleanly: knowing that label is wrong only rules out it being apples-only, but it could still be either oranges-only or the actual mixed box, and pulling out one orange doesn't distinguish between those two remaining possibilities. The exact same ambiguity applies to drawing from the box labeled 'Oranges' instead. Claiming it can't be done with one draw ignores that the mixed-labeled box is the single spot in this setup where the wrong-label rule pins the contents down to only one real possibility.

Source: Classic mislabeled-boxes fruit puzzle; the draw-from-the-mixed-label-box solution is cross-verified against GeeksforGeeks, 'Puzzle: Label The Fruit Boxes,' and Mind Your Decisions, 'Can You Solve The Apples And Oranges Riddle?'

Correct answer: .

Let x stand for the guest's age right now. 'In exactly 3 years' means an age of x + 3, and 'exactly 3 years ago' means an age of x - 3. The statement says the first amount equals twice the second: x + 3 = 2(x - 3). Expanding the right side gives x + 3 = 2x - 6, and moving all the x terms to one side gives 3 + 6 = 2x - x, so x = 9: the guest is 9 years old right now (in 3 years, 12; three years ago, 6; and 12 really is twice 6). The answer of 6 comes from mishandling the 'twice as old' relationship -- treating the two time-shifted ages as simply differing by a fixed gap rather than one being exactly double the other, which loses track of what 'twice' actually requires once you plug the age back in. The answer of 12 mistakes the FUTURE age (x + 3) for the CURRENT age, forgetting that the riddle describes an age 3 years from now, not today. The answer of 15 overshoots by folding the 3-years-ago shift into the current age in the wrong direction, adding both time shifts together instead of solving the doubling relationship the sentence actually describes.

Source: Standard algebraic age-riddle reasoning (translating 'X years from now' and 'X years ago' statements into a linear equation and solving for present age), a common word-problem type in introductory algebra curricula.

Correct answer: .

Cleo already owns the dog, which leaves only the cat and the fish available for Ana and Ben to split between them. Suppose Ben owned the fish: the fish owner is stated to love blue, so Ben would have to love blue -- but Ben is explicitly ruled out from loving blue, so that possibility collapses immediately, leaving only one option: Ben owns the cat. Since the cat owner is stated to love red, Ben's favourite colour is forced to be red, which also comfortably satisfies the separate rule that Ben doesn't love blue. From there the rest falls into place on its own: Ana takes the remaining pet, the fish, so she loves blue by the same fish-owner rule, and Cleo, holding the dog, is left with the one remaining colour, green. The option where Ben owns the fish and loves blue is exactly the possibility that gets eliminated first, since it directly contradicts the rule that Ben never loves blue. The option putting Ben with the dog is impossible outright, since Cleo -- not Ben -- is the one confirmed to own the dog. The option pairing Ben's cat with green ignores the direct rule that whoever owns the cat loves red, not green.

Source: Original constraint-satisfaction ('logic grid') puzzle constructed for this question; the unique solution was verified by directly checking every possible pet/colour assignment for Ana, Ben, and Cleo against all four stated clues and confirming only one assignment satisfies them simultaneously.

Correct answer: .

Lighting a rope from both ends at once always burns it out in exactly half its one-end time, no matter how unevenly it burns along its length: wherever the two flames happen to meet is irrelevant, because between them they've always consumed the rope's full 60-minute fuel supply in just 30 minutes. So lighting the first rope from both ends, together with lighting only one end of the second rope, guarantees the first rope finishes at the 30-minute mark; at that same instant, the second rope -- burning from only one end so far -- has exactly 30 minutes of burn time left in it, regardless of which half of it has burned faster or slower. Lighting that remaining unlit end at the 30-minute mark makes the second rope burn from both ends too, and burning through its last 30 minutes' worth of fuel from both ends finishes it in 30 / 2 = 15 more minutes, for a total of 30 + 15 = 45 minutes. Lighting one end of each rope and only adding a second flame once the first rope is already fully spent wastes the whole first 60 minutes doing nothing useful, and doesn't produce a 45-minute mark at all. Lighting both ends of both ropes from the very start finishes everything in just 30 minutes, not 45, since every piece of fuel present is being burned from two directions simultaneously. Cutting each rope in half by length doesn't reliably cut its burn time in half at all, since these ropes are stated to burn unevenly -- a rope's length-wise midpoint is not guaranteed to be its time-wise midpoint, so a cut half could burn out in almost no time or in nearly the full 60 minutes.

