Riddle 7 - Glass balls

Here we go again - audience participation time! Yes, it's time for a new riddle. As usual, we will first review the solution to the riddle before. The solution is very basic - you just need to count the numbers from left to right... Each time we simply write down the number of appearances of the digit and the digit itself. For example, let's take the one before last row:

4 2 1 3 1 1

The first digit is 4, how many appearances? 1. Therefore we write 1 4
Now 2 which appears once, therefore 1 2
Now 1 which appears 3 times, 3 1 and last 1 3. So the result is:

1 4 1 2 3 1 1 3

The next row will be:
4 1 1 4 1 2 2 3, which is almost what Alex has wrote...


Riddle 7 - glass balls



Imagine you have a single glass ball and a 100 stories high building. You want to know what is the highest story from which you can throw the ball and still it will not shatter. Since you have only one ball, you must start from the first floor and go up one by one, or else you might find yourself with a shattered ball and no idea of the exact floor from which you can throw balls (if you had any...) and still admire the view...


So if I ask you how many trials you need in the worst case, the answer is 100. You can't do better.

But what if you had 2 balls? What is the minimal number of trials, worst case, that you need to do to find the exact spot?

What about 3 balls? 4 balls? Is there a pattern here?

Riddle 6 - Numbers numbers... hint

A hint to the sixth riddle - count!

As I said, any small child who knows to count can solve this puzzle...

Riddle 6 - Numbers numbers...

And here we go again - another riddle.

But first - what's the solution to the last riddle http://omer-behindthelines.blogspot.com/2008/08/riddle-5-doorway-to-heaven.html? The answer is that you should always pick the door that was left behind and not the door you originally picked. This is because there was a 33% chance for you to be right in the first place, meaning that there was a 66% that the right door is one of the other two doors. Since the gnome has helped you to throw away one of these two doors, all the 66% chance went to the door he left behind. So there is a bigger chance that that door is the correct one.

Riddle 6 - what is the next number?
Look carefully at the following list of numbers and please write down the next one to follow:
1
1 1
2 1
1 2 1 1
3 1 1 2
1 3 2 1 1 2
3 1 1 3 2 2
2 3 2 1 2 2
4 2 1 3 1 1
1 4 1 2 3 1 1 3
???

One small hint - no need for mathematics at all, a 8 year old kid can solve this...

21

Yesterday I saw the movie "21" with Kevin Spacey. In the beginning of the movie he asks a student of his a riddle, which is exactly the same as riddle #5 (only with a different story). The answer gave in the movie is correct - so you can copy it if you'de like.

P.S. I didn't like the film - everything was too obvious...

Riddle 5 - doorway to heaven

It's time for a new riddle, but let's first solve the last one.

How many times do you need to break the chocolate bar to get 24 sqauares? Well I am going to show you that it doesn't matter how you break it, it will always be 23 times. How come? the simple fact is that every time you break a piece of the bar, you actually add another piece to your collection (breaking a bar creates two parts of that bar). So, if you start with 1 piece - the whole bar - then after exactly 23 breaks you will have 24 pieces - exactly the 24 squares of chocolate we wanted.

Doorway to heaven

You are standing in front of three doors.

You know that one door leads to heaven, while the other two lead to hell. You have no way of knowing which is which, and running out of time, you pick one (doesn't matter which one) and go towards it.

Suddenly there appears a gnome.

He sees that you were going to open the door you picked and says:

"I know which door leads to heaven and which don't. Believe me when I say that this door is a door to hell" and while he is saying that, he is pointing to one of the other two doors (and not to the door you chose).

The gnome tells the truth (it doesn't matter which door you picked in the first place, he will always have a door that he can point out that leads to hell), so the door he pointed out is out of the question.

The riddle is: which door should you choose now? The door you chose in the first place, or the door that the gnome has left out of the two?

There is no right door or wrong door, but maybe there is a difference in the probability between the two?

Riddle 4 - a piece of chocolate

It's time for the next riddle, but before that, let's examine the last one (http://omer-behindthelines.blogspot.com/2008/07/riddle-3-ants-everywhere.html). Alex suggested that it will not take more than 2 minutes for the ants to fall off. The fact is that it will take 1 minute, maximal, for all the ants to fall. The way to prove this is to take a look at what happens when two ants meet (see illustration below). As I said, each ant turns around instantly and goes back to the direction it came from. However, since ants have no names and ant #1 is, for the sake of the riddle, just the same as ant #2, we can imagine that instead of turning around, each ant replaces places with the ant that it met and continues in its journey. Taking this into account, it is obvious that each ant, no matter how many ants it meets in its way, will go in one direction until it drops off - meaning that all the ants will drop after at the most 1 minute.

Riddle 4

In this riddle we have a bar of chocolate. This bar has 6 columns and 4 rows of chocolate, making 24 squares or pieces of chocolate.

The question is - how many times, minimal, do I need to break the bar until I get 24 separate pieces of chocolate (assuming that each break is along a line that separates between two rows or columns and there is no additional breaks)? Please explain why your answer is the lowest number of breaks that can actually give the wanted result.

Riddle 3 - a hint

A hint for the ants riddle - an ant has no name.
Think about it...

Riddle 3 - ants everywhere

It's time for a new riddle, but first let's take care of the last one (http://omer-behindthelines.blogspot.com/2008/07/riddle-2-grand-slam.html).

The key to this riddle is the fact that after each game exactly one player leaves the tournament. Since we must have a single player that is left in the tournament and wins it, all the other players must lose, each one in a different game. Therefore, if we have 8192 players, it will take 8191 games to lose all but one players. As you can see, the math needed here was very very simple, as opposed to adding up the number of games in each round...

Riddle 3

In this riddle we have a stick. This stick is exactly 1 meter long. It is a known fact that an ant walks on this stick at exactly a speed of 1 meter per minute. It is also known that once an ant starts to walk in a given direction (left or right) it will continue to do so until she drops of the stick or until she meets another ant.

So far we can say that if we drop one ant on the stick, no matter where we drop it and to what direction, the ant will fall from the stick after 1 minute or less.

In the case of an ant meeting an ant, each one jumps around to the opposite direction and continues to walk (see diagram below). This meeting and jumping does not take anytime what so ever.

The question is, if we drop 1000 ants on the stick and each ant randomly picks a direction from its random dropping point, how much time (maximal) do we need to enable all the ants to fall off?

P.S. again - this is a logic question and there is no need to get into complicated math.

Riddle 2 - Grand slam

Well it's time for our second riddle. Before we go into detail, let's review our last riddle (
The question is how many matches do you need to hold in a grand slam, to declare the winner, when the number of starting players is 8192? To answer this question you don't need to use any advanced math and not even a calculator. The less math you use and the more logic you apply - the easier and faster the answer gets...

Riddle 1 - chess board

Let's say you have a normal chess board, like in the picture to the right. It is obvious that you can use domino tiles to fully cover the whole board, where each tile covers two squares of the chess board.

But what if I take down two squares - a8 and h1 (top-left and bottom-right)? Can the board still be fully covered with domino tiles (that do not co-exist on any square)? If so - give me your solution. If not - prove that it can't be done.