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Read Eckel Thinking in C++ vol. 2, Sutter, Joshi's book is good too, read Meyers books, they are thin and you can read them quickly. These are all very good books.

Questions that I got for an entry level position: 1) Explain how a forward contract is priced 2) How does the volatility affect an option price? Can the derivative be negative? Give an example. Explain different options, which one is more/less expensive, explain geom. BM, Black-Scholes, all...

Then you should re-word your solution. Think about it this way: Take a random kid "1". If he has only one grandpa, then everybody is connected to him via that gradpa => solved. Hence he(and all other kids) have two grandpas. "1" has 2 grandpas, call them A and B. Everybody else has A or B as...

Sure, prove by contradiction. Well, that's wrong. It should be "who is connected to at most 12 cousins". Anyway, the question is trivial: first you assume there exits two grandpas A and B that have a grand kid in common, then you argue that there could be at most one extra grandpa...

Deutsche bank trading floor position questions(I wanna be asked something like this, they are so simple): 1.A subway station. There are trains going in each direction. For each direction there is a train every 3 min. Stops are fast, say 1 sec. One direction is to work, the other one is...

Sure, I have it, but it is 60 pages shorter ;)

Hi, would somebody be so kind to send me a copy of the guide to vladimir.zhuravlev@utoronto.ca? I sent my CV to Dominik, but did not get the guide... Thank you ;)

quantyst: yep, we tried this method too.

Closed form is then you get (x_n=f(n)).

kevin, your 0.999... question was trivial. I wrote something just b/c I got a new idea (of taking out 9) :D Here is a cute puzzle(I know several solutions): Consider a 3x3x3 cube of 27 1x1x1 cubes glued together. You are allowed to make straight cuts and place pieces close to each other so...

Well, I know several proofs, but just now I got this idea, completely elementary: 0.999... = 9 * 0.111... = 9 * (1/9) = 1. The identity 1/9=0.1111 can be proven easily using, for example, long division. Normally I just solve the questions in this topic in the order of their appearance(when I...

Well, since you saw my post it would be more productive if you tried to find a mistake in my solution, your solution, or did something else. Like write a quick Monte Carlo check to see what the answer is approximately. Good luck!

Agreed, this one is hard. I tried many methods too. It looks like the answer(if it exists, which I doubt) is really ugly.

The best way to solve it is as follows: For n=1 this is of course possible. If n=2(or any other even number) the product of all numbers is negative. When you flip all numbers in a column you effectively multiply by (-1)^n=1, hence the product is always negative => the answer is no. For...

Here is a nice and simple question: There is a nxn square board, each square contains "-1", except that in one corner there is a "+1". On each turn you pick a row or a column and flip all signs in it. Is it possible to make all numbers +1?

This can be solved as follows: Denote by f_n(t) the probability that X_1+...+X_n<t, where 0<t<1. f_1=t. Then you can do induction by the number of random variables. (f_{n+1}(t)=\int_0^t f_n(t-x)\, dx) You get f_n(t)=t^n/n!

By the intermediate value theorem.

I did not see anybody posting the answer to this question, so I will do it. Answer: (\frac{50^5-49^5}{50^4}\approx 4.8) The idea is to calculate this by induction on the number of candies. Denote P_k(m) the probability to have m varieties out of k candies and E_k the expected number of types...

Hint: the answer is (\frac{k}{k+1}(N+1)). I proved this using some formula from "Heard on the street questions..." book.

So, we solved the first one. The second one has a similar solution(I suppose...), but you need to know that (\frac{20^{19}-1}{19^2}=14523213296398891966759) factors into two primes as (14523213296398891966759 = 192696104561\times 75368484119 ). The third one was solved in 2007, see...

A couple of updates to my last post: First of all, if n satisfies the question, then it has to be divisible by 17, and 17 is the smallest prime dividing it(the proof is very cute, you need to use Fermat's little theorem). Second, $n=\frac{18^{17}-1}{17^2}=7563707819165039903$ is a prime...

Well, I cannot find the smallest such number. But I can find one. $n=\frac{18^{17}-1}{17}$ works. Let me explain why. All numbers that we consider are natural $(\mathbb{N})$ . Lemma. Suppose that A mod k = 1. Then $A^k-1$ is divisible by\ (A-1)k\). Proof. This is easy. Just factor the...

Here is a solution I know: Denote (N_k) the expected number of coin flips to get "H" k times in a row. We are going to find a connection between $N_k$ and $N_{k-1}$. Here is the equation: $N_k=\frac12(N_{k-1}+1)+\frac12(N_{k-1}+1+N_k)$ Consider a sequence that ends in k-1 head for the...

This is done in one line using Bayes theorem. Suppose there are x white socks. A -- the event of adding a white sock, B -- event of picking up a black sock. Apply Bayes' theorem...

I think this is impossible. You will have one answer yes/no(2 possibilities) and you need to choose one out of three. Picking a non-random computer is easy though. Maybe there was a twist, like make the computer crash with a tricky question because it does not know how to answer... By the...

Huh? It is obvious that 18^n-1 is divisible by 17, but why would n^2 be divisible by 17?

Here are some more interview questions: 1. Calculate $\int_0^{\pi/6} \sec x \, dx$. 2. A bag contains 2 black socks and some white socks. One sock (black or white) is added to the bag and after that one sock is selected randomly. The selected sock happens to be black. What is the...

Anybody knows how to solve this?

Here is a question: There is a bag with N balls numbered 1,...,N. You pick out k balls, X_1, ..., X_k. What is the expected value of the maximum, E[max X_i, 1<=i<=k] ?

I would pour water or just pee in the hole. Can't think of anything else.

I have no idea what you wanted to say here. Could you reword it?

Here I got (\sum_{k=0}^{50} C^k_{100-k}). Can this be simplified? I have no idea. The reasoning goes as follows: Suppose you have k 2-steps, (0\le k\le 50). Then you have 100-2k 1-steps and 100-k steps in total. You need to decide the positions of the k 2-blocks. You have (C^k_{100-k})...

Ok, this one is hard. Here is how I did it: First do it not with 100, but with N passengers for small N, N=2 and 3. Observe that the probability is 1/2 in both cases. Then one can use math. induction: Suppose we have N passengers now and we proved for all k<N that the probability is...

If I were interviewing in #1 and #6 I would also ask to prove that this number exists (just 1st year calculus...). It's easy anyways. In #4 I got 3/4 and in #5 1/4 (did both by calculating appropriate integrals). The anwer in #5, of course, depends on how you interpret the question. In what...

Here is one like this from me: Make this III=I an identity without moving ANY matches at all. :D Good luck :D