Investment Banking puzzles, solved step by step

Puzzles: 100
Traced to a firm: 36
Topics: 12
Hard: 29

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002A 10 x 10 x 10 cube is built from small unit cubes and the outside surface is painted green. How many small cubes have green paint on them, and how many have none?Mental maths and countingCoreTD SecuritiesNew York · 2024
Try it first
Quick instinct: how many small cubes are never painted?
Show the worked solution
488 small cubes are painted and 512 are not. The unpainted ones form a hidden core with one layer peeled off every side, so the core is 8 x 8 x 8, which is 512. The whole cube is 1,000, so 1,000 minus 512 leaves 488 with at least one painted face.
Why count the cubes you cannot see?
Picture a mango with its skin peeled. Measuring the skin piece by piece is fiddly; measuring the fruit left inside and subtracting from the whole is easy. The unpainted cubes form one clean block, so counting them and subtracting is faster and harder to get wrong than adding up the painted surface. Peeling one layer from both sides of each dimension takes 10 down to 8.
Each face of the big cube shows 8 corner cubes with three painted faces, 96 edge cubes with two and 384 face-centre cubes with one. They add to 488 painted cubes, and the 8 x 8 x 8 core of 512 cubes is never painted.
How do you check 488 the other way?
Count the surface directly, taking care not to double count. There are 8 corners, 12 edges each carrying 8 cubes once the corners are removed, and 6 faces each carrying an 8 x 8 centre. That gives 8 plus 96 plus 384, which is 488. Two methods agreeing is exactly what the interviewer wants to hear.
The relationship
$\text{painted} = n^3 - (n-2)^3 = 1000 - 512 = 488$
n cubes along one edge, here 10
(n-2)^3 the hidden core after peeling one layer from each side
What it says in wordsThe painted cubes are everything minus the core you cannot see.
Where candidates lose it
The fast wrong answer is 600: six faces of 100. It counts every edge cube twice and every corner cube three times. Candidates who start from the surface usually end up correcting on the fly and lose the room.
Start from the inside. Say the core formula, give 512 and 488, then offer the corner, edge and face split as a check.
What the interviewer asks next
How many small cubes have exactly two painted faces?
What size cube has as many unpainted as painted small cubes, roughly?
Now paint only the top and bottom. How many cubes are painted?
Asked at TD Securities, Generalist, New York, 2024 (Wall Street Oasis): If I gave you a 10x10x10 cube and I painted the surface green, how many cubes are green?
015Without a calculator, what is 9 to the power of 6?Mental maths and countingCoreNomuraTokyo · 2025
Try it first
Before computing anything, which size is right?
Show the worked solution
531,441. Rewrite 9 to the 6th as the square of 9 cubed. 9 cubed is 729. Then square 729 by splitting it into 700 plus 29: 700 squared is 490,000, twice 700 times 29 is 40,600, and 29 squared is 841. They add to 531,441. As a check, it should be a little over half a million, because 0.9 to the 6th is about 0.53.
How do you keep every step in your head?
Think of carrying a heavy box up six floors: you take it in two flights of three with a rest between, not six lifts in a row. Rewriting a power as the square of a smaller power cuts the work to one easy cube and one square you can split. 9 squared is 81, and 81 times 9 is 729, so 9 cubed is 729. Then 9 to the 6th is 729 squared, and squaring a number near a round one is the most practised move in mental maths.
Nine cubed is 729, and squaring 729 as 700 plus 29 gives a 490,000 block, two strips of 20,300 and an 841 corner, which add to 531,441.
The relationship
$9^6 = (9^3)^2 = 729^2 = (700+29)^2 = 490{,}000 + 2(700)(29) + 29^2 = 531{,}441$
9^3 729, nine cubed
(700 + 29)^2 the square split into a round part and a remainder
2(700)(29) the two strips, 20,300 each and 40,600 together
What it says in wordsSquare a number by splitting it into a round part and a remainder; the cross term counts twice.