Source: Classic rope-burning measurement puzzle; the both-ends/one-end solution is cross-verified against the Wikipedia article 'Rope-burning puzzle' and GeeksforGeeks, 'Puzzle: How to Measure 45 Minutes Using Two Identical Wires.'

Correct answer: .

Splitting the coins into three even groups of four -- four on the left pan, four on the right pan, and four set aside untouched -- is the only first move that gets this harder, direction-unknown version of the puzzle down to a small enough set of possibilities to finish in exactly two more weighings. If the two groups of four balance, the fake coin is guaranteed to be among the four set-aside coins, and those four (now known to include the only suspect, alongside three coins already proven genuine by the very fact that the first weighing balanced) can be resolved for both identity and direction in two more weighings. If the two groups of four tip one way, the fake is narrowed down to those eight coins with a known 'which side was heavier' clue attached, and the untouched four set-aside coins become a trusted set of known-genuine coins that the next two weighings can test suspects against. Weighing all 12 coins at once as two groups of six leaves no known-genuine coins in reserve at all, since every coin was already placed on a pan; without a trusted reference group, the remaining two weighings have no reliable genuine coins to compare suspects against, which is exactly why a held-out group is necessary. Testing coins one pair at a time can't guarantee success either, since a single pair might happen to be two genuine coins that balance, wasting a weighing, and there's no way to guarantee narrowing down to one specific coin out of twelve within only three such attempts. Splitting into groups of five, five, and two is uneven in a way that doesn't cleanly divide down to a resolvable set of suspects in the following two weighings, unlike the balanced four-four-four split.

Source: Classic 'twelve coins, unknown heavier or lighter, 3 weighings' balance puzzle; the four-versus-four first weighing is cross-verified against cut-the-knot.org, 'Weighing 12 Coins, an Odd Ball puzzle,' and the Wikipedia article 'Balance puzzle.'

Correct answer: .

Since the camel can carry at most 1,000 bananas, moving the pile forward while more than 2,000 bananas remain takes three loaded trips plus two empty return trips for every stretch covered, meaning 5 kilometres of total camel-walking, and therefore 5 bananas eaten, for every 1 kilometre of real progress. That harsh 5-bananas-per-kilometre burn rate only lasts until the pile has shrunk enough to fit in just two loads of 1,000 each, which takes (3,000 - 2,000) / 5 = 200 kilometres, leaving exactly 2,000 bananas at that point. From there, only one loaded trip plus one empty return trip is needed per stretch, so the burn rate drops to 3 bananas eaten per kilometre; that continues until the pile shrinks to fit in a single 1,000-banana load, which takes a further (2,000 - 1,000) / 3 = 333 1/3 kilometres, bringing the trader to the 533 1/3-kilometre mark with exactly 1,000 bananas left. From there, one final single loaded trip (no more return trips needed) covers the remaining 466 2/3 kilometres, eating 1 banana per kilometre along the way, and delivers 1,000 - 466 2/3 = 533 1/3 bananas to market -- 533 whole bananas. Guessing 1,000 ignores that the camel eats bananas the entire way and can't simply carry a full load the whole distance without ever needing extra relay trips for the other 2,000 bananas left behind. Guessing 2,000 badly underestimates how many bananas the repeated back-and-forth relay trips actually consume over the full 1,000 kilometres. Guessing that nothing arrives ignores that a well-planned relay strategy, which reduces the number of loads needed as the pile shrinks, gets a meaningful amount of the original stockpile all the way to market.

Source: Classic 'Camel and Bananas' transportation-optimization puzzle (a version of the jeep problem); the 533-banana optimal-relay solution is cross-verified against GeeksforGeeks, 'Puzzle: Camel and Banana Puzzle,' and Tim Hesterberg's published solution walkthrough (timhesterberg.net).

Correct answer: .

