Riddles about money

logicmath

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.
78.86 %
51 votes
logicmath

A women walks into a bank to cash out her check. By mistake the bank teller gives her dollar amount in change, and her cent amount in dollars. On the way home she spends 5 cents, and then suddenly she notices that she has twice the amount of her check. How much was her check amount?
The check was for dollars 31.63. The bank teller gave her dollars 63.31 She spent .05, and then she had dollars 63.26, which is twice the check. Let x be the dolars of the check, and y be the cent. The check was for 100x + y cent He was given 100y + x cent Also 100y + x - 5 = 2(100x + y) Expanding this out and rearranging, we find: 98y = 199x + 5 Which doesn't look like enough information to solve the problem except that x and y must be whole numbers, so: 199x ≡ -5 (mod 98) 98*2*x + 3x ≡ -5 (mod 98) 3x ≡ -5 ≡ 93 (mod 98) This quickly leads to x = 31 and then y = 63 Alternative solution by substitution: 98y = 199x + 5 y = (199x + 5)/98 = 2x + (3x + 5)/98 Since x and y are whole numbers, so must be (3x + 5)/98. Call it z = (3x+5)/98 so 98z = 3x + 5, or 3x = 98z - 5 or x = (98z - 5)/3 or x = 32z-1 + (2z-2)/3. Since everything is a whole number, so must be (2z-2)/3. Call it w = (2z-2)/3, so 3w = 2z-2 so z = (3w+2)/2 or z = w + 1 + w/2. So w/2 must be whole, or w must be even. So try w = 2. Then z = 4. Then x = 129. Then y = 262. if you decrease y by 199 and x by 98, the answer is the same: y = 63 and x = 31.
78.13 %
55 votes
logicclean

An archeologist claims he found a Roman coin dated 46 B.C. in Egypt. How much should Louvre Museum pay for the coin? Note: Roman coins can really be found in Egypt
Nothing. That coin is as phony as a three dollar bill. In 46 B.C., they wouldn't have known how many years before Christ it was.
78.00 %
66 votes
logicsimple

This is a newspaper headline: Workers Strike - Want to Make Less Money! What is going on?
The workers work at the mint and are tired of being overworked. They want to work less, which is making less money, since money is made at the mind!
77.76 %
54 votes
simplecleanlogicinterview

You are blindfolded and 10 coins are place in front of you on table. You are allowed to touch the coins, but can't tell which way up they are by feel. You are told that there are 5 coins head up, and 5 coins tails up but not which ones are which. How do you make two piles of coins each with the same number of heads up? You can flip the coins any number of times.
Make 2 piles with equal number of coins. Now, flip all the coins in one of the pile. How this will work? lets take an example. So initially there are 5 heads, so suppose you divide it in 2 piles. Case: P1 : H H T T T P2 : H H H T T Now when P1 will be flipped P1 : T T H H H P1(Heads) = P2(Heads) Another case: P1 : H T T T T P2 : H H H H T Now when P1 will be flipped P1 : H H H H T P1(Heads) = P2(Heads)
77.53 %
42 votes
logiccleancleverstory

Once upon a time, in the West Lake village, a servant lived with his master. After service of about 30 years, his master became ill and was going to die. One day, the master called his servant and asked him for a wish. It could be any wish but just one. The master gave him one day to think about it. The servant became very happy and went to his mother for discussion about the wish. His mother was blind and she asked her son for making a wish for her eye-sight to come back. Then the servant went to his wife. She became very excited and asked for a son as they were childless for many years. After that, the servant went to his father who wanted to be rich and so he asked his son to wish for a lot of money. The next day he went to his master and made one wish through which all the three (mother, father, wife) got what they wanted. You have to tell what the servant asked the master.
The servant said, "My mother wants to see her grandson swinging on a swing of gold."
76.46 %
82 votes
123
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