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Bravo, sir.

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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

1st: just conditional probability:
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.

gStyling:

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

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

MonkeyToBe88:

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

feenans:
MonkeyToBe88:

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?

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?

Indeed:

Of course! Gotta know those primes and perfect squares if you're going to spread comps.

If I remember correctly, I think the most difficult question I was asked that was actually within the realm of possibility was to give the angle formed by the hands of a clock when it is 1:47.

Didn't get it exactly right. I was close, however, and still got the offer.

I generally don't ask candidates heavily quantitative brainteasers. If I'm going to hammer somebody it's usually on accounting or financial topics, particularly if they start to get all cocky.

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?

NickTW:

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)

spacejam22:

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

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?

krypt:

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.

krypt:

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?

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...

krypt:

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?

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?

Dealem41:

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.

Dr Joe:

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...

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

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

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?

krnguy228:

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

krnguy228:

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 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

qweretyq:

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.

theATL:

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)

theATL:

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)

A Quarter and Nickle

This better be good,
+Hammy

theHam1:

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

NickCarraway:

A Quarter and Nickle

what a dumbass, r u really that stupid nickcarraway?

gsmsml:
NickCarraway:

A Quarter and Nickle

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.

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.

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?

thinker481:

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.

Jimbo:

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

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=?

chron:

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."

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"

regicide:

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"

The information is enough. Think carefully.

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

b:

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

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...

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.

Two groups of zero pennies ldo.

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...