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Just wanted to hear from different monkeys on examples of brain teasers you faced on interviews...

1111 x 1111

"What we can, we must; and because we can, we must"

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1111 x 1111 = 1111000 + 111100 + 11110 + 1111 = 1234321

Notice the pattern:
11 x 11 = 121
111 x 111 = 12321
1111 x 1111 = 1234321
11111 x 11111 = 123454321
...

rominet:

1111 x 1111 = 1111000 + 111100 + 11110 + 1111 = 1234321

Notice the pattern:
11 x 11 = 121
111 x 111 = 12321
1111 x 1111 = 1234321
11111 x 11111 = 123454321
...

Nice try, but there is no way you would be able to come up with a patern like that in your head while stressed out interviewing, plus adding up 1111000 + 111100 + 11110 + 1111 in your head is pretty challenging.

The way I did it in my head was like I would do it on a piece of paper and it took me about 5 seconds.

1111

x 1111

----- 1111
--- 1111
- 1111
1111

and just add the numbers up vertically to get 1234321

They never allowed me to use a pen or paper on a brainteaser.
I got 47*38 on my interview. Got the answer in 15 seconds.

Walk into an interview and the first thing that happens is that one of the partners throws me a quarter. He says "if you stack quarters one on top of another, how many will you need to reach the top of the Sears (Willis now) Tower?" Follow up to that, "if you take all those quarters to cover the area of this paper, can you reach the floor of this room to the ceiling?" Followed by how many fish are in the world?

• 1

^ what he said.

You have 50 white balls, 50 black balls, and two buckets. How should you allocate the balls (you have to use them all) so that you have the best probability of pulling a white ball out of a random bucket?

pantherdb26:

You have 50 white balls, 50 black balls, and two buckets. How should you allocate the balls (you have to use them all) so that you have the best probability of pulling a white ball out of a random bucket?

I'd put 25 black balls in the bottom of each barrel and 25 white balls on top of the black ones. Maybe not what you were looking for, but it would guarantee you a 100% chance of grabbing a white ball, no matter the barrel you picked from (assuming that you didn't dig into the bottom and that the circumference of the barrel isn't too large).

pantherdb26:

You have 50 white balls, 50 black balls, and two buckets. How should you allocate the balls (you have to use them all) so that you have the best probability of pulling a white ball out of a random bucket?

Put 1 ball in the first bucket, then the rest of the balls in another. Your probability for picking a white ball will in this way be maximized :)

• 1

This last one is a trick question isn't it? if you are picking out of a "random bucket" then you might as well be picking from a huge bucket with all the balls in it. no matter what you do your chances will be 50/50

right?

wrong. you're assuming that you have to split up all the balls uniformly.

Will throw some poo for silver. Just send me a PM.

1. Do this by induction. Suppose the chessboard is 1x1. Then there is clearly 1 square in total.
Now, 2x2. here you have 1 square that has dimension 2x2, and 4 squares that have dimension 1x1. Making 1+4 = 5 in total.
Now 3x3. you have 1square that is 3x3, 4 that are 2x2 and 9 that are 1x1. Making 1+4+9 = 14.

Noticing a pattern?

If you have a nxn chessboard. You will have 1 nxn square, 4 (n-1)x(n-1), 9 (n-2)x(n-2) .....

Or more concisely, for i = 1 to n, you have i^2 squares of size (n+1-i).

So to find the total number of squares in an nxn chessboard, just add the first n squares.

In this case, it's 1^2 + 2^2 ........+ 8^2.

1. The question is too vague. What operations are you allowed?

2) 1.3^2-.01

1. Ask if you can draw it on a piece of paper with 8 square by 8 squares since it makes it a lot easier to visualize.

start with the smallest square combo so you instantly have 64. I started with 8 after, but you can easily do it with 1,2 and the pattern should be just as easy to recognize. 8 block squares = 1 (the entire board), 7 block squares = 4, 6 block squares = 9...at this point you need to tell the interview that using the pattern you would assume that 5 block squares = 16, 4= 25, 3=36, 2=49, 1 =64 (which I already did). Add them up and you get 204 squares.

