## Pages

Ill go first:

1) What is 49 x 49?
2) If I have a cube which is 10x10x10, and I take off one layer. How many squares are left?
3) What is the sum of all numbers between 0-100?

1) 49^2 = (50-1)^2 = 2500-100+1 = 2401
2) 8^3 = 648 = 512
3) 100(101)/2= 50
101= 5050

you have three guys, 1 girl, and two condoms. you all want to have heterosexual sex separately with her except you don't want to contract aids (meaning you never actually exchange fluids with either other men or the other woman). How is this possible?

There was a pretty long brainteaser topic a few weeks ago. Over 80 responses. Here's a link.

condom answer: first guy puts on both condoms (unsafe, buts its the only way to do it). second guy puts on the condom that was on top of the first guy when he wore two, because that inside has only touched the other condom. then the third guy turns the last condom inside out and uses that.

Grinder86:

condom answer: first guy puts on both condoms (unsafe, buts its the only way to do it). second guy puts on the condom that was on top of the first guy when he wore two, because that inside has only touched the other condom. then the third guy turns the last condom inside out and uses that.

lol....that's just terrible :( so much for sex ed these days

for number 2) isn't it just 9 cubed....oh shit, never mind, haha

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

I can't believe you're actually asked a condom question. I've NEVER ever been asked any sexual questions during interviews. Prolly since I'm a girl. It's so ridiculous. The girl has three holes right? One can stick it in her pussy, one in the ass and one in the mouth. PRESTO! AIDS schmaids.

********"Babies don't cost money, they MAKE money." - Jerri Blank********

********"Babies don't cost money, they MAKE money." - Jerri Blank********

jesus atropolation... you have a sick sick mind!

and grinder, close but you're not right!

its simple
1 guy fucks her with condom
2nd guy fuckers with other condom
3rd guy inverts one condom puts it on his D, takes the second condom and wraps up his inverted condom with one thats been used.
Vegas Baby

well first guy uses 2 condoms = outside condom (condom 1) has girl fluids and inside condom (condom 2) has guy fluids

second guy uses condom 1 such that he has the unused part on himself and the girl on the side with her own fluids.

the last guy turns condom 2 inside out, puts the 2 condoms together such that the 1st and 2nd guy's fluids are together in the middle while he is wearing the fresh part with the girl fluids side of the condom on the outside again.

to summarize - 2 condoms as follows
girl/unused unused/guy 1
girl/guy 2
girl/guy 2 guy 1/ unused

presto.

pillz is the only one with the correct answer; nothing else works

45mins

So if I'm reading this correctly someone will have to walk the i-phone back after each crossing:

MD and analyst walk across in 30 min followed by the MD crossing the bridge in his lonesome = 31 min

MD and associate cross in 9 min with the MD walking back in 1 min = 10 min

MD and VP cross in 4 min = 4 min

Total = 45 minutes

Making money is art and working is art and good business is the best art - Andy Warhol

MD and VP walk across together, 4
MD walks back with iPhone = 1
MD and AS walk across together = 9
MD walks back with iPhone = 1
MD and AN walk across together = 30

45.

edit: Oh, man. Nicely done sb_guru.

MD and VP walk together = 4 minutes
MD returns back with iPhone = 1 minute
Associate and analyst walk = 30 mins
VP returns back with iphone and joins MD = 4 minutes
MD and VP walk and join Associate and analyst = 4 minutes

Total time = 43 mins

• 2

Wow! But I think others will still get the job :D

sb_guru:

MD and VP walk together = 4 minutes
MD returns back with iPhone = 1 minute
Associate and analyst walk = 30 mins
VP returns back with iphone and joins MD = 4 minutes
MD and VP walk and join Associate and analyst = 4 minutes

Total time = 43 mins

sb_guru:

MD and VP walk together = 4 minutes
MD returns back with iPhone = 1 minute
Associate and analyst walk = 30 mins
VP returns back with iphone and joins MD = 4 minutes
MD and VP walk and join Associate and analyst = 4 minutes

Total time = 43 mins

Win.

sb_guru:

MD and VP walk together = 4 minutes
MD returns back with iPhone = 1 minute
Associate and analyst walk = 30 mins
VP returns back with iphone and joins MD = 4 minutes
MD and VP walk and join Associate and analyst = 4 minutes

Total time = 43 mins

Credit given where credit is due. (What was I thinking?)

