Puzzle: A Visit to the Casino!

There is a casino and it has 4 gates, let say A, B, C and D. 

Now the condition is that every time you enter casino you have to pay $5 and every time you leave the casino, you again have to pay $5. Also, whenever you enter the casino whatever amount you have with you will get double.

Now you enter the casino through gate A and come out through gate B, again you go inside casino from gate C and come out of gate D, at the end of this process you should be left with no money? 

So calculate how much money you should carry with you when you enter the Casino?

THIS should be the amount that you need to carry... 

Solution: Amount Needed for a Casino Visit

What is the puzzle?

Let's suppose you have amount 'x' initially in your wallet.

1. On paying $5 for entry at gate A amount left is x - 5 .

2. After entry into the casino it double and becomes 2 ( x - 5 ) = 2x - 10.

3. For exit at gate B you pay $5 and the amount left with you is 2x - 10 - 5 = 2x - 15.

4. Again, for the entry at gate C, you pay $5 more. So the amount with you will be  - 
    2x - 15 - 5 = 2x - 20.

5. The amount is doubled to become 2(2x - 20) = 4x - 40 after the entry into casino.

6. Now, you have to pay $5 one more time to have exit via gate D. Hence, the amount 
left will be 4x - 40 - 5 = 4x - 45.

7. As per given condition, the amount that you should have on exit must be 0.

Hence, 4x - 45 = 0 i.e. 4x = 45. Therefore, x = 11.25. 

So you should carry $11.25 before you enter into the casino to have $0 after exit out of the casino.

Let's verify this one with the amount at each of stages above.

1. Amount = 11.25 - 5 = 6.25

2. Amount = 2 x 6.25 = 12.50

3. Amount = 12.50 - 5 = 7.50

4. Amount = 7.50 - 5 = 2.50

5. Amount = 2 x 2.50 = 5 

6. Amount = 5 - 5 = 0

7. Amount = 0
    

The Greek Philosophers

One day three Greek philosophers settled under the shade of an olive tree, opened a bottle of Retsina, and began a lengthy discussion of the Fundamental Ontological Question: Why does anything exist?

After a while, they began to ramble. Then, one by one, they fell asleep.
 
While the men slept, three owls, one above each philosopher, completed their digestive process, dropped a present on each philosopher's forehead, the flew off with a noisy "hoot." Perhaps the hoot awakened the philosophers.

As soon as they looked at each other, all three began, simultaneously, to laugh.

Then, one of them abruptly stopped laughing. Why?

Interesting reason behind it! 

Source 

Theory Of The Smartest Philosopher

What's the story behind? 

The one who stopped laughing was the smartest one among! Read how he was the smartest.

We need to think from the smartest Philosopher's point of view. Let's name 2 other Philosophers as A & B.

Now here is what the smartest Philosopher would think.

"If I had nothing on my head then A & B must have been laughing after looking each other's head. And at least one of them is smart enough to realize that the other is laughing only after looking at him. That means A (or B) would have realized that the some thing is on his head too as B (or A) is laughing after looking him (not me if I had nothing on my head). Hence one of them would have stopped laughing. Since they are not stopping to laugh, I too must have something on my head."

Hence the smartest Philosopher stopped laughing after realizing that the fact.

  
A brilliant puzzle based on the similar logic is here!