## Pages

- What is the largest possible number you can write using only 2 numbers - just 2 numbers, no other mathematical symbols?

2.If, having only one match, on a freezing winter day, you entered a room which contained a lamp, a kerosene heater, and a wood burning stove, which should you light first.

- What belongs to you, but is mostly used by others?
- How many birth days does the average man have?
- Two planes take off at the same exact moment. They are flying across the Atlantic. One leaves New York and is flying to Paris at 500 miles per hour. The other leaves Paris and is flying to New York at only 450 miles per hour ( because of a strong head wind ). Which one will be closer to Paris when they meet?
- A 10 foot rope ladder whose rungs are one foot wide is hanging over the side of a boat. At this time there is only 1 foot of ladder under the water. It now begins to rain heavily, if it rains at the rate of 1 foot every two hours, how many hours will it take to cover half of the ladder?

As simple as these may sound, a lot of people do not get it in investment banking interviews.

*Throwback Thursday, this was originally posted almost 9 years ago! 4/13/08 *

## Comments (914)

Bravo, sir.

<><><><><><><><><><><><><><>

Once more into the breach, dear friends.

Side-by-side comparison of top modeling training courses + exclusive discount through WSO here.

well done

that's impressive.

I don't get it though. Why does starting lower than 14 prevent you from getting to 100? Say if you start at 11, 22, 33, 44, 55...99

Salam -

11, 22, 33... wont beat 14 because if you make it to 99, you'll have made 9 drops and still have 9 drops left (90-98).

The "14 drops" case is the most efficient iterative test. You could also ask yourself "if I add the series 1+2+3+4+5...., on which number will I first cross 100?" The answer is 14.

Interestingly, a lot of people have just as much trouble answering the following question:

"Why do you need two balls?"

How long would you get to answer a question like the one aachimp solved?

A few minutes. The interviewer would want you to talk out your reasoning as you tried to solve it.

My friend (now an Econ PhD at Princeton) heard this one at DE Shaw as an undergrad:

"A bunny rabbit is standing at the bottom of a flight of N stairs. The bunny is capable of moving up either one step or two steps per 'hop'. In terms of N, what is the maximum number of hop permutations the bunny can take to get to the top? Here is a pad of paper and a pencil. There are no shortcuts -- you will use real math. Good luck."

Ouch

idea is backward recursion,

V(k)- # of permutations with k stairs to go.

V(0)=0, V(1)=1, V(2)=2, for any k>2, conditioning on the first hop:

V(k)=1+V(k-1)+1+V(k-2)=2+V(k-1)+V(k-2); hence

V(N)=2+V(N-1)+V(N-2),

the actual # as a functioon of N is a bit messy , get back after sleep...

"A bunny rabbit is standing at the bottom of a flight of N stairs. The bunny is capable of moving up either one step or two steps per 'hop'. In terms of N, what is the maximum number of hop permutations the bunny can take to get to the top? Here is a pad of paper and a pencil. There are no shortcuts -- you will use real math. Good luck."

Correction, hope that works:

idea is backward recursion,

V(k)- # of permutations with k stairs to go.

V(0)=0, V(1)=1, V(2)=2, for any k>2, conditioning on the first hop:

V(k)=V(k-1)+V(k-2); hence

V(N)=V(N-1)+V(N-2),

it is almost a Fibonacci sequence, but the starting conditions are different:

in Fibonacci it's: 0,1,1 ,2,3,5, 8 ....

here it's: 0,1,2,3,5,8,....

Hence, V(N) is (N+1)th element of the Fibonacci sequence, hence the closed form solution for that is the solution for the Fibonacci sequence:

V(N)=F(N+1)=[phi^(N+1)-(1-phi)^(N+1)]/sqrt(5);

where phi is the golden ratio.

Well, I don't think that remembering golden-ratio formula is required, but relating to Fibonacci sequence would be welcome.

Although, one crazy ...ck once asked me just that over the phone... I didn't remember it back then.

..t happens.

you don't have to remember the golden ratio, but you can derive it by solving the linear homogeneous recurrence relation. Total= f(n+1), but solving it in terms of fn, you get

x^n=x^(n-1)+x^(n-2), dividing by x^(n-2) you get

x^2-x-1=0

roots of x are (1+-sqrt(5))/2,

general form of equation is c1x1^n+c2x2^n, where x1, x2 are the roots

using initial conditions f0=0,f1=1

you have c1+c2=0

c1x1+c2x2=1

c1=1/sqrt(5), c2=-1/sqrt(5),

with f(n+1)

you get what you wrote down

alternatively you can write a combinatorial formula in terms of n: (n choose 0) +(n-1 choose 1)+...+(n-k+1 choose k),

you can see this intuitively by thinking of k either equalling the number of 1 step hops or 2 step hops

wow, these are good guys

I'll name a couple of mine:

Start of with 2 coins. One double-headed, one normal. You RANDOMLY pick one from a hat. Flip it 10 times and realize u get all 10 heads.

Find the probability that you HAD CHOSEN the double headed coin. (from my Wall st interview. I eventually got it, but I didn't end up getting a job because they said I was just a sophomore...oh well)

Another one:

if a CD is priced at $5 today. Tomorrow it is EITHER $6 or $1 (and nothing in between). Find the prob of it being $6. (no more info given, this was from another wall st bank)

However, just as a future reference

be sure to provide your answer and ur thought processes eventually...

1st: just conditional probability:

probability that you HAD CHOSEN the double headed coin.

P(Double|10Heads)=1

0.5/(10.5+0.5^10*0.5)=1/(1+0.5^10)=approx=1-0.5^10=approx=0.999.

2nd.

if a CD is priced at $5 today. Tomorrow it is EITHER $6 or $1 (and nothing in between). Find the prob of it being $6. (no more info given, this was from another wall st bank)

is it just 80% ?

$5=E[X]=p

$6+(1-p)$1->p=0.8.

"Prove mathematically that a call option on a stock does not depend on the stock's drift rate"

after having derived the BS PDE:

"Derive the solution to this for a call option"

gstyling what level and role were you applying for that they made you derive the pde? i'd think the order for those questions would be reverse...the black scholes pde does not include a drift term.

Grad level - quant structuring. The fact that the PDE that the process follows doesn't depend on mu implies that the solution doesn't contain the drift term (that was my argument, and he then asked me to solve it). I wasn't able to solve it but still got the job.

That's a poor argument. The reason the drift term cancels is obvious when you derive the PDE.

Another i was asked in the same interview (not very hard but has a nice solution):

You have three pizzas, sized L, M and S. P(L)=P(M)+P(S), where P(A) is the price of pizza sized A. With nothing but a knife, how can you determine whether it's better value buying 1 L or [1 M and 1 S].

In terms of quantity or quality?

