A Door Of Fate Or Logics?

A prisoner is faced with a decision where he must open one of two doors. Behind each door is either a lady or a tiger. They may be both tigers, both ladies or one of each.

If the prisoner opens a door to find a lady he will marry her and if he opens a door to find a tiger he will be eaten alive. Of course, the prisoner would prefer to be married than eaten alive :).

Each of the doors has a statement written on it. The statement on door one says, “In this room there is a lady, and in the other room there is a tiger.”

The statement on door two says, “In one of these rooms there is a lady, and in one of these rooms there is a tiger.”
 
The prisoner is informed that one of the statements is true and one is false.

Which door should the Prisoner open?

This should be his choice!

Logical Choice Of Door

What were the choices?

For a moment, let's assume that first statement is true. The lady is behind the Door 1 and tiger is behind the Door 2. But this makes statement 2 also true where it says there is tiger behind one of these door & lady behind one of these doors. Hence, the statement 1 can't be true.

Hence, statement 2 must be true.

Only possibilities left are -

Door 1 - Tiger
Door 2 - Tiger

Door 1 - Lady
Door 2 - Lady

Door 1 - Tiger
Door 2 - Lady.

Since, the true second statement is suggesting there is lady behind 1 & tiger behind the other door, the possibilities of both tigers or ladies are eliminated.

That's why behind Door 1 is tiger & behind Door 2 is lady.

 

One More CryptArithmatic Problem

What three digits are represented by X, Y, and Z in this addition problem?

  XZY

+XYZ
______
  YZX

Here is the solution!

Solution Of One More CryptArithmatic Problem

What was the problem?

Here is the equation rewritten.

  XZY
+XYZ
______
  YZX
 

Let's start with the tens place. Z + Y = Z is there. That means either Y = 0 or 9 if 1 is carry from ones place.

Since Y is at hundred's place it can't be 0. Hence, Y = 9.

At hundred's column, now we have, X + X = 9. That's only possible if X = 4 and carry 1 forwarded from tens place. So X = 4.

Now, finally, at ones place, we have, 9 + Z = 4. Hence, Z must be 5 with carry 1 being forwarded to the next place.

To sum up, X = 4, Y = 9 and Z = 5.