1. I am not really sure of all the rules but would 1 +2/3 +.02 be a possibility---using all digits and uses 2 twice.

^That's what I was thinking too, but technically it's 1.3^2 - 0.01, which doesn't satisfy the condition.

Sorry, I was referring to the post by streetlegend.

gekko, I also thought that yours might work, but it's not an exact answer, if we are told that we can round to the nearest hundredth, then it would be 1 +2/3 +0.01, which has the same problem I outlined earlier.

everythingsucks:

Sorry, I was referring to the post by streetlegend.

gekko, I also thought that yours might work, but it's not an exact answer, if we are told that we can round to the nearest hundredth, then it would be 1 +2/3 +0.01, which has the same problem I outlined earlier.

let me help you out right there, they really don't give a fuck what the answer is. The purpose of the question(s) is to see how your mind works...in that problem's case, do you have simple problem solving skills and can you make numbers work work to your advantage. As you can see there are multiple ways to get the same answer, which makes this probably the best brainteaser related to IBD I have ever seen,,,I mean that is what a banker does, takes numbers that normally equal 6 and makes them look like 7.65. Don't get hung up on the technicals of rounding too much. I am sure that those two above answers would have been acceptable to the interviewer, but you can ask him about rounding.

for questions two, all operation are allowed but not sure 0.02 (.02) count as one 0 or two 0s....

Anyone with a better answer for question 2?

lol, thanks gekko. Coming from a pure math background, I need a reality check every once in a while.

wkc207 as gekko pointed out, both answers are fine. Just state your assumptions first (rounding to nearest hundredth or by .02 you mean 0.02 etc.)

1 + 1 - .32 = 1.68

progolfer...forgot to add the zero..i'll take credit for the whole thing

cclow:

progolfer...forgot to add the zero..i'll take credit for the whole thing

You're supposed to subtract it.

Here is a more detailed answer to #1: http://puzzles.nigelcoldwell.co.uk/twentyseven.htm

1)

For a NxN Chessboard:
1. You get NxN 1-squres.
2. You get (N-1)x(N-1) 2-squres--going left to right you get (N-1) 2-squares, and likewise going down.
3. You get (N-2)x(N-2) 3-qaures--same logic; just draw a picture and check.
...
N. You get 1 N-square.

So it's simply 1+2^2+...+7^2+8^2. There's a formula for the sum of squares but I'm sure the interviewer will be satisfied enough with that expression.

put 1 white ball in one bucket and the rest in the other bucket.

Will throw some poo for silver. Just send me a PM.

PooSlinger:

put 1 white ball in one bucket and the rest in the other bucket.

Pretty sure that no matter how you divide them, you will get the same 50/50 split.

(1/50).5 + (49/50).5 = .5
(20/50).5 + (30/50).5= .5

That's why I just said put the 25 black balls on the bottom, then put the 25 white balls on top of the black ones. I could be wrong on this one, but it makes sense to me.

AverageGuy:
PooSlinger:

put 1 white ball in one bucket and the rest in the other bucket.

Pretty sure that no matter how you divide them, you will get the same 50/50 split.

(1/50).5 + (49/50).5 = .5
(20/50).5 + (30/50).5= .5

That's why I just said put the 25 black balls on the bottom, then put the 25 white balls on top of the black ones. I could be wrong on this one, but it makes sense to me.

Who said you have to put 50 balls in each bucket? I agree with 1 white in one bucket and the 99 others in the other.

LeBron.James:
AverageGuy:
PooSlinger:

put 1 white ball in one bucket and the rest in the other bucket.

Pretty sure that no matter how you divide them, you will get the same 50/50 split.

