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:

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:

Hussey should have Amy Hussey pull the trigger again without spinning.

We know that the first chamber Amy Hussey fired was one of the four empty chambers. Since the bullets were placed in consecutive order, one of the empty chambers is followed by a bullet, and the other three empty chambers are followed by another empty chamber. So if Hussey has Amy Hussey pull the trigger again, the probability that a bullet will be fired is 1/4.

If Amy Hussey spins the chamber again, the probability that she shoots Hussey would be 2/6, or 1/3, since there are two possible bullets that would be in firing position out of the six possible chambers that would be in position.