Debt Capital Markets puzzles, solved step by step

Puzzles
100
Traced to a firm
16
Topics
13
Hard
30
Topic
All topicsLeverage, coverage and cash flow9Mental maths and numeracy8Estimation and market sizing7Logic and brainteasers8Cost of capital and valuation riddles7Bond pricing and yield7Compounding, PIK and fees6Issuance and refinancing arithmetic8Credit spreads and default probability8Duration and convexity8Capital structure and recovery8Probability and expected value10Yield curve and forward rates6
Level
AnyWarm upCoreHard
Source
AnyReported at a firmStandard
Showing 1–3 of 3 · filtered from 100Clear filters
  1. 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.

    Only perfect squares have an odd number of divisors, so only they stay open123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100Door 12: toggled once per divisor1 and 122 and 63 and 46 toggles, an even number: ends closedDoor 16: one divisor pairs with itself1 and 162 and 84 alone5 toggles, an odd number: ends openOpen at the end: 10 doors1, 4, 9, 16, 25, 36, 49, 64, 81, 100
    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
    toggles(n)=d(n)d(n) odd  ⟺  n=m2#{m2≤100}=⌊100⌋=10\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^2a perfect square
    floor of root 100how 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?
  2. 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.

    Each weighing has three outcomes, so split the nine into three groups of threeWeighing 1: ABC against DEFLeft side heavyforgery is in A, B or CThey balanceforgery is in G, H or IRight side heavyforgery is in D, E or FWeighing 2: G against HG heavyG is forgedBalanceI is forgedH heavyH is forgedThen: A against Bheavier is forged; if level, CThen: D against Eheavier is forged; if level, F3 outcomes x 3 outcomes = 9 cases: exactly enough to single out one of 9
    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
    certificates you can search=3w32=9\text{certificates you can search} = 3^{w} \qquad 3^{2} = 9
    wthe number of weighings allowed
    3outcomes 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?
  3. 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.

    Lighting both ends halves whatever burn time is left0 min15 min30 min45 min60 minRope Aboth endstwo flames meet: 30 mingoneRope Bone end, then bothone flame: 30 min used15 minA goes out: light B's other end45 minutes
    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
    tboth ends=tone end230+302=45t_{\text{both ends}} = \frac{t_{\text{one end}}}{2} \qquad 30 + \frac{30}{2} = 45
    t one endburn time with one flame
    t both endsburn 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.