The minute hand moves 6 degrees per minute (360 degrees in 60 minutes), so at 40 minutes past the hour it sits at 6 x 40 = 240 degrees from the 12. The hour hand moves only 0.5 degrees per minute in addition to its hourly position (30 degrees per hour, since 360/12 = 30), so at 3:40 it sits at (30 x 3) + (0.5 x 40) = 90 + 20 = 110 degrees from the 12. The angle between the two hands is the difference, |240 - 110| = 130 degrees, and since that is already 180 degrees or less, it is already the smaller of the two possible angles between the hands, with no further adjustment needed. The option giving 110 degrees is simply the hour hand's own position measured from the 12 -- it never actually compares the two hands at all, so it cannot be the angle between them. The option giving 150 degrees comes from freezing the hour hand exactly on the '3' mark at 90 degrees and forgetting that the hour hand keeps creeping forward as the minutes pass; using 90 instead of the true 110 produces |240 - 90| = 150, a plausible-looking but incorrect shortcut. The option giving 160 degrees results from a small arithmetic slip while combining the two hands' rates, landing close to but short of the exact combined-rate calculation that produces 130.

Source: Clock angle problem (combined hand-rate formula, angle = |30H - 5.5M|); consistent with Wikipedia's 'Clock angle problem' entry and standard secondary-school geometry treatments of the problem.

Correct answer: .

Suppose the sign is true: then what it claims -- that it is false -- must actually hold, which makes it false, contradicting the assumption that it is true. Suppose instead it is false: then its claim ('this statement is false') fails to hold, meaning the statement is not false after all, which makes it true, again contradicting the assumption just made. Either starting assumption flips into its own opposite, so no consistent truth value can ever be assigned to the sentence. This is the classic liar paradox, one of the oldest puzzles in logic and philosophy, traceable at least to the ancient Greek philosopher Eubulides of Miletus in the 4th century BCE. The option about signs being incapable of lying misses the point entirely -- the paradox concerns whether the sentence's content can be consistently assigned a truth value, not whether the physical object stating it has some special capacity for dishonesty; the exact same contradiction appears regardless of who or what states the sentence. The option invoking how often everyday statements turn out false brings in an irrelevant frequency argument that has nothing to do with this specific self-referential sentence's logical structure. The option declaring it 'simply true' because a sign lacks a truth value dodges the exercise -- the question explicitly treats it as a meaningful statement making a truth claim, and denying it a truth value does not make the underlying contradiction disappear, it just refuses to engage with it.

Source: The liar paradox, traced to Eubulides of Miletus (4th century BCE); consistent with standard treatments in Wikipedia's 'Liar paradox' entry and the New World Encyclopedia's 'Liar paradox' entry.

Correct answer: .

Zeno correctly notices that the chase can be broken into infinitely many smaller catch-up steps, but he wrongly concludes that infinitely many steps must therefore take an infinitely long time (or cover infinite distance). In fact each successive step's distance and duration shrinks geometrically -- say the tortoise starts 2 metres ahead, and Achilles' catch-up steps cover 1 metre, then 1/2 metre, then 1/4 metre, and so on. Mathematically this is the geometric series 1 + 1/2 + 1/4 + 1/8 + ..., and using the standard convergent-series formula (first term divided by one minus the common ratio, here 1 / (1 - 1/2)) gives exactly 2, a finite total, not an ever-growing sum. Infinitely many terms can still add up to a finite number, which is exactly the insight that later calculus formalized. The option crediting Achilles' speed alone is circular: being faster does not by itself explain why the infinite subdivision resolves, and the identical convergent-series resolution applies no matter how much faster Achilles is. The option about the tortoise stopping invents a detail nowhere in Zeno's actual setup, and the paradox resolves the same way even if the tortoise never stops at all. The option dismissing infinite sequences as impossible in reality actually concedes to Zeno's mistaken conclusion rather than refuting it -- modern mathematics and physics both accept that infinitely many events can be packed into a finite stretch of time or distance, which is precisely the resolution long accepted by mathematicians.

Source: Zeno of Elea's paradox of Achilles and the Tortoise, as reported in Aristotle's Physics; resolution via convergent geometric series, consistent with standard calculus and mathematical-analysis treatments of the paradox.

Correct answer: .