If its only about size, then 1/2 P(L) = 1/2 P(M) + 1/2 P(S). Slice M and S into half and put them on top of L. If smaller, less value.

Or what is the point in this question?

Quantity - should have also specified that all pizzas are circle shaped, and have equal thickness (i.e. look at area).

Yeah, so if the two halves of M and S are each smaller than L (one half of each put on top of an unsliced L), then buy L. If any half is as big, buy M+S.

Let A(X) be the area of size X. The fact that A(L)>A(M)/2+A(S)/2 doesn't tell us anything.

you get the radius of each pizza (assuming you know where the middle is) and then try to form a right angle triangle from your 3 radii with the radius of L being the hypotenuse. If your "right-angle" (or where the right angle should be if L=S+M) is greater than 90, you know that L is bigger than S+M and if the "right angle" is less than 90 degrees you want to take S+M.

that is smart

I was thinkng in terms of squares, but i never thought about using the pythag for pizza radii.

good job

the coin question's answer was

1024/1025

Think of it as binomial trees and count up the combinations-hint hint

The CD pricing one is indeed 80% and the logic u used is correct. However, I still really believe that it was a weird question.

exactly , that what I wrote 1/(1+0.5^10)=1024/1025

not that weird of question...basically using spot prices to back out the market's expectation of future event.s

I dunno. Most people would overthink it and not realize that the same incremental price inc/drop would be the same likelyhood.

It does make sense, and is really simple math. However, I would prob get it wrong at first just because I would never think of it that way unless they tell me.

back to this question:

Problem:

You are standing outside a 100-story building holding two identical glass spheres. You are told that either sphere, if dropped from the roof, would shatter upon hitting the earth, but that it would not necessarily break if dropped from the 1st story. Your task is to identify the lowest possible floor that you can drop a ball from and break it. QUESTION: In the general case, what is the smallest number of drops required to guarantee that you have identified the lowest story?

Notes:

- Both balls have the same minimum breakage story.

- You only have two balls to use. If one breaks, it cannot be used for the rest of the experiment.

14 is correct for the case with 2 balls, but how about when you have 3 balls instead of 2?

You have an object of K weight where K is an integer. You have an unlimited number of weights of each integer value from 0 to K on one side of a room. You have a balance scale (you put the object of K weight on one side and weights on the other side until it balances out) on the other side of the room. What is the fewest number of weights you would need to bring over to the scale to be guaranteed that you can correctly identify the weight of the object?

1) Assume that K = 100 and give me your solution.

2) Now give me the general solution for any K.

Got this one in an MS S&T interview.

www.gottamentor.com

50, 25, 13, 7, 3, and 1?

double post

I think the general case for all k between 2^n (inclusive) and 2^n+1 (noninclusive) should be n weights, each weight representing each of 2^1, 2^2, ... 2^n. There is no need for 1 weights since all odd D's can be identified based on the inequality between two even weights.

Not the hardest but my favorite:

You're betting on the world series and your bookie will only take a $60,000 maximum bet. You want to bet $100,000 that the Yankees will win the world series. How do you split this betting over the 7 possible games?

What was the answer to the clock question?

I assume it is 252 degrees.

[360 x 47/60] - [ 360 x 1/12] = 252

Close, but remember that your hour hand is going to be moving as well, and should be approx 47/60th of the way to 2 o'clock at 1:47... so [360 x 1/12] should be ~ [360 x 1.783/12], right?

OHHHH ! Yeah, I just realized your point. You're right. That does make things harder though, it can be solved mentally through simplification.

{360 x 47/60 = 6 X 47 = 282... degree of minute hand relative to 12] - [(30 ... the degree of the hour hand at one) + (30 x 47/60=47/2=23.5 ... 47/60th of the way between one and two)]

282 - [ 30+23.5]

= 228.5

I suppose this could be done mentally but it would take 45s to a minute or so. How long would they give you for a question like that ?

at 1:47

= (18/60)

360 + (47/60)30 = 131.5 degrees^ That is coming from the opposite end that you had done.

For very small elite boutique:

Find the square root of 72! to the nearest integer (2 minutes)

let me guess, the elite boutique is Gridley and Co...

^^ not hard - i've had sht like that ...

square root of 0.5

inassessable -- not square root of 72, square root of 72! (factorial)

Spacejam, I would love to see an elegant way to estimate the 51-digit answer to the nearest integer in 2 minutes,,,

Is this a joke?

ha! abacus is for pussies. the cool kids just -pretend- they have an abacus and wave fingers in the air

http://www.youtube.com/watch?v=wIiDomlEjJw

Back when you guys took the SAT, you still had the analogy section; so here is one for you:

Is Gridley & Co. to Spacejam22 as Jefferies is to everyone else?

Like every good satarist, it is hard to tell if he is joking.

wow gonna try this, let me get my abacus

Here's one...

On a magic island, there are 99 lions and 1 sheep. Lions want to eat the sheep, but will settle for grass. Sheep only eat grass. If a lion eats a sheep, he will turn into a sheep. Assume the lions behave logically and each will look out own survival (ie. each lion acts independently). How many lions and sheep will be left?

Classic story of backwards induction. Assuming there is a sequential process, 98 Lions and 1 sheep will be left behind.

A rational lion looking out for his own survival would never eat a sheep, as if that is a rational action for him, it is a rational action for another lion, guaranteeing that he will in turn be eaten. Therefore, a lion will never eat a sheep, and you'll end up with 99 lions and 1 sheep. You can prove it with game theory if you want...

The answer is there is no such thing as magic.

No but seriously I think we need more information; how much grass is there? If the lions will settle for grass, there is no reason to choose to eat the sheep over the grass (is there a limited supply?)

Also, for the factorial question, there is no way to get the exact integer except by manual calculation... they are probably just testing your thought process...

you would just start by calculating the number of zeros which is quite easy, then take the power of two, and probably if you have time left go with the next primes... if you get that far I think you are good

I suck at brainteasers, how do you get better? Just practice and try to understand the logic behind each one?

im with you! i enjoy them but havent every been taught or spent the time to get these gnarly ones.

There will be one sheep left?

How do you figure? The way I see it, the 1st sheep will be eaten immediately, and then none of the lions will want to eat grass, since if they do they will turn into sheep and be eaten. So the lions start starving, and do so until the cost-benefit analysis of eating the grass makes sense (die of hunger in 1 second vs eat grass and be eaten by the other lions in 10 seconds.) Assuming continuous time, the no two lions will come to this conclusion at the same time, so one at a time they will eat the grass, become sheep, and be eaten. This continues until there is only one lion left, who can then eat the grass and live happily ever after as a sheep.

You turn in to a sheep by eating other sheep, not eating grass.

For the question about the bunny hops, you have to use generating functions.

