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Career Resources
1111 x 1111
1111 x 1111 = 1111000 + 111100 + 11110 + 1111 = 1234321
Notice the pattern: 11 x 11 = 121 111 x 111 = 12321 1111 x 1111 = 1234321 11111 x 11111 = 123454321 ...
Nice try, but there is no way you would be able to come up with a patern like that in your head while stressed out interviewing, plus adding up 1111000 + 111100 + 11110 + 1111 in your head is pretty challenging.
The way I did it in my head was like I would do it on a piece of paper and it took me about 5 seconds.
1111 x 1111 ----- 1111 --- 1111 - 1111 1111and just add the numbers up vertically to get 1234321
Walk into an interview and the first thing that happens is that one of the partners throws me a quarter. He says "if you stack quarters one on top of another, how many will you need to reach the top of the Sears (Willis now) Tower?" Follow up to that, "if you take all those quarters to cover the area of this paper, can you reach the floor of this room to the ceiling?" Followed by how many fish are in the world?
//www.wallstreetoasis.com/forums/answering-brainteasers //www.wallstreetoasis.com/forums/brainteasers //www.wallstreetoasis.com/tag/brain-teasers //www.wallstreetoasis.com/forums/brain-teasers
^ what he said.
You have 50 white balls, 50 black balls, and two buckets. How should you allocate the balls (you have to use them all) so that you have the best probability of pulling a white ball out of a random bucket?
I'd put 25 black balls in the bottom of each barrel and 25 white balls on top of the black ones. Maybe not what you were looking for, but it would guarantee you a 100% chance of grabbing a white ball, no matter the barrel you picked from (assuming that you didn't dig into the bottom and that the circumference of the barrel isn't too large).
Put 1 ball in the first bucket, then the rest of the balls in another. Your probability for picking a white ball will in this way be maximized :)
Need answers to brain teasers, please help! (Originally Posted: 11/01/2010)
1: Chess board, 8 by 8, how many squares in total?
2: 0,1,2,3 can use all digit once and one digit twice to get 1.68 as final answer (can use 2 digit at once e.g. 12)
Thanks guys
This last one is a trick question isn't it? if you are picking out of a "random bucket" then you might as well be picking from a huge bucket with all the balls in it. no matter what you do your chances will be 50/50
right?
wrong. you're assuming that you have to split up all the balls uniformly.
Noticing a pattern?
If you have a nxn chessboard. You will have 1 nxn square, 4 (n-1)x(n-1), 9 (n-2)x(n-2) .....
Or more concisely, for i = 1 to n, you have i^2 squares of size (n+1-i).
So to find the total number of squares in an nxn chessboard, just add the first n squares.
In this case, it's 1^2 + 2^2 ........+ 8^2.
2) 1.3^2-.01
start with the smallest square combo so you instantly have 64. I started with 8 after, but you can easily do it with 1,2 and the pattern should be just as easy to recognize. 8 block squares = 1 (the entire board), 7 block squares = 4, 6 block squares = 9...at this point you need to tell the interview that using the pattern you would assume that 5 block squares = 16, 4= 25, 3=36, 2=49, 1 =64 (which I already did). Add them up and you get 204 squares.
^That's what I was thinking too, but technically it's 1.3^2 - 0.01, which doesn't satisfy the condition.
Sorry, I was referring to the post by streetlegend.
gekko, I also thought that yours might work, but it's not an exact answer, if we are told that we can round to the nearest hundredth, then it would be 1 +2/3 +0.01, which has the same problem I outlined earlier.
let me help you out right there, they really don't give a fuck what the answer is. The purpose of the question(s) is to see how your mind works...in that problem's case, do you have simple problem solving skills and can you make numbers work work to your advantage. As you can see there are multiple ways to get the same answer, which makes this probably the best brainteaser related to IBD I have ever seen,,,I mean that is what a banker does, takes numbers that normally equal 6 and makes them look like 7.65. Don't get hung up on the technicals of rounding too much. I am sure that those two above answers would have been acceptable to the interviewer, but you can ask him about rounding.
for questions two, all operation are allowed but not sure 0.02 (.02) count as one 0 or two 0s....
