Debt Capital Markets puzzles, solved step by step

Puzzles: 100
Traced to a firm: 16
Topics: 13
Hard: 30

Showing 1–8 of 8 · filtered from 100Clear filters

004What is the angle between the hour hand and the minute hand of a clock at 3:15?Logic and brainteasersWarm upSyndicate desks
Try it first
Answer inside ten seconds.
Show the worked solution
7.5 degrees. The minute hand at 15 minutes points exactly at the 3, which is 90 degrees from 12. The hour hand moves half a degree every minute, so at 3:15 it sits at 3 times 30 plus 15 times 0.5, which is 97.5 degrees. The gap is 97.5 minus 90, or 7.5 degrees.
Why is zero the wrong answer?
Zero comes from picturing a clock as two independent pointers that jump. The hour hand moves continuously: it covers 30 degrees between one number and the next over 60 minutes, so it moves half a degree every minute. At quarter past, it has done a quarter of its journey from 3 to 4. Think of a train between two stations: fifteen minutes into an hour long run, you are not still standing on the first platform.
At 3:15 the minute hand points exactly at the 3, 90 degrees from 12, while the hour hand has moved a quarter of the way to the 4, to 97.5 degrees, leaving a 7.5 degree gap between them.
What is the general method, so any time works?
Measure both hands from 12 in degrees. The minute hand sits at 6 degrees times the minutes; the hour hand sits at 30 degrees times the hours plus half a degree times the minutes. Subtract, and if the result is over 180, take 360 minus it to get the smaller angle. Saying the method before the number is what the interviewer is listening for, because the next question will be a harder time.
The relationship
$\theta = \left|\,30h + 0.5m - 6m\,\right| = \left|\,30h - 5.5m\,\right| = \left|\,90 - 82.5\,\right| = 7.5^\circ$
h the hour, here 3
m the minutes, here 15
5.5 how many degrees a minute the minute hand gains on the hour hand
What it says in wordsThe minute hand gains 5.5 degrees a minute on the hour hand, starting 30 degrees behind for every hour on the clock.
Why does a capital markets desk ask this?
It is a speed and care test, not a clock test. The question checks whether you notice that two things are moving when the obvious reading has only one moving. The same slip in a desk setting is quoting a yield as if the price had not moved since the morning, or pricing accrued interest as if the settlement date were today. After the answer, one sentence on the 5.5 degrees a minute rule shows you can generalise.
Where candidates lose it
Zero is the whole trap. It comes from answering the picture in your head rather than the mechanism, and it is said fast because the question sounds too easy to need thought.
The second loss is the follow up. Candidates who guessed 7.5 without the half degree a minute rule freeze on 9:45 or 2:20. Learn the one line formula and the times take seconds.
What the interviewer asks next
What is the angle at 9:45?
How many times a day do the hands overlap exactly, and why is it not 24?
At what exact time after 3:00 do the hands first overlap?
025There are 100 closed doors in a row. On pass 1 you open every door. On pass 2 you toggle every second door, on pass 3 every third door, and so on, until pass 100 toggles only door 100. How many doors end open, and which ones?Logic and brainteasersCoreSyndicate desks
Try it first
Quick instinct: how many doors are open at the end?
Show the worked solution
Ten doors end open: 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100. Door n is toggled once on each pass whose number divides n, so it is toggled as many times as n has divisors. Divisors come in pairs, 2 with 6 for 12, so most numbers have an even count and end closed. Only a perfect square has a divisor paired with itself, giving an odd count, so only the squares stay open.
What decides whether one door ends open?
Think of a light switch pressed by everyone who walks past: whether the light ends on depends only on whether an odd or even number of people pressed it. Door n is toggled on pass k exactly when k divides n, so the number of toggles equals the number of divisors of n, and the door ends open only if that number is odd. This changes a hundred passes into one question about numbers.
Why do only perfect squares have an odd number of divisors?
Divisors come in pairs that multiply to the number: for 12 they are 1 and 12, 2 and 6, 3 and 4, six in all. Every divisor has a partner, so the count is even, unless a divisor is its own partner, which happens only when the number is a perfect square. For 16, the pairs are 1 and 16, 2 and 8, and 4 on its own, five in all. Squares up to 100 run from 1 times 1 to 10 times 10, so ten doors.
Of the 100 doors only the ten perfect squares end open, because door 12 is toggled six times by its three divisor pairs and closes, while door 16 is toggled five times, since 4 pairs with itself, and stays open.
The relationship
$\text{toggles}(n) = d(n) \qquad d(n) \text{ odd} \iff n = m^2 \qquad \#\{m^2 \le 100\} = \lfloor \sqrt{100} \rfloor = 10$
d(n) the number of divisors of n
m^2 a perfect square
floor of root 100 how many squares fit below 100, which is 10
What it says in wordsToggles equal divisors; the count is odd only for perfect squares, and there are ten of those up to 100.
How do you show your working in the room?
