Solution:

The main idea for getting the solution for this puzzle is that the pirate should know that he will be getting the treasure for the acceptance of the proposal. And if he gets to know that the treasure he is getting is less than the others then he will not accept the proposal. If the last one who is left is the pirate 1 then he will be the one who will be getting all the golden coins.


If pirate 1 would be the only one left, he would get all the golden coins. If only pirates 1 and 2 would be left, pirate 2 would die for sure, since pirate 1 is bloodthirsty and will reject all proposals of pirate 2 (since he will get all coins anyway). So:
| Pirate 1 | Pirate 2
----------------+------------+------------
Number of coins | 1000 | 0
----------------+------------+------------
Pirate agrees | No | Yes
When also pirate 3 would still be alive, he needs the agreement of one of the other two. Pirate 2 will agree with every proposal since, as we have seen, he would die if he did not. Therefore, pirate 3 should propose to keep everything for himself.
| Pirate 1 | Pirate 2 | Pirate 3
----------------+------------+------------+------------
Number of coins | 0 | 0 | 1000
----------------+------------+------------+------------
Pirate agrees | No | Yes | Yes
When we have four pirates, pirate 4 should make two other pirates agree with his proposal. Therefore, he proposes to give one coin to pirate 1, one coin to pirate 2, and the rest to himself. Pirates 1 and 2 will accept the proposal, since they are greedy, and if they would not accept, they would get less.
| Pirate 1 | Pirate 2 | Pirate 3 | Pirate 4
----------------+------------+------------+------------+------------
Number of coins | 1 | 1 | 0 | 998
----------------+------------+------------+------------+------------
Pirate agrees | Yes | Yes | No | Yes
However, as we know, there are five pirates. If pirate 5 gives both pirate 1 (or 2) and pirate 3 one coin more than in the previous case, they are willing to accept the proposal. Then a majority (three out of five) of the pirates will support the proposal, and pirate 5 can keep the rest of the treasure to himself.
| Pirate 1 | Pirate 2 | Pirate 3 | Pirate 4 | Pirate 5
----------------+------------+------------+------------+------------+------------
Number of coins | 2 | 0 | 1 | 0 | 997
----------------+------------+------------+------------+------------+------------
Pirate agrees | Yes | No | Yes | No | Yes
Conclusion: Pirate 5 should propose to give two coins to pirate 1 (or 2), one coin to pirate 3, and the remaining 997 coins to himself.

Solution:

Can be found on the pic below.

Solution:

2100 miles.

Each vowel equivalent to 500 and each consonant equivalent to -200

Solution:

Colors:
Y B R G
G R B Y
B Y G R
R G Y B


Signs:
x + - /
- / x +
/ - + x
+ x / -

Solution:

Let us denote the number of boys as 'p' and the number of girls as 'q'.

The first statement says that each boy has equal numbers of brothers and sisters
(p - 1) = q
=> p - q = 1 ------- (a)

The second statement says that every girl has twice as many brothers as that of sisters
2 (q - 1) = p
=> 2q - p = 2 ------- (b)

By solving (a) and (b)
We get q as 3 and p as 4.

Therefore, there are 4 boys and 3 girls in the family.

Solution:

I will be thrown at lions. Now if Akbar threw him at lions, Birbal’s statement will stand true and he will have to bury him alive. But if he bury him, the statement will emerge as false. Thus he had no choice left

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:

192

Explanation:
I started with the number 192.
Adding the reversed 291 to 192 gave me 483.
Similarly,
483 + 384 = 867
867 + 768 = 1635
1635 + 5361 = 6996

Solution:

2 minutes (duration between 9:59 and 10:01)

Solution:

C7 => B8
Exchange pawn for a knight

Check-Mate

GameOver

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