Anyone with a better answer for question 2?
lol, thanks gekko. Coming from a pure math background, I need a reality check every once in a while.
wkc207 as gekko pointed out, both answers are fine. Just state your assumptions first (rounding to nearest hundredth or by .02 you mean 0.02 etc.)
1 + 1 - .32 = 1.68
progolfer...forgot to add the zero..i'll take credit for the whole thing
You're supposed to subtract it.
Here is a more detailed answer to #1: http://puzzles.nigelcoldwell.co.uk/twentyseven.htm
For a NxN Chessboard: 1. You get NxN 1-squres. 2. You get (N-1)x(N-1) 2-squres--going left to right you get (N-1) 2-squares, and likewise going down. 3. You get (N-2)x(N-2) 3-qaures--same logic; just draw a picture and check. ... N. You get 1 N-square.
So it's simply 1+2^2+...+7^2+8^2. There's a formula for the sum of squares but I'm sure the interviewer will be satisfied enough with that expression.
1) I would say 8*8=64, if you don't remember a chess board is 8 by 8 you can count the number of different chess pieces when you set it up
2) I did 2-(1/3)+.02 or 2-(1/3)+.01 if they get mad about the rounding
put 1 white ball in one bucket and the rest in the other bucket.
Brain teaser preparation guide? (Originally Posted: 12/08/2011)
Hi guys:
i have an interview for fixed income trading coming up, which is going to include a brainteaser component. can someone recommend a good book for brainteasers? the ones i found on vault guide wasn't very good....
thank you so much!
Pretty sure that no matter how you divide them, you will get the same 50/50 split.
(1/50).5 + (49/50).5 = .5 (20/50).5 + (30/50).5= .5
That's why I just said put the 25 black balls on the bottom, then put the 25 white balls on top of the black ones. I could be wrong on this one, but it makes sense to me.
Who said you have to put 50 balls in each bucket? I agree with 1 white in one bucket and the 99 others in the other.
You have eight weights and a balance (scale that determines which side weighs more). Exactly one weight is heavier than the others. What is the minimum number of times you need to use the balance to determine which weight is heaviest?
I was asked a variation of this brain teaser by an MD during a FT interview. He asked how one could find the heaviest weight using the scale only twice. Hadn't heard it before, but I worked through it and got the right answer. I ended up getting an offer, and I'm pretty sure that correctly answering this brain teaser made the difference.
For Illini Programmer's brainteaser:
Is the answer 3?
1) 4 on each side, heavier side has heavier weight, choose those 4 weights 2) 2 on each side, heavier side has heavier weight, choose those 2 3) 1 on each side, heavier side is heavier weight
easy, its 2 times.
divide into 3 groups. weigh two groups, and you'll find out which group the heavy weight is in. eliminate all other groups. weigh 2 of the weights in the heavy group and voila.
damn i wish i got some some these questions...they sound pretty easy to me
search you lazy b*******
.
wso sells one
http://www.ibankingfaq.com/category/interviewing-brainteasers/
Brain Teasers - Quick read (Originally Posted: 12/27/2011)
From google.
http://online.wsj.com/article/SB100014240529702045523045771130037050897…
Its just sad that instead of this, we continue to have the MD, VP, Associate and Analyst cross the bridge with one flashlight.
For the question about the buckets and balls I'd put 49 white balls and 50 black balls in one bucket and one white ball in the other bucket. I could be wrong about that. I never got any probability questions in my interviews but got a few things like "how much does a 747 weigh? How many people voted in X City last year? How many boxing gloves were sold in X state last year?"
I don't really see it as sad. Working in IBD doesn't necessarily require the amount of intellect as someone working in software development at Google does.
2-4 are extremely doable. 1 and 5 are retarded.
I like 2-4. Lol if you can get #1 and wtf at #5
2 is high school science 3 is in every puzzle book and 4 is middle/high school math - anyone who cannot immediately get them should be dinged 1 is retarded 5 involves "lateral thinking" which most companies explicitly choose not to test because it is stupid to test for.
Have you personally recieved brainteasers at your interviews? (Originally Posted: 03/11/2013)
if so, how many? and which interview?
