Brainteasers interview questions

The minute hand is at $90^\circ$. The hour hand has moved a quarter of the way from 3 to 4: $90^\circ+\tfrac14\cdot30^\circ=97.5^\circ$. Difference $7.5^\circ$.

The minute hand gains one full lap on the hour hand every $\tfrac{12}{11}$ hours, so it laps it 11 times in 12 hours.

Locker $n$ is toggled once per divisor of $n$. It ends open iff it has an odd number of divisors, i.e. $n$ is a perfect square: $1,4,\dots,100$, ten lockers.

Count factors of 5: $\lfloor100/5\rfloor+\lfloor100/25\rfloor=20+4=24$.

Five heats, then a race of the five winners. Only five horses can still be in the top three: the second and third of the winners' race, the second and third from the overall winner's heat, and the second from the runner-up's heat. A seventh race among those five settles 2nd and 3rd.

With $d$ drops you can cover $d+(d-1)+\dots+1=\tfrac{d(d+1)}2$ floors. The smallest $d$ with $\tfrac{d(d+1)}2\ge100$ is 14: drop from 14, 27, 39, ….

There are 24 possibilities and each weighing has 3 outcomes, so at least 3 are needed ($3^3=27\ge24$), and a careful 4-4-4 scheme achieves it.

$\binom{10}{2}=45$.

There are $(9-k)^2$ squares of side $k$: $\sum_{k=1}^{8}k^2=204$.

Choose 2 of the 9 horizontal lines and 2 of the 9 vertical lines: $\binom92^2=36^2=1296$.

Pair $1+100,\ 2+99,\dots$: 50 pairs of 101.

Last digits of powers of 7 cycle $7,9,3,1$. $2026\equiv2\pmod4$, so the last digit is 9.

$b+(b+1)=1.10$ gives $b=0.05$. The intuitive answer, 10p, would make the bat £1.10.

It doubles to full on the last day, so it was half the day before.

$\dfrac{11!}{4!\,4!\,2!}=34{,}650$ (I, S four times each; P twice).

$\dfrac{n(n-3)}{2}=\dfrac{10\cdot7}{2}=35$.

Label bottles in binary with 10 bits ($2^{10}=1024\ge1000$). Strip $i$ tastes every bottle whose bit $i$ is 1. The pattern of positive strips spells the poisoned bottle's number.

Swapping labels, a collision is the same as the ants passing through each other. So each "ant" just walks straight off, taking at most 1 minute.

They meet after 1 hour; the fly flies for 1 hour at 75 mph. No series needed.

Ten times in the units place (7, 17, …, 97) and ten in the tens place (70–79).

Draw from the box labelled Mixed. It must be all one fruit, say apples. Then the box labelled Oranges cannot be oranges or apples, so it is Mixed, and the last box is Oranges.

Pigeonhole: with two colours, any three socks include two of the same colour.

Four could all differ; the fifth must repeat a suit.

12 months, so 13 people force a repeat.

$255=2^8-1=11111111_2$.

$(n-2)\cdot180^\circ=4\cdot180^\circ$.

$2^{10}5^8=2^2\cdot10^8=4\cdot10^8$.

Exactly-two means the middle of an edge: one per edge, 12 edges.

The unpainted core is $2\times2\times2$.

The last prisoner calls the parity of the black hats in front (their own survival is a coin toss). Every other prisoner can then deduce their own hat from that parity and the calls behind them.

$f(n)=f(n-1)+f(n-2)$ with $f(1)=1,f(2)=2$: Fibonacci, giving $f(10)=89$.

A geometric series: $2^{10}-1$.

Any shortest path is 4 rights and 4 ups in some order: $\binom84=70$.

Every match eliminates exactly one player, and 63 must be eliminated.

After 7 days and nights it is at 7 m; on day 8 it climbs the last 3 m before it can slip.

For the Josephus problem with every second person removed, write $n=2^m+\ell$; the survivor is $2\ell+1$. Here $100=64+36$, so $2(36)+1=73$.

Weigh 3 against 3. The heavy ball is in the heavier group, or in the third group if they balance. One more weighing of 1 against 1 settles it.

Count factors of 5: $\lfloor50/5\rfloor+\lfloor50/25\rfloor=10+2$.

The minute hand is at $180^\circ$; the hour hand is halfway between 9 and 10, at $285^\circ$. The difference is $105^\circ$.

For coprime $a$ and $b$ the largest unreachable amount is $ab-a-b=15-8=7$. Every amount from 8 upwards can be made.

