# A very difficult logic puzzle

• Jun 23rd 2009, 07:58 AM
jashansinghal
If 100 people are playing a game namely 1,2,3------100
in the first game 1 wins
in the second game 2 wins
in the third game 3 wins and so on till 100
so in total 100 games are played.
if a person wins a game he has to give some money to the other 99 players
this amount of money which he has to give is the same amount as the other person has in his pocket.for eg:If 1 wins he has to give the other 99 players money.TO 2 he will give the same amount of money which 2 has it in his pocket.to 3 he will give the same amount of money which 3 has in his pocket.
At last after the 100 th player wins the amount of money which everone has is same and is "X"
Calculate in terms of"X" the value of money which 63rd player had in beginning.

If anyone has not understood anything ,he can ask me
• Jun 23rd 2009, 08:49 AM
TheAbstractionist
The total amount of money possessed by all 100 players is $100X$ and this is constant throughout the games. Suppose the 63rd player has $Y$ amount of money at the start. After 62 games, he will have $2^{62}Y$ money. The total amount of money of the other players is $100X-2^{62}Y.$ Thus, after the 63rd game, the 63rd player gives this amount of money in total to the other players, and he is left with $2^{62}Y-(100X-2^{62}Y)=2^{63}Y-100X$ in his pocket. The remaining 37 games are then played, after which the 63rd player has $2^{37}(2^{63}Y-100X)=2^{100}Y-2^{39}25X.$

Hence

$2^{100}Y-2^{39}25X\ =\ X$

$\implies\ \fbox{Y\ =\ \dfrac{(1+2^{39}25)X}{2^{100}}}$
• Jun 24th 2009, 01:09 AM
jashansinghal
wow its simply amazing