For the mental arithmetic ones you guys should look at vedic mathematics. My grandfather taught it to me growing up and it means I can pretty much do any simple mathematical problem in a few seconds. Just an example:
45^2 always ends in 25 and starts with 4 times one more than 4 so 4 x 5 = 20. Put them together and you get 2025
35^2 = 1225 (3x4)
65^2 = 4225
etc
I got one during interview (successfull) in one of the Big4 firms. The clock shows 3-15 time. What is the angle (exactly, without rounding, of course) between hour and minute hands.
Not a tough one but I liked it a lot (coincidence?).
Had 3.
Clock @ 3:15 2 glasses with orange & water mixed together 9 coins, find the odd one out with a set of scales
had 2.
1.5 chickens, 1.5 eggs, 1.5 days. how many eggs does one chicken lay in one day. 3:15 degree one
I had 4 for my technical interview.
Try reading "Heard on the Street"
Had on average 5/superday for BB S&T SA.
S&T, hedge funds, and AM I have received brainteasers. Either Superdays or 1st rounds. They just want to see how you think and communicate your thought process.
Is it better to be up front and say you don't know if a brainteaser really baffles you, or better to try even if you get it totally wrong?
You are given 13 balls and a scale. Of the 13 balls, 12 are identical and 1 weighs slightly more. How do you find the heavier ball using the scale only three times?
Pretend you have a bucket of water, a 5 liter container and a 3 liter container. How do you come up with exactly 4 liters of water?
During consulting interview, interviewer said I could pick one and usually people don't get it right. Told him I knew both, and explained them in like a minute each. Pretty easy in my opinion.
Hardest brainteaser ever? (Originally Posted: 11/16/2011)
More mathy brainteasers! So if A^n+B^n=C^n , we all know there are many cases for n=2 (right triangles...) but for n>2, there are no solutions.
Prove it.
Anyone that knows anything about math/this problem: don't say anything. Just curious as to how people approach this problem/whether it causes funny arguments.
hardest brainteaser i got was find the square root of sum of all prime numbers from 1-1000 (for boutique IBD). tricky because a lot of people thought 1 was prime and screwed up...
Anyone for this question??
philliefan: is that a joke?
I'm sure it isn't. The correct answer (according to Excel) is 275.9112176.
lol didnt read your last sentence. deleted.
Are you kidding me? Who the fuck gives this at an interview? I'd punch them in the face and walk out.
After solving it of course.
Trivial by inspection
I think this one may be too difficult for non-maths/cs people
I have discovered a truly marvelous proof of this, but it's too long for this comment box.
lol, well done.
true wasn't wile's proof like 200 pages or something, gl getting through that in an interview
HA ha ha ah ah......I'm sure NO ONE got asked this in an interview.
The proof is like 100 pages long, it's probably the most argued theorem in math, and way beyond even some PhD's .
+1 SB to Joe Monkey for quoting the person who first claimed to solve this interview brainteaser.
Yeah, this was a social experiment. Didn't expect this many people to be familiar with Fermat's last theorem/wiles' proof on WSO.
And yeah, you need really advanced graduate level math to even begin understanding the proof. You have to admit, with the mathematical ineptitude displayed in some of the brain teaser threads, it wasn't that farfetched to think most people here wouldn't recognize it on sight.
This logic puzzle, on the other hand, actually is very hard. It's not conceptually a brain teaser, per se, but requires really deep logical thought. You practically need to do everything based on symbolic logic.
"Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are da and ja, in some order. You do not know which word means which. " http://en.wikipedia.org/wiki/The_Hardest_Logic_Puzzle_Ever
To answer your original questions:
http://math.stanford.edu/~lekheng/flt/wiles.pdf
Have fun!
You have an unlimited supply of quarters, dimes, nickels, and pennies. What is the greatest value you can reach such that the coins that you have chosen cannot be rearranged to equal exactly $1.00?
solution: imagine there is a bowl in front of you. take the bowl, your coins, and carry out the following steps:
[step 1] pick a denomination, d, that has not yet been selected. if there are no denominations left to choose from, stop. otherwise, go to step 2.
[step 2] determine the maximum number, n, of coins of denomination d that you can put in the bowl such that you cannot, even with the change in bowl from any previous steps, make change for a dollar. then, go to step 3.