The four empty chambers form a block, and you are on one of them. The next chamber is loaded only if you are on the last empty chamber before the bullets, which is 1 case in 4. So not spinning survives with probability $\tfrac34$, while spinning survives with $\tfrac46=\tfrac23$. The click told you something about where you are in the cylinder, and spinning throws that away.

$2^{10}=1024$ and $1+0+2+4=7$.

Up to order: $(2,5,5)$, $(3,4,5)$ and $(4,4,4)$. Others such as $(2,4,6)$ and $(1,5,6)$ fail the triangle inequality.

Unfold the room into a flat net and the shortest walk becomes a straight line. The naive route — straight down the wall, along the floor, up the far wall — measures $11+30+1=42$. But a net that crosses one end wall, a side wall, the ceiling, the other side wall and the far end wall lays the two points out at a horizontal separation of 40 and a vertical separation of 0, giving exactly $40$. The lesson is that "shortest path on a surface" is a geodesic, and you find it by flattening, not by guessing which faces to use.

Lighting a rope at both ends consumes it in 30 minutes whatever the burn profile, because the two flames together always finish the whole rope. When the first rope is gone, 30 minutes have passed and the second rope has 30 minutes of burn left in it. Light its other end now and it finishes in 15 more. Total $45$. The trick is that "both ends" halves the time without needing to know anything about where the rope burns fast.

The greedy route — the fastest person ferries everyone — costs $2+1+5+1+10=19$. Better to send the two slowest together so their times overlap. Send 1 and 2 across (2), return 1 (1), send 5 and 10 together (10), return 2 (2), send 1 and 2 (2): total $17$. Pairing the two slow walkers means you pay 10 once instead of 10 and 5 separately.

Opposite corners share a colour, so removing them leaves 32 of one colour and 30 of the other. Every domino covers one square of each colour, so a full cover of 31 dominoes would need 31 of each. It is impossible. The most you can place is $30$, leaving two same-coloured squares uncovered. Counting an invariant — here the colour balance — settles the question without trying a single arrangement.

Nine different answers drawn from $\{0,1,\dots,8\}$ must be exactly those nine numbers. The person who shook 8 met everyone but their spouse, so their spouse is the 0. Remove that pair: in the remaining party of eight everyone has shaken one fewer of the people still present, and the same argument pairs 7 with 1, then 6 with 2, then 5 with 3. The pairs are $(8,0),(7,1),(6,2),(5,3)$, leaving 4 unpaired — and the only person not asked is the host, so 4 is the host's wife. She shook $4$ hands.

Take 1 coin from the first stack, 2 from the second, and so on to 10 from the tenth: 55 coins, which would weigh 550 g if all were honest. Weigh the lot at once. The excess in grams names the stack — 3 g over means stack 3. One weighing suffices because a digital scale returns a number, not a comparison, and you have encoded the answer into the sample.

Fill the 5 (1), pour into the 3 leaving 2 in the big jug (2), empty the 3 (3), move the 2 across (4), fill the 5 again (5), top up the 3 — which takes just 1 litre — leaving 4 in the 5-litre jug (6). Every reachable amount is a combination $3a+5b$, and since 3 and 5 are coprime every whole number of litres up to 5 is reachable; the puzzle is only ever about the route.

While more than 2000 bananas remain, moving the stock forward costs 5 bananas per km (three trips out, two back). Run that for 200 km: $3000-1000=2000$ bananas at the 200 km mark. Now two loads remain, costing 3 per km. Run for 333 km: $2000-999=1001$, and drop 1 to make it a single load of 1000 at 533 km. The last 467 km costs 1 per km, delivering $1000-467=533$. The structure is that the cost per kilometre drops each time the stock falls below a multiple of the carrying capacity.

Imagine the ascent and the descent happening on the same day, as two monks starting at opposite ends of the path at 6 a.m. They are on the same path moving towards each other, so they must meet. Formally, the difference between the two position functions is continuous, negative at 6 a.m. and positive at dusk, so it vanishes somewhere: the intermediate value theorem. No information about pace is needed, and none is available.

Its centre travels a circle of radius $2r$, of circumference $4\pi r$, while the coin's own circumference is $2\pi r$ — suggesting one rotation, which is the trap. Rolling contributes one turn, but the centre also revolves once around the fixed coin, and in a fixed frame those add: the answer is $2$. This is the coin rotation paradox, and it is the same bookkeeping that makes a sidereal day differ from a solar one.