(1/50).5 + (49/50).5 = .5
(20/50).5 + (30/50).5= .5

That's why I just said put the 25 black balls on the bottom, then put the 25 white balls on top of the black ones. I could be wrong on this one, but it makes sense to me.

Who said you have to put 50 balls in each bucket? I agree with 1 white in one bucket and the 99 others in the other.

Wasn't paying attention and thought I read that they had to be split evenly. Yes, if you can put a different amount of the balls in the two barrels the answer is 1 white ball in one barrel and the other 49 white balls along with the 50 black balls in the other barrel.

PooSlinger:

put 1 white ball in one bucket and the rest in the other bucket.

I'm pretty sure this answer is correct, assuming the odds of picking any given bucket is 50% and not dependant on the number of balls in the bucket.

You have eight weights and a balance (scale that determines which side weighs more). Exactly one weight is heavier than the others. What is the minimum number of times you need to use the balance to determine which weight is heaviest?

search you lazy b*******

IlliniProgrammer:

You have eight weights and a balance (scale that determines which side weighs more). Exactly one weight is heavier than the others. What is the minimum number of times you need to use the balance to determine which weight is heaviest?

I was asked a variation of this brain teaser by an MD during a FT interview. He asked how one could find the heaviest weight using the scale only twice. Hadn't heard it before, but I worked through it and got the right answer. I ended up getting an offer, and I'm pretty sure that correctly answering this brain teaser made the difference.

For Illini Programmer's brainteaser:

1) 4 on each side, heavier side has heavier weight, choose those 4 weights
2) 2 on each side, heavier side has heavier weight, choose those 2
3) 1 on each side, heavier side is heavier weight

No - you only need two tries.

1) Weigh three on each side:

If all are equal you know the heavy ball is one of the remaining two,

2) weigh the pair to figure out which ball is heaviest

or

When weighing 3vs3 and one side is heavier take those three balls,

2) weigh 1vs1 to find the heaviest; if both are equal then the ball you left out is the heavier one

GOT BEAT TO IT!

IlliniProgrammer:
PooSlinger:

put 1 white ball in one bucket and the rest in the other bucket.

I'm pretty sure this answer is correct, assuming the odds of picking any given bucket is 50% and not dependant on the number of balls in the bucket.

You have eight weights and a balance (scale that determines which side weighs more). Exactly one weight is heavier than the others. What is the minimum number of times you need to use the balance to determine which weight is heaviest?

easy, its 2 times.

divide into 3 groups. weigh two groups, and you'll find out which group the heavy weight is in. eliminate all other groups. weigh 2 of the weights in the heavy group and voila.

damn i wish i got some some these questions...they sound pretty easy to me

.

wso sells one

For the question about the buckets and balls I'd put 49 white balls and 50 black balls in one bucket and one white ball in the other bucket. I could be wrong about that. I never got any probability questions in my interviews but got a few things like "how much does a 747 weigh? How many people voted in X City last year? How many boxing gloves were sold in X state last year?"

I don't really see it as sad. Working in IBD doesn't necessarily require the amount of intellect as someone working in software development at Google does.

2-4 are extremely doable. 1 and 5 are retarded.

I like 2-4. Lol if you can get #1 and wtf at #5

2 is high school science
3 is in every puzzle book and 4 is middle/high school math - anyone who cannot immediately get them should be dinged
1 is retarded
5 involves "lateral thinking" which most companies explicitly choose not to test because it is stupid to test for.

For the mental arithmetic ones you guys should look at vedic mathematics. My grandfather taught it to me growing up and it means I can pretty much do any simple mathematical problem in a few seconds. Just an example:

45^2 always ends in 25 and starts with 4 times one more than 4 so 4 x 5 = 20. Put them together and you get 2025

35^2 = 1225 (3x4)

65^2 = 4225

etc

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I got one during interview (successfull) in one of the Big4 firms.
The clock shows 3-15 time. What is the angle (exactly, without rounding, of course) between hour and minute hands.

