Two trains are traveling toward each other on the same track, each at 60 miles per hour. When they are exactly 120 miles apart, a fly takes off from the front of one of the trains, flying toward the other train at a constant rate of 100 miles per hour. When the fly reaches the other train, it instantly changes directions and starts flying toward the other train, still at 100 miles per hour. It keeps doing this back and forth until the trains finally collide.
If you add up all the distances back and forth that the fly has travelled, how much total distance has the fly travelled when the trains finally collide?
The fly has travelled exactly 100 miles. We can figure this out using some simple math. Becuase the trains are 120 miles apart when the fly takes off, and are travelling at 60 mph each, they will collide in exactly 1 hour. This gives the fly exactly 1 hour of flying time, going at a speed of 100 miles per hour. Thus, the fly will travel 100 miles in this hour.
You are visiting NYC when a man approaches you.
"Not counting bald people, I bet a hundred bucks that there are two people living in New York City with the same number of hairs on their heads," he tells you.
"I'll take that bet!" you say. You talk to the man for a minute, after which you realize you have lost the bet.
What did the man say to prove his case?
This is a classic example of the pigeonhole principle. The argument goes as follows: assume that every non-bald person in New York City has a different number of hairs on their head. Since there are about 9 million people living in NYC, let's say 8 million of them aren't bald.
So 8 million people need to have different numbers of hairs on their head. But on average, people only have about 100,000 hairs. So even if there was someone with 1 hair, someone with 2 hairs, someone with 3 hairs, and so on, all the way up to someone with 100,000 hairs, there are still 7,900,000 other people who all need different numbers of hairs on their heads, and furthermore, who all need MORE than 100,000 hairs on their head.
You can see that additionally, at least one person would need to have at least 8,000,000 hairs on their head, because there's no way to have 8,000,000 people all have different numbers of hairs between 1 and 7,999,999. But someone having 8,000,000 is an essential impossibility (as is even having 1,000,000 hairs), So there's no way this situation could be the case, where everyone has a different number of hairs. Which means that at least two people have the same number of hairs.
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.
A farmer lived in a small village. He had three sons. One day he gave $100 dollars to his sons and told them to go to market. The three sons should buy 100 animals for $100 dollars. In the market there were chickens, hens and goats. Cost of a goat is $10, cost of a hen is $5 and cost of a chicken is $0.50.
There should be at least one animal from each group. The farmer’s sons should spend all the money on buying animals. There should be 100 animals, not a single animal more or less! What do the sons buy?
They purchased 100 animals for 100 dollars.
$10 spent to purchase 1 goat.
$45 spent to purchase 9 hens.
$45 spent to purchase 90 chickens.
An infinite number of mathematicians are standing behind a bar. The first asks the barman for half a pint of beer, the second for a quarter pint, the third an eighth, and so on. How many pints of beer will the barman need to fulfill all mathematicians' wishes?