Start by tracking one small door aloud, say door 6: passes 1, 2, 3 and 6, four toggles, closed. Moving from a single example to the rule about divisor pairs is what the interviewer is really scoring, more than the final count. Then generalise: with N doors, the open ones are the squares up to N, so the count is the whole part of the square root of N. For 1,000 doors, 31 stay open. The limit is simply that the puzzle is pure logic; its value on a desk is the habit of reducing a big process to one property.
Where candidates lose it
The common wrong answers are 50, guessing that alternate doors survive, and the primes, because primes feel special in divisor puzzles. Primes have exactly two divisors, so they end closed.
The other loss is simulating pass by pass and running out of time. Track one door, find the rule, then count.
What the interviewer asks next
With 1,000 doors, how many end open?
Which doors are toggled exactly twice, and what are they called?
If you only did passes 1 to 50, how many of the 100 doors would be open?
037You have a 3-litre and a 4-litre bottle and unlimited water. How do you measure exactly 2 litres and exactly 5 litres, and which whole-litre amounts up to 7 can you make?Logic and brainteasersWarm upNomuraNew York · 2026
Try it first
How many of the amounts 1 to 7 litres can you measure, counting water held across both bottles?
Show the worked solution
For 2 litres: fill the 3, pour it into the 4, fill the 3 again and top up the 4; exactly 2 litres stay in the 3-litre bottle. For 5: fill the 4, pour into the 3 to leave 1, empty the 3, move the 1 into it, then fill the 4, for 1 plus 4. Every whole amount from 1 to 7 is possible, because 3 and 4 differ by 1.
What moves are you actually allowed?
Three: fill a bottle to the top, empty it, or pour from one into the other until the first is empty or the second is full. The only amounts you can know for certain are full bottles and what is left after a pour stops at a full bottle, so every measurement is built from 3s and 4s. Think of it as making change with only Rs 3 and Rs 4 coins, where you are also allowed to hand coins back.
Four moves leave 2 litres in the 3-litre bottle after the 4-litre bottle is topped up, and five moves put 1 litre in the 3-litre bottle beside a full 4-litre bottle, which together hold exactly 5 litres.
Why can you reach every amount from 1 to 7?
Because 4 minus 3 is 1, and once you can make 1 you can make anything by adding bottles. The amounts you can measure are exactly the combinations of 3 and 4 that fit in the bottles, and since the two sizes share no common factor, every whole litre up to their total of 7 is reachable. One litre: fill the 4 and pour into the 3. Three and four: fill one bottle. Five: 1 plus a full 4. Six: 3 in each. Seven: both full.
The relationship
$\gcd(3, 4) = 1 \;\Rightarrow\; 3a + 4b \text{ can equal any whole number}$
gcd the greatest common divisor, the largest number dividing both sizes
a, b how many times you add or remove each bottle's volume, positive or negative
What it says in wordsWhen the bottle sizes share no factor, their combinations reach every whole number.
The same rule tells you when a puzzle has no answer. With a 4-litre and a 6-litre bottle, every amount you can make is even, so 5 litres is impossible, and you can say so without trying a single pour. Interviewers like that sentence more than the pouring itself, because it shows you found the structure rather than a lucky sequence.
Where candidates lose it
Candidates start pouring at random and lose track of the state, which in a phone interview is fatal because the interviewer cannot see your paper. Say each state as a pair, litres in the 3 then litres in the 4, after every move.
The second miss is solving 2 litres and freezing on 5, which cannot fit in either bottle. The question is asking for water held across both bottles, and saying that out loud is half the answer.
What the interviewer asks next
With a 5-litre and a 7-litre bottle, what is the fewest number of moves to measure 1 litre?
Can you measure 5 litres with a 4-litre and a 6-litre bottle? Prove it either way.
How does this relate to what bond sizes you can build from fixed lot sizes?
Asked at Nomura, Equity Capital Markets, New York, 2026 (Wall Street Oasis): How much water can you fill using 1 3liter and 1 4liter bottle using each other?
041K investors each send a sorted list of n orders by limit yield. You need one sorted order book. How many comparisons does a naive merge take against a min-heap merge, and why is the heap the right tool as K grows?Logic and brainteasersCoreCitadelNew York · 2026Citadel SecuritiesNew York · 2026
Try it first
Merging 64 lists of 1,000 orders: roughly how many comparisons does scanning every list's front order each time take, against a heap?
Show the worked solution
A naive scan takes about n K (K minus 1) comparisons; a min-heap takes about n K log2 K. For 64 investors with 1,000 orders each, that is about 4.03 million against 0.384 million, roughly 10 times fewer. The heap holds only each list's current best order, so finding the next order costs a few steps down one branch rather than a look at every list.
What is the naive way, and where does it waste effort?