Making money is art and working is art and good business is the best art - Andy Warhol

Its 43 min. Good Job

brain teasers are starting to become old news for ib interviews.

Depends entirely on the firm you interview at. At one of the places I interviewed at, they asked me 3 brain teasers, all of which you couldn't necessarily prepare for. At a bunch of others, I didn't get any brain teasers.

Often times, they just want to see how you think through a problem under pressure, not necessarily if you get it "right" or "wrong."

"If you travel at 80 miles per hour, how long does it take to travel 80 miles? "

1. I'm gunna assume these prisoners are champs and wana get their drink on. So you arrange the bottles in a cube and each prisoner drinks 100 bottles which will be either a on the x, y, z plane and from there you play battle ships to find the bottle which killed three prisoners.

EDIT: i googled it, i'm wrong.

Maybe somebody else can actually explain it better/more accurately/in detail but you can have 10 prisoners and assign them all to a certain combination of the 1000 bottles, and 10 people is enough to cover slightly over 1000 different combinations (to cover all 1000 bottles), so that when a certain combination of the 10 people dies, you know exactly which bottle is the only bottle that combination of people drank from. It's that damn thing on the calculator with nCr...

• 1
BlackHat:

Maybe somebody else can actually explain it better/more accurately/in detail but you can have 10 prisoners and assign them all to a certain combination of the 1000 bottles, and 10 people is enough to cover slightly over 1000 different combinations (to cover all 1000 bottles), so that when a certain combination of the 10 people dies, you know exactly which bottle is the only bottle that combination of people drank from. It's that damn thing on the calculator with nCr...

Yes, the power set of a set of 10 people has 1024 elements.

BlackHat:

Maybe somebody else can actually explain it better/more accurately/in detail but you can have 10 prisoners and assign them all to a certain combination of the 1000 bottles, and 10 people is enough to cover slightly over 1000 different combinations (to cover all 1000 bottles), so that when a certain combination of the 10 people dies, you know exactly which bottle is the only bottle that combination of people drank from. It's that damn thing on the calculator with nCr...

you're trying to match a 10-bit binary and the slaves' lives are the bits.

Can anyone elaborate on the answer of BlackHat? I don't really get it..

Haha basically. You have the 1st slave drink like bottle #1, 3, 5, 7, 9, 11... 2nd slave does like 3, 6, 9, 12... and then it ends up such that every bottle has a unique combination of slaves that would die if that was where the poison was. It's also time efficient so it makes the 24 hours period.

10 slaves

500
250
125
63
32
16
8
4
2
1

EDIT: Apparently right answer for the wrong reason. Forgot about the time

Just looked up the answer, a hard one to work out on the spot.

let every slave drink a bit from a bottle.

Wait for 20 hours.

Replace the dead with a prisoner.

brainteasers are most popular in S&T interviews. IB and ER also have them but I don't think as many as S&T.

1100

break it out this way:
each year you get a perpituity of \$10
so in year 1 you get 10 year 2 you get a new 10 + old 10 so forth
value of the first perpituity is 100, second is 100/1.1 third is 100/1.21

1001/1.1^0 + 1001/1.1^1 + 100* 1/1.1^2...+ 100*1/1.1^infinity

this is an infinite geom series and r = 1/1.1 a = 100
100/(1-(1/1.1)) = 1100

no this not the good answer

Bankerph: breakinginnew's answer is correct - its \$1,100, assuming first payment is at period end, and there is no principal recovery (which is implict given the coupon is paid out in perpetuity).

To ride the other brothers' camel.

• 1
SECIB:

To ride the other brothers' camel.

lol or this ^

SECIB:

To ride the other brothers' camel.

And the SB goes to....

He tells them that if they wait around too long either the king or the red jasmine will be dead before they get back, if the flower is dead they failed their mission and if the king is dead then they will never know who will become king.

