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Member Since 18 Feb 2013
Offline Last Active May 22 2015 04:54 PM

#341084 Friendly Mathematicians

Posted by BMAD on 18 November 2014 - 12:05 AM

At a certain mathematical conference, every pair of mathematicians are either friends or strangers. At mealtime, every participant eats in one of two large dining rooms. Each mathematician insists upon eating in a room which contains an even number of his or her friends. Prove that the number of ways that the mathematicians may be split between the two rooms is a power of two.

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#340876 Quick Fun Geometric Puzzle

Posted by BMAD on 31 October 2014 - 01:53 PM


Place the remaining numbers from 4 to 10 in the seven divisions of the above figure so that the outer divisions total 30 and each geometric figure totals 30.

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#338807 minimum distance

Posted by BMAD on 29 June 2014 - 06:24 PM

I got the same figure.

But, with plasmid, I don't know where the cables are.


Assumption (from post 1): We are to minimize PA + PB + P


yes. PA + PB + PC = L



Spoiler for Corrected using mathematica

I am not sure what 1.407 is referring to but i get a different answer.

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#338062 lamp algorithm

Posted by BMAD on 15 April 2014 - 05:47 AM

N lamps are set in a circle, and for each integer M you have a tool that can toggle the state (on/off) of any set of M consecutive lamps.
Find a possible N which satisfies the following statements:
The sum of its digits is less than 10.
By applying the tool for M=105 several times, we can toggle a single lamp.
If we remove one lamp and start from a random initial setting for the remaining N-1 lamps, the probability that there exists a way to apply the tool for M=32 several times and switch all the lamps off is positive and less than 0.0011%.
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#338061 foreign country currency

Posted by BMAD on 15 April 2014 - 05:30 AM

In a certain country 1/2 of 5 = 3. If the same proportion holds, what is the value of 1/3 of 10?
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#338049 Probability: equal sequences

Posted by BMAD on 14 April 2014 - 02:58 AM

Try out (or imagine) the following simple game with a friend:
You each toss a fair coin as many times as necessary to get a required sequence. 
Your required sequence is H T H 
Your friend's required sequence is H T T. 
Do this several times and each time write down the number of tosses you needed
The winner is the player whose average number of tosses is the lowest. For example:
Game 1:
You toss: H T T H T H
Friend tosses: H T H H H T H H T T
Your 'score' is 6 and your friend's is 10
Game 2:
You toss: T T H T T H H T H
Friend tosses: T T H H T H T T
Your 'score' is 9 and your friend's is 8 
Game 3:
You toss:   T T H H T H
Friend tosses:   H H T H T T
You both 'score' 6.
At this point your average is 7 and your friend's average is 8.
The conundrum: Assuming you played the game many times, which of the following outcomes would you expect?
A) You win (i.e. your average is lower than your friend's).
B) Your friend wins (your average is higher than your friend's).
C) You tie (your average is about the same)

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#338043 more fun with coins

Posted by BMAD on 11 April 2014 - 07:51 PM

Suppose John tosses a coin 250 times and Eric tosses a coin 251 times, what is the probability that john's coin has more heads than Eric?  What is the probability that Eric has more heads than John?

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#338003 Broken keyboard

Posted by BMAD on 06 April 2014 - 02:00 AM

   Imagine you have a special keyboard with the following keys: 
  1. A 
  2. Ctrl+A 
  3. Ctrl+C 
  4. Ctrl+V 
    where CTRL+A, CTRL+C, CTRL+V each acts as one function key for “Select All”, “Copy”, and “Paste” operations respectively. 
    If you can only press the keyboard for N times (with the above four keys), please write a program to produce maximum numbers of A. If possible, please also print out the sequence of keys. 
    That is to say, the input parameter is N (No. of keys that you can press), the output is M (No. of As that you can produce). 
For N = 8 the answer is M = 9, where S = { A, A, A, CTRL+A, CTRL+C, CTRL+V, CTRL+V, CTRL+V }. 
For N = 9 the answer is M = 12, where S = { A, A, A, CTRL+A, CTRL+C, CTRL+V, CTRL+V, CTRL+V, CTRL+V }.

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#337994 alternative to Pachinko

Posted by BMAD on 04 April 2014 - 08:47 PM

This is a simple game like Pachinko and in a similar vein as my "coin with a conscience" problem.


