# Puzzle Archives :: Alcatel-Lucent

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- Question :: 1
There are 100 doors they are all closed.

A Person walks through these 100 doors 100 times. Each time he toggles some of the doors. ( Toggles - close if open, open if close )

In the first walk, he will toggle all the doors.

In the second walk, he will toggle every second door, i.e., 2nd, 4th, 6th, 8th and so on

In the third walk, he will toggle every third door, i.e. 3rd, 6th, 9th and so on,

So after 100th Walk, what all door will be open?

- Question :: 2
# 100 hat puzzle

100 prisoners stand in line, one in front of the other. Each wears either a red hat or a blue hat. Every prisoner can see the hats of the people in front – but not their own hat, or the hats worn by anyone behind. Starting at the back of the line, a prison guard asks each prisoner the color of their hat. If they answer correctly, they will be pardoned. Before lining up, the prisoners confer on a strategy to help them.

What should they do?

- Question :: 3
You have a 100 coins laying flat on a table, each with a head side and a tail side. 10 of them are heads up, 90 are tails up. You can't feel, see or in any other way find out which side is up. Split the coins into two piles such that there are the same number of heads in each pile.

- Question :: 4
**Lazy people need to be smart**Four glasses are placed on the corners of a square Lazy Susan (a square plate which can rotate about its center). Some of the glasses are upright (up) and some upside-down (down).

A blindfolded person is seated next to the Lazy Susan and is required to re-arrange the glasses so that they are all up or all down, either arrangement being acceptable (which will be signaled by say ringing of a bell).

The glasses may be rearranged in turns with the subject to the following rules: Any two glasses may be inspected in one turn and after feeling their orientation the person may reverse the orientation of either, neither or both glasses. After each turn, the Lazy Susan is rotated through a random angle.

The puzzle is to devise an algorithm which allows the blindfolded person to ensure that all glasses have the same orientation (either up or down) in a finite number of turns. (The algorithm must be deterministic, i.e. non-probabilistic )