Solution:

The five cards must contain at least one card of the same suit.
Let's assume Rhodes placed one of the two cards of the same suit at the end and the second card was given to Daniel.
In that case, Daniel will know the suit of the card.

Now the real question is how Rhodes made sure that Daniel also knows the number of the card.

Daniel got four cards and one of them will determine the suit, so we are left with 3 cards.
3 cards can be arranged in 3! ways i.e. 6 ways but our card numbers can vary from 1 to 13.
Since we have two cards of the same suit we will make sure that we don't pick the King (13). This leaves us with 12 numbers.

We can also distinguish card from upside down position, therefore, we now have 6 * 2 ways to arrange these card.
"TADA.... We cracked the magic trick".

Let us assume that the smallest card is C1U (upwards position) and C1D (downwards position). Similarly, we have C2U, C2D, C3U and C3D

According to that, the 12 possible arrangements are:
C1U C2U C3U => NUMBER 1
C1U C3U C2U => NUMBER 2
C2U C1U C3U => NUMBER 3
C2U C3U C1U => NUMBER 4
C3U C1U C2U => NUMBER 5
C3U C2U C1U => NUMBER 6
C1D C2D C3D => NDMBER 7
C1D C3D C2D => NDMBER 8
C2D C1D C3D => NDMBER 9
C2D C3D C1D => NDMBER 10
C3D C1D C2D => NDMBER 11
C3D C2D C1D => NDMBER 12

Please note that king (number 13) is not possible as discussed already.

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:

== Mathematical Answer ==
Mathematical Answer = number of seconds in a minute * number of minute in an hour * number hour in a day * number of day in year
60 * 60 * 24 * 365 = 31536000
60 * 60 * 24 * 365.25 = 31557600 (if you want account for leap year)


== We are looking for trick answer (which is 24) ==
Every month contains only 2 second like
A) 2nd January ("second" January)
B) 22nd January (twenty"second" January)

Therefore 2 * 12 months = 24.

Solution:

Fifteen minutes can easily be measure using these two hour glasses.

Step 1: He will start both the hourglass.

Step 2: The moment the eleven minute hourglass is empty, he will invert it.

Step 3: When the thirteen minutes hourglass is empty, he will invert the eleven minute hourglass.
In step 3, we will have counted thirteen minutes. Since we inverted the eleven minute hourglass in step 2, it started from fresh and was inverted just for two minutes (13-11=2). In this manner when it is reversed when the thirteen minute hourglass is finished, it will have two minutes of sand left. This time when the sand finishes, he will have measured fifteen minutes. (13+2=15)

Solution:

Four

As the window was closed it is obvious that the six candles that were not extinguished were lit till they melted away entirely. Thus only four candles that were extinguished remain at the end.

Solution:

You will ask, I have a number in my mind that i.e. either 2 or 3. Is the number that I have in my mind smaller or equal to the number that you are thinking of?

If Alan replies with no, he is thinking of the number 1.
If Alan replies with yes, he is thinking of the number 3.
If Alan replies with don't know, he is thinking of the number 2.

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:

Lucky days are here again.

Explanation:
Lucky + Day + s + R + H + Ear + Again (Flower represent lucky symbol)

Solution:

It can be done as shown in the picture below.

Solution:

XIX is the answer which refers to nineteen in roman numeric system. On taking out one i.e. I, we will get XX that is equal to twenty.

Solution:

10

Explanation:
Equation 1 + 9 + 11 = 1 can be derived from
One (o) + nine (n) + eleven (e) = one => 1

Similarly for equation,
12 + 11 + 9
Twelve (t) + eleven (e) + nine (n) => ten (10)