They mount each others camels. Thus each has incentive to reach there fastest, as if one reaches there first, his camel actually reaches there last. Hence, that person succeeds.

Switch camels and race

you assholes are really playing out my brainteaser posts

to switch camel's?

I liked this one. More please..

It's opposite day bitches.

HarvardOrBust:

It's opposite day bitches.

Haha +1.. If I was interviewing and you answered with this I'd be like "yup, that was it" and just move on

According to my excel model that solves all problems.

It's pretty quick doing it out with a payout tree cause the sum of the expected values converge to \$2.00/3. It'd be a pain in the ass to not be able to use pen and paper though.

KillerMike:

It's pretty quick doing it out with a payout tree cause the sum of the expected values converge to \$2.00/3. It'd be a pain in the ass to not be able to use pen and paper though.

how did you solve it thanks

Ha you're both wrong! You're playing with pounds and pence, not dollars and cents. When one converts your answers into the correct currency, you lose!

Just draw an expected value tree. Fair price would just be the expected payout

how is expected payout 2/3, the expected value increases from 2/3 because the pot increases.

Probability of getting \$1 is (1/3)(1/2), probability of getting \$2 is (1/3)(1/2)^2, probability of getting \$3 is (1/3)*(1/2)^3

so it's like (1/3)*(1/2+1/2+3/8+2/+5/32 ... ) which I just ballparked to converge to \$2

killer mike ain't give a damn what you think about him
last good debate he had was w/michael eric dyson

33.3 cents, you convert it you bloody brit

sorry, that should be 66 and 2/3 cents.

oldhat - can you send me your excel model that solves all problems please

Probability of game going on = Probability of getting 1,2,3 = 1/2

Probability of game ending = Probability of getting 4,5,6 = 1/2

Probability of the game ending with \$1 in the pot is (1/2)(1/2) = 1/2^2
Probability of the game ending with \$2 in the pot is (1/2)
(1/2)(1/2) = 1/2^3
...
Probability of the game ending with \$n in the pot is 1/2^(n+1)
...
Expected value of the amount in the pot when the game ends is E = 1
1/2^2 + 21/2^3 + ...+n1/2^(n+1)+...

E-1/2E = 1/2^2 + 1/2^3 + ...+1/2^n+... = 1/2

Thus, E = 1

Money in the pot can be taken only when the game ends with 4 or 5 being rolled.

The probablity of the game ending with 4 or 5 being rolled = Probabliy of 4,5 out of 4,5,6 = 2/3

So, the expected return of playing the game is 1*2/3 = 2/3

Sorry for putting those annoying numbers and formula here. My english is not good, so it's hard for me to explain it with just words.

Can we resell the tokens? If yes, infinite amount. If no, 0 tokens.

Xdeatel:

Can we resell the tokens? If yes, infinite amount. If no, 0 tokens.

Unless there is very good prospect of infinite demand tomorrow, this approach seems foolish. Also, conceptually impossible.

Assuming no other depreciation, your ROIC is (2/1.5) -1 = 33% which is higher than the WACC (10%). Depending on 1) Amount of capital you have, 2) Time horizon, 3) Method of financing, you can optimize your economic profit (DCF).

Sounds fair to me, seems like a silly question or not enough info provided

Thank you all for your comments. I too believe that this question has some missing information, just wanted to get some second opinions. I found it as one of the questions previously asked in interviews which one of the WSO members posted on here. The said the answer was calculated as follows: "(1.1)^3 = 1.33 which means you should get 3 years of token supply" - an answer that makes absolutely no sense to me.

ceres0305:

Thank you all for your comments. I too believe that this question has some missing information, just wanted to get some second opinions. I found it as one of the questions previously asked in interviews which one of the WSO members posted on here. The said the answer was calculated as follows: "(1.1)^3 = 1.33 which means you should get 3 years of token supply" - an answer that makes absolutely no sense to me.

1) you are only willing to make an investment that EQUALS your required rate of return, can't exceed
2) the rate of token appreciation is 33% over the year even though only a daily appreciation was given

Poorly worded question

25%

I eat success for breakfast...with skim milk

100%. You didnt propose that they weren't true.

