**Level 1 – Easy – 10 points**

*The first problem is something which is somewhat elementary.*

*From a biased coin(a coin where probability of heads is not be 1/2) how can you generate two events which are equally likely.(same probability).*

To restate it suppose there is a chocolate cake.There are two persons Nilasha and Diganta,both of them want to eat the cake.They decide both of them should have the same probability of eating the cake.But they have a biased coin.How should they design the toss so that both of them are equally likely to eat the cake.(What it means is that you need to find two events with same probability with a biased coin)

**Level 2 – Medium – 20 points**

The next problem was asked to me in a mock interview by some seniors which was actually a very nice application of a well know theorem.

*Prove that there exists a series of 100 consecutive natural numbers with exactly 13 primes between them.*

**Hint-I** could have given any number less than 26 in place of 13 in the above problem

**Level 3 – Hard – 20 points**

This problem was given in MOP 2005 and none of the students could answer this one.Then two former gold medalists along with a hint from the proposer of the problem, solved this seemingly easy yet intense problem.

So here goes the problem,

*There are n co-linear points on a straight line.No two distances between any two points can appear more than twice.(That is the same length cannot be repeated more than twice).Prove that there are at least [n/2] distances which appear only once.(Here [] means the greatest integer function.*

Seems easy right? Intuition say induction might kill the problem.Well try hard. The solutions will be out by next week.

P.S-And with this I will commence a new section called *problems of the month.*

The full solutions are to be given at officialbhattacharya@gmail.com and the top scorer of MARCH and APRIL will receive a book from us.You may discuss your observations in the forum section

**Best of Luck**

I have tried the answer of the no. 1 problem.The answer of this problem is that Nilasha and Diganta will do 8 tosses so that both of them are equally likely to eat the cake.

Now,it is said that they have a biased coin.They can’t toss 2 tosses (which is the least number of tosses) and divide it equally to eat the cake.So my design for the 8 tosses is to half these 8 tosses,which means they will first toss the first 4 tosses.Then they will use the biased coin.The result will be at least 1 head and 3 tails.Now they also have some remaining 4 tosses. They will use the biased coin and will get a result of 3 heads and 1 tail.This way 4 heads and 4 tails match together and form 8.This way Nilasha and Diganta will have the same probability of the cake and they will equally divide it and will eat the cake.I have not given 4 as my answer because i can also half it and give 2 to 2 tosses. But they cannot use the biased coin as because the 2 tosses can have the least result as 1 head and 1 tail(which is making them equal). Also if they do the biased coin on the whole 8 tosses,it would give the result as 5 to 3 or 6 to 2 tosses which would not be possible to share the cake amongst each other.So the least number of tosses that they can made are 8.

In the answer of question number 1,

Nilasha and Diganta have to toss a biased coin N number of times. Let’s assume that the coin is biased towards Tail.

Then in the first N/2 tosses Nilasha will choose Tail and have higher chances of winning and in the next N/2 tosses Diganta will choose Tail and have higher chances of winning.

After total N number of tosses, Nilasha and Diganta will have probability of getting equal number of Tails or Heads ( in case the coin is biased towards Heads ).Then it is likely that Nilasha and Diganta could have the cake equally . Also N should be a reasonably higher number .