Logic & Brain Teasers

Why B? And why not the others?

Correct answer: B. 13 kids

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').

Why D? And why not the others?

Correct answer: D. 22

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).

Why A? And why not the others?

Correct answer: A. The hen

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').

Why C? And why not the others?

Correct answer: C. Nowhere -- the $27 the friends paid already includes the runner's $2, so adding the $2 to it a second time is what creates the fake gap, not a real missing dollar

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').

Why B? And why not the others?

Correct answer: B. 1, because adding two four-digit numbers can never carry a large enough amount to create more than one extra digit in front of the result

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.

Why D? And why not the others?

Correct answer: D. 17 minutes -- the two fastest hikers cross first, one brings the lamp back, the two slowest cross together, then the other fast hiker walks the lamp back one last time for the remaining fast hiker to rejoin

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.'

Why A? And why not the others?

Correct answer: A. "If I asked you whether the left road is the way out, would you say yes?" -- then take the left road if the guard says yes, and the right road if the guard says no

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.'

Why C? And why not the others?

Correct answer: C. Weigh 3 coins against 3 other coins, setting the remaining 2 coins aside

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.'

Why B? And why not the others?

Correct answer: B. Switch -- doing so wins the prize about 2 times out of 3, roughly double the odds of staying

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.'

Why D? And why not the others?

Correct answer: D. 9 -- the back-most student announces whether the number of red caps they can see is odd or even (accepting that this guess about their OWN cap might be wrong), and every student after them combines that one clue with all the guesses already given to work out their own cap color with certainty

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.'