"...all truth passes through three stages. First, it is ridiculed. Second, it is violently opposed. Third, it is accepted as being self-evident."

• Schopenhauer

B) 50%?

25% chance of choosing the correct answer, options A and D correspond to 25%. Hence 2/4 or 50%.

1/3

0

LOL

A. 25%
B. 50%
C. 60%
D. 25%

none of these literally say "correct."

if answer C had been "C. Correct" instead of "C. 60%", then the probability wouldve been 1/4. But you gave percentages as answer choices to throw us off.

Nobody knows what the correct answer choice is nor how many correct answer choices there are. The 4 answer choices could've been taken from an MCAT exam for all we know where the question states "Which is correct? You may mark more than one answer."

Since it's impossible to know the number of correct responses to a group of answer choices without its parent question, the only other thing you're asking of us is whether "correct" is among the answer choices to which I answer, "0/4" or "0%"

5 posts, 5 answers....interesting...and if you wish, you can swap the word "true" for "correct"...that isn't it...

Prim.<span class=keyword_link><a href=/finance-dictionary/equity-research-overview rel=nofollow><abbr title=equity research&#10;>ER</abbr></a></span>.ate:

5 posts, 5 answers....interesting...and if you wish, you can swap the word "true" for "correct"...that isn't it...

haha DAMN. i thought this was going to be a grammar-based brainteaser in disguise for once lol

The sentence below is false.
The sentence above is true.

I'm amazed other people have actually looked at reddit math

Reality hits you hard, bro...

MMBinNC:

I'm amazed other people have actually looked at reddit math

i never even knew reddit math existed haha. thanks for giving me something to look at at work now.

maybe i should start posting questions from my old college discrete math textbook on here

there are only 3 possible answers, SINCE A and D are the same thing...IF WE CONSIDER that there is only one correct answer, the odds would be of exactly 1/3...

1/3

its one way or the other: hate me or admire.

If these are asked at job interviews, then I am well prepared...kkkkkkkkkkkkk

Oh this is going to take a while...(the question's gone viral at the moment if that's any help)

Just Do It

Answer cannot be 1/3 because 25% is represented twice in the choices, so higher chance of randomly choosing it. Just like HT/TH case with coins... Same result overall, but got to account for probability of randomly landing tails or heads first.

Or have I got it completely wrong?

you have to count A nad D as 1 choice

it is impossible to know for sure....for all we know, 25% could be the answer...since A and D are 25%, the probability would be 2/4 or 1/2....

if the answer is NOT 25%, we need to put A and D as one single answer, giving the answer of 1/3...it is impossible to know for sure.

I got ~38%. Is the answer an actual answer choice? Where are people getting 0%?

Logic (probably wrong): If choose A or D, 25% of the time you have a 50% chance of being correct. Else, 25% chance if B or C.

XPJ:

I got ~38%. Is the answer an actual answer choice? Where are people getting 0%?

Logic (probably wrong): If choose A or D, 25% of the time you have a 50% chance of being correct. Else, 25% chance if B or C.

the choices are answers to the question. you have a .5 chance of picking 25%, so 25% is not the correct answer; you have .25 chance of picking 50%, so 50% is not the answer.

42.....but seriously its 37.5% or 6/16

It's not a paradox. If the choices were

A. 25%
B. 50%
C. 0%
D. 25%

then yes, it would be a paradox.

it's not a paradox, 0% works perfectly

jec:

it's not a paradox, 0% works perfectly

Maybe I'm missing something, but I still don't see the logic behind 0%.

For example:

If I choose a random answer from among those below, what is the probability that the answer is correct?

A. Blue
B. Red
C. Yellow
D. Blue

Assuming that either Red, Blue or Yellow is the correct answer, I don't see how you get 0%. The %'s are thrown in as a mind-fuck. Unless, it's wrong to assume that R, B or Y are the 3 possible answers in which case this is a retarded question.