[step 3] put n coins of denomination d in the bowl. then, go to step 1.
so, if i go dimes, quarters, nickels, then pennies, we get: 9 dimes + 1 quarter + 0 nickels + 4 pennies. no matter which denomination you start with, the max is $1.19
Didn't wiles present at some conference and it took various hours to go through it all?
I don't think it's unreasonable to ask questions that are clearly too hard to solve in an interview. Depending on the question, the candidate can still make some progress, demonstrate their ability to think about really difficult math/problem solving, and show that they are not overconfident.
Off the top of my head this problem isn't a good one for an interview, but my point is just that being (nearly) unsolvably difficult doesn't rule out a problem as a good interview question.
"Follow up to that, "if you take all those quarters to cover the area of this paper, can you reach the floor of this room to the ceiling?""
I got asked something like this once. Guy asked me, if there was a stack of quarters the height of the building, what market i would make in the percentage of the office we'd need to hold them all.
Brainteaser - Unsure if I got the answer (Originally Posted: 06/04/2012)
I'm a longtime reader and first time user for WSO. Last Thursday I got interviewed at a small $300 - 350 million hedge fund in San Francisco. In the interview I was asked a brainteaser that I am unsure if I got the answer to and it has been bothering me all weekend. The reason I don't know the answer is because my interviewer, one of the two PMs, just grinned when I gave the answer and moved on to the next question.
Anyways the question is: You have a deck of cards with 50 cards.
Four of the cards are Aces.
Four of the cards are Kings.
The rest of the cards are blank. (42 cards blank)
After the cards are shuffled they are all laid out next to one another. In order to win the game an Ace needs to be next to a King. What is the probability of winning?
I was allowed to use a pen & paper, but no calculator.
Umm... 32 in 1225 or 2.6%?
That's a nasty interview question. I had a question like it in an algorithms class.
This professor gives a good overview of the reasoning behind it (adapt it for 50 cards): http://recursed.blogspot.com/2010/01/neat-problem-on-card-arrangements…
Was this Grandmaster Capital? It sounds like a question they would give, given the Clarium heritage and Patrick Wolff's genius level mind.
That awkward moment when you realize that professor, is your professor... (Ian Gouldon not the other fellow)
edit
Get some sleep. I bet the PMs couldn't have gotten the answer themselves. I think they just wanted to see your thought process.
The correct answer is to grab a sharp pen and stab your interviewer in the jugular.
Assuming the cards are in a circle instead of a line so every card will have two other cards next to them. Their are four aces, which can be taken as a given, 8 cards will be next to the aces (which could include another ace, a king, or a blank card).
I would think of it as what are the odds of it not happening. You get 8 trials, where each trial incorporates one less card to pull from, increasing (slightly) the odds that each next card is a king next to the ace.
1- ((45/49)x(44/48)x(43/47)x(42/46)x(41/45)x(40/44)x(39/43)x(38/42)) = 52.2%
Is that what you got?
Simulating it with 1,000,000,000 trials I got 50.1%
When I started it on paper I also modeled it as a circle. I don't think your method is correct, though, because the number of places where an ace can be next to a king depends on the particular distribution of aces; i.e., given four aces placed there may be anywhere from two to five places where a king may be placed to satisfy the requirement (in the circular model) or one to five (in the linear model).
This was my approach:
1) How many ways are there to place four aces in fifty locations? (50 nCr 4) 2) How many of those have 2, 3, 4, 5 "sweet spots"? 3) What is the probability of missing all of those "sweet spots" with your four kings, for each number of sweet spots?
It basically comes down to the weighted sum of the complements of missing all of the sweet spots.
Also note that in the circular arrangement there are slightly more ways to get the win.
Thanks for the solution and the question I asked to you guys was the simplified version of the question. In the real question my interviewer asked:
Q = interviewer D = me
Q:Pick two cards. (King, Queen, Ace...) D: Ace and Jack Q: You like blackjack? D: Yea, more of a poker player, but still fun from time to time. Q: Well, you aren't playing poker or blackjack. You are playing the probability game and here it is:
Following this he asked the most difficult question:
I then started talking about how all the other 42 cards are irrelevant. Hence the blank cards. I think the rest of my explanation sounded alright, but who knows. After this question he then switched it up about asking me about something on my resume.