Not a tough one but I liked it a lot (coincidence?).

Clock @ 3:15
2 glasses with orange & water mixed together
9 coins, find the odd one out with a set of scales

1.5 chickens, 1.5 eggs, 1.5 days. how many eggs does one chicken lay in one day.
3:15 degree one

I had 4 for my technical interview.

Try reading "Heard on the Street"

Had on average 5/superday for BB S&T SA.

S&T, hedge funds, and AM I have received brainteasers. Either Superdays or 1st rounds. They just want to see how you think and communicate your thought process.

Is it better to be up front and say you don't know if a brainteaser really baffles you, or better to try even if you get it totally wrong?

"Do whatever it takes to keep the legend of Wall Street as it was truly intended live on. When you think back on investment banking of the early 21st century, remember the heat--remember the passion. But mostly, remember the titans. " - LSO

Soul_Reaper:

Is it better to be up front and say you don't know if a brainteaser really baffles you, or better to try even if you get it totally wrong?

If you don't know, but have any semi-intelligent thoughts, just tell your interviewer(s) that you aren't quite sure but this is how you're thinking about it...... They might give you a hint that will help you solve it.

You are given 13 balls and a scale. Of the 13 balls, 12 are identical and 1 weighs slightly more. How do you find the heavier ball using the scale only three times?

Pretend you have a bucket of water, a 5 liter container and a 3 liter container. How do you come up with exactly 4 liters of water?

During consulting interview, interviewer said I could pick one and usually people don't get it right. Told him I knew both, and explained them in like a minute each. Pretty easy in my opinion.

7,548 questions across 469 investment banks. The WSO Investment Banking Interview Prep Course has everything you'll ever need to start your career on Wall Street. Technical, Behavioral and Networking Courses + 2 Bonus Modules. Learn more.

hardest brainteaser i got was find the square root of sum of all prime numbers from 1-1000 (for boutique IBD). tricky because a lot of people thought 1 was prime and screwed up...

Anyone for this question??

"What we can, we must; and because we can, we must"

philliefan: is that a joke?

iluvhermione:

philliefan: is that a joke?

I'm sure it isn't. The correct answer (according to Excel) is 275.9112176.

Are you kidding me? Who the fuck gives this at an interview? I'd punch them in the face and walk out.

After solving it of course.

Just Do It

Trivial by inspection

I think this one may be too difficult for non-maths/cs people

Best Response

I have discovered a truly marvelous proof of this, but it's too long for this comment box.

joe_monkey:

I have discovered a truly marvelous proof of this, but it's too long for this comment box.

lol, well done.

true wasn't wile's proof like 200 pages or something, gl getting through that in an interview

HA ha ha ah ah......I'm sure NO ONE got asked this in an interview.

The proof is like 100 pages long, it's probably the most argued theorem in math, and way beyond even some PhD's .

+1 SB to Joe Monkey for quoting the person who first claimed to solve this interview brainteaser.

Yeah, this was a social experiment. Didn't expect this many people to be familiar with Fermat's last theorem/wiles' proof on WSO.

And yeah, you need really advanced graduate level math to even begin understanding the proof. You have to admit, with the mathematical ineptitude displayed in some of the brain teaser threads, it wasn't that farfetched to think most people here wouldn't recognize it on sight.

This logic puzzle, on the other hand, actually is very hard. It's not conceptually a brain teaser, per se, but requires really deep logical thought. You practically need to do everything based on symbolic logic.

"Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are da and ja, in some order. You do not know which word means which.
"
http://en.wikipedia.org/wiki/The_Hardest_Logic_Puz...

