Derivatives Foundation puzzles, solved step by step

Puzzles: 100
Traced to a firm: 66
Topics: 12
Hard: 29

Showing 1–4 of 4 · filtered from 100Clear filters

001How many golf balls fit inside a 100-storey office tower? Talk me through your thought process; I care more about the structure than the final number.Mental maths and estimationCoreTower Research CapitalNew York · 2013

Try it first

Before any arithmetic: which single assumption will move your answer the most, once the building's size is fixed?

Show the worked solution

About 12 billion, with an honest range of 10 to 14 billion. A floor plate of 4,000 square metres at 3.8 metres a floor gives 1.52 million cubic metres over 100 floors; take half as usable, 0.76 million. A golf ball is 4.27 centimetres across, about 41 cubic centimetres, so 24,500 would fit in a solid cubic metre. Spheres leave gaps: at a packing factor of 0.64 that is 15,700 per cubic metre, and 0.76 million times 15,700 is 11.9 billion.

Why build a chain instead of guessing a number?

If someone asks how much rice a storeroom holds, you do not guess a tonnage. You measure the room, decide how much of it is shelving, and know how much a sack holds. An estimation question is marked on whether each link in the chain is stated, sized and defensible, not on the final figure. Here the chain is floors, floor plate, floor height, usable fraction, ball volume, packing factor. Say the six links before you fill any of them in, so the interviewer can follow and correct a single link rather than the whole answer.

The chain runs from 100 floors, a 4,000 square metre plate and 3.8 metre floors to 1.52 million gross cubic metres and 0.76 million usable, then from a 40.8 cubic centimetre ball to 24,531 balls per solid cubic metre, and the packing factor turns that into 10.4, 11.9 or 13.8 billion balls, a spread of about a third from the packing assumption alone.

Where does the packing factor come from, and why does it matter so much?

Pour marbles into a jar and shake it: they settle at roughly 64% of the jar's volume filled, with the rest air. Stack them by hand in the tightest possible pattern and you reach about 74%. Tip them in gently without shaking and you can be as low as 56%. The floor plate and height are things you can look up or pace out, but the packing factor is a physical assumption you must own, and it moves the answer from 10.4 to 13.8 billion on its own. Say which one you are using and why: balls poured into a building are a shaken random pile, so 0.64 is the defensible middle.

The relationship

N = \frac{100 \times 4{,}000 \times 3.8 \times 0.5}{40.8 \times 10^{-6}} \times 0.64 \approx 1.19 \times 10^{10}

100 x 4,000 x 3.8	floors times floor plate times floor height, the gross volume in cubic metres
0.5	the usable fraction after the core, structure and services
40.8 x 10^-6	one golf ball in cubic metres
0.64	the packing factor for a shaken random pile of spheres

What it says in wordsUsable volume divided by the volume of one ball, scaled down for the gaps between balls.

What do you say when the interviewer pushes on the usable fraction?

Lifts, stairs, structural columns, ducts and ceiling voids take a large slice of a tower, and nobody knows it to the percent. Give a range and show which way it pushes the answer: 40% usable takes you to about 9.5 billion, 60% to about 14 billion, so the final figure is somewhere between 10 and 15 billion whatever you assume. A candidate who reports a single number to three figures has missed the point of the question. A candidate who says 12 billion, give or take a third, with the reasons, has answered it.

Where candidates lose it

The common failure is to fix on one number early, such as the number of balls in a room, and then multiply by a guessed room count without ever stating the building's volume. The interviewer cannot follow it, cannot correct it, and marks it as a guess.

The second loss is forgetting that spheres do not fill space. Dividing the volume by the ball's volume overstates the answer by about half. Saying the words packing factor, and a number for it, is what separates a trader's estimate from a schoolchild's.

What the interviewer asks next

Now the balls are tennis balls, 6.7 centimetres across. Roughly how does your answer change?
If I told you the real answer was 20 billion, which of your assumptions would you revisit first?
Make me a market on it: give me a bid and an offer in billions, and tell me how wide and why.

