Solution:

89

Solution:

McLaughlin

Solution:

Peter will have to use a lock box that can hold two locks. He will put the ring in a box and will mail it to Mary. Upon receiving the box, Mary will put her own lock box inside and will send back to Peter. Peter can then-then take off his lock and send it off to Mary again.

Solution:

The color of the hat which the man A is wearing is the white color hat.
It is clear with the question and the clue itself that neither B nor C has information regarding this and therefore they do not answer. For the explanation, let's start from C. the man C can see man A's hat and also man B's hat. He must be seeing at least one white hat on any one of them. If he sees two black hats, then he clearly knows that he is wearing a white colored hat. But the man B cannot give an answer. He might see a white hat on man A. if the man B saw a black hat on A, he would know that he is wearing a white one. The question itself says that the man A answers to the question and therefore, the man A shouts that he is wearing a white one.

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:

She has misplaced 119 pens in total.

Solution:

There can be more than one possible solutions. We are sharing one with you:
Anabelle lost to Betty, but won every other game for a total of 3 points.
Betty beat Annabelle, but tied in all other games for a total of 2.5 points.
Christie beat Dora and tied with Betty and Evelyn for 2 points.
Dora beat Evelyn and tied only with Betty for 1.5 points.
Evelyn tied with Betty and Christie for 1 point.

Solution:

The best way is to divide the batteries in three groups. Take three batteries in two groups each and two batteries in the third group.

In the groups with three batteries, you can have three possible combinations in each group. In the group with two batteries, you can have just one possible combination.

If you add the combinations: 3 + 3 + 1 = 7, you get the least number of pairs you must check.

Solution:

43

Explanation:
Here is the thing. You can buy any number of wings that can be divided evenly by three except for three by using the combinations of boxes of 6 and boxes of 9. (You must use at most the box of 9 and then multiple boxes of 6). Now, if there is a number that can't be divided by 3, use a box of 20. After purchasing the box of 20, if the number that remains is divisible by 3, then you are good to go. On the other hand if it is not, you will have to use a second box of 20. And the remaining number will definitely be divisible by 3.

Thus if we calculate the largest number that can't be bought is the one that requires two 20 pieces boxes before the remainder can be thinned down to a number divisible by 3. We know that 3 is the number that can't be purchased and can evenly be divisible, the largest possible number of wings that can't be bought are:
3 + 20 + 20 = 43.

Solution:

Let us denote the four people with A, B, C and D.
A takes 1 minute to cross, B takes 2, C takes 5 and D takes 10.

A and B cross first spending 2 minutes.
A comes back with torch taking 1 minute.
C and D cross taking 10 minutes.
B comes back with torch taking 2 minutes.
Finally, A and B cross the bridge taking 2 minutes.

2 + 1 + 10 + 2 + 2 = 17 minutes

Thus, this is the way they all managed to cross that bridge that night.