Four colors are always enough, and sometimes genuinely necessary: a map can easily contain four regions that all border one another (for example, one central region touching three outer regions that also all touch each other), which forces four distinct colors at once, so fewer than four can never be guaranteed to work for every possible map. The question was first posed in 1852 by Francis Guthrie while coloring a map of England's counties, and it took over a century to settle: an 1879 proof by Alfred Kempe was widely accepted for a decade until Percy Heawood found a flaw in it in 1890 -- though Heawood's own work did manage to prove that five colors are always enough. It was not until 1976 that Kenneth Appel and Wolfgang Haken, at the University of Illinois, finally proved that four really is the tight minimum, using a computer to mechanically check thousands of reduced map configurations -- making it one of the first major mathematical theorems proved with essential computer assistance, and one so lengthy that no human has ever verified the whole thing by hand. The option naming three colors fails on exactly the four-mutually-bordering-regions example above, so three is demonstrably not always sufficient. The option naming five colors is a genuinely correct and provable upper bound, proved decades before the full four-color proof, but it is one more color than the actual tight minimum. The option naming six colors is an even looser, still-earlier bound, further still from the true minimum.

Source: Four Color Theorem, conjectured by Francis Guthrie (1852) and proved by Kenneth Appel and Wolfgang Haken (1976); consistent with Encyclopaedia Britannica's 'four-color map problem' entry and the MacTutor History of Mathematics archive (University of St Andrews).

Correct answer: .

The puzzle has a recursive structure: to move n disks to another peg, you first move the top n-1 disks to the spare peg (whatever the minimum number of moves that sub-task takes), then move the single largest disk across in one move, then move those same n-1 disks back on top of it (the identical sub-task again). That gives minimum(n) = 2 x minimum(n-1) + 1, starting from minimum(1) = 1 move for a single disk. Working it up: 1 disk takes 1 move, 2 disks take 3, 3 disks take 7, and 4 disks take 2 x 7 + 1 = 15 moves -- which also matches the closed-form shortcut 2^n - 1 (here, 2^4 - 1 = 16 - 1 = 15). The option giving 7 moves is actually the correct minimum for a 3-disk stack, not 4 -- an easy slip if you forget you added an extra disk and reuse the previous stack's count. The option giving 14 moves is one move short of the true minimum, missing the single dedicated move that carries the largest disk itself across to the target peg partway through the solution. The option giving 16 moves is one move too many, effectively double-counting by treating the final placement as needing an extra shuffle that the recursive solution never actually requires.

Source: Tower of Hanoi, invented by Edouard Lucas (1883); minimum-moves formula 2^n - 1, consistent with Wikipedia's 'Tower of Hanoi' entry and standard recursion and combinatorics references.

Correct answer: .

Reading 111221 aloud run by run gives 'three 1s, then two 2s, then one 1' -- and writing that down as a count immediately followed by its digit, run by run, literally produces 312211 (3, 1, 2, 2, 1, 1), which is exactly the correct next term. The option adding an extra digit at the end double-counts the final run, claiming two 1s close out the string when 111221 actually ends in only a single 1. The option that comes out five digits long undercounts the very first run, treating the three leading 1s in 111221 as if there were only two of them, which drops a digit the string actually contains. The option opening with a 4 overcounts that same leading run, claiming four 1s start the string when there are only three. Only a term that opens with the true count of the first run (three), followed immediately by the digit it's counting, and continues the same way for every run left to right, actually reproduces what reading 111221 aloud sounds like.

Source: Look-and-say sequence, studied by John Conway (1986); OEIS A005150 and Wikipedia's 'Look-and-say sequence' entry confirm the 111221 -> 312211 step and the general read-aloud, run-length rule.

Correct answer: .

Working backward from the winning number 21, the numbers that actually matter are each exactly 4 apart -- 21, 17, 13, 9, 5, 1 -- because 4 is one more than the biggest jump either player is allowed to make in a single turn (3). Once a player calls out 17, the other player is forced to call 18, 19, or 20 next, leaving a gap of exactly 1, 2, or 3 back up to 21 -- a gap the first player can always close in a single turn no matter which of the three the opponent picks. Calling out 20 instead hands the win away entirely: on the very next turn, the opponent only needs to call a single number, '21,' to win immediately. Calling out 16 is just as risky, since the opponent can simply call '17' themselves on their next turn and seize the checkpoint instead. Calling out 19 is similarly unsafe, since the opponent can jump straight from 19 to 21 in one move by calling out two numbers at once. Only locking in 17 -- and, earlier in the game, the matching checkpoints 13, 9, 5, and 1 -- guarantees control of the count all the way to 21.

