Geekery

Riddles: The King, The Emperor, and The Magician

Posted on 07. Oct, 2009 by Jake in brain teasers

Life or Death? The Emperor’s Proposition
You are a prisoner sentenced to death. The Emperor offers you a chance to live by playing a simple game. He gives you 50 black marbles, 50 white marbles and 2 empty bowls. He then says, “Divide these 100 marbles into these 2 bowls. You can divide them any way you like as long as you use all the marbles. Then I will blindfold you and mix the bowls around. You then can choose one bowl and remove ONE marble. If the marble is WHITE you will live, but if the marble is BLACK… you will die.”

How do you divide the marbles up so that you have the greatest probability of choosing a WHITE marble?

Knights of the Round Table
King Arthur, Merlin, Sir Lancelot, Sir Gawain, and Guinevere decide to go to their favorite restaurant to share some mead and grilled meats. They sit down at a round table for five, and as soon as they do, Lancelot notes, “We sat down around the table in age order! What are the odds of that?”

Merlin smiles broadly. “This is easily solved without any magic.” He then shared the answer. What did he say the odds were?

Solutions for The Really, Really, Really Big Number:

The Really, Really Big Number

Solution:
1

Explanation:
This one is so sneaky.

First, consider 100 divided by 11.  The remainder here is 1.  Now consider the remainder when 100×100 is divided by 11.  Don’t do it on your calculator or on paper.  Rather, consider that you have one hundred hundreds, and each of them has a remainder of 1 when divided by 11.  So, go through each of your hundred hundreds and divide it by 11, leaving remainder 1.  Then collect up your remainders into a single hundred, and divide it by 11, leaving a remainder of 1.  This process can be extended to dividing 100×100x100 by 11, and indeed, to dividing any power of 100 by 11.

The Unkindest Cut of All, Part 1 of 2

Solution:
12.5

Explanation:
I am especially fond of these two ways to solve this problem:

1. Draw the right triangle whose hypotenuse is the square’s diagonal, and whose two legs are two sides of the square.  Then use the Pythagorean Theorem (a^2 + b^2 = c^2) to solve for the length of each side.  Since two sides are equal, we get (a^2 + a^2 = c^2), or (2(a^2) = c^2) ).  Since c is 5, 2(a^2) = 25, making a^2 equal to 25/2, or 12.5.  Since the area of the square is a^2, we’re done: it’s 12.5.

2. Tilt the square 45 degrees and draw a square around it such the the corners of the original square just touch the middles of the sides of the new, larger square.  The new square has sides each 5 units long (the diagonal of the smaller square), and it therefore has area 25.  However, a closer inspection reveals that the area of the larger square must be exactly twice that of the smaller.  Therefore the smaller square has area 25/2, or 12.5.

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