Investment Banking puzzles, solved step by step

Fill the 3-litre bottle, pour it into the 4, fill the 3 again and pour until the 4 is full: 2 litres are left in the 3-litre bottle. That is four moves. Because 3 and 4 share no common factor, fills and pours can build every whole number of litres, so any total from 1 to 7 is possible across the two bottles.

What does each move actually measure?

Think of paying for something with only Rs 3 and Rs 4 coins. You can pay Rs 2 by handing over two Rs 3 coins and taking a Rs 4 coin back as change. The bottles work the same way. Every fill adds a full bottle and every pour into a full bottle sets one bottle's worth aside, so the volumes you can make are sums and differences of 3s and 4s. Two litres is 3 plus 3 minus 4: you draw two small bottles from the tap, the big bottle swallows 4 of those 6 litres, and 2 are left.

Is there a second route, and which volumes are out of reach?

Start from the big bottle and it still works, only slower. Fill the 4, pour into the 3 to leave 1, empty the 3, move the 1 litre across, fill the 4 again and top up the 3, which takes 2 litres and leaves 2 in the big bottle. That is six moves, 4 plus 4 minus 3 minus 3. Saying both routes and choosing the shorter one shows the interviewer you searched the problem rather than remembered it. Write each state as a pair, small bottle first, so neither of you loses track.

For the second half of the question, use BezoutA result in number theory: whole-number combinations of two numbers produce exactly the multiples of their highest common factor, and nothing else.'s rule: two bottle sizes can measure exactly the multiples of their highest common factor. The highest common factor of 3 and 4 is 1, so every whole number of litres can be built, limited only by capacity: up to 4 in one bottle and 7 across both. Contrast a 4 and a 6: their common factor is 2, so 3 litres is impossible however long you pour.

7 races. Race five heats of five, then race the five heat winners: the winner of race 6 is the fastest overall. Now strike out every horse that has three horses known to be faster. Only five can still be second or third: the second and third from the winner's heat, the first two from the runner-up's heat and the winner of the third-placed heat. Race them; the top two are second and third overall.

What does each race actually tell you?

Without a stopwatch, a race tells you the order only among the horses in it. Think of five classrooms each running their own sprint: you know the fastest in each room, but nothing about how one room's second best compares with another room's winner. Every horse must race at least once, so five heats are unavoidable, and they give five separate rankings that a sixth race between the heat winners stitches together. Name the heats A to E in the order their winners finished race 6, so A1 is the fastest horse of all.

Which horses can you strike out after race 6?

Any horse with three horses known to be faster cannot finish in the top three. Heats D and E go entirely, because D1 and E1 already finished behind A1, B1 and C1, and everything in those heats is slower still. In heat C only C1 survives: C2 trails C1, B1 and A1. In heat B, B1 and B2 survive, but B3 trails B1, B2 and A1. In heat A, A2 and A3 survive and A4 trails A1, A2 and A3. That leaves exactly five: A2, A3, B1, B2 and C1.

Race 7 puts those five on the track, and its first two finishers are second and third overall. A1 sits out, because it is already known to be the fastest. Could six races ever be enough? No: the five heats alone cannot name the fastest horse, and the race that does name it leaves five horses still unranked for second and third. That is the reasoning to say out loud, because it shows the 7 is a minimum rather than just a method that works.

Light rope A at both ends and rope B at one end at the same moment. When A burns out, 30 minutes have passed: light B's other end, and B burns out 15 minutes later, at 45 minutes. Lighting a rope at both ends halves whatever burning time it has left, however unevenly it burns, so B's remaining 30 minutes take 15.

Why can you not just cut a rope into pieces?

Because uneven burning means length tells you nothing about time. Think of a candle with a thick base and a thin top: half its height might burn in ten minutes or in forty. The only thing you know for certain about each rope is its total burning time, 60 minutes, so every step must work with time, never with length. Cutting, folding or marking a rope uses length, which is why every answer built on them fails.

Why does lighting both ends halve the time?

Picture the rope as a row of short segments, each with its own burning time, adding up to 60 minutes. Two flames eat into the row from both ends at the same moment and meet somewhere. When they meet, each has burned for the same length of time, and between them they have consumed all 60 minutes of segments. Two flames sharing 60 minutes of burning finish in 30, wherever along the rope they happen to meet. The same holds for a rope with any amount of burning time left: lit at both ends, it lasts half that.

How does that build 45 minutes?

Light A at both ends and B at one end at time zero. A is gone at 30 minutes, which is your signal. B has burned for 30 minutes from one end, so exactly 30 minutes of burning time remain in it, wherever the flame has reached. Light B's other end at that moment and those 30 minutes burn in 15. B goes out at 30 plus 15, which is 45 minutes, and no step relied on length. Say the answer as three events with clock times, 0, 30 and 45, so the interviewer can follow each one.

22 times. The minute hand goes round 12 times in 12 hours and the hour hand once, so the minute hand gains 11 laps and passes the hour hand 11 times. The overlaps come every 12/11 hours, about 65 minutes 27 seconds apart, and the one you expect between 11 and 12 is the one at 12:00 itself. Two 12-hour cycles make 22, counting midnight once.