Asked at Tower Research Capital, Assistant Trader, New York, 2013 (Wall Street Oasis): How many golf balls fit in the empire state building? Explain thought process and detailed solution

013How many people work in a large bank's 45-storey London headquarters tower? Give me a number and the assumptions behind it.Mental maths and estimationCoreHSBCCentral · 2026

Try it first

Before building anything: which route gives an estimate the interviewer can check link by link?

Show the worked solution

About 8,400 people, with an honest range of roughly 5,558 to 12,994. Of 45 floors, take 40 as ordinary office floors after plant rooms, lobby and trading floors. A floor plate of 3,000 square metres with 70% usable gives 2,100 square metres of desk space per floor. At 10 square metres per person that is 210 people a floor, and 40 floors give 8,400. The widest assumption is the space per person, 8 to 12 square metres, which alone moves the answer by half.

Why build from the floor rather than from the bank?

If you wanted to know how many people a wedding hall holds, you would not guess from the size of the family; you would pace the hall and think about chairs per row. An estimate is only as good as the link the listener can check, and anyone who has worked on an office floor has a feel for how many desks it holds. So the chain is office floors, floor plate, usable share, square metres per person. Say the four links before any number, so the interviewer can argue with one of them rather than with the whole answer.

Forty office floors of 3,000 square metres at 70% usable and 10 square metres a person give 8,400 people, while taking every assumption at its low end gives 5,558 and at its high end 12,994, and the space per person is the link with the widest range, a ratio of 1.5 between its ends.

The relationship

N = \frac{\text{floors} \times \text{plate} \times \text{usable}}{\text{sq m per person}} = \frac{40 \times 3{,}000 \times 0.70}{10} = 8{,}400

floors	office floors out of 45, after plant, lobby and other uses
plate	gross floor area in square metres
usable	the share of a floor that holds desks rather than lifts, cores and meeting rooms
sq m per person	the desk plus a share of corridors and kitchens

What it says in wordsPeople equals total desk area divided by the area each person uses.

Which assumption should you spend your time on?

The one with the widest range. Floors run 38 to 42, a ratio of 1.11; the plate 2,700 to 3,300, a ratio of 1.22; the usable share 65% to 75%, 1.15; but space per person runs 8 to 12 square metres, a ratio of 1.5, so it moves the answer most. Check it against a floor you know: a trading floor packs people at 6 to 8 square metres, a floor of meeting rooms and offices spreads them at 15 or more. If the interviewer gives you one fact, ask for that one.

What would you add about occupancy?

That desks and people are different counts. With hot-desking, a floor of 210 desks might be home to 250 or 300 people who are not all in on the same day, so a question about who works in the building can give a larger answer than a question about who is in it. Say which one you are answering. The limitation to state: the usable share and the space per person are guesses from general experience, not measurements, and a real number would come from the building's floor plans and the badge-in data.

Where candidates lose it

The common failure is to answer with a bare number, often a round 10,000, and then be unable to defend any part of it. The number is not what is marked; the chain is.

The second loss is spending the time on floors and plate, which are tight, and waving at the space per person, which is loose. Put the effort where the range is.

What the interviewer asks next

Now estimate how many lifts the tower needs to get everyone in between 8 and 9 in the morning.
How many taxis operate in a city's central business district on a weekday morning? Build the chain.
If the building's badge data showed 6,000 entries a day, which assumption would you revisit first?

