Solution:

Suppose if there is just one unfaithful husband, then his wife will kill her husband as she must be knowing that all other men are faithful but the queen declared that one of them all is unfaithful.

If there are two unfaithful husbands, then both their wives will believe there is just one unfaithful husband, that too the other one whom she knows. Thus both of them will expect the above case to happen at midnight. But when no one is shot, they will realize that there are two unfaithful husbands and since they know about everyone, they will know that the other unfaithful man is their own husband on the next day.

If there are three unfaithful husbands, each of their wives will be knowing about two other unfaithful men so they will be expecting the above case and will be waiting for the gunshots on the second day. When that does not happens, they will realize that there are more than two and since they all will know about the rest, they will realize that their own husband is unfaithful and they will kill him on the third day.

Thus if we talk about general terms and suppose that there are n unfaithful husbands, their wives will believe that there is n-1 unfaithful husbands and will expect the gunshot on n-1 day. When they don’t hear that, they will realize their own husband is the nth.

Solution:

1. Fill 5 liters jar ( 5l-jar:5, 3l-jar:0)
2. Transfer to 3 liters pail (5l-jar:2, 3l-jar:3)
3. Empty 3 liters jar ( 5l-jar:2, 3l-jar:0)
4. Transfer 2q from 5 pail to 3 pail (5l-jar:0, 3l-jar:2)
5. Fill 5 liters pail(5l-jar:5, 3l-jar:2)
6. Transfer 1q from 5 pail to 3 pail(5l-jar:4, 3l-jar:3)

Solution:

2003.

Explanation:
Its a series of prime numbes.

Solution:

Move the coins as given in the figure and you will get the required result.

Solution:

The mules is moving in circular path and hence outside legs travels more distance than inner legs.

Solution:

One Night Stand.

Explanation:
Number 'One' + Knight(chess piece) + st + and(symbol &)

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:

Rhea will open just one present.
Cindy|Y - Y is the 25th alphabet
Duk|E - E is the 5th alphabet
Joh|N - N is the fifteenth alphabet
Rhe|A - A is the first alphabet

Solution:

Slumdog Millionare = Slum + Dog + Million + air

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:

Firstly: Policeman should take the thug across the river thames and escort the boat back with him. (Thug across, everyone else on this side)
Then, Policeman will take one cat along, leave it there, and on the way back, bring back the thug (since thug cannot be left with anyone without policeman's supervision).
Then Mrs. Mother will take the second cat on the other side. Drop it there and bring the boat back. For the 4th trip, Mrs. Mother And Mr. Father Will travel to the other side, and Mrs. Mother will stay there, and MR. Father will bring the boat back. (Now Mrs. Mother and 2 cats are on the other side.)
5th trip, the policeman takes the thug across. Both of them stay there and Mrs. Mother will come back. Then for the next trip, Mr. Father and Mrs. Mother will go on the boat, and Mrs. Mother will stay there, and Mr. Father will return back.
Then for the 7th trip, Mr. Father will travel with one dog. On the return, the policeman and the thug will return. Then in the next trip, the Policeman will take the second dog. (Now Mrs. Mother and her two cats; Mr. Father and his two dogs and the Policeman is here).
Now finally the policeman takes the boat on the other side and brings back the thug.
Tough one, right ?