Turn switch A on for ten minutes, then off. Turn B on and go in. The lit bulb is B, the dark but warm bulb is A, and the dark cold bulb is C. The puzzle is only hard while you assume the bulb carries one bit — on or off. Heat is a second channel, and two channels distinguish three states.

Work backwards. With two left, the senior takes all 100, since his own vote is half. With three, the junior-most knows he gets 0 in the two-pirate case, so 1 coin buys him: $99,0,1$. With four, the proposer needs one more vote and buys the pirate who would get 0 in the three-case: $99,0,1,0$. With five, he needs two more votes, and buys the two who get 0 under the four-pirate split: $98,0,1,0,1$. The senior keeps $98$. Backward induction, and the fact that a pirate's price is exactly one coin more than his fallback.

Place the first coin exactly at the centre, then mirror every opponent move through the centre point. The table is symmetric about its centre, so if the opponent's move was legal, the reflected square is free and the reply is legal too. The first player therefore always has a move and cannot be the one who runs out. Strategy stealing by symmetry: the opening move destroys the symmetry that would otherwise favour the second player.

Cut the square into four quarters of side $\tfrac12$. With five points and four quarters, some quarter holds two of them, and two points in a square of side $\tfrac12$ are at most its diagonal $\tfrac{\sqrt2}{2}$ apart. So $d=\tfrac{\sqrt2}{2}\approx0.707$ and $d^2=\tfrac12$. The bound is tight: four points at the corners and one at the centre sit exactly that far apart.

Track the fraction of the band behind the ant rather than its distance, because stretching carries the ant along and leaves that fraction unchanged. In second $n$ the band is $n$ metres long and the ant adds $\tfrac{1}{100n}$ of it. The total progress after $N$ seconds is $\tfrac1{100}\sum_{n=1}^{N}\tfrac1n$, a harmonic sum, which diverges. So the fraction reaches 1 and the ant arrives — after about $e^{100}$ seconds, a time beyond absurd, but finite. Divergence of the harmonic series is doing all the work.

Rearranging cannot help. Consider the central unit cube: it has six faces, none on the surface of the original block, and each of those faces must be produced by a distinct cut, since one plane cut can create only one of them. So at least 6 cuts are needed, and 6 obviously suffice — two in each direction. The invariant is the innermost cube, and it caps the answer regardless of how cleverly you stack the pieces.

Triples with product 36 and their sums: $(1,1,36)=38$, $(1,2,18)=21$, $(1,3,12)=16$, $(1,4,9)=14$, $(1,6,6)=13$, $(2,2,9)=13$, $(2,3,6)=11$, $(3,3,4)=10$. Only the sum 13 is ambiguous, so the gate reads 13. "The eldest" implies a unique oldest child, ruling out $(1,6,6)$ and leaving $(2,2,9)$. The eldest is $9$. The census taker's failure is itself the decisive piece of information.

Use 1, 3, 9 and 27. Allowing a weight in either pan or neither gives each one three roles — $+1$, $-1$, $0$ — so four weights express every integer in balanced ternary from $-40$ to $40$. For instance 5 is $9-3-1$, meaning 9 in one pan against the object plus 3 and 1 in the other. Powers of 3, not 2, because the second pan gives you subtraction for free.

Not 45. Average speed is total distance over total time, and the slow leg occupies twice the hours of the fast one, so it dominates. For distance $d$ each way the time is $\tfrac d{30}+\tfrac d{60}=\tfrac{d}{20}$ for $2d$ miles, giving $40$ mph — the harmonic mean of 30 and 60. Whenever equal distances are travelled at different speeds, the harmonic mean is the right average.

Look at the counts modulo 3. They start at $13,15,17\equiv1,0,2$ — all three residues distinct. A meeting reduces two counts by 1 and raises the third by 2, which shifts every count by the same amount modulo 3, so the differences between counts are invariant mod 3. Ending in one colour means counts $45,0,0$, whose residues are $0,0,0$ — all equal. Distinct residues can never become equal, so it cannot happen. Almost every "can this state be reached" puzzle is an invariant in disguise.

The nine numbers total 45, so each line sums to 15. Add the four lines through the centre — two diagonals, the middle row, the middle column: they cover every cell once except the centre, which they cover four times. So $4\times15=45+3c$, giving $c=5$. The centre is forced before you place a single other number.

Positions that are multiples of 4 are losing for whoever faces them: whatever you take, your opponent restores the multiple. 21 is not a multiple of 4, so take $1$ to leave 20, then always take $4-k$ when your opponent takes $k$. You hand over 16, 12, 8, 4 and finally 0. Every subtraction game of this shape reduces to arithmetic modulo one more than the largest legal move.

