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Problem: A telegraph machine can transmit only lines and dots; it takes 2 seconds to transmit a line, but only 1 second to transmit a dot. We generally want to transmit texts containing letters of the English alphabet, and digits (so we have N<=36 symbols in total). Therefore, a prefix-free...

Solution as posted on http://pratikpoddarcse.blogspot.in/2012/11/math-olympiad-problem-simple-and.html Since the final state matches the initial state, we can imagine this process going on continuously. Consider the dwarf whose cup contains the smallest amount of milk just before he begins...

Very interesting problem. When two ants A and B meet, they change directions. Lets view this as when two ants A and B meet, they continue in the same direction but their "original point" gets changed by the reflection property by the radius at the point of intersection as the mirror. Now...

Source: Nick's Mathematical Puzzles Problem:Let n be a positive integer, and let \(S_n = {n^2 + 1, n^2 + 2, ... , (n + 1)^2}\). Find, in terms of n, the cardinality of the set of pairwise products of distinct elements of \(S_n\) For example, \(S_2 = {5, 6, 7, 8, 9},\) 5 × 6 = 6 × 5 = 30, 5 × 7...

Very interesting problem. Thank You. Interesting discussion on the blog: http://pratikpoddarcse.blogspot.com/2012/01/lazy-walking-strategy-puzzle.html

solutions anyone?

Thanks AlexandreB

This is another of those problems I have not been able to solve since over an year. Fit disks of the following diameters into a circular cavity of size 12.000: 2.150 2.250 2.308 2.348 2.586 2.684 2.684 2.964 2.986 3.194 3.320 3.414 3.670 3.736 Write a program or give a general algorithm to...

Lots of quant problems: http://www.pratikpoddarcse.blogspot.com

I got this problem from Peter Winkler's Puzzle Book. I have not been able to solve it since 2 years now. In a room stand n armed and angry people. At each chime of a clock, everyone simultaneously spins around and shoots a random other person. The persons shot fall dead and the survivors spin...

Nice problem. Both solutions (by @albertino and @peterruse) are interesting and correct. Thanks.

Lots of quant problems: http://www.pratikpoddarcse.blogspot.com

Another solution posted on the blog link above: Number of ways in which we can have 52 runs is same as number of ways in which we can have 2 runs. Similar result holds for 52-k and k+2. This can be explained using the following construction : Suppose wlog we always start with red (this will...

Looks correct to me. Good one.

Answer: 27 Solution: http://pratikpoddarcse.blogspot.com/2011/05/coloured-run-of-cards.html

Another interesting solution at http://pratikpoddarcse.blogspot.com/2011/04/smallest-number-in-decreasing-sequence.html

Right. Thanks. I was expecting that would mean that the integrals to the further right should be zero. But that is not the case. Thanks.

I think it should be \(\int_0^{1-nh}\int_{x_1+h}^{1-nh}\cdots\int_{x_{n-1}+h}^{1-nh}1dx_ndx_{n-1}\cdots dx_1\) Right?

Nice problem

agree with @peterruse Can you please elaborate on this calculation?

http://www.pratikpoddarcse.blogspot.com

Which method of differential equation are you referring to?

You are supposed to keep the cards side by side. Vertically is not allowed. The only information the other guy should have is the permutation. Nothing else. -- http://www.pratikpoddarcse.blogspot.com

This is very interesting. I saw this in Peter Winkler's book. Just sharing an interesting thought. At the first look, one feels that its not possible because permutation of 4 cards represent 24 different outcomes and the fifth card can be any of the 48 cards. So, this should not be possible. But...

Thanks for the solution Bob

Nice post. Thanks

I am intrigued. Can someone please find a flaw in koupparis solution?

Koupparis, The probability of getting even has to be more. P(getting even | Game does not get over in first two chances) = P(getting odd | Game does not get over in first two chances) But game can get over in first two chances. And then the sum is even. Hence, prob of getting even > 0.5...

For the expectation over time, the solution given by peteruse is correct. Let expected number of rolls be T. T = 5/6(1+T) + 1/6(1+1) So, T= 7 There are similar very interesting problems in coins space. http://pratikpoddarcse.blogspot.com/2009/10/lets-say-keep-tossing-fair-coin-until.html

Very interesting problem. You can get the solution at http://pratikpoddarcse.blogspot.com/2010/07/differing-views.html Cheers!

Well 79 is prime. So, one of the 4 numbers had to be 79, 158, 237, 316, 395, 474... So 711-(that number) would be: 632, 553, 474, 395, 316, 237... and 711000000/(that number) would be 90000, 45000, 30000, 22500. 18000, 15000... Since all the multiples have a lot of zeroes in them, I am more...

Information: a*b*c*d = 7.11 a+b+c+d = 7.11 a>b>c>d>0 100*a, 100*b, 100*c, 100*d are integers To find a. 711=3*3*79 (100a)*(100b)*(100c)*(100d) = 711000000 = 3*3*79*100*100*100 (100a)+(100b)+(100c)+(100d) = 711 Possible solution for (100a, 100b, 100c, 100d): (316, 150, 125, 120) So...

7 races 5 races for 25 horses. Eliminate last 2 in all the races. So, we are left with A1, A2, A3 B1, B2, B3 C1, C2, C3 D1, D2, D3 E1, E2, E3 Race between A1, B1, C1, D1, E1 and without loss of generality, say the result is A1>B1>C1>D1>E1 Then the candidates for top 3 are A1, A2, A3, B1, B2...

Very interesting problem, I think! Classic example of Simpson's Paradox -- Pratik Poddar http://www.pratikpoddarcse.blogspot.com

Equal Probability of P and Q P(Q winning) = 1/64 + (6 choose 1)*1/128 + (7 choose 2)*1/256 + (8 choose 3)*1/512 = 1/64 + 6/128 + 21/256 + 56/512 = 130/512 = 0.253 Probability of P=7/12 P(Q winning) = (5/12)^6 + (6 choose 1)*(5/12)^6*(7/12)^1 + (7 choose 2)*(5/12)^6*(7/12)^2 + (8 choose...

Generalized Semicircle Covering Points Problem A nice generalization of this problem can be found here If (n) points are drawn randomly on a circle, the probability of them being on the same semi-circle is (\frac{n}{2^{n-1}})