Why is it not once an hour?

Think of two runners on a track, one doing 12 laps while the other does 1. The faster runner passes the slower one only 11 times, because the slower runner's single lap is subtracted from the faster one's total. Overlaps count the laps the minute hand gains on the hour hand, and it gains 12 minus 1, which is 11, every 12 hours. The hour hand keeps creeping forward, so each catch-up takes a little longer than an hour.

When exactly do the hands meet?

The minute hand moves 6 degrees a minute and the hour hand half a degree, so the minute hand gains 5.5 degrees a minute. A full lap of 360 degrees takes 360 over 5.5, which is 65 5/11 minutes, so the meetings come at 12:00, about 1:05:27, about 2:10:55 and so on. The eleventh gap lands exactly back on 12:00, which is why there is no meeting between 11 and 12: the hands meet at about 10:54:33 and next at 12:00.

What do you say about the edges?

State how you count midnight. Counting from one midnight up to but not including the next gives 22; counting both midnights gives 23. That one sentence shows you saw the edge case rather than stumbling into it. The same lap method answers the follow-ups in seconds: the hands point in exactly opposite directions 22 times a day as well, and form a right angle 44 times, because a right angle happens twice per lap gained.

Two weighings. Put three bars on each side and leave two off. If the scale balances, the light bar is one of the two left off, and weighing them against each other finds it. If one side is lighter, the light bar is among those three: weigh one against one, and if they balance it is the third. A weighing has three outcomes, so two weighings separate up to 3 x 3 = 9 bars.

Why is halving the wrong instinct?

Think of guessing a number with questions that can be answered higher, lower or spot on. Each answer splits the possibilities three ways, not two. A balance scale gives three outcomes, left lighter, right lighter or level, so each weighing should split the suspects into three groups, not two. Halving, four against four, uses only two of those outcomes and needs 3 weighings for eight bars.

Why can it not be done in one weighing?

One weighing has only three possible results, and any of the eight bars could be the light one. Three outcomes cannot point to eight different answers, so one weighing is never enough, and two weighings, with 3 x 3 = 9 outcome paths, are the fewest that can cover eight bars. That counting argument is the proof the interviewer wants: it shows two is a floor, not just a method that happened to work.

What does the puzzle teach beyond the scale?

The same counting sits behind any search. The fewest questions you need is set by how many answers each question can give, not by how clever the questions are. With nine bars, two weighings still suffice; with ten, you need a third. Saying where the method breaks shows the interviewer you understand why it works, which is the point of asking it.

Number the bottles 1 to 1,000, write each number in 10-digit binary, and have tester k sip from every bottle whose kth digit is a 1. After 24 hours, write a 1 for each tester who is ill and a 0 for each who is not. That 10-digit binary number is the poisoned bottle. It works because 10 yes-or-no results can tell apart 2 to the 10th, 1,024 cases, which is more than 1,000.

Why does giving each tester a separate batch fail?

Picture 10 friends each tasting their own 100 bottles. One friend falls ill and you know the poison is among their 100, but not which one, and the day is over. When testers taste separate batches, each result answers one question about one batch, so 10 testers can single out at most 10 groups, not 1,000 bottles. The fix is to let testers overlap, so that every bottle is sipped by its own unique combination of testers.

How does the binary code name the bottle?

Give each tester a place value: tester 1 is worth 512, tester 2 is worth 256, and so on down to tester 10, worth 1. Bottle 357 equals 256 + 64 + 32 + 4 + 1, so testers 2, 4, 5, 8 and 10 sip from it. If exactly those five fall ill, adding their place values gives back 357, and no other bottle has the same set of tasters. Every bottle number is a different set of testers, so the pattern of illness points to one bottle only.

What is the general lesson?

Each tester is one yes-or-no question asked of every bottle at once. Ten yes-or-no answers can tell apart 1,024 possibilities, so the right design makes each answer carry one binary digit of the bottle's number. The same idea finds one faulty item among many with very few tests, and it is why 2,000 bottles need only one more tester. With two days, each tester has three outcomes, ill on day one, ill on day two, or never, and 7 testers would be enough.

42. The gaps between terms are 4, 6, 8 and 10, rising by 2 each time, so the next gap is 12 and 30 + 12 = 42. The rule underneath is that the nth term is n x (n + 1): 1 x 2, 2 x 3, 3 x 4, 4 x 5, 5 x 6, and next 6 x 7, which is 42.

What should you do first with any number series?

Imagine a taxi meter that charges a little more for each extra kilometre than it did for the one before. The fares look irregular, but the jumps between them tell the story. Write the differences between terms underneath the series before guessing, and if they are not constant, take differences again. Here the first differences are 4, 6, 8 and 10, and the second differences are a steady 2, which tells you the rule is a quadratic: something times something.

How do you check the answer, not just find it?

A second route that lands on the same number is your proof. Each term splits into two neighbouring whole numbers: 2 = 1 x 2, 6 = 2 x 3, 12 = 3 x 4, 20 = 4 x 5, 30 = 5 x 6. Two methods agreeing, differences and a formula, turn a guess into an answer. The formula also lets you jump ahead: the 10th term is 10 x 11 = 110 without writing out the terms in between.