Imagine 64 queues at a bank, each already in order of arrival, and you must call people one at a time in overall order. The naive clerk walks along all 64 queue fronts every time to find the earliest. Every time one order leaves the book, the naive merge re-compares all K front orders, even though only one of them changed. That is K minus 1 comparisons for each of n K orders: 64,000 orders times 63 is 4,032,000 comparisons.
The other naive route is to merge lists one at a time: merge list 1 and 2, then merge in list 3, and so on. Each merge re-reads everything merged so far, which costs about n times K squared over 2, here about 2.08 million. Better than scanning, but it still grows with the square of K.
A min-heap keeps each investor's best remaining order, with the lowest yield at the top, so each step costs about log2 K comparisons; merging 64 lists of 1,000 orders then takes about 0.384 million comparisons against 4.03 million for scanning every front order.
Why does a heap fix it?
A min-heapA tree in which every parent is smaller than its children, so the smallest item is always at the top and can be removed and replaced in a number of steps equal to the tree height. keeps the K front orders only partly sorted: the best is always at the top, and the rest are arranged so that fixing the tree after a change touches one path from top to bottom. Taking the best order and inserting that investor's next one costs about log2 K comparisons instead of K, which is 6 instead of 63 at K of 64. Total work becomes n K log2 K, about 384,000 comparisons.
The relationship
$\text{scan: } nK(K-1) \approx 4.03\text{m} \qquad \text{heap: } nK\log_2 K = 64{,}000 \times 6 = 384{,}000$
n orders per investor list, 1,000
K number of investor lists, 64
\log_2 K height of the heap, 6 for 64 lists
What it says in wordsBoth methods output every order once; the heap makes each output cost the height of a small tree instead of a scan of every list.
Say where the heap does not matter. With four or five lists, scanning is about as fast and simpler to code, and the orders arrive as fast as a person can read them anyway. The heap earns its place when K is large or the lists do not fit in memory, which is the version in the reported question: arrays read from disk, where only the front of each list is held at once. A careful heap counts about two comparisons per level on the way down, so treat log2 K as the order of the cost, not an exact count.
Where candidates lose it
The common miss is proposing to concatenate all the lists and sort them. It works, but costs about n K log2 of n K and throws away the fact that each list is already sorted, which is the whole hint in the question.
The second loss is naming a heap without saying what sits in it. Say clearly: one entry per list, the current front order, plus which list it came from so you know where to fetch the next one.
What the interviewer asks next
What else does each heap entry need to store besides the yield?
How would you merge the lists if they were too large to fit in memory at once?
Two orders have the same yield. How do you keep allocation fair in the merged book?
Asked at Citadel, Equity Capital Markets, New York, 2026 (Wall Street Oasis): I was asked to implement K-way merge of K sorted arrays
Asked at Citadel Securities, Equity Capital Markets, New York, 2026 (Wall Street Oasis): and the cadidate was expected to use a min heap
053Nine bond certificates look identical, but one is a forgery printed on slightly heavier paper. With a balance scale and only two weighings, how do you find the forgery?Logic and brainteasersCoreSyndicate desks
Try it first
What should the first weighing be?
Show the worked solution
Weigh three certificates against three, then one against one inside the suspect group. If the first weighing tips, the forgery is on the heavy side; if it balances, it is among the three set aside. Take that group of three and weigh one against another: the heavier one is the forgery, and if they balance, the third is. Each weighing has three outcomes, so two weighings separate nine cases.
Why split into thirds rather than halves?
Think of a quiz where each answer can be yes, no or maybe, instead of just yes or no. Every question now splits the possibilities three ways, so you get to the answer in fewer questions. A balance scale is a three-answer question: left heavy, right heavy or level, and a good weighing uses all three answers. Splitting in halves throws the level answer away, which is why it needs more weighings.
The first weighing of three against three sends each of its three outcomes to a group of three suspects, and the second weighing of one against one inside that group sends each outcome to a single certificate, so two weighings cover all nine cases.
How do you prove two weighings is the minimum, and the limit?
Count the outcomes. One weighing has 3 outcomes and two weighings have 3 x 3 = 9, so two weighings can pick out at most 9 certificates, and nine is exactly what you have. One weighing cannot do it, because 3 outcomes cannot separate 9 suspects. The same count tells you the scale for any number: three weighings handle up to 27, four handle up to 81.
The relationship
$\text{certificates you can search} = 3^{w} \qquad 3^{2} = 9$
w the number of weighings allowed
3 outcomes per weighing: left heavy, right heavy, level
What it says in wordsEach weighing multiplies the cases you can tell apart by three.