37.5%

a/a
a/b
a/c
a/d
b/a
b/b
b/c
b/d
c/a
c/b
c/c
c/d
d/a
d/b
d/c
d/d

6/16 or 37.5%

1750 equal distribution to 4 pirates including himself? That way he locks up 4 votes and gets a quarter of the loot?

Or maybe now I think about it 1001 for 3 lowest ranking pirates and 3997 for the captain, since they know they'd get more than the 1/7 share, and they'd have a ways to go to get to be the captain.

That's not how I would do it.

Solve by induction.

Edit: you wanted the number, propose 6997.

• 1

.

6997 for captain
1 for fifth
1 for third
1 for last

It's the usual MBP "brainteaser" where it all boils down to the one vs. one situation
where the next to last pirate can take all 7000 because he needs only his own vote.
So it would be enough to offer the last 1 because otherwise he would receive 0.

1. cap 7000
2. 0
3. 0
4. 0
5. 0
6. 0
7. 0

aye! come at me bros

Unforseen:

1. cap 7000
2. 0
3. 0
4. 0
5. 0
6. 0
7. 0

aye! come at me bros

This man deserves a SB hahaha you just made my day

pirates -> captain's proposal (voting breakdown)

1 -> 7000 (1)
2 -> 7000 (0 1)
3 -> 6999 (1 0 1)
4 -> 6999 (0 1 0 1)
5 -> 6998 (1 0 1 0 1)
etc.

dabanobo:

# pirates -> captain's proposal (voting breakdown)

1 -> 7000 (1)
2 -> 7000 (0 1)
3 -> 6999 (1 0 1)
4 -> 6999 (0 1 0 1)
5 -> 6998 (1 0 1 0 1)
etc.

You got it.

If pirate 1 is the only pirate left he gets 7000 coins. This case will never happen.

If Pirate 1 and 2 are left, pirate 2 will get 7000 coins because he can give himself half of the votes.

If 3 pirates are left, pirate 1 gets 1 coin, and pirate 3 gets 6999 coins, pirate 2 gets zero. Pirate 1 will vote for this distribution because if he doesn't, pirate 3 will die and all the coins will go to pirate 2.

If 4 pirates are left, Pirate 4 will give 1 coin to pirate 2. This secures is vote because he gets one more coin than the 3 pirate scenario. Pirates 1 and 3 get nothing, Pirate 4 gets 6999.

If 5 pirates are left, Pirate 5 will offer Pirates 1 and 3 two coins each in order to secure >50% of the vote. Pirates 2 and 4 get nothing. Pirate 5 gets 6998 coins.

If 6 pirates are left, Pirate 6 will prevent mutiny by giving one coin to Pirates 2 and 4. Pirates 1,3, and 5 get nothing, and Pirate 6 takes 6998 coins.

If 7 pirates are left Pirate 7 needs to earn 3 votes besides his own to prevent mutiny. He offers 1 coin each to pirates 1, 3 and 5. None to pirates 2, 4, 6. He takes home 6997 coins.

The key is to offer just enough coins to secure at least half the votes. You secure a vote by offering more coins than the pirate would get in a scenario with one less pirate.

• 1

If pirate 1 is the only pirate left he gets 7000 coins. This case will never happen.

If Pirate 1 and 2 are left, pirate 2 will get 7000 coins because he can give himself half of the votes.

If 3 pirates are left, pirate 1 gets 1 coin, and pirate 3 gets 6999 coins, pirate 2 gets zero. Pirate 1 will vote for this distribution because if he doesn't, pirate 3 will die and all the coins will go to pirate 2.

If 4 pirates are left, Pirate 4 will give 1 coin to pirate 2. This secures is vote because he gets one more coin than the 3 pirate scenario. Pirates 1 and 3 get nothing, Pirate 4 gets 6999.

If 5 pirates are left, Pirate 5 will offer Pirates 1 and 3 two coins each in order to secure >50% of the vote. Pirates 2 and 4 get nothing. Pirate 5 gets 6998 coins.

If 6 pirates are left, Pirate 6 will prevent mutiny by giving one coin to Pirates 2 and 4. Pirates 1,3, and 5 get nothing, and Pirate 6 takes 6998 coins.

