Six friends go to a casino. They sit around a circular table, and we call them 1,2,3,4,5,6 in that order around the table. The dealer deals coins to them in the following fashion :
He selects a person, say X.
X receives some even number of coins, say 2n.
The person to the left of X, and to the right of X gets ncoins each.
Initially no one had any coins.
After a while the number of coins they had were like this :
1 : 349
2 : 342
3 : 345
4 : 368
5 : 338
How many coins does the 6th person have?
Comments on Question 1
Notice that if a even numbered person is selected by dealer, his account increases by 2n, and two odd places get neach. Consider two numbers, the sum of coins in the even positions =Eand the sum of coins in the odd positions =O.
By the property of the dealings, E=Oalways.
According to the problem, O = 349 + 345 + 338 = 1032.
If the 6th person has xcoins then E=710 + x = O = 1032