Solution:

he was genius.

Officials must have thought he was jumping with a hostage, they would never risk giving him a faulty parachute.

Solution:

10 prisoners must sample the wine. Bonus points if you worked out a way to ensure than no more than 8 prisoners die.

Number all bottles using binary digits. Assign each prisoner to one of the binary flags. Prisoners must take a sip from each bottle where their binary flag is set.

Here is how you would find one poisoned bottle out of eight total bottles of wine.

Bottle 1 Bottle 2 Bottle 3 Bottle 4 Bottle 5 Bottle 6 Bottle 7 Bottle 8
Prisoner A X X X X
Prisoner B X X X X
Prisoner C X X X X
In the above example, if all prisoners die, bottle 8 is bad. If none die, bottle 1 is bad. If A & B dies, bottle 4 is bad.

With ten people there are 1024 unique combination so you could test up to 1024 bottles of wine.

Each of the ten prisoners will take a small sip from about 500 bottles. Each sip should take no longer than 30 seconds and should be a very small amount. Small sips not only leave more wine for guests. Small sips also avoid death by alcohol poisoning. As long as each prisoner is administered about a milliliter from each bottle, they will only consume the equivalent of about one bottle of wine each.

Each prisoner will have at least a fifty percent chance of living. There is only one binary combination where all prisoners must sip from the wine. If there are ten prisoners then there are ten more combination where all but one prisoner must sip from the wine. By avoiding these two types of combination you can ensure no more than 8 prisoners die.

One viewer felt that this solution was in flagrant contempt of restaurant etiquette. The emperor paid for this wine, so there should be no need to prove to the guests that wine is the same as the label. I am not even sure if ancient wine even came with labels affixed. However, it is true that after leaving the wine open for a day, that this medieval wine will taste more like vinegar than it ever did. C'est la vie.

Solution:

prinka is 10, in a system that awards five for each vowel.

Solution:

fIvE thOUsAnd(5000)

Solution:

Think deeply about the door number 56, people will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1, the 1st person will open the door; pass 2, 2nd one will close it; pass 4, open; pass 7, close; pass 8, open; pass 14, close; pass 28, open; pass 56, close.
Thus we can say that the door will just end up back in its original state for each pair of divisor. But what about the cases in which the pair of divisor has analogous number for example door number 16? 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is recurrent because 16 is a perfect square, so you will only visit door number 16, on pass 1, 2, 4, 8 and 16… leaving it open at the end. So only perfect square doors will remain open at the end.

Solution:

Fred Odea won.

Just look at the announcement lines. The first sentence in 'Ok now everybody!' If you take out just the first letter of every word, it will form 'ONE'. In the same manner the second sentence 'The winner of…..' will give form 'TWO' and the third sentence 'The Hardest Riddle Ever Event.' Will form THREE.

Thus the final sentence must form FOUR. There is only one candidate who can suffice with the first two letter and he is Fred Odea. The final sentence of the mastermind must be, 'Fred Odea: Ultimate Riddler!'

Solution:

In the attached picture, you can see that the formation allows 2 castles to be protected by the walls. Congrats!

Solution:

Sucre : 1 cupcake
T-Bag : 2 cupcakes
Burrows: 3 cupcakes
Scofield : 5 cupcakes

Explanation :
It's obvious if the prisoner has eaten 5 or more cupcakes than he will not ask this question "did you had more cupcakes than me?" as it will make him the person who had eaten maximum cupcakes.
A) Sucre : Must not have eater 5 cupcakes, therefore he has eaten anywhere between 1-4 cupcakes.

B) T-bag : Since sucre asked him the question T-bag knows sucre have eaten between 1 - 4 cupcakes, and therefore he must have eaten 2 to 4 cupcakes(it can't be 1 else he will reply answer as "No".)

C) Burrows : Replied with the answer as "I don't know and based on logic B) he knows T-bag must have eaten at least 2 and he replied no which implies that he must be eaten 3 or 4.

D) Scofield replied he know all, which indicates he must have eaten 5.
Therefore cupcake division is as below:
Sucre : 1
T-Bag : 2
Burrows: 3
Scofield : 5

Solution:

73

Explanation:
It's a very famous puzzle known as "Josephus Problem Puzzle". It can be solved in various ways but the trick to solve this problem is illustrated below.

Binay number of 100 is: 1100100
complement pf 100 is 0011011, whose decimal equivalent is 27.
The desired answer will be 100 - 27 = 73

Solution:

Rashid's son-1: 1 11 21 31 41 51 61 71 81
Rashid's son-2: 2 12 22 32 42 52 62 72 73
Rashid's son-3: 3 13 23 33 43 53 63 64 74
Rashid's son-4: 4 14 24 34 44 54 55 65 75
Rashid's son-5: 5 15 25 35 45 46 56 66 76
Rashid's son-6: 6 16 26 36 37 47 57 67 77
Rashid's son-7: 7 17 27 28 38 48 58 58 78
Rashid's son-8: 8 18 19 29 39 49 59 69 79
Rashid's son-9: 9 10 20 30 40 50 60 70 80