I was wondering why the deck only had 50 cards...
Jokers brah
To make sure the OP couldn't just recall the solution from memory
P (AK or KA) in slots 1 and 2 is 2(4/504/49) = 0.0130612245 probability for slots 2-3, 3-4, ... 49-50 is 490.0130612245 = 0.64
0.64
don't know if this is right, but i wouldve approached it using combinatorics.
So you have 8 kings and aces, and you can use PIE to try to count how many combos there are that let you win. You could then break into the event that you have at least 1 pair, then subtract event that you have at least 2, add event that you have at least 3, then subtract that you have at least 4. And then for each event multiply by the number of possible ways to arrange.
So for example the #combos that you have at least 1 pair is 4C14C1 * 49!. You choose a king and an ace and stick them together and view them as 1 card. Since you would then only have 49 "cards" (its 50 - 1 since you have 1 pair) to distribute, there are 49! ways to place everything. But this will double count when you have 2 pairs, you have to apply PIE and subtract out the #combos of at least 2 pairs which is 4C24C2*48!, then add back the 3 pairs, and then subtract the 4 pairs. Then divide that number by 50!.
so you get something like (4C14C149! - 4C24C248! + 4C34C347! +4C44C446!)/50! cancel stuff out and you have 16/50 - 36/(5049) + 16/(504948) -1/(50494847)~15/50 since the last 2 terms are really small
You win if you have at least one "pair" -- the two, three, four pair cases include the at least one pair case which satisfies the victory condition.
Agree with the above, the chance to win is rare. =424140.....1/504948*.....=0.0000000000000003453=0
The cards are not laid out in a circular manner. I asked him this during the interview.
I'm not going to work this out, but for future reference you should always try to use some form of induction. Whether that's induction on the total number of cards or the number of kings/aces depends on what you feel more comfortable attempting first. Once you can solve it for some base cases, you generally start seeing patterns on how to generalize to an arbitrary set.
Given the wide range of answers here (with ample time and no pressure), it's safe to assume that they were more interested in your thought process and whether you looked compose rather than looking like you took a shit in your pants. If you got it right tho, then hat-tip to you sir.
Thanks for the responses. I received an email today saying I would hear back at the latest by Friday. If I get the internship I'm gonna ask my interviewer for the answer. ;)
wow some of these are pretty intense...chim chim, how did you respond to the fish question?
Brain teaser questions. Just wanted to check... (Originally Posted: 01/28/2012)
Thanks All. Got it sorted. Cheers.
Question 2 is a perfect answer, great job. Will definitely show the interviewer (IF YOU GET THE QUESTION) that you can first do the rational thing and then think outside the box when the answer looks funky.
Question 1 do you really think you'll be able to do all that math in your head? And isn't the answer the 59th minute?
If the bacteria doubles every minute and takes one hour to fill the pond then:
1 hour = 1.0 (full pond)
59 min = 1.0 / 2.0 = 0.5 (Half full)
58 min = 0.5 / 2.0 = 0.25 (Quarter full)
so on...
Hopefully you realize though that it is just pure luck when it comes to brainteasers. Obviously you can read and try to memorize a bunch of them but the main thing you want to get from reading / looking at solutions to brainteasers is the critical thinking element. They all have a common way of approaching a problem and the more you do the more you'll learn to take different factors into consideration, etc...
Never thought of qn1 that way..
thanks prospectivemonkey!.
(0.55/25) + (0.55/25) = 0.2 ...
prospective monkey is right for Q1.
get heard on the street... all these are in there...
haha thanks monkeymark. i was using the ms calc. while i was writing this post.
a clarification for q1 is that the pond is exactly full by minute 60.
Can someone post the original questions so other can learn? Kinda the point of a forum...
Now that I've been studying for the GMAT, a lot of the 'brainteasers' are actually really simple mathematic concepts/functions; though not really common in subjects studied during college such as calculus.
I'd look over some of the more talked about topics for GMAT math as they help go over the basic concepts that seem to be asked in brainteasers e.g. combinatorics, exponential growth, rates/work, algebraic translations. It's basically math you did in high school, but forgot about.
