Solution:

Forty.

The number of decks is irrelevant; the answer is the same if one or one-hundred decks are used.

Any card drawn will be a A,2,3,4,5,6,7,8,9,10,J,Q, or K, so there are 13 possibilities each time a card is drawn.

The fastest way to draw a four of a kind is if the first four cards all have the same 'value.' The slowest way, which provides the solution, is to first draw 13 three of a kinds, and then one more card.

Since 13 x 3 + 1 = 40, if 40 cards are drawn it is guaranteed that those forty cards contain at least one four of a kind.

Solution:

The man asked the king that he wanted to die of old age.

Solution:

NONE, holes are empty.

Solution:

31

Explanation:
Ramanujan lighting the candles in binary.
11110 =30
11111 =31

Solution:

Sam : 2
Dean : 5
Castiel : 7

Age diff
7 - 2 = '5' is prime
7 - 5 = '2' is prime
5 - 2 = '3' is prime

Solution:

The numbers that have been written over the calendar are:
12, 4, 8 and 3

A few numbers have been circled under particular months. The curious thing is that the numbers that have been written are the same number of corresponding months under which the dates have been marked.

What we can judge by the overall calendar is that the dates that have been marked may refer to the letters at that position in the month's name. In addition, the written numbers can be the position of those letters. Let's simplify below:

12th month is December
The number marked under December = 5
The fifth letter of December = M

4th month is April
The number marked under April = 4
The fourth letter of April = I

8th month is August
The number marked under August is 6
The sixth letter of August = T

3rd month is March
The numbers marked under March are 4 and 5
The fourth and fifth letter of March = C and H

When you combine the letters, you get the name as MITCH.
Therefore, Mitch is the killer.

Solution:

So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8.

There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second one possibility is to weigh 9,10,11 against 1,2,3

(1) They balance, in which case you know 12 is the different coin, and you just weigh it against any other to determine whether it is heavy or light.
(2) 9,10,11 is heavy. In this case, you know that the different coin is 9, 10, or 11, and that that coin is heavy. Simply weigh 9 against 10; if they balance, 11 is the heavy coin. If not, the heavier one is the heavy coin.
(3) 9,10,11 is light. Proceed as in the step above, but the coin you're looking for is the light one.

That was the easy part.

What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.

Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. The direction of the tilts can tell us whwther the offending coin is heavier or lighter.

Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin.

For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.

Solution:

The answer is 2. First, divide the coins into 3 equal piles. Place a pile on each side of the scale, leaving the remaining pile of 3 coins off the scale. If the scale does not tip, you know that the 6 coins on the scale are legitimate, and the counterfeit is in the pile in front of you. If the scale does tip, you know the counterfeit is in the pile on the side of the scale that raised up. Either way, put the 6 legitimate coins aside. Having only 3 coins left, put a coin on each side of the scale, leaving the third in front of you. The same process of elimination will find the counterfeit coin.

Solution:

2003.

Explanation:
Its a series of prime numbes.

Solution:

20
Just count the nines in the numbers: 6, 16, 26, 36, 46, 56, 76, 86, 96, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69.
Note: 66 contains TWO nines

Solution:

This can be done if you draw a five pointed star and place the balls on the corners and intersection points. See the figure to understand better.