Solution:

10 Bulbs

For this tricky puzzle, you must check how many light bulbs in the row are having an odd number of factors. The first one surely has odd number of factors, the second has even, four has odd. Thus the bulb four and one will remain lit. The bulbs that are going to remain lit are perfect squares as they have an odd number of factors - 1, 4, 9, 16.

Since there are 100 passes, you can go up to 10 times 10 i.e. the square of 10. There are 10 perfect squares available to you - one, two, three, four, five, six, seven, eight, nine and ten. They corresponds to the bulb number 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100. All of them will remain lit and thus total ten bulbs will remain lit after 100 passes.

Solution:

If you are thinking of how to cut in such a way that you are able to make seven equal pieces, you are thinking in the wrong direction. You should give more stress on how to make the transaction with the worker.

You have to break the gold bar two times so you have the following sizes of the bar:

1/7 ---- Bar A
2/7 ---- Bar B
4/7 ---- Bar C

When the day 1 ends:
Give Bar A (You remain with Bar B and Bar C; Worker possess - Bar A)

When the day 2 ends:
Give Bar B and take the Bar A back (You remain with Bar A and Bar C; Worker possess - Bar B)

When the day 3 ends:
Give Bar A (You remain with Bar C; Worker possess - Bar A and Bar B)

When the day 4 ends:
Give Bar C and take back Bar A and Bar B (You remain with Bar A and Bar B; Worker possess - Bar C)

When the day 5 ends:
Give Bar A (You remain with Bar B; Worker possess - Bar A and Bar C)

When the day 6 ends:
Give Bar B and take back Bar A (You remain with Bar A; Worker possess - Bar B and Bar C)

When the day 7 ends:
Give Bar A (You remain with no bar; Worker possess - Bar A, Bar B and Bar C)

Thus by the end of seven days, you have taken care of everything.

Solution:

The question can practically disarm you. But we promise by the time you read the solution, you will find that it was way easy than you were thinking.

Let us name the boxes with a number - Box1, Box 2..., Box10. Now you must be familiar with the weights of the balls precisely. By saying that we are not implying that you will have to take one ball from every box and judge the weight of every ball.

What you have to do is pick one ball from Box 1, 2 balls from Box 2, 3 balls from Box 3, ...., 10 balls from Box 10. Thus after taking balls from every box in such a manner, you will finally have 55 balls in total. There comes the time to use the weighing machine.

If all the balls were weighing adequate, the combined weight should be 55 x 10 grams = 550 grams. But since one of the box has defected balls, the weight will be less than that. Here is the tricky part and explains why we took different number of balls from each box.

If the total weight is less than 1 gram, then the defective box is the Box 1 since we took 1 ball from that box. If the Box 2 is defective, the total weight will be less than 2 grams. In similar fashion you can identify the defective box by analyzing the total weight that is calculated by the weighing machine.

Solution:

Were you thinking of using the fastest person who takes one minute to travel to and fro till everyone has reached the other end? Yes that can be done but it will take 10 + 1 + 7 + 1 + 2 = 21 minutes in total.

If we pay attention towards finding a way to merge the one taking 7 minutes and the one taking 10 minutes, we will have to acknowledge the fact that one of them will also have to return back which will take much more time than 21 minutes.

What if we use the person taking 2 minutes to escort the one taking 1 minutes across? Let's come to the conclusion:

1 and 2 cross the bridge
2 returns with the torch
7 and 10 cross the bridge
1 comes back
1 and 2 cross the bridge

In such a manner, the total time taken will be
2 + 2 + 10 + 1 + 2 = 17 minutes.

Solution:

Fill the 3 quart bucket (5 quart bucket has 0 quarts and 3 quart bucket has 3 quarts)

Pour it entirely in the 5 quart bucket (5 quart bucket has 3 quarts and 3 quart bucket has 0 quarts)

Fill the 3 quart bucket (5 quart bucket has 3 quarts and 3 quart bucket has 3 quarts)

Pour the contents to fill the 5 quart bucket (5 quart bucket has 5 quarts and 3 quart bucket has 1 quart)

Empty down the 5 quart bucket (5 quart bucket has 0 quarts and 3 quart bucket has 1 quart)

Pour the water from 1 quart bucket to 5 quart bucket (5 quart bucket has 1 quart and 3 quart bucket has 0 quart)

Fill the 3 quart bucket (5 quart bucket has 1 quart and 3 quart bucket has 3 quarts)

Pour the contents from 3 quart bucket to 5 quart bucket (5 quart bucket has 4 quarts and 3 quart bucket has 0 quarts)

We are done and it took us eight steps to measure exactly 4 quarts of water.