How do you check it before saying it?
Three quick checks. An even power of 9 ends in 1, because 81 does, and 531,441 ends in 1. Any power of 9 is divisible by 9, so its digits must sum to a multiple of 9: 5 + 3 + 1 + 4 + 4 + 1 is 18. And the size is right: 9 is 10% below 10, and six 10% trims leave about 53% of a million. A digit check and a size check together catch nearly every slip in the cross term, which is where most mental errors happen.
A second route, if you prefer it: 9 to the 6th is also 81 cubed. 81 times 81 is 6,561, and 6,561 times 81 is 6,561 times 80 plus 6,561, which is 524,880 plus 6,561, giving 531,441 again. Saying one route and confirming with the other is worth more than speed.
Where candidates lose it
The usual loss is multiplying 9 by itself six times in a row and dropping a digit somewhere in the five-digit middle steps. By the time you reach 59,049 times 9, the interviewer has watched you struggle for a minute.
The other loss is saying the number without a check. Give the answer, then the two-second test: it ends in 1 and its digits sum to 18.
What the interviewer asks next
What is 11 to the power of 4?
What are 99 squared and 999 squared?
Is 2 to the power of 20 bigger or smaller than a million, and by how much?
Asked at Nomura, Generalist, Tokyo, 2025 (Wall Street Oasis): I think one of the math questions was 9 to the power of 6
036You must staff a three-person deal team from eight analysts, but two of them refuse to work together. How many different teams are possible?Mental maths and countingCoreBulge bracket IBMiddle market IB
Try it first
Quick instinct: how many teams?
Show the worked solution
50 teams. Choosing any 3 of 8 analysts gives 8 x 7 x 6 / 6 = 56 teams. The forbidden ones contain both analysts who refuse to work together, and the third seat can go to any of the other six, so there are 6 forbidden teams. 56 minus 6 leaves 50. Count everything, then subtract the cases the rule forbids.
Why count the teams you do not want?
Think of seating guests at a dinner when two of them have fallen out. Listing every acceptable table is slow; counting all possible tables and removing those that seat the two together is quick. When a rule forbids a small set of cases, count everything and subtract the forbidden ones, because the forbidden set is usually the easy one to count. Here a forbidden team is the two feuding analysts plus a third person, and there are only 6 choices for the third.
Of the 56 possible three-person teams from eight analysts, only the 6 that contain both A and B break the rule, so 50 remain: 20 with neither of the pair and 30 with exactly one.
How do you get 56 without a formula sheet?
Fill the team one seat at a time: 8 choices, then 7, then 6, which is 336 ordered picks. Each team of three turns up in 3 x 2 x 1 = 6 different orders, so divide by 6 to get 56 distinct teams. That is the combinationA selection where order does not matter. The number of ways to choose k from n is written n choose k. count, 8 choose 3. Subtract the 6 forbidden teams and 50 remain.
The relationship
$\binom{8}{3} - \binom{6}{1} = 56 - 6 = 50$
8 choose 3 every team of three from eight analysts
6 choose 1 teams holding both feuding analysts: one choice of third member from the other six
What it says in wordsAll teams, less the teams that contain the forbidden pair.
How do you check 50 a second way?
Count the allowed teams directly, split by how many of the feuding pair they include. Teams with neither of the two are 6 choose 3, which is 20; teams with exactly one are 2 times 6 choose 2, which is 30; together 50. Two methods agreeing is the answer the interviewer remembers, and the second method is the one you will need when a follow-up adds a second rule.
Where candidates lose it
The common wrong answer is 20: candidates take both feuding analysts out of the pool and count teams of three from the remaining six. That throws away the 30 perfectly good teams containing one of them.
The other slip is subtracting the feuding pair from 56 as if it were one team. The pair plus a third person can be formed 6 ways, so 6 teams go, not 1.