Sarah, was that at Goldmans?
Easy Brainteaser (Originally Posted: 05/07/2012)
As penance for liberally beating a dead goat in the other brainteaser thread, here is a fresh one that should be easy for many of you:
Does .9 repeated = 1?
Please prove mathematically and explain in English. SB for the first to do both.
A Geometric series is probably easier.
just construct a cauchy sequence
No.
You can explain this without any fancy math, using the definition of numbers alone.
x = 0.9 10x = 9.9 10x - x = 9
.9 repeated has a limit value of 1, but is never actually 1.
If abs (x) 1, the infinite series sum(x^n) converges to 1/(1-x) let x = (1/10)
finite sum = 1/(1-.1)
this gives you 1 + .1 + .01 + .001 .... subtracting 1 and multiplying by 9 gives you .9 + .09 + .090 ....
solution = [1/(1-.1) -1] *9 = [10/9 - 9/9] * 9 = 1
This statement makes no sense at it stands. If you mean to say constructing a sequence where each additional term is made by appending another 9, then yes, this sequence has a limit of 1. If you mean the number .9 repeating, well, it's a constant, so its "limit value" must be itself. By the way, that is most definitely 1.
x = 0.9
10x = 9.9
10x - x = 9
9x = 9
x = 1
.
I don't think your answer is correct.
10*.9=9
not 9.9, so in your equation x does not equal .9 like you said.
the short answer is no it does not equal 1
Wtf is this shit...
....
...
You just saw this on reddit didn't you
Yes - my simplest 'explain in english' is that .3 repeating is generally known to equal one third, and 3 times one third equals 1
That's a good way to explain it. Also if 1/3 = .3 repeating 2/3= .6 repeating
Add them together and you get
3/3=.9 repeating 1=.9 repeating
Let me also do it with math:
0.999 repeating equals x. 10x equals 9.999 repeating equals 9 + x Subtract x from both sides 9x = 9 therefore x=1
Assuming you mean 10* .999 repeating = 9.999 repeating, this step is only true if the series converges (i.e it equals 1). In order to finish your proof you essentially assumed that what you were proving was true.
Are you retarded?
Oh Cola Coca got the answer first. Everyone who's disagreeing with him isn't taking the time to think about what it means for the decimal to be repeating indefinitely
Here's my answer:
Yes. When you want a repeated decimal, you just take that number and divided it by 9. 0.7777.....=7/9 0.5555.....=5/9
Thus, 0.999999.....should theoretically equal 9/9, and 9/9 = 1.
Do I get the SB?
Gave it to Dimon, wasnt thinking of it that way but it makes sense.
You can prove this another way using negation
how can one number equal another number? even if a limit, it still doesn't technically equal it...
plz splain
This was how I thought of it:
If a > b, x exists where a x b or a > x > b. If not, a=b
The answer is yes. There is a long mathematical proof that you learn in calc 2.
But the easy straightforward answer is...
1/3 = .3 Repeated 2/3 = .6 Repeated
Add the two previous
3/3 = .9 Repeated
Wikipedia has a pretty cool proof with (Dedekind) cuts too! I didn't know that one since my analysis class skipped the construction of real numbers, but it was nice. I liked it!
Assume that 9.999.../=10
Then you can list a number that is greater than 9.999... and less than 10.
Since you can't, 9.999...=10
1/3 + 1/3 + 1/3 = 3/3 = 1 and 1/3 is 0.33... so 0.3333..... + 0.3333..... + 0.3333.... = 0.9999 = 1
Weird corollary imo: every terminating number has 2 equivalent decimal representations
One terminating and one non-terminating?
Models and bottles!
Did I get it right?
I know why our financial system failed. Because there are too many guys who think 0.9 is 1.
There are several ways of showing this. An alternate version is using infinite series.
Let S(i) = i_Sum_k=1 .9*10^(-k)
And so .99 repeated is just Lim_(i--> infiniti) S(i)
But S(i) = .9* (1-10^(-i-1))/(1-1/10)
So Lim_(i--> infiniti) = .9*(1/(1-1/10)) = 1
Numbers are concepts. If the number isn't the same number, then it's not the same number. The idea of .9 repeating is that its as close as possible to 1 without actually being 1. Otherwise it would be 1. So I'm gonna ruin all the fun and say it's not possible based on the idea of what numbers are.