Solution:

This puzzle is also known as Monty Hall Puzzle. If Michael choose the box on the table, the probability of finding the pearl inside will be 2/3.

Let us consider the same thing with more boxes. If Sarah placed 100 boxes on the table with one box that contains the pearl and Michael is allowed to pick random, 99 boxes will remain on the table. Now at least 98 boxes in the table will be empty. Sarah will remove these 98 boxes and now Michael will have to choose from the box on table or the box in his hand. If he chooses the box below, the probability of finding the pearl in the box will be 99/100.

Now you must be convinced why he should pick the box on the table.

Solution:

In the first wave, let us make an assumption that there is only 1 red hat and other 8 hats are blue. The student who can see all the other 8 blue hats on others' head will certainly be clear that there are 8 blue hats and 1 red hat since the professor told them that there is at least 1 red hat among the nine. But no one was able to answer after the first half an hour. Thus this can't be possible and our assumption is wrong.

Now let us assume that there are 2 red hats and 7 blue hats. The student who is having a red hat on his head can definitely see 7 blue hats and one red hat. He can deduce that he is wearing a red hat. But nobody was able to answer in the second interval as well. Thus our assumption is again wrong.

In the last wave of final five minutes, let us assume that there are 3 hats and 6 blue hats. The student with a red hat can see 2 other red hats and 6 blue hats and he may deduce that he is wearing a red hat. In this interval, everyone was able to answer which states that our assumption is correct.

Thus there are 3 red hats and 6 blue hats.

Solution:

Seven

Dividing the horses in a group of five, we will have five different groups of horses. A race will be conducted separately for each group which will help us find the fastest three horses in every group. We have used five different races till now.

Now we can have a race among the winners from every race which will help us determine the fastest of them all. We have used the sixth race.

The second fastest horse can be the one who came second in the sixth race or the second fastest horse in the finals winners' first race.

If we look at the first case, the third fastest horse can be the third fastest horse in the sixth race or the second fastest horse in either the fastest or second fastest horse first race.
If we look at the other case, the third fastest horse may be the second fastest in the sixth race or the third fastest in the final winners' first race.

If you calculate our assumptions, we have five more horses to race (seventh race) and determine the second fastest and the third fastest horse.

Solution:

Let us first assume that the number of drops required are N.

If the egg breaks at maximum number of tries, we will have N - 1 drops till it does not break. Thus we must drop the first egg from the height N. Now if the first drop of the first egg does not break the egg, we can have N - 2 drops for the second egg if the first egg breaks in the second drop.

Let us put that into a valid example for better understanding. Suppose we need 16 drops. Now let us drop the egg from the sixteenth floor, if it breaks, we will try all the floors below sixteen. Suppose it does not breaks, then we have 15 drops left and we will drop it from 16+15+1 = 32nd floor. This is because if it breaks at 32nd floor, we can try all the way down to 17th floor in fourteen tries making the total tries to be 16.

Let us assume that 16 is the correct answer

1 + 15 16 if breaks at 16 checks from 1 to 15 in 15 drops
1 + 14 31 if breaks at 31 checks from 17 to 30 in 14 drops
1 + 13 45.....
1 + 12 58
1 + 11 70
1 + 10 81
1 + 9 91
1 + 8 100 we can easily do in the end as we have enough drops to complete the task.

Seeking the above case, we must note that we have achieved it in 14 or 14 drops but we still need to find out the optimal one. We know from above that the optimal one will require 0 linear trials in the last step.

Thus we can say that

(1+p) + (1+ (p-1)) + (1+ (p-2)) + .........+ (1+0) >= 100.

Let 1+p=q which is the answer we are looking for

q (q+1)/2 >=100

q=14

Thus the optimal number of drops required are 14.

Drops will be tried at floor number 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100. How? Just move up 14, 13, 12 .... Floor till it breaks or does not.

Solution:

We know you must be having numerous thought about it. But trust us that when you know the solution to the problem you will regret for not thinking in that direction.

You just have to ask the person, "Which way is your city?" Just follow the path he directs you to and you will reach Truthcity. How?

Well if the person belongs to Liarcity, he will speak lie and will tell you the direction of the Truthcity and if the person belongs to Truthcity, he will tell the truth and will lead you to the Truthcity.

Isn't it simpler than you thought?