You're right, I misread the question, although I think it's more interesting the way I misread it...

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There is an unlimited supply of grass, but the lions WANT to eat the sheep.

**Answer below***

Start backwards, what if you had 1 lion and 1 sheep? Lion would eat sheep, turn into sheep, and he has ensured his survival because there are no lions left to eat him.

Now, 2 lions and 1 sheep, if lion #2 decides to eat the sheep, he turns into a sheep, and immediately the remaining lion #1 eats him because there will be no lions left eat him in turn. Therefore, lion #2 is logical and will not eat the sheep because he knows lion #1 will be waiting.

3 lions and 1 sheep, lion #3 will eat the sheep because he knows that #2 will not eat the sheep.

Follow this reasoning, all odd number lions will eat and even numbers will not... 98 lions and 1 sheep will remain.

Your solution assumes that the lions are shortsighted and only look at the decision the lion before them will make. This is not "rational" if they are looking out for their own survival.

Any given lion will not eat sheep if there is even 1 other lion around.

But yes, I understand your logic from a mathematical perspective.

nope try again

I've got a tough investment banking interview question:

"My friend bought a company and recently sold it for the same price. He claimed to have made a killing.. how is this possible?"

Can anyone provide insight on how they would respond?

He purchased it with debt, and paid down the debt with the company's cash flow. Aka an LBO

Is this answer valid:

He bought shares of a public company which had a split and proceeded to increase in value. Therefore, the selling price was the same as the buying price but he sold twice as many shares therefore making a killing.

You have a ten digit number.

The sum of the ten digits is 10.

The leftmost digit is equal to the number of digits which equal 0. The second to the left is equal to the number of digits which are ones, ect.

Given this, lay out a proof which demonstrates all possible values for the ten digit number.

Were you actually asked for the proof or just the number?

One number is pretty easy: 6210001000

I think its the only one but I may be wrong

Proof

how would you value an oil/gas company?

qweretyq,

I don't think that answer is valid because by "price" I assume that they mean price of the company (ie - valuation or market cap) so the sale price in the scenario you mention would be twice that of the original purchase price.

The only answer I can think of is leverage. So use cash flows to pay down debt (or pocket them...).... and you can take it from there.

Leverage is the answer they were looking for. Other correct answers would include deflation (eg, same nominal P, increased real P), stripping of the company's assets prior to resale of the company itself, or if the company paid dividends (including a dividend recap scenario)

Leverage is the answer they were looking for. Other correct answers would include deflation (eg, same nominal P, increased real P), stripping of the company's assets prior to resale of the company itself, or if the company paid dividends (including a dividend recap scenario)

Riddle(Originally Posted: 04/12/2007)What two coins make 30 cents? One of them cannot be a nickle.

A Quarter and Nickle

So...what's the answer?!

This better be good,

+Hammy

lol, as NickCarraway said...the answer is a quarter and a nickel. One of them (the quarter) isn't a nickel.

Oh, LMAO...I see.

That was worth a chuckle.

Thanks,

+Hammy

what a dumbass, r u really that stupid nickcarraway?

why is he stupid?

You just watched Scrubs, didn't you. :)

Yes, yes I did. Damn you all.

E-L

not L-E

yes a i am.

those stanford kids...thats why its HYP and not HYPS

KIDDING. calm down stanford kids.

brain teasers - consulting interviews?(Originally Posted: 08/08/2007)It's been a while since college math classes so I need some warmup before my HF/consulting interviews...anyone have any good ones from their interview experiences?

Please refrain from providing the answers on this forum for at least a day or two after it has been posted...

using the search function might help =]

Man that's a hard teaser. Post the answer to that one because I just can't figure it out:)

If the amount of lillys in a lilly pond double every minute, after how many minutes will there be half the amount of lillys at 59 minutes?

58 minutes???

i say 58 as well.

it's 29 minutes, please guys.

why 29?? it doubles every minute, so i would think the total ponds for every min. would be 2 to the power of that minute, i.e. the 59th minute would have 2^59 ponds. Half of that would be 2^59 divided by 2 which equals 2 ^58, so the 58 minutes???

i think thats correct

was that a real brain teaser?

I don't really think so

Its your birthday in a couple of days and your gf has promised you a day you wont forget. You have been trying to convince her that taking it from behind can be a interesting experiece. You have a good feeling that your birthday wish will come true.

You have been out with her for the night celebrating something special and she seems like she is in the mood for you to show her a goodtime.

Do you

a) Go for it as you undress her, knock her legs apart bend her over and gor for it. Taking the risk that it will pay off leaving you with something even more interesting for your birthday that might finally involve her and her hot best friend.

or

b) Don't take the risk, let it on what you would like for your birthday experience. Giving up the chance of a possible special prize of a threesome.

There is no right or wrong answer I am interested in the way you think

At dawn on Monday a snail fell into a bucket that was 12 inches deep. During the day it climbed up 3 inches. During the night it fell back 2 inches. On what day did the snail finally manage to climb out of the bucket?

because 29 is a mysterious number

it would basically crawl an inch a day until the 9th day when it crawls up the 3 inches and gets out before it can fall back. That's assuming it doesnt get lazy and not crawl over the top and lose ground...stupid snail...good one though random99, I was keen to say Saturday (12 days) but that little extra thought makes all the difference.

btw, I just got a call back for another round with the HF job so you know I'm gonna need some more of these :-)

have four guys, they all need to cross a bridge, bridge can only fit two people and they must carry the one flashlight they have across the bridge. the bridge is going to collapse in 18 minutes and have to get across before it falls (meaning n less than 18). each of them have different times A=1 min B=2 C=5 D=10. when two people cross they take the bigger time AB=2 minutes. How do you get them across?

AB cross, A returns. 3 minutes have elapsed.

CD cross, B returns. 15 total have elapsed

AB cross together. 17 minutes total.

if you need to wait for your birthday to do that with your girlfriend, find a new one.

Clearly you don't follow my threads. That was a riddle for you guys benefit. i don't have such problems.

3 guys crash at a hotel. they each have 10 bucks and the clerk charges them 30 for the room, they all pitch in and then go to their room.... later the clerk realizes the rate for the room is only 25bucks, not 30, so he goes back to return the change. On his way he realizes he can't split 5 bucks in 3 ways so he pockets 2 bucks and then gives each of the guests one dollar.

So essentially each guy gave 9 bucks right? the clerk kept two. Here's the thing though, 3*9=27 +2=29.... what happened to the extra dollar???

They paid $25 for the room, tipped $2 to the clerk, and got back $3. Hence they paid $9 each. There is no extra dollar. The equation above is incorrect because it double counts the $2 clerk fee but never mentions the $3 they got back.