Now say why a DCM interviewer asks it. The puzzle rewards the habit of asking how much information each step gives before choosing the step. Due diligence on a bond issue works the same way: the best question to ask a management team is the one whose possible answers split the risks most evenly, not the one whose answer you already expect. The analogy is loose, so keep it to one sentence.
Where candidates lose it
Most candidates start with four against four because halving feels natural. If the scale tips, four suspects remain, and one weighing cannot finish the job, so they need three. The interviewer is waiting to see whether you notice the level outcome is information too.
The second loss is solving it but not proving two is the minimum. Have the counting argument ready: 3 outcomes per weighing, 3 x 3 = 9, and one weighing gives only 3.
What the interviewer asks next
You now have 12 certificates and do not know whether the forgery is heavier or lighter. How many weighings do you need?
What is the largest number of certificates you could search with three weighings?
If two of the nine are forged, can you still find them in two weighings?
066Five creditor classes, ranked by seniority, must agree how to split 100 units of recovery. The most senior class still at the table proposes a split. If at least half of the classes at the table vote for it, it passes; otherwise the proposer leaves with nothing and the next most senior class proposes. Every class wants the most units and votes no when indifferent. What should the most senior class propose?Logic and brainteasersHardRestructuring
Try it first
How many units can the most senior class keep?
Show the worked solution
The senior class proposes 98 for itself, 0 for class 2, 1 for class 3, 0 for class 4 and 1 for class 5, and the plan passes. Work backwards. With two classes left, class 4 takes all 100, because its own vote is half. With three, class 3 buys class 5 with 1 unit. With four, class 2 buys class 4 with 1. With five, class 1 needs two more votes and buys classes 3 and 5, who would otherwise get nothing, with 1 unit each.
Why start from the end of the game?
When you must catch a train that leaves at a fixed time, you plan from the departure backwards to when you leave home. A bargaining game played in turns is solved the same way: the last stage has an obvious answer, and each earlier proposal only has to beat what each voter would get one stage later. The end here is two classes, 4 and 5. Class 4 proposes 100 for itself and votes yes, and one yes out of two is half, so the plan passes.
With two classes left class 4 takes 100; with three, class 3 keeps 99 and buys class 5; with four, class 2 keeps 99 and buys class 4; with five, class 1 keeps 98 and buys classes 3 and 5 with one unit each, because they are the classes that get nothing one stage later.
How does each proposer buy the cheapest votes?
With three classes left, class 3 needs two of three votes. Class 5 gets nothing if class 3 is removed, so one unit buys it: 99, 0, 1. At every stage the proposer buys the classes that would get nothing at the next stage, because one unit beats nothing. With four left, class 2 needs two of four votes and class 4 gets nothing in the three-class outcome, so the split is 99, 0, 1, 0. With five, class 1 needs three votes, its own plus classes 3 and 5, who get nothing in the four-class outcome.
Classes left Proposer Class 1 Class 2 Class 3 Class 4 Class 5 Votes needed
2 Class 4 out out out 100 0 1 of 2
3 Class 3 out out 99 0 1 2 of 3
4 Class 2 out 99 0 1 0 2 of 4
5 Class 1 98 0 1 0 1 3 of 5
Units of recovery at each stage. The proposer keeps everything it does not need to spend on votes, so the final split is 98, 0, 1, 0, 1 for classes 1 to 5.
What does this teach about a real restructuring?
Bargaining power comes from what each party gets if the deal fails, not from its size or its rank on paper. The senior class keeps 98 because it can see what every other class would get if its plan failed, and it pays only for the votes it needs. Real restructurings do not run like this: priority rules, court oversight and class voting thresholds all change the outcome, and junior classes often hold out for a share because a fight costs everyone. Treat the puzzle as a lesson in reasoning from the fallback, not as a model of a court process.
Where candidates lose it
Most candidates try to be fair, or try to buy class 2 because it looks most powerful. Class 2 is the most expensive vote on the table: it becomes the proposer if class 1 is removed and would keep 99.
The second loss is getting the three-class and four-class stages wrong by forgetting that a tie passes. State the voting rule at the start, at least half including the proposer, and check each stage against it.
What the interviewer asks next
What changes if a plan needs a strict majority rather than at least half?
How does the answer change with seven classes?
If indifferent classes vote yes, what does the senior class propose?
085You have two ropes. Each takes exactly 60 minutes to burn from one end to the other, but they burn unevenly, so half a rope does not take 30 minutes. With a lighter and nothing else, how do you measure exactly 45 minutes?Logic and brainteasersCoreSyndicate desks
Try it first
What is the one move that makes this possible?
Show the worked solution