Ask either guard: "If I asked the other guard whether the left road leads to the city, would he say yes?" Both guards give the same answer, and it is the opposite of the truth — the liar lies about the truth-teller's honest answer, and the truth-teller reports the liar's lie faithfully. So take the road the answer denies. Routing the question through both guards makes exactly one lie enter the chain whichever guard you ask, which is what removes your ignorance of who is who.

Books on a shelf face left to right, so volume 1's front cover is on the *right* of that volume — adjacent to volume 2. Likewise volume 3's back cover faces left. The worm therefore passes only volume 1's front cover (0.5), all of volume 2 (0.5 + 4 + 0.5), and volume 3's back cover (0.5), totalling $0.5+5+0.5=6$ cm. Reading the shelf correctly rather than the volumes is the whole puzzle.

Let $f(n)$ count tilings of a 2 by $n$ strip. The leftmost column is covered either by one vertical domino, leaving $f(n-1)$, or by two horizontal ones stacked, leaving $f(n-2)$. So $f(n)=f(n-1)+f(n-2)$ with $f(1)=1$, $f(2)=2$: the Fibonacci numbers. $f(10)=89$. The recursion comes from asking only what happens at the edge, which is the standard way into almost every tiling count.

Two is impossible: any face touches four others, which between them touch each other. Three works — give each pair of opposite faces its own colour, and since opposite faces never share an edge, no conflict arises. So $3$. The graph of a cube's faces is the octahedron's vertex graph, and its chromatic number is 3.

Treat 1 to 99 as two-digit strings 00 to 99. Each of the digits 0 to 9 appears ten times in the units place and ten times in the tens place, so each place contributes $10\times45=450$, giving 900. The number 100 adds 1. Total 901.

Remainder 1 by each of 2 to 6 means $n-1$ is a multiple of $\operatorname{lcm}(2,3,4,5,6)=60$, so $n=60k+1$. Modulo 7, $60\equiv4$, so we need $4k+1\equiv0$, i.e. $k\equiv5\pmod7$. The smallest is $k=5$, $n=301$.

The minute hand gains $5.5^\circ$ a minute on the hour hand, so the angle between them sweeps through $360^\circ$ 22 times in 24 hours. Each sweep passes through $90^\circ$ once and $270^\circ$ once, giving 44. The guess of 48, two per hour, fails because the hands do not complete a relative lap every hour.

Each digit has 5 odd choices. One-digit: 5. Two-digit: 25. Three-digit: 125. Total 155. Leading zeros cannot sneak in, because 0 is even.

$\dfrac1{n(n+1)}=\dfrac1n-\dfrac1{n+1}$, so the sum telescopes to $1-\tfrac1{100}$.

With 1 to 5 points you get 1, 2, 4, 8, 16 regions, and 6 points break the pattern with 31. Count by Euler's formula, or directly: every interior crossing comes from choosing 4 of the points ($\binom64=15$), and each chord and each crossing adds a region, so the count is $1+\binom62+\binom64=1+15+15=31$. The puzzle is a standard warning against trusting a pattern from its first five terms.

The $k$-th cut can cross each of the previous $k-1$ cuts once, passing through $k$ existing pieces and splitting each in two. So it adds $k$ pieces, and $n$ cuts give $1+(1+2+\dots+n)=1+\tfrac{n(n+1)}2$. For $n=7$, 29.

Induct on the number $n$ of blue-eyed people. With $n=1$, that person sees no blue eyes, knows it must be them, and leaves on night 1. With $n=2$, each sees one blue-eyed person and waits; when nobody leaves on night 1, each concludes there must be a second, namely themselves, and both leave on night 2. In general all $n$ leave on night $n$. The visitor told everyone something they already knew, but made it common knowledge, and that is what starts the count.

Each glass ends with the volume it started with. Whatever wine is missing from the wine glass has been replaced by exactly that volume of water, and that missing wine is now in the water glass. So the two amounts are equal, however much you stirred.

Every square sits inside an axis-aligned bounding square of side $b$ (from 1 to 4). A bounding square of side $b$ contains exactly $b$ inscribed squares with corners on its edges, including itself. It can be placed in $(5-b)^2$ positions. So the total is $\sum_{b=1}^4 b(5-b)^2=16+18+12+4=50$. Counting only upright squares gives 30; the tilted ones are the 20 people miss.