Funny enough, this one stumped me a few hours ago when my recruiter gave it to me during our interview prep. Are there really so few brain teasers that everyone uses the same ones or at least some version of those ones. If so, can I just memorize the underlying theme of as many as I can over the weekend and just pretend I have never heard them before in the interview? I feel this is my best chance cuz I suck at these. Will these keep me from getting the job?

you got it... and yes I think there's a pool of brainteasers interviewers ussually pick from

There are ten pennies 5 heads up and 5 tails up. Create 2 group (any number of pennies) so that the heads in one group is equal to the heads in the other.

Oh and yeah, You are blindfolded and cannot see.

you have two threads each of which burns for an hour. The threads don't burn uniformly and they are not identical. how will you measure 45 minutes using the threads.

just split them into two groups and then flip over all the pennies in either one of the groups

the rationale is that no matter how you divide the pennies initially, heads in one group equals tails in the other so when you flip one of the groups, you now have equal numbers of heads (and tails) in both groups

light one thread on both ends, the other on just one end, when the first is completely burned, light the second end of the second thread. 45 minutes has elapsed when the second thread is completely burned away.

if you light the first thread on both ends, it'll be burned at 30 minutes. the secnd thread lit only on one end will also have burned away 30 minutes worth (and have 30 minutes worth). by lighting the second end, you're cutting the duration of this burn in half (15 minutes). 30 + 15 = 45

if you interview at quant groups this is a must

more please!

Jack's birthday is M/D (month/date) and he tells Bill and Sam that M/D is one date in the following ones:

3/1, 3/5, 3/8, 6/4, 6/7, 9/1, 9/5, 12/1, 12/2, 12/8

Bill says "If I don't know M/D, Sam won't know it either."

Sam says "I didn't know it just now, but now I know it."

Bill say "Then I think I get it too."

So M/D=?

I am going to say 12/2, although i'm not totally sure. the only way they could both know is if there was one date that is unique in some way. so first i eliminated all the dates that share the same day as another (3/1,9/1,12/1; 3/5,9/5; 3/8,12/8). now we have 6/4, 6/7 and 12/2. the first two share a month and so are eliminated. so 12/2 is the answer.

There's definitely not enough information to solve that problem.

you are on top of a 200 foot wall and need to descend. you have two ropes, 100 ft and 50 ft long.

there is a hook (to tie the rope to) at the top and a platform with a hook 100 feet below.

how can you get down without dying?

You tie the two ends of the 100 foot rope together, making a circle. Tie the 50 ft rope to that circle at any point. Hook the circle made from the 100ft rope over the hook at the top, and throw the tail end down towards the platform. The 100 ft circle becomes approx 50ft long and the other 50 ft rope hangs from the bottom (6:00 position) of the circle you made, essentially making a 100ft scalable rope. Scale down the first 100 feet. Once on the platform unhook the circle 100 feet above by holding the end of the 50 ft strand and making a wave motion in the rope; picutre a streamer. Hopefully that will come unhooked if it is truely a hook on the top and if you hold on to the rope you can repeat the same process for descending the second 100 feet.

Looking through some brainteasers during lunch and found this old one with no response, anyone follow my logic?

"They key to having it all is the realization that you already do"

You tie the two ends of the 100 foot rope together, making a circle. Tie the 50 ft rope to that circle at any point. Hook the circle made from the 100ft rope over the hook at the top, and throw the tail end down towards the platform. The 100 ft circle becomes approx 50ft long and the other 50 ft rope hangs from the bottom (6:00 position) of the circle you made, essentially making a 100ft scalable rope. Scale down the first 100 feet. Once on the platform unhook the circle 100 feet above by holding the end of the 50 ft strand and making a wave motion in the rope; picutre a streamer. Hopefully that will come unhooked if it is truely a hook on the top and if you hold on to the rope you can repeat the same process for descending the second 100 feet.

Looking through some brainteasers during lunch and found this old one with no response, anyone follow my logic?

"They key to having it all is the realization that you already do"

The information is enough. Think carefully.

You may exclude possibilities step by step from the 2 guys' conversation.

oh I apologize I did forget something...

Jack also told Bill M and Sam N and the nthe 2 guy had that conversation

Now this problem can be solved...

Two groups of zero pennies ldo.

Post your favorite brainteaser!(Originally Posted: 11/29/2007)Let's play a game. I'll post my favorite brainteaser. The next person solves my problem and then posts his/her favorite. So on and so forth. This thread will indubitably get sidetracked by people arguing over the answer to some particular problem, but try to stay focused. =)

brainteasers need not be from ib interviews. I think IB candidates/employees think in a manner where brainteasers are fun anyway... so let's enjoy!

Question:

Three men check into a B&B, sharing a room for the night. The receptionist tells the men that the room costs $30, so each man ponies up $10. Later, the manger informs the receptionist that there was a special going on that night; the room rate was only $25. The receptionist gives the bellboy $5 to return to the three men. The bellboy, however, concludes that the three men could not evenly split $5, so he pockets $2 and returns each man $1. In total, each man paid $9 ($10 minus the $1 refund). $9 times three men is $27... plus the $2 the bellboy took comes out to $29. The men initially paid $30; what happened to the missing dollar?

I have the general idea- just my wording wasn't the greatest in the world.

no you idiot... they paid a total of 27 and that includes the $2 that the bell boy got. So the room rate 25 + bell boy $ 2 is equal to 27, and the rest 3 dollars are the ones that they got back.

Eek... that is the exact wording of the brainteaser, not b's conclusion that there was actually a missing dollar. Unnecessarily strong wording...

I might be missing something (since I have almost no brainteaser experience), but I don't think baloogafish's explanation makes sense.

The room rate at this incredibly shitty hotel was $30.00 (which is equivalent to the hourly cost of a weeknight hotel stay at a low-end hotel in Manhatten), so each paid $10.00 . Since the actual rate should have been $25.00, each man should have paid $8.33 (not $10.00) and were thus each entitled to a $1.67 refund. However, the bellboy only distributed $1 to each of the men, thus stiffing each by the residual amount ($9.00 - $8.33 = $0.67) and pocketing the additional $2.00 ($0.67 x 3 = $2.00) in change. The brainteaser portion of this question, as I see it, comes from the assumption that distributions should sum to $30, which I do not believe to be correct. Rather, the cash distributions should sum to the post-adjustment hotel room rate of $25. Each man paid $9.00 for a grand total of $27. Because the bellboy stiffed them, he pocketed their residual "true-up" on the room rate adjustment for a total of $2. Once distributed, total net proceeds sum to the adjusted room rate of $25.00 ($9.00 x 3 - $2 = $25).

Perhaps I'm way off on this, so people should feel free to correct me. In any case - nice brain teaser b. I've never really done these but it is a very nice break from the painful LBO deal that I've been working on for more than 20 straight hours at this point (waiting on feedback from the Latham guys on a few things, and they are slow as usual in responding).

