Solution:

The deer hiding in the trees.

Solution:

Yes, some people are going get really tiny pieces but you were allowed for that right. Also, there is another way of doing it. Can you think of it?

Solution:

13 numbers

Explanation:
Since we are looking to find odd reminder which implies that the remainder will be either 1 or 3
Therefore number must follow below two conditions:
A) We can list number 5n + 1 or 5n + 3. (where n = 0,1,2,3.....)
B) The number must end in 1 or 3.

List of such numbers less than 100 are:
1, 3, 11, 13, 21, 23, 31, 33, 41, 43, 51, 53, 61, 63, 71, 73, 81, 83, 91, 93

Let's remove the nonprime ones which are 1, 21, 33, 51, 63, 81 and 91

So, I will be left with 13 numbers which fit the bill.
3, 11, 13, 23, 31, 41, 43, 53, 61, 71, 73, 83 and 93

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:

Yes, you should switch the envelope. In the beginning when you picked up the envelope, you had a 1/3 probability of finding an offer letter in the envelope. There was 2/3 chance that the letter was there in the two envelopes on the table.

If you keep your selected envelope, you still have a 1/3 chance of finding an offer letter in that. However, since the interviewer has removed one empty envelope from the table, if you switch, you have a probability of 2/3 that the offer letter is inside that.

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:

The answer is very simple. All she had to do is take the fifteen cards from the top and reverse them. This would make another pile out of that and there will be two piles - one of 15 cards and one of 37 cards. Also both of them will have the same number of inverted cards.

Just think about it and if the mathematical explanation will help you understand better, here it is.
Assume that there were p inverted cards initially in the top 15 cards. Then the remaining 37 cards will hold 15-p inverted cards.
Now when she reverses the 15 cards on the top, the number of inverted cards will become 15-p and thus the number of inverted cards in both of the piles will become same.

Solution:

Now there can be two solutions to the given situation because there can be two situations:
Suppose if B and C have same colors, A will know that his color is the other one and he will be able to speak up the first.
Now if B and C do not have same colors, A will stay silent. This will tell B that his and C's colors are different. Thus he will speak up first.

So either A or B will speak up first depending on the situation.

Solution:

If two widows, each having a son married the son of the other widow, and then by the consummation of marriage, both the couples had a daughter, all the aforementioned relationships will turn to be true.

Solution:

Just 2

Day One: You make your first cut at the 1/7th mark and give that to the worker.
Day Two: You cut 2/7ths and pay that to the worker and receive the original 1/7th in change.
Day three: You give the worker the 1/7th you received as change on the previous day.
Day four: You give the worker 4/7ths and he returns his 1/7th cut and his 2/7th cut as change.
Day Five: You give the worker back the 1/7th cut of gold.
Day Six: You give the worker the 2/7th cut and receive the 1/7th cut back in change.
Day Seven: You pay the worker his final 1/7th.

Solution:

One Night Stand.

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