Question 1

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

Solution

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

So x=322.