"Life all comes down to a few moments. This is one of them." - Bud Fox

You have an unlimited supply of quarters, dimes, nickels, and pennies. What is the greatest value you can reach such that the coins that you have chosen cannot be rearranged to equal exactly \$1.00?

al865149:

You have an unlimited supply of quarters, dimes, nickels, and pennies. What is the greatest value you can reach such that the coins that you have chosen cannot be rearranged to equal exactly \$1.00?

solution: imagine there is a bowl in front of you. take the bowl, your coins, and carry out the following steps:

[step 1] pick a denomination, d, that has not yet been selected. if there are no denominations left to choose from, stop. otherwise, go to step 2.

[step 2] determine the maximum number, n, of coins of denomination d that you can put in the bowl such that you cannot, even with the change in bowl from any previous steps, make change for a dollar. then, go to step 3.

[step 3] put n coins of denomination d in the bowl. then, go to step 1.

so, if i go dimes, quarters, nickels, then pennies, we get: 9 dimes + 1 quarter + 0 nickels + 4 pennies. no matter which denomination you start with, the max is \$1.19

Didn't wiles present at some conference and it took various hours to go through it all?

I don't think it's unreasonable to ask questions that are clearly too hard to solve in an interview. Depending on the question, the candidate can still make some progress, demonstrate their ability to think about really difficult math/problem solving, and show that they are not overconfident.

Off the top of my head this problem isn't a good one for an interview, but my point is just that being (nearly) unsolvably difficult doesn't rule out a problem as a good interview question.

"Follow up to that, "if you take all those quarters to cover the area of this paper, can you reach the floor of this room to the ceiling?""

I got asked something like this once. Guy asked me, if there was a stack of quarters the height of the building, what market i would make in the percentage of the office we'd need to hold them all.

Umm... 32 in 1225 or 2.6%?

That's a nasty interview question. I had a question like it in an algorithms class.

This professor gives a good overview of the reasoning behind it (adapt it for 50 cards): http://recursed.blogspot.com/2010/01/neat-problem-...
Was this Grandmaster Capital? It sounds like a question they would give, given the Clarium heritage and Patrick Wolff's genius level mind.

West Coast rainmaker:

That's a nasty interview question. I had a question like it in an algorithms class.

This professor gives a good overview of the reasoning behind it (adapt it for 50 cards): http://recursed.blogspot.com/2010/01/neat-problem-...
Was this Grandmaster Capital? It sounds like a question they would give, given the Clarium heritage and Patrick Wolff's genius level mind.

That awkward moment when you realize that professor, is your professor...
(Ian Gouldon not the other fellow)

edit

Robert Clayton Dean: What is happening?
Brill: I blew up the building.
Robert Clayton Dean: Why?
Brill: Because you made a phone call.

• 1

Get some sleep. I bet the PMs couldn't have gotten the answer themselves. I think they just wanted to see your thought process.

The correct answer is to grab a sharp pen and stab your interviewer in the jugular.

Assuming the cards are in a circle instead of a line so every card will have two other cards next to them. Their are four aces, which can be taken as a given, 8 cards will be next to the aces (which could include another ace, a king, or a blank card).

I would think of it as what are the odds of it not happening. You get 8 trials, where each trial incorporates one less card to pull from, increasing (slightly) the odds that each next card is a king next to the ace.

1- ((45/49)x(44/48)x(43/47)x(42/46)x(41/45)x(40/44)x(39/43)x(38/42)) = 52.2%

Is that what you got?

SirPoopsaLot:

Assuming the cards are in a circle instead of a line so every card will have two other cards next to them. Their are four aces, which can be taken as a given, 8 cards will be next to the aces (which could include another ace, a king, or a blank card).

I would think of it as what are the odds of it not happening. You get 8 trials, where each trial incorporates one less card to pull from, increasing (slightly) the odds that each next card is a king next to the ace.

1- ((45/49)x(44/48)x(43/47)x(42/46)x(41/45)x(40/44)x(39/43)x(38/42)) = 52.2%

Is that what you got?