What the interviewer asks next
One senior analyst must be on every team, and the feud still applies. How many teams now?
What if the team is four people instead of three?
Two separate pairs refuse to work together. How many three-person teams are possible?
068How many zeros are at the end of 100 factorial, written out in full?Mental maths and countingCoreBulge bracket IBMiddle market IB
Try it first
Before you count: how many trailing zeros?
Show the worked solution
24 trailing zeros. Each zero at the end is a factor of 10, which is a 2 paired with a 5. Among the numbers 1 to 100 there are 97 factors of 2 but far fewer 5s, so the 5s decide it. The 20 multiples of 5 contribute one 5 each, and the four multiples of 25 contribute a second, giving 20 + 4 = 24. There is no multiple of 125 below 100, so there is no third layer.
Why is the answer about fives and not about tens?
Think of making pairs of shoes from a pile of lefts and rights. If you have 97 left shoes and 24 right shoes, you can make 24 pairs; the extra lefts do not help. A trailing zero is a pair of one 2 and one 5 multiplied together, and in 1 to 100 the 2s vastly outnumber the 5s, so the number of 5s sets the number of zeros. Counting multiples of 10 misses the zeros built from a 5 in one number and a 2 in another, such as 5 x 2 = 10 or 15 x 4 = 60.
The 20 multiples of 5 from 5 to 100 each contribute one factor of 5, and the four multiples of 25 contribute a second one, so 100 factorial contains 24 fives, each paired with one of its 97 twos to make 24 trailing zeros.
Where do the extra four come from?
A number like 50 is 2 x 5 x 5: it carries two fives, not one. Every multiple of 25 hides a second five, and 25, 50, 75 and 100 are the four of them below 100, so the count is 20 from the multiples of 5 plus 4 from the multiples of 25. The next layer would be multiples of 125, which carry a third five, but 125 is beyond 100, so the count stops at 24. The pattern continues for larger factorials: divide by 5, then 25, then 125, and add the whole-number parts.
The relationship
$Z = \left\lfloor \tfrac{100}{5} \right\rfloor + \left\lfloor \tfrac{100}{25} \right\rfloor + \left\lfloor \tfrac{100}{125} \right\rfloor = 20 + 4 + 0 = 24$
Z the number of trailing zeros
floor the whole-number part of the division
5, 25, 125 each power of 5 adds a layer of fives
What it says in wordsAdd up the whole-number parts of 100 divided by 5, by 25 and by 125, because each layer counts one more five per number.
How do you check it without writing out 100 factorial?
Test the method on a factorial you can write. 10 factorial is 3,628,800, with two trailing zeros, and the rule gives 10 over 5 = 2, plus 10 over 25 = 0, which matches. Showing the rule on a small case you can verify is the fastest way to earn trust for the big case you cannot. Then say the general version: the number of trailing zeros of n factorial is the whole-number sum of n over 5, n over 25, n over 125 and so on, and for 1,000 factorial that is 200 + 40 + 8 + 1 = 249.
Where candidates lose it
The fast wrong answers are 10, from counting multiples of 10, and 20, from counting multiples of 5 once. The second is close enough that candidates say it with confidence and lose the point on the four multiples of 25.
The other loss is explaining it as a memorised formula. Say why fives are the scarce factor first; the formula then sounds like reasoning rather than recall.
What the interviewer asks next
How many trailing zeros does 1,000 factorial have?
How many factors of 2 are in 100 factorial, and why does that number not matter here?
What is the smallest n for which n factorial ends in exactly 30 zeros, and is every count reachable?
088Aptitude test style: you convert Rs 5 lakh to dollars at Rs 83.50 per dollar, and the bank first deducts a 1.5% fee on the rupee amount. How many dollars do you receive, to the nearest dollar?Mental maths and countingCoreBulge bracket IBMiddle market IB
Try it first
Pick the answer the test is looking for.
Show the worked solution