Mass_banker, you forgot to post a brainteaser.

I'm no expert but his logic is correct. But I didn't get it :(

One I had in an interview a while back:

You've got a hose and 2 buckets - one a 3 gallon bucket and one a 5 gallon bucket. How do you get exactly 4 gallons of water?

I had this one in high school but its pretty intuitive...

You fill the 3 gallon bucket and pour it into the 5 gallon bucket. You then re-fill the 3 gallon bucket and pour it into the 5 gallon bucket (which already has 3 gallons in it)until it is completely full. You now have 1 gallon of water in the 3 gallon bucket. You pour out the 5-gallon bucket and then pour in the the 1 gallon of water from the 3 gallon bucket into the 5 gallon bucket. Then you fill the 3 gallon bucket completely and pour into the 5 gallon bucket (which already has 1 gallon of water) giving you exactly 4 gallons of water.

Here's one I got in an interview -

You have two pieces of fuse - each burns end to end for exactly one hour. How can you time EXACTLY 45 minutes?

- Capt K -

"Prestige is like a powerful magnet that warps even your beliefs about what you enjoy. If you want to make ambitious people waste their time on errands, bait the hook with prestige." - Paul Graham

Light both ends of one fuse and one end of the other and then light the second end of the second piece just as the first piece burns away completely. (Assuming the fuses burn at uneven rates along their length).

Sticking with volume...

If you have a 1 gallon jug and it is half full (50% full of liquid) with Jack and coke plus 4 ice cubes (1 cubic inch each). What percentage of the 1 gallon jug will be full when all 4 ice cubes melt?

You have 9 identitical looking marbles and one balance scale. 8 weigh the same, 1 weighs slightly more than the rest. What is the minimum # of weighings needed to find the one marble that is different (example - if u weighed one marble vs. all the rest one by one, it would take 8 weighings of 1 vs. 1 to ensure that you find the heavier one)

2 weigh-ins.

First, you take 6 of the marbles and weigh 3 on each side of the balance scale. If 1 side is heavier, take 2 of the marbles from the heavier side and weigh them and that will tell you which one is heavier.

If neither side is heavier, take 2 of the remaining marbles (that you haven't weighed yet) and weigh them.

but what is the probability that one of the two selected is the heavier one?

I know that is not tha answer you are looking for, which is two.

Atmost 2 measures it enuf to spot the faulty one.

Step:1 :Split the 9 stones into 3 pockets of 3 each.Take any 2/3 and measure.After the measure,you can spot-out the 3 having faulty one.

Step2: split the 3 into 1 each.Take any 2 and place on either side of the scale-pan and the next moment u can identify the faulty one.

-Deepak.

The Classic: 9 weights, each identical looking, 8 weigh the same, 1 weighs a different amount then the others. Using only twice a "balance scale" (think statue of justice) how can you determine which of the 9 balls is the abnormal one?

The impossible:

There are ten gnomes. They are in the dungeon. Their captor has given the gnomes a chance of survival. Here is the offer:

He lines the gnomes up in a single-file row. This means that the tenth gnome sees the back of the person in front of him, and there is no gnome behind the tenth gnome. The ninth gnome has the tenth gnome behind him and the eighth gnome directly in front of him, and so on. Finally, the first gnome has the second gnome directly behind him, but there is no one in front of the first gnome.

The captor has a bag full of many black hats and many white hats. There is not necessarily the same number of black hats as white hats. (important) He randomly reaches into his bag and places a hat on each of the gnomes. This means that the tenth gnome can see everyone's hat except his own, the ninth gnome can see everyone's hat except his own and the tenth gnome's hat, and so on. The first gnome can see no one's hat.

The captor then takes out his gun and puts it to the temple of the tenth gnome. He asks the gnome, "What color is your hat?" If the gnome answers correctly, he lives and gets freed from the dungeon. If he does not, he dies. He continues up the line in this progression.

However, before placing the hats on the gnomes, he allows the gnomes to meet as a group and discuss a strategy to save as many of the gnomes as possible. How many gnomes can you guarantee to save, and with what strategy?

REMEMBER: When it is your turn to say the color of your hat you must ONLY say "white" or "black." If you say anything else, the king will shoot you and all of the remaining gnomes.

wow, two in a row of the same one - but it's a lot better if you don't tell that it's two!!!

i think that guy posted as i was typing mine! haha.

anyway...the second one i posted is for real.

I can't solve it, but here's a start on the gnomes -

Because there are an unknown number of black vs. white hats, without any extra information (or even with information about the proportion in the random sample of hats on the gnomes heads), a guess about your own hat color is a toss up.

Therefore, the 10th gnome needs to use his "guess" (white or black) to convey some kind of information to the other nine gnomes. If the "guess" matches his hat color, he's lucky and lives. If not, he dies, but has communicated some kind of important info to the other nine.

This probably isn't the best solution but here it is - the 10th gnome guesses the color that he sees the most of in front of him (ie. if there are 5 white and 4 black, he guesses white). That way, you are assured that at least 5 gnomes survive.

Can anyone do better than 5?

- Capt K -

"Prestige is like a powerful magnet that warps even your beliefs about what you enjoy. If you want to make ambitious people waste their time on errands, bait the hook with prestige." - Paul Graham

I should clarify - the 10th gnome guesses white, so all the other 9 know that there is higher than a 50% chance they are wearing a white hat, so they will all guess white as well.

- Capt K -

"Prestige is like a powerful magnet that warps even your beliefs about what you enjoy. If you want to make ambitious people waste their time on errands, bait the hook with prestige." - Paul Graham

that the gnomes cannot attempt to communicate with each other outside of the words "black" or "white," (i.e. hitting the person in front of you to signal to them that they are wearing a black hat) the gnomes could use the pitch of their voice to signal in a similar manner. For example, using an extremely high pitched voice to signal "black" and low pitch to signal "white" to the next person. This would guarantee 9 and give a 50/50 at 10. (assuming there is a notable difference in pitch between high and low)

I agree with five, but my method is a little different...I would have the tenth gnome call out the hat color of the first, and the ninth call out the color of the 2nd, all the way through to the sixth gnome. Then the fifth through first would be guaranteed to know their colors.

that the gnomes cannot attempt to communicate with each other outside of the words "black" or "white," (i.e. hitting the person in front of you to signal to them that they are wearing a black hat) the gnomes could use the pitch of their voice to signal in a similar manner. For example, using an extremely high pitched voice to signal "black" and low pitch to signal "white" to the next person. This would guarantee 9 and give a 50/50 at 10. (assuming there is a notable difference in pitch between high and low)

*sorry if this posted twice

I think that the path will be the 10th gnome should say the color on the 1st gnome, 9th say the color on the 2nd, and etc. In this fashion, the first 5 gnomes will be saved.

as for gnomes 6~10, I'm going to assume that black and white are proportional to each other, so theoretically 2.5 will survive on random odds. So my answer saves 5 + any number between 0 and 5 with the long run average of 7.5 gnomes saved...

make sense? or am i totally off

Please keep with the system to ensure that all brain teasers get answered and there are not a bunch of unanswered questions floating around. Posting a brainteaser is a privilege obtained from answering the previous one. =)

I got the gnome question in a comp sci interview once.