Source: Classic '21 counting game' (a subtraction game equivalent to Nim); checkpoint strategy of multiples-of-4-plus-1 cross-verified against Riverbend Community Math Center's 'Nim Games' lesson materials and Beast Academy's published Nim worksheet.

Correct answer: .

Euler's key insight was to stop thinking about the actual geography and instead count how many bridges touch each landmass. Walking through a landmass uses up two of its bridges at a time -- one to arrive, one to leave -- so every landmass gets 'used up' in pairs, except possibly the landmass where the walk starts and the one where it ends, which can each get away with one odd bridge left over. In Konigsberg, every single one of the four landmasses had an odd number of bridges touching it, which is two more odd landmasses than any single walk can ever tolerate, so no route -- regardless of starting point -- could cross all seven bridges exactly once. The option about starting on the busiest island misunderstands the problem as being about total bridge count rather than the odd-or-even count at each individual landmass; a busier island is still just as odd as the others. The option blaming the plain odd total of seven bridges confuses the overall bridge count with the count of bridges at each landmass, which is the number that actually decides the outcome. The option about avoiding same-direction repeats misses the point entirely, since the puzzle already forbids recrossing any bridge at all, in either direction, so that restriction changes nothing about the four odd landmasses.

Source: Leonhard Euler, 'Solutio problematis ad geometriam situs pertinentis' (1736); consistent with Encyclopaedia Britannica's 'Konigsberg bridge problem' entry and the MacTutor History of Mathematics archive (University of St Andrews).

Correct answer: .

Working through the circle step by step: the count first eliminates kid #2, then kid #4, then kid #6, leaving #1, #3, #5, and #7 still standing. The count doesn't reset after that first lap -- it simply keeps going, so the very next kid counted is #1, who gets eliminated next, leaving only #3, #5, and #7. Continuing around from there eliminates #5, leaving just #3 and #7, and the following count lands on #3, eliminating #3 and leaving #7 as the sole survivor. This elimination pattern -- counting out every second person around a circle until one remains -- is a classic puzzle called the Josephus problem, widely studied in computer science because the eventual survivor's position follows a predictable pattern based on how the circle's size compares to the nearest smaller power of two. The option naming kid #5 describes someone who survives several eliminations but is still the second-to-last kid removed, not the final survivor. The option naming kid #1 describes the very first kid in the original count, which carries no special protection and is in fact the first kid eliminated once the counting continues past the opening lap. The option naming kid #3 describes the very last kid eliminated, removed one step before the circle is down to the sole survivor.

Source: Josephus problem (circular counting-out elimination puzzle, step size 2); the n=7 survivor position cross-verified against Wikipedia's 'Josephus problem' entry and University College Dublin School of Mathematics and Statistics course notes on the problem.

Correct answer: .

Work backward from the smallest stage first: if it ever gets down to just Bo and Cal, Bo's proposal only needs 1 yes vote out of 2, and Bo's own vote already satisfies that, so Bo would simply keep all 100 coins and give Cal nothing -- meaning Cal is actually staring down 0 coins if Ada's proposal fails. Knowing that in advance, Ada needs just one more yes vote beyond her own to reach the required 2 out of 3, and the cheapest way to buy it is to offer Cal exactly 1 coin, since even 1 coin beats the 0 coins Cal would get if the matter were left to Bo and Cal alone -- a perfectly logical Cal votes yes for that 1 coin. That leaves Ada free to keep 99 coins for herself while giving Bo nothing at all, since Bo's vote was never needed. The option splitting the coins close to evenly ignores that Ada doesn't need to win anyone's goodwill beyond the bare minimum number of votes required, so there's no reason to pay more than necessary. The option giving everything away to Bo and Cal misreads the voting rule entirely -- Ada's own vote already counts toward the needed total, so Ada has no reason to keep nothing. The option that instead pays off Bo misjudges which pirate is cheap to buy: Bo stands to collect all 100 coins if the matter falls to the two-pirate stage, so a single coin would never be enough to win Bo's vote, while Cal -- who'd get nothing at that stage -- accepts that same single coin gladly.

Source: Pirate gold-division game-theory puzzle, solved by backward induction; three-pirate base case cross-verified against CEMC (Centre for Education in Mathematics and Computing, University of Waterloo) 'Game Theory' solution materials and standard game-theory treatments of the pirate puzzle.