5,898 dollars. The fee is 1.5% of Rs 5,00,000, which is Rs 7,500, leaving Rs 4,92,500 to convert. Divided by 83.50 that is about 5,898.2. A quick route: 83.5 x 6,000 is 5,01,000, so Rs 5 lakh buys about 5,988 dollars before the fee, and taking 1.5% off 5,988 gives the same 5,898.
What is the question really testing?
These questions sit in the timed numerical section of an online assessment, and the arithmetic is easy. What they test is whether you apply each adjustment to the right base under time pressure. Think of a restaurant bill with a 10% service charge: it is 10% of the food, not of the food plus tax, and getting the base wrong gives a number that looks almost right. The test designers put the almost-right numbers in the options, so the base you apply the fee to decides whether you pick the correct one.
Deducting the Rs 7,500 fee from Rs 5,00,000 and converting the remaining Rs 4,92,500 at 83.50 gives 5,898 dollars, while forgetting the fee gives 5,988 and dividing by 1.015 gives 5,900, both of which a test will list as options.
Does it matter whether you take the fee before or after converting?
Not for a percentage fee. Taking 1.5% off the rupees and then converting, or converting and then taking 1.5% off the dollars, multiplies by the same two numbers in a different order: 5,988 x 0.985 is also 5,898. Order matters only for a flat fee or a fee charged on a different base; a percentage fee on the same amount commutes with the conversion. Knowing this lets you pick whichever order makes the mental maths easier.
The relationship
$\$ = \frac{5{,}00{,}000 \times (1 - 0.015)}{83.50} = \frac{4{,}92{,}500}{83.50} \approx 5{,}898$
5,00,000 the rupee amount handed over
0.015 the fee, 1.5% of the rupee amount
83.50 rupees per dollar
What it says in wordsDollars received equal the rupees left after the fee divided by the rupee price of one dollar.
Where does 5,900 come from? Dividing by 1.015 answers a different question: how much could you convert if the 1.5% were charged on top of the converted amount. That base is smaller than Rs 5 lakh, so the fee comes out as about Rs 7,389 instead of Rs 7,500. On a test with options two dollars apart, that is a wrong answer.
Where candidates lose it
The trap is speed. Candidates divide 5,00,000 by 83.5, see 5,988 in the options and move on, or remember the fee and divide by 1.015 because it feels like the same thing. Both are listed because both are common.
Read what the fee is charged on, write the base down, and check the order of magnitude: about six thousand dollars, a little under.
What the interviewer asks next
What if the bank also charged a flat Rs 500 per transaction?
What exchange rate are you effectively getting after the fee?
How would you convert the dollars back, and what do the two fees cost you in total?

n	cubes along one edge, here 10
(n-2)^3	the hidden core after peeling one layer from each side

9^3	729, nine cubed
(700 + 29)^2	the square split into a round part and a remainder
2(700)(29)	the two strips, 20,300 each and 40,600 together

8 choose 3	every team of three from eight analysts
6 choose 1	teams holding both feuding analysts: one choice of third member from the other six

Z	the number of trailing zeros
floor	the whole-number part of the division
5, 25, 125	each power of 5 adds a layer of fives

Try it first

Why count the cubes you cannot see?

How do you check 488 the other way?

Where candidates lose it

What the interviewer asks next

Try it first

How do you keep every step in your head?

How do you check it before saying it?

Where candidates lose it

What the interviewer asks next

Try it first

Why count the teams you do not want?

How do you get 56 without a formula sheet?

How do you check 50 a second way?

Where candidates lose it

What the interviewer asks next

Try it first

Why is the answer about fives and not about tens?

Where do the extra four come from?

How do you check it without writing out 100 factorial?

Where candidates lose it

What the interviewer asks next

Try it first

What is the question really testing?

Does it matter whether you take the fee before or after converting?

Where candidates lose it

What the interviewer asks next