Solution:

The first step that they will take will be a leap of logic. What it means is that they will deduce that every color must appear twice at least. Why? Because the master logician has assured them that the puzzle will not be impossible for anyone of them. And if a color appears only once in the circle, the person wearing it will have no clue about that color which will not be fair for him.

Then the logicians will follow the same and look for all the colors of hats in the circle. If one of them sees a color just once, he can safely assume that he is also wearing the hat of the same color as by leap of logic, no color can appear just once. Thus when the bell is rung, he will leave.

In the similar fashion, if anyone sees another color just once, he can determine that he is wearing the hat of the same color and will leave when the bell rings or will be disqualified and sent home. Unvaryingly, if a color is seen twice, they will be eliminated after the first bell. Hence, there must be at least three hats of any of the remaining color.

Assume that you are sitting in the circle and you don’t see a color once but see it twice. Then if they were the only two hats of the same color, the logicians must have left at the first bell already. But they did not. Which means that there are three hats of that color and you are wearing one. Thus you will leave after the second bell.

Solution:

4 6 5 3 (from magician perspective)
3 5 6 4 (viewer perspective)

Detail:

From user perspective:
3 5 6 4 => 5 6 4 3 => 6 5 4 3

For magician perspective
3 4 5 6 (ascending order as)

Solution:

All Digits [0-9] follow the following pattern: There is one occurrence of all digits in every 10 numbers Example. Digit-9 occurred one time between 10 and 19 in 9, between 20 and 29 in 29
However, for tens(91,92,93...99) and hundreds (900,901.... 999), the occurrence is much more. Since we have included 1000, Therefor 1 is the most occurred digit.
Note: Despite three '0s' in 1000, 0 is not most occurred digit as they don't have tens and hundreds like other digits.

Solution:

Bond was suspicious of the women as one never knocks its own door at the hotel.

Solution:

Now there can be two solutions to the given situation because there can be two situations:
Suppose if B and C have same colors, A will know that his color is the other one and he will be able to speak up the first.
Now if B and C do not have same colors, A will stay silent. This will tell B that his and C's colors are different. Thus he will speak up first.

So either A or B will speak up first depending on the situation.

Solution:

Hockey Stick $49.50 & ball $0.50

Solution:

Flyball should pay Lavy $11 instead of accepting his counter offer as in that case the club will end up paying almost double the amount.

Solution:

To do it, the marbles must be arranged as the vertices of an octahedron. If you are unsure, check the image below.

Solution:

Jeff Killed Mike

1. Jack is not the murderer, because he is the brother of the murderer.
2. Dan can't be the murderer since he ran a marathon, and the murderer recently had his leg amputated, and wouldn't be running a marathon of any magnitude that quickly.
3. Ben is not the murderer if he just met Jack, since Jack and the murderer grew up together.
4. This leaves Jeff and Mike.

Since Jeff is still alive (he wants to install a new computer next week, present tense) he must be the murderer. Mike also didn't grow up with Jack. It has been determined that Jack, Dan and Jeff are all alive. Ben must also be alive since Jeff plans to install Ben's computer next week. This means that Jeff killed Mike.

Solution:

There are two things to keep in mind:
Firstly there is at least one red hat. (There can be two or three as well).
Secondly the competition is fair for everyone.

Thus if there is only one red hat, that person will see two green hats on other heads and will be able to deduce his own color as red. However the other students will see one red and one green hat and can never be sure. In such manner, the competition will prove to be partial for one student.

Suppose if there are two red hats. Then the students who are wearing red hats will see one red and one green hat on others. Now they must have deduced that there can’t be just one red hat. Thus they will know that they are also wearing a red hat. But the one who is wearing a green hat will see two red hats and can never be sure of his own color. In this case as well, the competition will not be fair.

Thus the only possible and fair means is if all of them are wearing a red hat. The one who is able to deduce the situation first, will raise his hand and will tell the correct answer.

Solution:

It can surely be true because it is not mentioned that it was raining.