The guru says that she can see only one blue eued person, then if there is only one blu eyed person he would have to leave. Note that according to the quesrion, a person cant see his own eye colour, but he can see others eye colour. Then he looks around and see that noone else have blue eyes, so according to question, then, he would have to leave, because he would understand that guru is talking about him. So, if there is one blue-eyed person, he leaves the first night.

If there are two blue-eyed people, they will each look at the other. They will both think that if I don't have blue eyes, then other one should be the only person wuth blue eyes.. And if he's the only blue eyed person, then he will leave tonight. They wait and see, and when no one leaves and understand that their guess was would. They would think "I must have blue eyes." And both of them will leave in night $2$ .So, If there are two blue-eyed people on the island, they will each leave the 2nd night.

Now, if there are three blue -eyed person. I may explain this with an example, and this is a little bit tricky. Consider Person 1 = Myself
Person 2 = @Rishabh Sood
Person 3 = @Nihar Mahajan

I look at Nihar and Rishabh, that I dont have blue eyes, then the other two who I see should leave the island two nights hence. The exact same thinking would go in the minds of Rishabh and Nihar, but when noone leaves the island after 2 nights, then each of us would realize that "what I guessed was wrong, I should have blue eyes". So, I, Rishabh and Nihar would leave the third night.

This logical formula would continue till the 100th day, when everyone with blue eyes will think that the other 99 people whom I see with blue eyes would leave on the 99th night, but when none of them leaves,they would all leave on the 100th night realizing that they have blue.

---

So, the answer is $100+100+10 = 210$ .

---

This question is really very logical, it took me around 3 hours to solve it. Nice question Rishabh. Thanks for the question.

Source

Here are some observations that help simplify the problem:

No one without blue eyes will ever leave the island, because they are given no information that can allow them to determine which non-blue eye color they have. The presence of the non-blue-eyed people is not relevant at all. We can ignore them. All that matters is when the blue eyed people learn that they actually are blue-eyed. There are two ways in which blue-eyed people might leave the island. A lone blue-eyed person might leave on the first night because she can see that no one else has blue eyes, so the guru must have been talking about her. Or an accompanied blue-eyed person can leave on a later night, after noticing that other blue-eyed people have behaved in a way that indicates that they have noticed that her eyes are blue too. The problem is symmetrical for all blue-eyed people, so this means they will either all leave at once or all stay forever.

Theorem: If there are N blue-eyed people, they will all leave on the Nth night.

Dual Logic.

Blue eyed people leave on the 100th night. If you (the person) have blue eyes then you can see 99 blue eyed and 100 brown eyed people (and one green eyed, the Guru). If 99 blue eyed people don't leave on the 99th night then you know you have blue eyes and you will leave on the 100th night knowing so.

Intuitive Proof.

Imagine a simpler version of the puzzle in which, on day #1 the guru announces that she can see at least 1 blue-eyed person, on day #2 she announces that she can see at least 2 blue eyed people, and so on until the blue-eyed people leave.

So long as the guru's count of blue-eyed people doesn't exceed your own, then her announcement won't prompt you to leave. But as soon as the guru announces having seen more blue-eyed people than you've seen yourself, then you'll know your eyes must be blue too, so you'll leave that night, as will all the other blue-eyed people. Hence our theorem obviously holds in this simpler puzzle. But this "simpler" puzzle is actually perfectly equivalent to the original puzzle. If there were just one blue-eyed person, she would leave on the first night, so if nobody leaves on the first night, then everybody will know there are at least two blue-eyed people, so there's no need for the guru to announce this on the second day. Similarly, if there were just two blue-eyed people, they'd then recognize this and leave on the second night, so if nobody leaves on the second night, then there must be a third blue-eyed person inspiring them to stay, so there's no need for the guru to announce this on the third day. And so on... The guru's announcements on the later days just tell people things they already could have figured out on their own.

It's obvious that our theorem holds for the "simpler" puzzle, and this "simpler" puzzle is perfectly equivalent to the original puzzle, so our theorem must hold for the original puzzle too.

Formal Proof.

To prove this more formally, we can use mathematical induction. To do that, we'll need to show that our theorem holds for the base case of N=1, and we'll need to show that, for any given X, if we assume that the theorem holds for any value of N less than X, then it will also hold for N=X. If we can show both these things, then we'll know the theorem is true for N=1 (the base case), for N=2 (using the inductive step once), for N=3 (using the inductive step a second time) and so on, for whatever value of N you want.

Base case: N=1. If there is just one blue-eyed person, she will see that no one else has blue eyes, know that the guru was talking about her, and leave on the first night.

Inductive step:

Here we assume that the theorem holds for any value of N less than some arbitrary X (integer greater than 1), and we need to show that it would then hold for N=X too. If there are X blue-eyed people, then each will reason as follows: "I can see that X-1 other people have blue eyes, so either just those X-1 people have blue eyes, or X people do (them plus me). If there are just X-1 people with blue eyes, then by our assumption, they'll all leave on night number X-1. If they don't all leave on night number X-1, then that means that there is an Xth blue-eyed person in addition to the X-1 that I can see, namely me. So if they all stay past night number X-1, then I'll know I have blue eyes, so I'll leave on night number X. Of course, they'll also be in exactly the same circumstance as me, so they'll leave on night number X too."

This suffices to prove our theorem. The base case tells us the theorem holds for N=1. That together with the inductive step tells us that it therefore holds for N=2, and that together with the inductive step again tells us that it holds for N=3, and so on... In particular, it holds for the case the original puzzle asked about, N=100, so we get the conclusion that the 100 blue-eyed people will leave on the 100th night

Hence your answer will be 100+100+10=210.

Cheers!