They discuss a strategy. The gnome in the very back counts to see whether there is an even or odd number of black hats. If even, he will say black, if odd, he will say white. He has a 50/50 chance of getting it right.

From this, the 9th gnome can deduce the color of his own hat. For example, if the 10th gnome saw an even number of black hats, and the 9th gnome sees an odd number, he knows that his own hat is black. If he sees an even number, he knows his own hat is white. After each "black" is called out, the system switches, as "black" now represents odd and "white" represents even number of black hats. So on and so forth.

This ensures a 9/10 survival rate; the 10th gnome has a 50% chance at survival.

Question:

A zebra has a stack of 300 bananas. He wants to transport as many bananas as possible across a field 100 miles long. He can carry at most 100 bananas at a time. He has to eat 1 banana for each mile he walks (both backwards and forwards). Assume that unattended bananas will not get stolen, lost, or otherwise affected. What is the maximum number of bananas he can get across the field?

I have a string which stretches all the way around the world. I want to raise it off the ground by 1 foot. how much longer does the string need to be?

there is a rail track 1 mile long, and bolted down at the ends. one particularly hot day the track expands by 1 foot. assuming it can only bend upwards (not sideways at all) how far will the track raise off the ground after the expansion.

Isn't it x = 2pi(r+1) - 2pi*r, x=2pi feet?

good work hoya

For Jimbo's rail track question, I would think you can simply derive this using basic geometry, but I'm probably missing something.

If the track length is 5280 ft (1 mile) long and expands by 1 ft to 5281 ft, it seems like the height would be:

(((5280 + 1)/2)^2 - ((1/4)*(5280)^2)))^(1/2) = ~51.4 feet

This does not seem intuitive to me, and I was most certainly not a math major, so any corrections or comments would be welcome here.

yup that's close enough...i like the counterintuitive nature....

If you have a 1 gallon jug and it is half full (50% full of liquid) with Jack and coke plus 4 ice cubes (1 cubic inch each). What percentage of the 1 gallon jug will be full when all 4 ice cubes melt?

Anyone know the answer?

231 cubic inches to a gallon

-ice cubes are only about half under surface level

-1 gal= 231 in3

-before: 115.5 in3

-after: 115.5+2=117.2 in3

50.736%

In reality most ice cubes are much more than just half under surface level,

so it would be even less. Pretty much stay at 50%

that was not a brainteaser though

It doesn't matter whether, and to what degree, the ice cubes are submerged. You already know that there is 1/2 gallon of liquid and 4 in^3 of ice. Submerging the ice cubes will simply displace the volume of the liquid -- not replace it.

In my previous post, I'm assuming that the jug is "(half of liquid) plus four ice cubes" and not "half full of (liquid plus four ice cubes)".

Confusing wording to the question.

it is an exercise in obscure weights and measures.

how many cubits to the mile?

here's a real teaser. It should make you think:

You are imprisoned in a chamber with two doors as the only exit. One door leads to death by cancer, filled with complications and malpractice; the other door leads to riches of jewelry, money and fine clothing for the rest of your life. There are two guards standing before you: one guard always lies; the other always tells the truth. Of course, you don't know their identities. You can ask only one question to save your life. WHAT WOULD YOU ASK?

"Which way would the other guy tell me to go?" -- And then go the opposite way.

... quite an old puzzle.

Sorry it is old,

how about this one, hopefully its not as known:

You leave your camp and travel a 100 miles south, then a 100 miles east, then a 100 miles north. You find yourself back at your camp site and tell your colleague that you saw a bear on your travels. What is its color?

White,

Your coordinates would only make sense if you're in the north pole. Given only polar bears live in the northpole, it has to be white.

When I hear a brainteaser...I just start reciting poetry "Two paths diverged.... " until the interviewer laughs and forgets he asked me the initial question.

As no one seems to be following the system that I laid out (and people seem to be posting brain teasers that I've heard 10 years ago) I'm going to go ahead and post some of my favorites. Have fun... I enjoyed these:

1.) There are 100 people in line to board a train. Each person is holding a ticket corresponding to both his place in line and his seat (i.e. Person 1 is supposed to sit in Seat 1, Person 2 is supposed to sit in Seat 2, etc.). All of the people in line are normal, except for Person 1, who is "crazy". When Person 1 boards the train, he will ignore his seat number and sit in a random seat. Following Person 1, each person will sit in his own seat, unless his seat is occupied. In this case, the Person who's seat is occupied becomes the new "crazy person" and will sit in a random seat. For example, if Person 1 sat in Seat 2, Person 2 will become the "crazy person" and sit in a random seat. What is the probability that Person 100 will end up in Seat 100?

2.) McNuggets come in boxes of 6, 9, and 20. What is the largest number of McNuggets that is not possible to obtain using some combination of these three box sizes?

3.) There are 100 closed windows and 100 people. The first person opens every window. The second person closes every 2nd window. The third person visits every 3rd window; if the window is closed, he opens it; if it is open, he closes it. So on and so forth. After the 100th person, how many windows are open?

4.) Three men are at a picnic. One man brought 3 loaves of bread, one man brought 4 loaves of bread. The third man did not bring any bread, but promised to pay the other 2 men the $14 he had in his pocket for a share of the bread. The men agreed, and each subsequently eats an equal share of the bread. Following the lunch, the men had a dispute as to how the $14 should be divided; what is the most fair way?

5.) Three men (A, B, C) are in a dual, where they take turns shooting. Person A gets to shoot first, Person B gets to shoot second, and Person C gets to shoot third. Person A and C are both amateur shots; Person A hits his target 1/3 of the time, person C hits his target 1/2 of the time. Person B is an expert and hits 100% of the time. Where should person A first shoot to maximize his chances of winning the game?

6.) There is a spider in the corner of a 10 ft x 10 ft x 10 ft room. The spider wants to get to the opposite corner of the room [i.e. he is in the bottom back left corner and wants to get to the top front right corner]. He can not fly; thus must stay along the wall at all times. What is the shortest distance to the opposite corner?

the answer is 1/2. I thought about it in smaller numbers and built up. here is the intuition.

2 person case (C2): 1st guy has 50/50 shot of sitting in 2nd guys seat. so odds are 1/2. C2 = 1/2.