Simulating it with 1,000,000,000 trials I got 50.1%

When I started it on paper I also modeled it as a circle. I don't think your method is correct, though, because the number of places where an ace can be next to a king depends on the particular distribution of aces; i.e., given four aces placed there may be anywhere from two to five places where a king may be placed to satisfy the requirement (in the circular model) or one to five (in the linear model).

This was my approach:

1) How many ways are there to place four aces in fifty locations? (50 nCr 4)
2) How many of those have 2, 3, 4, 5 "sweet spots"?
3) What is the probability of missing all of those "sweet spots" with your four kings, for each number of sweet spots?

It basically comes down to the weighted sum of the complements of missing all of the sweet spots.

Also note that in the circular arrangement there are slightly more ways to get the win.

Thanks for the solution and the question I asked to you guys was the simplified version of the question. In the real question my interviewer asked:

Q = interviewer
D = me

Q:Pick two cards. (King, Queen, Ace...)
D: Ace and Jack
Q: You like blackjack?
D: Yea, more of a poker player, but still fun from time to time.
Q: Well, you aren't playing poker or blackjack. You are playing the probability game and here it is:

1. What is the probability I will have one of your cards if I remove two of them?
This was fairly straightforward if you have some statistical background.
It's a straightforward combination so 1 - (44C2)/(52C2) = .287
-Before he told me whether it was right or wrong he drew two cards from the deck. They weren't a Jack or Ace if you were wondering. Next he told me my calculation was correct. (Obviously I had to explain why the combination and why I got the probability of it not happening.)

Following this he asked the most difficult question:

1. What is the probability that a Ace lays next to a Jack in this deck of cards? (show me deck of cards)

I then started talking about how all the other 42 cards are irrelevant. Hence the blank cards. I think the rest of my explanation sounded alright, but who knows. After this question he then switched it up about asking me about something on my resume.

Robert Clayton Dean: What is happening?
Brill: I blew up the building.
Robert Clayton Dean: Why?
Brill: Because you made a phone call.

• 1

I was wondering why the deck only had 50 cards...

SirPoopsaLot:

I was wondering why the deck only had 50 cards...

Jokers brah

bonks:
SirPoopsaLot:

I was wondering why the deck only had 50 cards...

Jokers brah

lol

febreeze:
bonks:
SirPoopsaLot:

I was wondering why the deck only had 50 cards...

Jokers brah

lol

Hahaha, I see this guy has never played cards before!

SirPoopsaLot:

I was wondering why the deck only had 50 cards...

To make sure the OP couldn't just recall the solution from memory

P (AK or KA) in slots 1 and 2 is 2(4/504/49) = 0.0130612245
probability for slots 2-3, 3-4, ... 49-50 is 49
0.0130612245 = 0.64

0.64

don't know if this is right, but i wouldve approached it using combinatorics.

So you have 8 kings and aces, and you can use PIE to try to count how many combos there are that let you win. You could then break into the event that you have at least 1 pair, then subtract event that you have at least 2, add event that you have at least 3, then subtract that you have at least 4. And then for each event multiply by the number of possible ways to arrange.

So for example the #combos that you have at least 1 pair is 4C14C1 * 49!. You choose a king and an ace and stick them together and view them as 1 card. Since you would then only have 49 "cards" (its 50 - 1 since you have 1 pair) to distribute, there are 49! ways to place everything. But this will double count when you have 2 pairs, you have to apply PIE and subtract out the #combos of at least 2 pairs which is 4C24C2*48!, then add back the 3 pairs, and then subtract the 4 pairs. Then divide that number by 50!.

so you get something like
(4C14C149! - 4C24C248! + 4C34C347! +4C44C446!)/50!
cancel stuff out and you have
16/50 - 36/(5049) + 16/(504948) -1/(50494847)~15/50 since the last 2 terms are really small

pig:

don't know if this is right, but i wouldve approached it using combinatorics.