.99999... = lim 9 x (1/10 + 1/10^2 + ... 1/10^n) = 1 n->∞ .9 repeating is indeed another representation of 1
this is a site Determined gave me
http://polymathematics.typepad.com/polymath/2006/06/no_im_sorry_it_.html
The comments on the bottom seem to make a more convincing argument.
1cent repeated is 1 mio.
Did you know that?
I know a way of figuring things out now, or how I would answer them. But given in an interview, I dont know how I would respond!
Brainteaser - Unlimited vases? (Originally Posted: 10/14/2008)
You have 10 blue balls, and 10 black balls. There are two vases. If you pick a blue ball you win, black ball you lose.What allocation gives you the highest probability of picking a blue ball. What about if there were unlimited vases (anywhere up to 20)?
put one blue ball in one vase and all the other balls in the other vase.
Same idea for the unlimited vases.
...
Put one blue in one vase. 100% chance of grabbing a blue ball. Put the other 19 into the other vase. 9/19 chance of grabbing blue ball.
Second scenario is to put 1 blue each in 10 vases and then put 10black into one vase.
You just think out loud and remain calm. That's really all they are looking for most of the time. I.e. Logic and confidence.
Job Interview Brainteaser (Originally Posted: 10/07/2010)
Got this at an interview at a reputable hedge fund and blew it:
You're walking up a hill at 6am and finish at 6pm at the summit. The next day you start walking back down at 6am and get to the bottom of the hill at 6pm. What's the probability that you cross the same point at the same time on both days?
did you answer 0?
I would say very very very close to zero but not zero.
i know what the answer is now but no it's not 0
nooo its 1 =(
One? No fucking way. As someone who hikes/climbs mountains quite frequently, I can confidently say the probability of you walking at a constant rate both up and down the mountain is close to zero.
But hey, lesson learned. You always have to ask clarifying questions when the information you're given is vague. I'm sure they would have said the walk was at a constant rate if you asked them.
My take: If you use the same route up and down (likely) and have the same speed (less likely), then it should be 1, as you will cross the same point at the same time halfway through your route. If you have different speeds up and down or use different routes, I don't see it happening at all.
i'd say EXTREMELY unlikely
sherminator -- you said you know the answer now? what is it?
The answer is 1
if you draw a line graph, with x-axis being hour of the day (from 6AM to 6PM) and y-axis being altitude from (base to summit), the two paths will always cross
^ Yeah he's right. Think about it this way:
if instead of you walking down the mountain the next day, your friend is walking down the mountain the same day. what's the probability that you two will hit the exact same point at the same time? so answer is 1.
also answered in question 17 here:
http://eklhad.net/funmath.html
This is probably overkill, but for those who have at some point studied Topology, this is a basic application of Brouwer's Fixed Point Theorem. And yes, the answer is, in fact, 1.
the answer is 1
Math Brainteaser - Type of Questions (Originally Posted: 10/29/2011)
What are some common math brainteaser questions?
And one of my interview questions is 153 * 67 should i come up with an exact answer? or an approximation is actually acceptable? like 150*70?
A wealthy man tells his two sons that they (the two sons) will race for his fortune. The two sons will race camels from town A to town B, and the son whose camel is the slowest to reach town B will win. After riding around in the desert for days, the two sons meet a wise man and ask for advice. The wise man tells the sons something, after which the sons get back on the camels and ride toward town B as quickly as they can.
What did the wise man tell the sons?
lol, he tells them to switch camels.
I used to think the answer was that the wise men told them that since neither son would ever want to win the race, thus rendering it impossible for the fortune to ever be inherited, it would be mutually beneficial to let one of them win and split the fortune. silly me
-
shouldn't take too long to get the exact number
15060=9000 1507=1050 360=180 37=21 add them up 10,251
so, no 150*70 is not an acceptable answer
you need to give an exact number, a faster way: (150+3) * (70-3) = 150 *70 - 3 * ( 150 -70 ) - 9 = 10500 - 249 = 10251
What is the exact question? Was this for a mostly Asian trading firm?
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