Asked at HSBC, Sales and Trading, Central, 2026 (Wall Street Oasis): How many employees in London hsbc building How many taxis are in HK central How many beds in the nyc hotel

028How many zeros are at the end of 1000! (1000 factorial)?Mental maths and estimationCoreJump TradingChicago · 2013
Try it first
Pick the count before you work it out.
Show the worked solution
249 trailing zeros. A zero at the end is a factor of 10, and a 10 is a 2 times a 5. Among 1 to 1,000 there are 994 factors of 2 but only 249 factors of 5, so the 5s decide. Multiples of 5 contribute 200, multiples of 25 add a second 5 each for 40 more, multiples of 125 add 8, and 625 adds 1: 200 + 40 + 8 + 1 = 249.
Why count 5s and not 10s?
Imagine packing gift boxes that each need one lid and one base, from a pile of 994 lids and 249 bases. You can make 249 boxes, and the extra lids are useless. A trailing zero is a box: it needs one factor of 2 and one factor of 5, and in the product 1 x 2 x 3 x ... x 1,000 the 5s run out long before the 2s. Counting multiples of 10 misses every zero that comes from pairing a 5 in one number with a 2 in another; 4 x 5 is 20, which ends in a zero though neither factor is a multiple of 10. So the question is simply: how many times does 5 divide into 1,000 factorial?
Multiples of 5 up to 1,000 contribute 200 factors of 5, multiples of 25 a further 40, multiples of 125 another 8 and 625 one more, which add to 249, while the 994 factors of 2 are never the constraint.
How do you count the 5s without missing the doubled ones?
Count in layers. Every multiple of 5 carries at least one 5: 1,000 divided by 5 is 200. Every multiple of 25 carries a second 5 that the first layer did not see: 1,000 over 25 is 40. Multiples of 125 carry a third, 8 of them, and 625 carries a fourth, once. Each layer counts only the extra 5 that the layer before it missed, which is why you add the plain quotients and never multiply. The sum is 249. The same method gives 24 zeros for 100 factorial, 20 plus 4, and the layering stops as soon as the power of 5 exceeds the number.
The relationship
$Z(n) = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \left\lfloor \frac{n}{625} \right\rfloor = 200 + 40 + 8 + 1 = 249$
Z(n) the number of trailing zeros of n factorial
floor of n over 5^k how many numbers up to n carry at least k factors of 5
What it says in wordsAdd the whole-number quotients of n by each power of 5 until the power is bigger than n.
What does the interviewer learn from how you say it?
The answer is a test of whether you decompose a question into its prime factors rather than reach for the obvious count. Say the pairing argument first, then the layered count, then check the shape: the answer is a little under n over 4, because 200 + 40 + 8 + 1 is a geometric series that sums towards 1,000 over 4. That check is useful on the follow-ups. For the number of factors of 2 the same layering gives 994; for 3s it gives 498, so in base 6, where a trailing zero needs a 2 and a 3, 1,000 factorial ends in 498 zeros. The limitation is worth a sentence: this counts zeros at the end, not zeros anywhere in the 2,568-digit number, which is a different and much harder question.
Where candidates lose it
The fast wrong answer is 100, one per multiple of 10, or 200, one per multiple of 5. The first misses every zero made by pairing a 5 with a 2 from a different number; the second misses the extra 5s inside 25, 125 and 625.
The other slip is counting 625 as four and 125 as three in the final sum, which double counts. Each layer adds only one new 5 per number, so the layers are added as plain quotients.
What the interviewer asks next
How many trailing zeros does 100! have?
How many trailing zeros does 1000! have when written in base 6? In base 12?
What is the smallest n for which n! ends in exactly 100 zeros, and is there an n that ends in exactly 5 zeros?
Asked at Jump Trading, Trading, Chicago, 2013 (Wall Street Oasis): How many 0's are in 1000! (factorial)?
079What are the last two digits of 4^3000?Mental maths and estimationCoreBelvedere TradingNew york · 2021
Try it first
What is the shape of the method, before any arithmetic?
Show the worked solution
