Solution:

Solution is done below.

(0! + 0! + 0! )! =6
(1! + 1! + 1! )! =6
2 + 2 + 2 = 6
3 * 3 - 3 = 6
4 + 4 -sqrt(4) = 6
5 + 5 / 5 = 6
6 + 6 - 6 = 6
7 - 7 / 7 =6
8 - sqrt(sqrt(8+8) = 6
(9 + 9) / sqrt(9) = 6

Solution:

The girl was born on 29 February. Thus her birthday came once in four years only while her sister was born on a normal day and celebrated her birthday every year. Thus the girl had a chance of depositing money only 10 times in 4 years through which she collected $2500 while her sister took $50 from the piggy bank every year making the total amount to be $2000. Thus when the piggy bank was opened, it had just $500.

Solution:

The guard and the prisoner cross the river
The guard comes back
The guard and the girl cross the river.
The guard and the prisoner comes back.
The mother and the girl cross the river.
Mother comes back
The mother and the father cross the river
Father comes back
The guard and the prisoner cross the river
Mother comes back
The father and the mother cross the river
Father comes back
The father and the boy cross the river
The guard and the prisoner comes back
The guard and the boy cross the river
The guard comes back
The guard and the prisoner cross the river.

Solution:

Upon reaching each room, he found out that the window facing the dead body was closed. He had to open the window himself. This tells that there was somebody else with him who closed the window. If he would have committed suicide, at least one window would have been open.

Solution:

Are you Thorgan

Explanation:
Below will be the reply of three brothers:
Eden: No (speaks truth)
Thorgan: No (speaks lie)
Kylian: Yes (speaks lie)

Therefore if its Kylian then only you will get a reply as "Yes" and then you ask for your money.

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:

Zouma is 3 and has two footballs.
Ake is 1 and has one football.
Kalas is 4 and has three footballs.
Aina is 2 and has four footballs.

Solution:

Costa

Explanation:
c + 1 characters-> d
o + 2 characters-> q
s + 3 characters-> v
t + 4 characters-> x
a + 5 characters-> f

=> dqvxf = costa

Solution:

Yes, some people are going get really tiny pieces but you were allowed for that right. Also, there is another way of doing it. Can you think of it?

Solution:

Ganpat had a black suit and a brown suit.
Harish had a brown suit and a pink suit.
Manoj had a white suit and a black suit.
Tarun had a pink suit and a white suit.

Solution:

13 numbers

Explanation:
Since we are looking to find odd reminder which implies that the remainder will be either 1 or 3
Therefore number must follow below two conditions:
A) We can list number 5n + 1 or 5n + 3. (where n = 0,1,2,3.....)
B) The number must end in 1 or 3.

List of such numbers less than 100 are:
1, 3, 11, 13, 21, 23, 31, 33, 41, 43, 51, 53, 61, 63, 71, 73, 81, 83, 91, 93

Let's remove the nonprime ones which are 1, 21, 33, 51, 63, 81 and 91

So, I will be left with 13 numbers which fit the bill.
3, 11, 13, 23, 31, 41, 43, 53, 61, 71, 73, 83 and 93