3 person case (C3): 1/3 chance 1st guy sits in his own seat, then odds of 3rd guy in his seat is 1. 1/3 chance of 1st guy in 3rd guys seat, then prob is 0. 1/3 chance 1st guy in 2nd guys seat, then 2nd guy sits in seat 1 or 3 randomly, so odds are 1/2.

C3 = (1/3)

(1) + (1/3)(0) + (1/3)((1/2)1+(1/2)*0) = 1/24 person case (C4): first two options for 1st guy same as C3, sits in own seat or 4th guys seat. If 1st guy sits in seat 3, then 2nd guy sits in his own seat and 3rd guy either sits in seat 1 or 4, so odds are 1/2. If 1st guy sits in seat 2, then 2nd guy is crazy and we are reduced to the same situation as case 3: three seats with first guy beign crazy. this had odds of C3=1/2.

C4 = (1/4)

(1) + (1/4)0 + (1/4)(1/2) + (1/4)C3 = 1/25 person case (C5): now you can see the pattern. the odds below are in order if 1st guys sits in seat 1,2,3,4,5

C5 = (1/5)

1 + (1/5)C4 + (1/5)C3 + (1/5)(C2) + (1/5)*0 = 1/2proof by math induction. We know that C(2) = 1/2, so base case is satisfied.

Now we prove that if C(j) = 1/2 for all j < n, then C(n+1) = 1/2. assume

C(n) = (1/n)

1 + (1/n))C(n-1) + ... + (1/n)C(2) + (1/n)0 = 1/2now you can extend this pattern, such that for n+1 people

C(n+1) = (1/(n+1))

1 + (1/(n+1))C(n) + ... + (1/(n+1))C(2) + (1/(n+1))0. since C(n) = 1/2 thenC(n+1) = 1/2. qed.

i'll take a stab at this one. you know each man consumes an equal amount of bread, 7/3 loaves. now since man #3 didn't bring any bread, he had to buy his shares from the other 2 guys.

man #1 brought 3, ate 7/3 for a leftover of 2/3. so he sold 2/3 to man #3.

man #2 brought 4, ate 7/3 for a leftover of 5/3. so he sold 5/3 to man #3.

so share paid to man #1 is (2/3)/(7/3)=(2/7).

man #1 gets (2/7)*14 = $4.

obviously man #2 gets the other $10.

Also correct. I'm surprised this wasn't answered earlier, as I didn't think this question was too difficult.

Also, no one ever solved this one:

A zebra has a stack of 300 bananas. He wants to transport as many bananas as possible across a field 100 miles long. He can carry at most 100 bananas at a time. He has to eat 1 banana for each mile he walks (both backwards and forwards). Assume that unattended bananas will not get stolen, lost, or otherwise affected. What is the maximum number of bananas he can get across the field?

0

I heard a similar one went something like this:

You are in a house, all four walls face south and all walls have a window, a bear walks by a window ...what color is the bear??

First, I don't think zebra's eat bananas, do they?!?

Second, well, what's the zebra's intention? Does he intend to carry the bananas across the field because he's going to move to the other side and live there forever? Or does he have to carry bananas across the field to make a delivery to someone else?

If he is making a delivery for someone else he can only get 100 bananas across the field. After he eats 100 on the way there and 100 on the way back. Then he's finished. No more bananas.

If he's moving to the other side forever... well... the answer is still 100. He carries 100, but needs to eats 100... so the remaining 100... well he's shit out of luck and can't go back and get the remaining 100. So he donates that 100 to the zebra friends he sadly left behind :-(

Am I right?

if it can only carry 100 at a time, and the field is 100 miles, and he eats 1 every mile...the zebra wont have any once he reaches each end of the field. hence, he can't get any all the way across the field.

50 white marbles in one jar, 50 black ones in another. how would you distribute the marbles in order to maximize your chances of pulling out a black one. what would be the chances you pull a black marble with that distribution?

Id just pull from the black jar

To aadpepsi (for some reason, her post is showing up below mine):

No, you're not right. He has to eat one banana for every mile he walks; I never said that he can walk 1 mile for each banana that he eats. Therefore, he can't fill up on 100 bananas at the start and walk across.

Answer is also not 0.

Anyone who's feeling intelligent want to give it a go?

Any chance the answer is ~44 bananas (I highly doubt this is correct, but wanted to toss out a solution better than 0).

Seems to me that one strategy would involve taking 100 banans for the first 1/3 of the trip, dropping 1/3 off and then consuming the remaining 1/3 on the trip back. Do this once more and now you've got ~66 bananas that are 33 miles from the start point, and a zebra at the starting point with 100 bananas left. He burns 33 of these getting to the starting point, but is then sitting 33 miles from the start point with 133 bananas (166 - 33). Since he can only carry 100 at a time, he takes 100 bananas roughly 11 miles (1/3 * 1/3 = 1/9 of 100 miles), drops 78 of his remaining 89 (since he burned 11 getting there), then uses the remaining 11 bananas to get back to his remaining 33 bananas that are sitting 33 miles from his starting point. He'll burn another 11 bananas getting out to his 44 mile point (where he dropped 78 bananas), and will thus end up with 100 bananas 44 miles from where he started. He burns 56 of these bananas on the remaining 56 miles and ends up with 44 bananas.

smuguy, very close, and proper approach, but answer is incorrect.

I figured as much. There's got to be an underlying purely mathematical approach, but I have no legit background here beyond AP Calc in high school and the very basic math that comes with working in PE.

Ah, but AP Calc in high school is more than enough math needed to solve this problem. =)

This problem was actually given to me in my entrance exam to a private high school to determine which students would be placed into an "honors class", thus it is actually solvable with solely pre-high school skills.

(1) Three men in a dual problem:

This is a classic quantum three person duel (truel). Each person will shoot that person who is the greatest threat to them. Thus, person A knows that person B and person C will target each other. If A tries to shoot C and hits, then he is dead because B will shoot him/her with 100% accuracy. If A tries to shoot B, there is a 1/3 probability of hitting B, in which case person C will then have one shot with 50% chance to hit A. Thus, the best option for person A is to:

shoot in the air (i.e. do not target either of the other two shooters). As a result, B will kill C with 100% probability, and A now has a 1/3 chance of killing B and being the last person standing - his highest probability.