$b=1+\tfrac b2$, so $\tfrac b2=1$ and $b=2$. The fast wrong answer, 1.5 kg, treats "half a brick" as half a kilogram.

Add four phantom coconuts. With $n+4$ in the pile, each night's step (remove one, take a fifth, keep four fifths) becomes exactly "take four fifths of the pile", because $n-1-\tfrac{n-1}5=\tfrac45(n+4)-4$. After five nights the pile holds $\left(\tfrac45\right)^5(n+4)-4$, and the morning division needs this to be a multiple of 5. So $n+4$ must be a multiple of $5^5=3125$. The smallest is $n=3121$, and checking forwards confirms it: 3121, 2496, 1996, 1596, 1276, 1020 in the morning, which is $5\times204$.

A position is lost for the player to move exactly when the binary XOR of the piles is 0. Here $3\oplus4\oplus5=2$, so the first player can win by moving to XOR 0. Changing pile $x$ to $x\oplus2$ requires $x\oplus2<x$: that holds for 3 (giving 1) but not for 4 (giving 6) or 5 (giving 7). So the only winning move is to reduce the 3-pile to 1.

A three-digit palindrome $aba$ is fixed by its first digit (1 to 9) and middle digit (0 to 9): $9\times10=90$.

Take any 10 coins and flip every one of them. If that group held $h$ heads, it now holds $10-h$. The other 90 coins hold the remaining $10-h$ heads. The two groups match whatever $h$ was.

With $d$ drops and $e$ eggs you can cover $\sum_{k=1}^{e}\binom dk$ floors, since each drop either breaks an egg (one fewer egg, one fewer drop) or not (one fewer drop). With three eggs: $d=8$ covers $8+28+56=92$ floors, too few; $d=9$ covers $9+36+84=129$. So 9. With two eggs the same formula gives the familiar 14.

The corners are 7 files and 7 ranks apart, 14 in total, and a knight move covers at most 3 of that, so at least 5 moves. But every knight move changes the colour of the square, and a1 and h8 are the same colour, so the number of moves is even: at least 6. Six is enough: a1–b3–c5–d7–f8–g6–h8.

Inclusion and exclusion with arithmetic series: multiples of 3 below 1000 sum to $3\cdot\tfrac{333\cdot334}2=166{,}833$, multiples of 5 to $5\cdot\tfrac{199\cdot200}2=99{,}500$, and multiples of 15, counted twice, to $15\cdot\tfrac{66\cdot67}2=33{,}165$. Total $233{,}168$.

Fix the number of 50p coins. None: $20b+10c=100$ gives $b=0,\dots,5$, six ways. One: $20b+10c=50$ gives three ways. Two: one way. Total 10. Fixing the largest coin first keeps the count honest.

The $k$-th cut is a plane, and the earlier cuts divide that plane into at most as many regions as $k-1$ lines divide a plane: $1+\tfrac{(k-1)k}2$. Each region splits one piece in two. Adding up: $1+\sum_{k=1}^{5}\big(1+\tfrac{k(k-1)}2\big)=\binom50+\binom51+\binom52+\binom53=26$. Lines give the lazy-caterer numbers; planes give these, the cake numbers.

Use the recursion $J(n)=(J(n-1)+3)\bmod n$ with $J(1)=0$, counting positions from 0. Iterating up to $n=41$ gives position 30, which is person 31. This is the original Josephus setting; with a count of 2 there is a closed form, but with 3 the recursion is the honest method.

Unfold two adjacent faces into a 1 by 2 rectangle; the two corners become opposite corners of that rectangle, and the straight line between them has length $\sqrt{1^2+2^2}=\sqrt5$. Going along edges costs 3, and crossing a face diagonal then an edge costs $\sqrt2+1\approx2.41$.

Cut the third link. That leaves pieces of 1, 2 and 4 links, and every total from 1 to 7 is a sum of some of them. Day 1 pay the 1; day 2 pay the 2 and take back the 1; day 3 add the 1; day 4 pay the 4 and take back the rest; and so on. Powers of two are the trick.

Split the board into sixteen 2 by 2 blocks. Any two squares in one block are adjacent, so each block holds at most one king: at most 16. Putting a king on the bottom-left square of every block achieves it.

All 32 squares of one colour work, because a knight always attacks the other colour. No more is possible: the board splits into 32 pairs of squares a knight's move apart (each 2 by 4 strip splits into four such pairs), and each pair can hold at most one knight.

Seven along the rank, seven along the file and seven along the long diagonal: 21. From a central square she attacks 27, the most possible.