Light rope A at both ends and rope B at one end at the same moment. Rope A burns out after exactly 30 minutes. At that instant, light the other end of rope B, which has 30 minutes of burn left. With two flames it finishes in 15 minutes, so rope B goes out at exactly 45 minutes.
Why does lighting both ends always give 30 minutes?
Two people painting a fence from opposite ends meet when the whole fence is painted, whatever the pace on each stretch; if the job is one hour for one painter, together they finish in half an hour. A rope's burn time behaves the same way: two flames consume the rope's 60 minutes of burn between them, so they must meet after 30 minutes, even though the meeting point is not the middle of the rope. Unevenness changes where they meet, never when.
Rope A, lit at both ends, burns out at 30 minutes; that is the signal to light rope B's other end, and the 30 minutes of burn left in rope B then finish in 15 minutes, at the 45 minute mark.
Why does rope B have exactly 30 minutes left?
Rope B has burned from one end for 30 minutes, so 30 of its 60 minutes of burn are gone and 30 remain, spread unevenly along whatever length is left. Lighting its other end applies the same halving trick to what remains: 30 minutes of burn with two flames takes 15. 30 plus 15 is 45. The trick works on any remaining burn time, not just a whole rope.
The relationship
$t_{\text{both ends}} = \frac{t_{\text{one end}}}{2} \qquad 30 + \frac{30}{2} = 45$
t one end burn time with one flame
t both ends burn time with two flames, always half
What it says in wordsTwo flames halve whatever burn time is left, so a half followed by a half of a half gives 45 minutes.
Say what the puzzle is testing, briefly: separating what you know (total burn time) from what you do not (where along the rope that time sits). A desk reading a bond's cash flows faces the same split: the total is fixed, the timing can be lumpy.
Where candidates lose it
The common wrong move is cutting a rope in half and assuming each half burns for 30 minutes. The question tells you the burn is uneven precisely to rule that out, and an interviewer will stop you there.
The second loss is lighting both ropes at both ends at once and then being stuck with two 30 minute timers. Keep one flame on rope B back until rope A tells you 30 minutes have passed.
What the interviewer asks next
How would you measure exactly 15 minutes with the same two ropes?
With only one rope, which times can you measure?
Using both ropes, list every time you can measure exactly.
098You have 1,000 bottles of wine and exactly one is poisoned. The poison shows its effect after 24 hours, and you have one day to find the bottle. What is the fewest testers you need, and how do you assign the bottles to them?Logic and brainteasersHardSyndicate desks
Try it first
How many testers do you need?
Show the worked solution
10 testers. Number the bottles 1 to 1,000 and write each number in binary with ten digits. Tester n drinks from every bottle whose n-th binary digit is 1. After 24 hours, the testers who fall ill spell out the poisoned bottle's number in binary. Ten testers give 1,024 possible patterns, enough for 1,000 bottles; nine give only 512.
Why is each tester worth a binary digit?
Think of the game twenty questions: every yes-or-no answer halves the possibilities, so twenty answers can pick out one thing in about a million. A tester is one yes-or-no answer: ill or fine. With one round of results, n testers can produce 2 to the power n different patterns, and you need at least as many patterns as bottles. 2^9 = 512 falls short of 1,000; 2^10 = 1,024 covers it. So ten is both enough and the minimum.
Each bottle number written in ten binary digits tells each tester whether to drink from it, and if bottle 613 is poisoned, testers 10, 7, 6, 3, 1 fall ill, whose place values add back to 613.
How do you read the answer back?
Give tester n the place value 2 to the power (n minus 1): tester 1 is worth 1, tester 2 worth 2, up to tester 10 worth 512. Add the place values of the testers who fall ill and you have the bottle number. If bottle 613 is poisoned, its binary form is 1001100101, so testers 10, 7, 6, 3, 1 fall ill, and 512 + 64 + 32 + 4 + 1 = 613. Numbering 1 to 1,000 fits in ten digits because 1,000 is below 1,024, and the pattern where nobody falls ill is left spare.
The relationship
$2^{9} = 512 < 1{,}000 \le 1{,}024 = 2^{10}$
2^n the number of distinct ill-or-fine patterns n testers can produce
What it says in wordsTen testers are the fewest whose outcomes can label every bottle.
The desk version of the lesson: design the test so that every outcome carries information. A tester who drinks from one bottle learns about one bottle; a tester who drinks from half of them splits the problem in two.
Where candidates lose it
The common answers are 1,000 testers, one bottle each, or some splitting scheme that needs several rounds. The 24 hour delay allows only one round, so the testers must be designed to answer in parallel.
The second loss is giving 10 without the proof that 9 is not enough. Say 2^9 is 512 in the same breath; it turns a remembered trick into an argument.
What the interviewer asks next
You have two days instead of one, and a tester who falls ill on day one is out. How many testers do you need?
Exactly two bottles are poisoned. Does the binary scheme still work?
How many testers would a million bottles need?