This can be quantified probabilistically, just google search for quantum truel, and you should be able to find plenty of papers discussing the mathematical theory behind the issue (obviously it can be attacked with traditional probability as well)

(2) McNugget problem:

This is an application of what is known in probability as the "coin problem." If you are given 3 natural numbers whose greatest common divisor is 1 (which is the case for 6, 9 and 20), then there exists a largest natural number which can not be expressed as a linear combination of the other three. This number is known as the Frobenius number. For this problem, the answer is 43. It is interesting to note that this problem, for arbitrary n, this problem is np-hard (np-complete is a subset of np-hard)

(3) 100 Windows Problem:

This one is also a classic, although it is phrased in many different ways (lights, gym lockers, hotel room doors, etc.). The key here lies in the fact that each number that has an even number of unique factors will end up closed. Thus, the only windows that will be open are those numbers which have an odd number of unique factors. Thus, the only windows that will be open are those numbers who are perfect squares (1, 4, 9, 16, 25, 36, 49, 64, 81, 100). For example, the factors of 36 are: 1, 36, 2, 18, 3, 12, 4, 9, 6... Thus, only an odd number of people will open/close this window and it will be left open at the end (since it starts off closed).

Chelsea, put 1 black marble in 1 jar and all the other marbles in the other jar.

If you reach into Jar #1, you have 1/1 or 100% chance of pulling a black marble.

If you reach into Jar #2, you have 49/99 chance of pulling a black marble.

Altogether, you have [99/99 + 49/99]/2 chance of pulling a black marble, or [148/99]/2 or 74/99 chance of pulling a black marble. Just under a 75% chance.

qwerty25, very good. Despite that I couldn't understand half the words in your post, you got all three problems correct.

going on 40 hours up straight, so it's bound to happen. I might get to some of the others if no one gets them. BTW, love the thread.

Ok, this is ridiculous. I don't need to be thinking of bananas. What's the zebra banana answer? I don't have time for this. I need instant gratification. What's the answer? WHAT'S THE ANSWER!

i think i got an answer, not sure if it is right

the zebra can move all 300 bananas 1 mile @ a loss of 5 bananas per mile, he can continue for 20 miles, until has only 200 bananas, then the loss per mile will be 3 bananas (makes only 2 trips instead of 3) until the 50mile mark. from there on he can stock up on 100 bananas and travel the rest of the distance at a loss rate of 1banana per mile.

so he can carry 50 bananas.

Close... right method/logic... wrong answer.

probably 52 +/- 1 banana.

i did the the work to figure out he can travel 53 miles with a reserve of 100, but didnt feel like working out the details.

plus im sure the zebra would probably be sick of doing all those repetitions, and would just haul it after the 50miles. i know i wouldnt go through the trouble of going through another repetition for 2 extra bananas.

What?!? Where'd you get the loss rate of 5 bananas per mile? Where did that come from?

Who posted this question? Will the original poster of the question please post the legitimate answer. It's driving me nuts.

here, i'll clarify

you move 100 bananas the first mile. eat up 1 banana, carry 1 banana for the trip back.

pick up another 100 bananas and travel another 1 mile. carry 1 banana for the trip back again.

repeat for final time. and the total bananas consumed to move all 300 bananas is 5.

when you have only 200 bananas, the loss rate is only 3 per mile.

Assuming the zebra travels in minimum increments of 1 mile, I believe the correct answer is 47 bananas.

smuguy... almost. I think you made a mistake when subtracting. You probably realized that the zebra had 47 miles left to travel, but he actually brings 53 bananas across (100-47).

Eat_My_SOX... bingo.

The exact answer is 53 1/3 bananas.

This was the easy problem... now try the one w/ the crazy guy on the train. That one is brutal. =)

Sure, all of you guys give answers but no real solid explanation. I had to Google it.

What I have concluded is that you fellas shamefully did the same. You googled for the answer and then posted a mish mash response here on the forum to make it look like it's your own authentic answer, but in truth it's not. You're faking that you know the answer when you really don't. Anyways, I think this link gives the most THOROUGH explanation, step-by-step for at least two popular variations of the brain teaser.

http://www.gamedev.net/community/forums/topic.asp?...

Sigh.

I don't think anybody is desperate enough to Google for these answers... this thread is purely for fun. I was the one who originally posted the question (and had solved it on my own 4 years ago), so I know that it's easily doable. A good saying:

"Never judge others by your own limitations."

Listen... one's limitations vary on any given day. On a more forgiving day, I would take the time to solve a brain teaser and it would be fun. Today, I just wanted immediate gratification. I wanted the ANSWER. I didn't want to work thru a solution.

Well I didn't google anything considering I still missed it. Actually, I was on an update call with one of our portfolio companies and chose to scribble out some possible solutions for this instead of actually listening to the call.

Perhaps some did look it up, but no need to accuse everyone just because you didn't get it and didn't want to think about it.

i did NOT!

anyways, if Wallstreetoasis had an upload feature, i'd scan and post my work.

plus,

the answers for spider in the corner would be 24.14ft.

the bread problem by abc (activity based costing) would be $4.67 to each of the two people.

and im going to take a random stab at the 100person question. by saying it is 1/100 chance that he'll get his seat. (didnt really solve, just felt like it might be a good answer)

Eat_My_SOXs - try again, all those answers are incorrect.

I believe the spider problem's answer should be 2*(125^.5) = 22.36 feet

yes.

ahhh, i just got back.

yeah i got 22.36 also.

realized what i did wrong.

well, thats all the the time i have today.

here's two good ones:

1) you have a shuffled pile of N cards, numbered 1,...,N. First, you pick the card on the top and you put it in a new pile called A. You draw the next card from the original pile, and you put it on top of pile A only if it is larger than the card currently on top of pile A; otherwise you throw it away. You continue like that until no cards remain in the original pile. What's the expected number of cards in pile A?

2) you have six horses, and you want to find the 3 fastest. You can only have races between 2 horses each time. What's the minimum number of races needed to find the three fastest?

My guesses (probably wrong):

1.) N/2

2.) 12

this one can be done by recursion. think about small N's first.

N=2: obviously expected number is (1/2)*(1 + 2)=A(2)=1.5.

N=3: if we pick 3 first, then A=1. if we pick 2 first, then A=2. if we pick 1 first, we have one card in pile A that is below both cards left. so, this is the same as the case for N=2, plus the one card already there. so A=A(2)+1. this is the key to the problem. so, summing up

A(3)=(1/3)*(1 + 2 + (A(2)+1)) = 1.8333

N=4: we can start to see the pattern

A(4)=(1/4)

(1 + 2 + (A(2)+1) + (A(3)+1)) = 2.0833(3we can substitute for the first three terms using the formula for A(3), so that we get

A(4)=(1/4)

A(3) + A(3) + 1) = (1/4)(4*A(3) + 1)now we can generalize.

A(N) = (1/N)

(NA(N-1) + 1) = A(N-1) + (1/N)A(N) = A(N-2) + 1/(N-1) + 1/N

...

A(N) = A(1) + sum(j=2..N, 1/j)

we know A(1)=1, obviously, so

A(N) = 1 + sum(j=2..N, 1/j)

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