In the land of Brainopia, there are three races of people: Mikkos, who tell the truth all the time, Kikkos, who always tell lies, and Zikkos, who tell alternate false and true statements, in which the order is not known (i.e. true, false, true or false, true, false). When interviewing three Brainopians, a foreigner received the following statements:
Person 1:
I am a Mikko.
Person 2:
I am a Kikko.
Person 3:
a. They are both lying.
b. I am a Zikko.
Can you help the very confused foreigner determine who is who, assuming each person represents a different race?
Person 1 is a Miko.
Person 2 is a Ziko.
Person 3 is a Kikko.
If,
Fernando + Alonso + McLaren = 6
Fernando x Alonso = 2
Alonso x McLaren = 6
Then,
McLaren x Fernando = ?
3 or 0.75
Explanation:
Rewriting the last 2 equations in terms of Alonso,
Fernando = 2/Alonso
McLaren = 6/Alonso
Replacing above values in equation "Fernando + Alonso + McLaren = 6"
2/Alonso + Alonso + 6/Alonso =6
(2 + Alonso^2 + 6)/Alonso = 6
8 + Alonso^2 = 6Alonso
Alonso^2 - 6Alonso + 8 = 0
(Alonso - 4) (Alonso - 2) = 0
Therefore;
Alonso = 4 or 2
Let's take value of Alonso as 2
Fernando = 2/2 = 1
McLaren = 6/2 = 3
Therefore;
McLaren x Fernando = 3 x 1 = 3
Let's take value of Alonso as 4
Fernando = 2/4 = 0.5
McLaren = 6/4 = 1.5
Therefore;
McLaren x Fernando = 1.5 x 0.5 = 0.75
You have a sock drawer.
It has 4 black socks, 8 brown socks, 2 white socks and 8 tan socks.
You need to pull out a matching pair of socks in the dark.
There is no light and you couldn't see the socks.
How many socks you should pull out in the dark to get one matching pair of socks?
Five. You have only four different colors of socks. If you pick 5, you can surely get one pair of matching socks.
There are n coins in a line. (Assume n is even). Two players take turns to take a coin from one of the ends of the line until there are no more coins left. The player with the larger amount of money wins.
Would you rather go first or second? Does it matter?
Assume that you go first, describe an algorithm to compute the maximum amount of money you can win.
Note that the strategy to pick maximum of two corners may not work. In the following example, first player looses the game when he/she uses strategy to pick maximum of two corners.
Example 18 20 15 30 10 14
First Player picks 18, now row of coins is
20 15 30 10 14
Second player picks 20, now row of coins is
15 30 10 14
First Player picks 15, now row of coins is
30 10 14
Second player picks 30, now row of coins is
10 14
First Player picks 14, now row of coins is
10
Second player picks 10, game over.
The total value collected by second player is more (20 + 30 + 10) compared to first player (18 + 15 + 14). So the second player wins.
Going first will guarantee that you will not lose. By following the strategy below, you will always win the game (or get a possible tie).
(1) Count the sum of all coins that are odd-numbered. (Call this X)
(2) Count the sum of all coins that are even-numbered. (Call this Y)
(3) If X > Y, take the left-most coin first. Choose all odd-numbered coins in subsequent moves.
(4) If X < Y, take the right-most coin first. Choose all even-numbered coins in subsequent moves.
(5) If X == Y, you will guarantee to get a tie if you stick with taking only even-numbered/odd-numbered coins.
You might be wondering how you can always choose odd-numbered/even-numbered coins. Let me illustrate this using an example where you have 6 coins:
Example
18 20 15 30 10 14
Sum of odd coins = 18 + 15 + 10 = 43
Sum of even coins = 20 + 30 + 14 = 64.
Since the sum of even coins is more, the first player decides to collect all even coins. He first picks 14, now the other player can only pick a coin (10 or 18). Whichever is picked the other player, the first player again gets an opportunity to pick an even coin and block all even coins.
As I was going to the mall I met a man with seven wives. Each wive held two bags, each bag held a mother cat, each mother cat had six babies,
How many people were going to the mall?
