### Month of January

### 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?**

Answer: 322

### 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.**