If 7 pirates are left Pirate 7 needs to earn 3 votes besides his own to prevent mutiny. He offers 1 coin each to pirates 1, 3 and 5. None to pirates 2, 4, 6. He takes home 6997 coins.

The key is to offer just enough coins to secure at least half the votes. You secure a vote by offering more coins than the pirate would get in a scenario with one less pirate.

You weren't first, but SB anyway for taking the time out to spell it out.

As the Captain, tell you team to sail to shore. Once your there, take the gold, kill them all, buy a new ship. Done. :)

In other words, if it is an even number of pirates, just give every pirate with an even number rank 1 gold coin.
If it is an uneven number if pirates, just give every pirate with an uneven number rank 1 gold coin.

Hi MBP.

• 1

.

Can one be rational AND selfish, assuming selfish means they want more for themselves than anyone else?

BlackHat:

Can one be rational AND selfish, assuming selfish means they want more for themselves than anyone else?

This is a good answer if you've stalked the interviewer and know they majored in philosophy :)

• 1

Anyone?

I've seen a couple of variations of this with 2 instead of 3. Working on it

You ask liar who he is-- he will say that he is Honest.
Then you ask Honest if Liar is lying.
Then I have no fucking idea :/?

dunno, but couldn't you just ask them who the person next to them is?

if the god says that the person next to him is honest then that person has to be random, if he says that the person next to him is random he must be honest, if he says that the god next to him is a liar you can't be sure on anything yet

at this point, if you already know who's random or honest, you can just ask the other two who that god is (the one you determined)... liar will call him a liar, honest will call him what he really is, and random will call him honest (if it's truly honest, the only other choice in this scenario, you know that liar wouldn't say that)

then you do the same idea again if needed...

I dunno, my explanation is kinda long, but the idea seems pretty easy (i'm looking at them as asking them left to right)

edit: gah, I know I must be missing something in my explanation, but wouldn't it work out even if you could only ask three questions to each? there's probably a way simpler explanation, but whatever, I tried -_-

If your dreams don't scare you, then they are not big enough.

"There are two types of people in this world: People who say they pee in the shower, and dirty fucking liars."-Louis C.K.

GODS - A/B/C

"Would you deny that A is RANDOM?"
1. If B answers 'ja', then
either B is RANDOM (and is answering RANDOMly ),
or B is not RANDOM and A is indeed RANDOM .
Either way, C is not RANDOM definitely
2. If B answers 'da' , then
either B is RANDOM (and is answering RANDOMly),
or B is not RANDOM and A is not RANDOM .
Either way, A is not RANDOM definitely

Go to the God who was identified as not being RANDOM definitely based on B's answer
by the previous question (either A or C) and find out whether he is GOD - TRUE or FALSE by asking the following question...

"If I asked you 'Are you TRUE-GOD?', would you say 'no'?

if he says NO then he is GOD-TRUE....if he says YES, then he is GOD-FALSE

Question No. 3 -

Ask the same GOD (who was identified as not being RANDOM) the question:
"If I asked you 'Is B RANDOM?', would you say 'no'?".
1. If the answer is 'no' then B is RANDOM;
2. if the answer is 'yes' then the GOD you have not yet spoken to is RANDOM.

The third GOD can be identified by elimination after we have identified first 2

SC911:

GODS - A/B/C

"Would you deny that A is RANDOM?"
1. If B answers 'ja', then
either B is RANDOM (and is answering RANDOMly ),
or B is not RANDOM and A is indeed RANDOM .
Either way, C is not RANDOM definitely
2. If B answers 'da' , then
either B is RANDOM (and is answering RANDOMly),
or B is not RANDOM and A is not RANDOM .
Either way, A is not RANDOM definitely

Go to the God who was identified as not being RANDOM definitely based on B's answer
by the previous question (either A or C) and find out whether he is GOD - TRUE or FALSE by asking the following question...