Why does a bank put this in an online test?

Numerical and logical screens are timed tightly and filter large numbers of applicants before anyone reads a CV. The skill being tested is spotting structure fast, the same skill you use when a line in a model grows by a changing amount each year. A revenue line whose yearly increase itself rises by a fixed amount has exactly this shape, and seeing it saves you from projecting the last increase forward as if it were fixed.

A always tells the truth and B always lies. Suppose A is a liar. Then A's statement is false, so neither of them is a liar, which means A tells the truth: a contradiction. So A is truthful, the statement is true, at least one of the two is a liar, and since A is not, B is. B never needed to speak.

Why does assuming A is a liar fall apart?

Picture someone who says, with a straight face, "I am lying right now." If that is a lie, they are telling the truth, and if it is the truth, they are lying; the sentence eats itself. A's statement is a milder cousin. A liar can never truthfully announce that there is a liar among us, because the announcement would then be true. So test the liar case first: if A lies, the statement is false, which means neither person is a liar, which means A is truthful. The assumption contradicts itself and is thrown out.

How does B get classified without saying a word?

Once A is truthful, A's sentence becomes a fact about the pair: at least one of them lies. A has been cleared, so the only person left who can be the liar is B, and B's silence changes nothing because the information came from A. The question is built this way to see whether you notice that evidence about a person can come from someone else. Say the two steps in order, A first and then B, and name the contradiction out loud rather than skipping to the answer.

Why do interviewers like this shape of puzzle?

It rewards the method you need for any logic test: assume a case, chase it to a contradiction, and move on. The same discipline is how you check a model: assume the circular reference resolves, trace it, and if it cannot, the assumption was wrong. Candidates who fail it usually try to argue from B's silence, which carries no information, instead of from A's sentence, which carries all of it.

The 10 perfect squares: lockers 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100. Locker n is toggled once by every student whose number divides n, so it ends open if n has an odd number of divisors. Divisors come in pairs, d and n over d, so the count is even unless one divisor pairs with itself, which happens only when n is a perfect square. Those ten lockers are open; the other 90 are shut.

What decides whether a locker ends open or closed?

Think of a light switch flicked by a series of people walking past. Only the number of flicks matters: an odd count leaves the light on, an even count leaves it off. Locker n is flicked by student d exactly when d divides n, so its final state depends only on how many divisors n has. Locker 12 is touched by students 1, 2, 3, 4, 6 and 12, six flicks, and ends closed. Locker 16 is touched by 1, 2, 4, 8 and 16, five flicks, and ends open.

Why do only perfect squares have an odd number of divisors?

Divisors arrive in pairs. If d divides n then so does n over d: for 12 the pairs are 1 and 12, 2 and 6, 3 and 4. Pairs give an even count, and the only way to break a pair is for a divisor to be its own partner, d equal to n over d, which means n is d squared. So 16 has the pairs 1 and 16, 2 and 8, and then 4 standing alone. Every perfect square has exactly one such unpaired divisor, and no other number has any, which is why the open lockers are exactly the squares.

How should you present it in the room?

Do not simulate 100 students. Work one small locker aloud, say the pairing rule, name the exception, and list the ten squares: that is a ninety second answer that shows structure rather than stamina. Then offer the check: the number of open lockers is the whole-number part of the square root of 100, which is 10, and for 1,000 lockers it would be 31. Interviewers use this puzzle to see whether you look for the rule behind a process instead of running the process, which is the same instinct a model builder needs when a spreadsheet has a thousand rows and one formula.

17 minutes. The 1 and 2 cross (2), the 1 returns (1), the 5 and 10 cross together (10), the 2 returns (2), and the 1 and 2 cross again (2). The obvious plan, the fastest person escorting each of the others, takes 19. The saving comes from pairing the two slowest so the 5 minute walk overlaps the 10 minute one.

Why does the escort plan feel right but lose two minutes?

Picture sharing taxis to the airport. Putting the two people with the latest flights in the same cab costs nothing extra for either; putting each in a cab with someone in a hurry makes every ride as slow as its slowest passenger. A pair always moves at the slower pace, so the 5 minute walker is free if he travels with the 10, and costs a full five minutes if he travels with anyone else. The escort plan pays for the 10, the 5 and the 2 separately.

How do you know 17 cannot be beaten?

Count what every plan must pay. Five crossings are needed: three trips over and two back with the torch. The 10 must cross at least once, costing 10. If the 5 crosses separately from the 10, that is another 5 on top. Pairing them leaves two forward trips and two returns to do with the 1 and the 2, and the cheapest way to do that is 2 + 1 + 2 + 2. The trade is paying the 2 minute walker an extra return in order to save the 5 minute walker's crossing, and 2 is less than 5 minus 1.

The rule is worth stating because it is not always pairing. If the walkers took 1, 4, 5 and 10 minutes, twice 4 is 8, more than 1 plus 5, so escorting wins: 21 minutes against 23 for pairing. Showing the interviewer you know when your trick stops working is better than the trick.