So you have 8 kings and aces, and you can use PIE to try to count how many combos there are that let you win. You could then break into the event that you have at least 1 pair, then subtract event that you have at least 2, add event that you have at least 3, then subtract that you have at least 4. And then for each event multiply by the number of possible ways to arrange.

So for example the #combos that you have at least 1 pair is 4C14C1 * 49!. You choose a king and an ace and stick them together and view them as 1 card. Since you would then only have 49 "cards" (its 50 - 1 since you have 1 pair) to distribute, there are 49! ways to place everything. But this will double count when you have 2 pairs, you have to apply PIE and subtract out the #combos of at least 2 pairs which is 4C24C2*48!, then add back the 3 pairs, and then subtract the 4 pairs. Then divide that number by 50!.

so you get something like
(4C14C149! - 4C24C248! + 4C34C347! +4C44C446!)/50!
cancel stuff out and you have
16/50 - 36/(5049) + 16/(504948) -1/(50494847)~15/50 since the last 2 terms are really small

You win if you have at least one "pair" -- the two, three, four pair cases include the at least one pair case which satisfies the victory condition.

Daedalus:

You win if you have at least one "pair" -- the two, three, four pair cases include the at least one pair case which satisfies the victory condition.

You have to account for these because of Principle of Inclusion and Exclusion (PIE) since you will overcount them. For example, if you look at the case of Ace of Clubs and King of Clubs, it will count the case where Ace of Hearts and King of Hearts are paired up, but will count the same arrangement again when it looks at the case when Ace of Hearts and King of Hearts is paired up.

Agree with the above, the chance to win is rare.
=424140.....1/504948*.....=0.0000000000000003453=0

Man in Black

The cards are not laid out in a circular manner. I asked him this during the interview.

Robert Clayton Dean: What is happening?
Brill: I blew up the building.
Robert Clayton Dean: Why?
Brill: Because you made a phone call.

• 1

I'm not going to work this out, but for future reference you should always try to use some form of induction. Whether that's induction on the total number of cards or the number of kings/aces depends on what you feel more comfortable attempting first. Once you can solve it for some base cases, you generally start seeing patterns on how to generalize to an arbitrary set.

Given the wide range of answers here (with ample time and no pressure), it's safe to assume that they were more interested in your thought process and whether you looked compose rather than looking like you took a shit in your pants. If you got it right tho, then hat-tip to you sir.

Thanks for the responses. I received an email today saying I would hear back at the latest by Friday. If I get the internship I'm gonna ask my interviewer for the answer. ;)

Robert Clayton Dean: What is happening?
Brill: I blew up the building.
Robert Clayton Dean: Why?
Brill: Because you made a phone call.

• 1

wow some of these are pretty intense...chim chim, how did you respond to the fish question?

Question 2 is a perfect answer, great job. Will definitely show the interviewer (IF YOU GET THE QUESTION) that you can first do the rational thing and then think outside the box when the answer looks funky.

Question 1 do you really think you'll be able to do all that math in your head? And isn't the answer the 59th minute?

If the bacteria doubles every minute and takes one hour to fill the pond then:

1 hour = 1.0 (full pond)

59 min = 1.0 / 2.0 = 0.5 (Half full)

58 min = 0.5 / 2.0 = 0.25 (Quarter full)

so on...

Hopefully you realize though that it is just pure luck when it comes to brainteasers. Obviously you can read and try to memorize a bunch of them but the main thing you want to get from reading / looking at solutions to brainteasers is the critical thinking element. They all have a common way of approaching a problem and the more you do the more you'll learn to take different factors into consideration, etc...

Never thought of qn1 that way..

thanks prospectivemonkey!.

(0.55/25) + (0.55/25) = 0.2 ...

prospective monkey is right for Q1.

get heard on the street... all these are in there...

haha thanks monkeymark. i was using the ms calc. while i was writing this post.

a clarification for q1 is that the pond is exactly full by minute 60.