"If I asked you 'Are you TRUE-GOD?', would you say 'no'?

if he says NO then he is GOD-TRUE....if he says YES, then he is GOD-FALSE

Question No. 3 -

Ask the same GOD (who was identified as not being RANDOM) the question:
"If I asked you 'Is B RANDOM?', would you say 'no'?".
1. If the answer is 'no' then B is RANDOM;
2. if the answer is 'yes' then the GOD you have not yet spoken to is RANDOM.

The third GOD can be identified by elimination after we have identified first 2

SC911:

GODS - A/B/C

"Would you deny that A is RANDOM?"
1. If B answers 'ja', then
either B is RANDOM (and is answering RANDOMly ),
or B is not RANDOM and A is indeed RANDOM .
Either way, C is not RANDOM definitely
2. If B answers 'da' , then
either B is RANDOM (and is answering RANDOMly),
or B is not RANDOM and A is not RANDOM .
Either way, A is not RANDOM definitely

Go to the God who was identified as not being RANDOM definitely based on B's answer
by the previous question (either A or C) and find out whether he is GOD - TRUE or FALSE by asking the following question...

"If I asked you 'Are you TRUE-GOD?', would you say 'no'?

if he says NO then he is GOD-TRUE....if he says YES, then he is GOD-FALSE

Question No. 3 -

Ask the same GOD (who was identified as not being RANDOM) the question:
"If I asked you 'Is B RANDOM?', would you say 'no'?".
1. If the answer is 'no' then B is RANDOM;
2. if the answer is 'yes' then the GOD you have not yet spoken to is RANDOM.

The third GOD can be identified by elimination after we have identified first 2

Lol, you totally didn't do this yourself. You even used the terminology of the original problem (although still simplified since in the original, you don't know whether ja means yes or no)

this is one of those things where it's gonna be hard to cover all possible scenarios before I stop giving a shit

If your dreams don't scare you, then they are not big enough.

"There are two types of people in this world: People who say they pee in the shower, and dirty fucking liars."-Louis C.K.

• 1

^BINGOOOOOOOOOO

well well well, looks like a little cheater had to come in here and ruin the fun... what happens to the SB now? I say you give it to the only person who actually gave it an honest shot before this rapscallion came in here and ruined our lives.

If your dreams don't scare you, then they are not big enough.

"There are two types of people in this world: People who say they pee in the shower, and dirty fucking liars."-Louis C.K.

• 1
wolverine19x89:

well well well, looks like a little cheater had to come in here and ruin the fun... what happens to the SB now? I say you give it to the only person who actually gave it an honest shot before this rapscallion came in here and ruined our lives.

Happy?

I did it! I did it!

did my explanation even make any sense to you? ha

If your dreams don't scare you, then they are not big enough.

"There are two types of people in this world: People who say they pee in the shower, and dirty fucking liars."-Louis C.K.

Just to figure it out as I go to google for most of my problems, #winning.

SC911:

Just to figure it out as I go to google for most of my problems, #winning.

I hope you look forward to the following interview exchange:
[INTERVIEWER]: How would you solve {insert brain teaser here}?
[SC911]: I'm not sure, but I am REALLY good at pressing ctl+c, ctl+v
[INTERVIEWER]: Get the fuck out.

Well, that's dumb.. Obviously what I would do/say in an interview.

These types of questions are so lame.

Easy, in two questions or less.

Label gods as A, B, and C. They are in a circle so that B is right of A, C is right of B and A is right of C.

If no one answers Random, move them into the order "BAC". Ask them the same question.

By now, one will have answered Random. This is "Honest", since random would never label someone else as random and since the liar labels everyone else as "liar". The person who has been labeled Random is random, and the remaining god is honest.

a towel

?

Family Guy?

99.

I feel so wise.

rufiolove:

When evaluating whether or not to post something on WSO, I think to myself, "would an idiot post this" and if the answer is yes, I do not post that thing...

9^9

GreenwichForLife is right.

^ doesn't count as a symbol or operation?

If you write 9 to the 9th power with a pen and paper, there would be no symbol.

RubiksCubeMath:

If you write 9 to the 9th power with a pen and paper, there would be no symbol.

Exponentiation is an operation.

couldn't you just say "nine hundred trillion to the nine hunded trillitionth (sp?) power" and you've used no digits and no symbols or operations?

