Easy riddles

what am I

Nowhere in tomorrow

I’m at the beginning of the end and the start of eternity, at the end of time and space, in the middle of yesterday but nowhere in tomorrow. What am I?
The letter "e".
93.84 %
41 votes



If money really did grow on trees, what would be everyone’s favorite season?
93.84 %
41 votes


Smartest kid in the world

Jake and his friend Paco had very famous challenge sessions at their school. One would suggest something they could do, and the other would prove it wrong somehow. One day, Jake surprised Paco by stating: "I can answer any question in the world." Sure that he would win the challenge, Paco accepted the task of proving it wrong. He wrote up a test full of impossible questions. After a while, Jake returned the test. Paco unbelievably lost the challenge and told Jake he could indeed answer any question. How did Jake win?
For all the impossible questions, Jake simply wrote "I don't know".
93.84 %
41 votes

shortwhat am I

I'm not easy to store

You have me today, Tomorrow you'll have more. As your time passes, I'm not easy to store. I don't take up space, But I'm only in one place. I am what you saw, But not what you see. What am I?
93.84 %
41 votes


2 Player and N Coin – Strategy Puzzle

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.
93.84 %
41 votes


Bowling Ball

If I were in Hawaii and dropped a bowling ball in a bucket of water which is 45 degrees F, and dropped another ball of the same weight, mass, and size in a bucket at 30 degrees F, them at the same time, which ball would hit the bottom of the bucket first? Same question, but the location is in Canada?
Both questions, same answer: the ball in the bucket of 45 degree F water hits the bottom of the bucket last. Did you think that the water in the 30 degree F bucket is frozen? Think again. The question said nothing about that bucket having anything in it. Therefore, there is no water (or ice) to slow the ball down...
93.70 %
40 votes