h	the hour, here 3
m	the minutes, here 15
5.5	how many degrees a minute the minute hand gains on the hour hand

d(n)	the number of divisors of n
m^2	a perfect square
floor of root 100	how many squares fit below 100, which is 10

gcd	the greatest common divisor, the largest number dividing both sizes
a, b	how many times you add or remove each bottle's volume, positive or negative

n	orders per investor list, 1,000
K	number of investor lists, 64
\log_2 K	height of the heap, 6 for 64 lists

Try it first

Why is zero the wrong answer?

What is the general method, so any time works?

Why does a capital markets desk ask this?

Where candidates lose it

What the interviewer asks next

Try it first

What decides whether one door ends open?

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

How do you show your working in the room?

Where candidates lose it

What the interviewer asks next

Try it first

What moves are you actually allowed?

Why can you reach every amount from 1 to 7?

Where candidates lose it

What the interviewer asks next

Try it first

What is the naive way, and where does it waste effort?

Why does a heap fix it?

Where candidates lose it

What the interviewer asks next

Try it first

Why split into thirds rather than halves?

How do you prove two weighings is the minimum, and the limit?

Where candidates lose it

What the interviewer asks next

Try it first

Why start from the end of the game?

How does each proposer buy the cheapest votes?

What does this teach about a real restructuring?

Where candidates lose it

What the interviewer asks next

Try it first

Why does lighting both ends always give 30 minutes?

Why does rope B have exactly 30 minutes left?

Where candidates lose it

What the interviewer asks next

Try it first

Why is each tester worth a binary digit?

How do you read the answer back?

Where candidates lose it

What the interviewer asks next