2

Is it just 3 or am I missing something?

Grab one sock, white or black, put it aside.

Grab another sock, possibly the other color.

If you grab a third no matter what you're going to have a pair.

"The only point in making money is so you can tell some big shot where to go."
-Bogie

MSFhopeful:

Is it just 3 or am I missing something?

Grab one sock, white or black, put it aside.

Grab another sock, possibly the other color.

If you grab a third no matter what you're going to have a pair.

Yeah.

3 ?

3 - after picking 3 you will either have 2Blacks and 1 white or 1black and 2 whites. either way there is a pair

Funniest

never said the pair needs to be the same color

• 2
melvvvar:

never said the pair needs to be the same color

you have been wearing this all along..havent you?

melvvvar:

never said the pair needs to be the same color

OP should have clarified.

^ ironic for a question that depends on reading the problem carefully

I don't need any goddamned socks. My pair's already swinging between my legs. Wild and free.

Is it not 2? Two is a pair?

2

This is your definition of fun?

youre taking the fun out of the brainsteasers

You must have a TON of friends.

For the first four, use one of the following three formulas:
(a+b)(a-b)=a^2-b^2
(a+b)^2=a^2+2ab+b^2
(a-b)^2=a^2-2ab+b^2

I'll leave it to you guys to think of what the "a" and "b" should be in each situation. That's a general trick for most "fast" multiplication problems people tend to ask: they usually fall into one of these patterns you can use. Except when they don't, and life is miserable (I had that happen to me on a phone interview and was pretty sad when I terribly failed).

For the last one... It's something annoying about squaring something with only 1s. It ends up being an ascending natural number sequence followed by the same thing in reverse.
Ex:
11=1
11
11=121
111*111=12321
...

These are just good tricks to keep in mind since I've heard bank bros occasionally ask these things.

canas15:

For the first four, use one of the following three formulas:
(a+b)(a-b)=a^2-b^2
(a+b)^2=a^2+2ab+b^2
(a-b)^2=a^2-2ab+b^2

I'll leave it to you guys to think of what the "a" and "b" should be in each situation. That's a general trick for most "fast" multiplication problems people tend to ask: they usually fall into one of these patterns you can use. Except when they don't, and life is miserable (I had that happen to me on a phone interview and was pretty sad when I terribly failed).

For the last one... It's something annoying about squaring something with only 1s. It ends up being an ascending natural number sequence followed by the same thing in reverse.
Ex:
11=1
11
11=121
111*111=12321
...

These are just good tricks to keep in mind since I've heard bank bros occasionally ask these things.

hmm, I just round to the nearest 10,
1921 = 2021-21 = 420-21= 399

did all but last one like this within 5-10 secs

I'd rather go through problems like ? ln(x) dx, but i doubt anyone here wants to go through integrals. Also, I am starting with relatively simple things. In the future I will add some more challenging ones. Problem solving skills cannot be learned by searching google for the answer to these questions and memorizing them, but rather building a foundation by solving problems.

RubiksCubeMath:

I'd rather go through problems like ? ln(x) dx, but i doubt anyone here wants to go through integrals. Also, I am starting with relatively simple things. In the future I will add some more challenging ones. Problem solving skills cannot be learned by searching google for the answer to these questions and memorizing them, but rather building a foundation by solving problems.

I would prefer integrals...

Best Response

The hardest brainteaser I ever got was a riddle:

It starts with the letter "S".
It's two words.
It's on every forum.
It rhymes with the phrase "Church Glutton".

S _ AR _ H
B_ _ _ ON

Solve the puzzle.

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

• 4
goodL1fe:

The hardest brainteaser I ever got was a riddle:

It starts with the letter "S".
It's two words.
It's on every forum.
It rhymes with the phrase "Church Glutton".

S _ AR _ H
B_ _ _ ON

Solve the puzzle.

:-D

+1

Money Never Sleeps? More like Money Never SUCKS amirite?!?!?!?

Let me search for that button to research this answer more throughly

@goodl1fe

I was beginning to solve your brain teaser until I slowly realized you were trolling OP. +1 SB for you.