76. Multiplying by 4 and keeping only the last two digits gives 04, 16, 64, 56, 24, 96, 84, 36, 44, 76, and then 76 x 4 = 304 returns to 04. The cycle has length 10 starting at 4^1, and 3000 is a multiple of 10, so 4^3000 ends like 4^10, in 76. Check: 76 x 76 = 5,776, so 76 reproduces itself under squaring, which is what a power of 4^10 must do.
Why do only the last two digits matter at each step?
When you work out what time it will be 3,000 hours from now, you do not count the hours; you note that the clock face has 24 positions and ask where 3,000 lands on it. Last two digits are a clock with 100 positions. The last two digits of a product are fixed by the last two digits of the factors alone, so the sequence of 4^n mod 100 can only visit 100 states and has to fall into a cycle. Walk it: 04, 16, 64, 56, 24, 96, 84, 36, 44, 76, and 76 x 4 = 304 brings you back to 04. Ten steps, then it repeats.
The last two digits of 4^n for n from 1 to 10 are 04, 16, 64, 56, 24, 96, 84, 36, 44 and 76, and 76 x 4 = 304 returns the ring to 04, so because 3000 is 300 full turns of ten, 4^3000 lands on the same box as 4^10, which is 76.
How do you reduce 3000 without miscounting the start of the cycle?
The cycle begins at 4^1 = 04, not at 4^0 = 01, because 01 is never revisited: once a power of 4 is a multiple of 4 it stays one, and 01 is not. So the positions n = 1, 11, 21 and so on share last digits 04, and the positions n = 10, 20, 30 and so on share 76. 3000 is a multiple of 10, so it sits in the same slot as 10, and 4^10 ends in 76. Candidates who start counting from n = 0 land one step off and say 44.
The relationship
$4^{3000} = (4^{10})^{300}, \qquad 4^{10} \equiv 76 \pmod{100}, \qquad 76^2 = 5776 \equiv 76$
mod 100 keep only the last two digits
76^2 = 5776 76 squared ends in 76, so every power of 76 ends in 76
What it says in wordsAny power of 4^10 ends in 76 because 76 reproduces itself whenever it is multiplied by itself.
A second route, worth one sentence: split 100 into 4 and 25. Any 4^n with n at least 1 is 0 mod 4, and 4^10 = 1,048,576 is 1 mod 25, so 4^3000 is 1 mod 25. The number under 100 that is 0 mod 4 and 1 mod 25 is 76. Two methods agreeing is the finish an interviewer wants. The limitation: the cycle trick is only this short because 4 is small; for a base like 7 the cycle mod 100 is 4 steps, for 3 it is 20, and you should check rather than assume.
Where candidates lose it
The fast loss is counting the cycle from the wrong end. People list ten residues, say the cycle is 10, then divide 3000 by 10 and read off the first entry, 04, or count from 4^0 and get 44. Decide whether your list starts at 4^1 and say so.
The slower loss is trying to use Euler's theorem with 4 and 100 not coprime, which it does not allow. Either walk the cycle or split 100 into 4 and 25.
What the interviewer asks next
What are the last two digits of 7^3000?
What is the last digit of 3^3000, and how long is that cycle?
Why does 76 reproduce itself under multiplication by any power of 4 above 4^9?
Asked at Belvedere Trading, Equities, New york, 2021 (Wall Street Oasis): Some basic number theory (4^3000 modulo 100), basic probability calculations, and combinatorics puzzles

Z(n)	the number of trailing zeros of n factorial
floor of n over 5^k	how many numbers up to n carry at least k factors of 5

mod 100	keep only the last two digits
76^2 = 5776	76 squared ends in 76, so every power of 76 ends in 76

Try it first

Why build a chain instead of guessing a number?

Where does the packing factor come from, and why does it matter so much?

What do you say when the interviewer pushes on the usable fraction?

Where candidates lose it

What the interviewer asks next

Try it first

Why build from the floor rather than from the bank?

Which assumption should you spend your time on?

What would you add about occupancy?

Where candidates lose it

What the interviewer asks next

Try it first

Why count 5s and not 10s?

How do you count the 5s without missing the doubled ones?

What does the interviewer learn from how you say it?

Where candidates lose it

What the interviewer asks next

Try it first

Why do only the last two digits matter at each step?

How do you reduce 3000 without miscounting the start of the cycle?

Where candidates lose it

What the interviewer asks next