You insert money into the machine at one metal ball per 10cents.  The balls drop one at a time onto discs.  The first ball drops and lands on a disc and then rolls and lands on the next disc, and then falls on two more (for a total of 4 discs).  If after landing on the fourth disc, the ball rolls off to the left, you get 20cents.  if the ball rolls off to the right, you earn nothing.  After inserting your money, you choose whether the 1st ball should aim for the left part of the first disc or the right part of the first disc (and the ball will always go down that particular side).  Once a ball goes down a particular side, the rest of the discs are rigged to bias the odds the ball falls the other way by a change of 5%.


For example--with 1 ball

  • aim for right side,  disc 1--(goes down right side), disc2-- 52.5% (50% * 1.05 = 52.5) likely to fall down left, disc 3-- 52.5% likely to fall down left, disc 4-- 52.5% likely to fall down left
  • ball reaches disc2 say it falls down the left side, 49.875% likely to fall down the right side of disc 3, 49.875% likely to fall down the right side of disc 4
  • ball reaches disc3, say it falls down the left side of disc3, 52.36875% chance it will fall down the right of disc4
  • say the ball rolls off the right side of disc4....darn, i lost 10cents.



Can you make money playing this game in the long run?  If so, how should you play it?  If not, why not?

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#337988 Minor pieces on a chessboard

Posted by BMAD on 03 April 2014 - 12:53 PM

Consider a an infinite chessboard.  


How many squares can a knight reach after precisely n moves?

How about a bishop?



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#337982 in terms of 5

Posted by BMAD on 02 April 2014 - 02:38 AM

4y = -y +5(x-2) -10z

Instead of the more traditional approaches of solving with respect to a variable, solve it with respect to 5. Prove your answer is correct by using substitution with your answer into the number 6.

E.g. since 5 + 5/5 = 6 then
Solution for 5 + solution for 5 ÷ 5
Should equal solution for 5 + 1
  • -1

#337917 hidden race order

Posted by BMAD on 28 March 2014 - 03:22 PM

The 400 metre dash sprinting event will be held at a field track with 5 lanes. 25 athletes will be participating in total, of which, obviously, only 5 can be running together at a time. Define the minimum number of dashes required to determine the 3 fastest athlets of all, so that they are awarded the gold, silver and bronze medal. Which athletes will be running in each dash?

We assume that each athlete performs exactly the same in each dash. The results of the event will be determined by the relative classification of the athletes and not by their exact times. We only need to determine the 3 fastest athletes and not to follow the exact procedure which usually is followed at such events. (Obviously we cannot use a stopwatch).

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#337899 A coin with a conscience

Posted by BMAD on 25 March 2014 - 02:46 PM

Suppose we have a coin that "tries" to be fair. To be more specific, if we flip the coin n times, and have X heads, then the probability of getting a heads in the (n+1)st toss is 1-(X/n).  
[The first time we flip the coin, it truly is fair, with p=1/2.] 
A short example is in order. Each row here represents the i-th coin toss, with associated probability of heads and its outcome: 
p=1/2: H 
p=0: T 
p=1/2: H 
p=1/3: T 
p=1/2: T 
p=3/5: T 
It certainly would seem that the expected value of X should be n/2. Is this the case? If so (and even if not), then think of this coin as following a sort of altered binomial distribution.  
How does its variance compare to that of a binomial [=np(1-p)]? 
Does the coin that tries to be fair become unfair in the process? Or does it quicken the convergence to fairness? 

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#337898 Buying more elevators

Posted by BMAD on 25 March 2014 - 02:38 PM

There is a building with 50 floors.  Every morning 100 guest, 2 on each floor, come to use a single elevator.  The elevator has a max capacity of 10 users.  The elevator begins on the 50th floor and travels down.  Once the elevator achieves maximum capacity, the elevator travels only one additional floor.  At this floor, everyone on the elevator exits and those waiting (who have not yet ridden in the elevator) get on the elevator, then the elevator continues with only these new riders.  Everyone above the 1st floor desires to make it to the first floor while those on the first floor wish to make it to the fiftieth floor.


How many additional elevators should this building have (assume all elevators begin on the fiftieth floor)?  Provide a quantifiable justification to support your answer (e.g. defend why 2 elevators are better than any other case, or 3, or 4 or...so on.)

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#337897 A deterministic probability game

Posted by BMAD on 25 March 2014 - 02:11 PM

Alice and Bob are going to play a game, with the following rules: 
1st Alice picks a probability p, 0 <= p < 0.5 
2nd Bob takes any finite number of counters B. 
3rd Alice takes any finite number of counters A. 
These happen in sequence, so Bob chooses B knowing p, and Alice chooses A knowing p and B. 
A series of rounds are then played. Each round, either Bob gives Alice a counter (probability p) or Alice gives Bob a counter (probability 1-p). The game terminates when one player is out of counters, and that player is the loser. 
Whom does this game favor?

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