Edit:
Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account. As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends. Of course, you can also enjoy our collection of amazing optical illusions and cool math games. If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top. If you have a website, we would appreciate a little link to BrainDen. Thanks and enjoy the Den :-) |
Posted by witzar on 13 December 2014 - 03:35 PM
Edit:
Posted by witzar on 14 November 2014 - 12:25 PM
Posted by witzar on 14 November 2014 - 12:04 PM
Posted by witzar on 14 November 2014 - 11:38 AM
Posted by witzar on 14 November 2014 - 11:24 AM
Consider a rook that can only move on black squares.
Let's call "reachable" the squares that can be reached by this rook
(in finite number of moves) from the bottom-left corner.
Only black squares in rows: 1 (bottom), 3, 5, 7 and 9 are reachable.
This accounts to 25 (5*5) reachable squares total.
Now observe that a quadomino, now matter how placed on the chessboard,
always covers an even (0 or 2) number of reachable squares.
But 25 is odd.
Posted by witzar on 02 November 2014 - 03:46 PM
Posted by witzar on 25 October 2014 - 12:41 PM
Here is the simple game I've invented (if someone invented it before, then I'm not aware of it):
A pawn is placed on every square of m*n chessboard.
Two players take alternate turns removing pawns.
On each turn, a player removes one or more pawns.
All pawns removed in a single turn have to be taken from the same row or the same column.
The player who cannot make a move loses (alternatively: the player who takes the last pawn wins).
Let's call the player who begins the "first player" and the other one the "second player".
Which player (the first or the second) has a winning strategy depending on the chessboard dimensions?
PS The game can be played on more interesting boards. Just draw some number of crossing lines
on a sheet of paper and place a pawn on each point of intersection of the lines.
The pawns removed in a single turn should come from a single line.
I've tried to play this game on the pentagram (the star: five lines, ten points/pawns) with my 9 years old daughter.
It was fun. Determining which player wins was even bigger fun.
Posted by witzar on 07 October 2014 - 05:47 PM
Asking about probabilities makes no sense unless there is a probability distribution.
Unfortunately there is no reasonable distribution on R^{+}.
But an obvious distribution exists on interval [0,x], where x is any positive real number.
One thing we can do in such case is to try to solve the problem for [0,x] and then see
if the solution has a limit when we go with x to infinity.
With this approach it's obvious that the probability we look for equals 0.5 for every x,
so obviously it's limit is also 0.5 when x goes to infinity.
Posted by witzar on 08 August 2013 - 05:54 PM
If BMAD decides to lie about a, then we'll notice errors in e,f,g.
If BMAD decides to lie about b, then we'll notice errors in e,f.
If BMAD decides to lie about c, then we'll notice errors in e,g.
If BMAD decides to lie about d, then we'll notice errors in f,g.
If BMAD decides to lie about e, f or g, then we'll notice a single error in e or f or g respectively.
If BMAD decides not to lie at all, we'll notice no error.
Observe that no mater what he does, the set of noticed errors will be unique,
thus letting us know what did BMAD lie about (if at all)
and letting us restore true values of abcd(efg).
Posted by witzar on 10 June 2013 - 04:37 PM
there are only three possibilities of this experiment after the cut.
1. Left peice is bigger
2. right peice is bigger
3. left peice = right peice
so the probability he calculated is 1/3
It would be true, if all three possibilities were equally probable.
Which is not the case.
Posted by witzar on 01 April 2013 - 01:59 PM
(Note: word 'obvious' is used as a joke.)
Posted by witzar on 14 February 2013 - 11:16 AM
CE|G|AFH|BD
CE|G|AFH|D|B
CE|G|AF|H|BD
CE|G|AF|H|D|B
CE|G|A|FH|BD
CE|G|A|FH|D|B
CE|G|A|F|H|BD
CE|G|A|F|H|D|B
CG|E|AFH|BD
CG|E|AFH|D|B
CG|E|AF|H|BD
CG|E|AF|H|D|B
CG|E|A|FH|BD
CG|E|A|FH|D|B
CG|E|A|F|H|BD
CG|E|A|F|H|D|B
G|C|E|AFH|BD
G|C|E|AFH|D|B
G|C|E|AF|H|BD
G|C|E|AF|H|D|B
G|C|E|A|FH|BD
G|C|E|A|FH|D|B
G|C|E|A|F|H|BD
G|C|E|A|F|H|D|B
Only the last one (found by k-man) has no nested dolls.
Posted by witzar on 08 February 2013 - 01:59 PM
Posted by witzar on 31 January 2013 - 05:19 PM
there are 3 paths for all square roll..G N R, which way you pass?
Posted by witzar on 05 July 2012 - 12:47 PM
the difference may be infinitesimally small but it exists...
Community Forum Software by IP.Board 3.4.7
Licensed to: BrainDen