A witch owns a field containing many gold mines. She hires one man at a time to mine this gold for her. She promises 10% of what a man mines in a day, and he gives her the rest. Because she is blind, she has three magic bags who can talk. They report how much gold they held each day, and this is how she finds out if men are cheating her. Upon getting the job, each man agrees that if he isn't honest, then he will be turned into stone. So around the witch's mines, many statues lay!
Now comes an honest man named Garry. He accepts the job gladly. The witch, who didn't trust him said, "If I wrongly accuse you of cheating me, then I'll be turned into stone."
That night, Garry, having honestly done his first day's job, overheard the bags talking to the witch. He then formulated a plan... The next night, he submitted his gold, and kept 1.6 pounds of gold. Later, the witch talked with her bags. The first bag said it held 16 pounds that day. The second one said it held 5 pounds. The third one said it held 2 pounds. Beaming, the witch confronted Garry. "You scoundrel, you think you could fool me. Now you shall turn into stone!" the witch cried. One second later, the witch was hard as a rock, and very grey-looking. How did Garry brilliantly deceive the witch?
Garry put 2 lbs. in bag #1. 3 lbs. were put in bag #2. 11 lb. were put into bag #3. He then put bag #2 into bag #3, and bag #1 into bag #2. The bags only felt the weight of the gold above it. Thus they inadvertently gave the message that 23 lbs. were taken. 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
First Player picks 14, now row of coins is
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:
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.logicmathshort
Five add six is eleven, but six add seven is one. How is that possible?
When you are looking at a clockmathshortwhat am I
A name of an animal minus the first three letters plus a flat piece of wood. What am I ?
(MON) KEY + board= Keyboardmathprobability
A girl was ten on her last birthday, and will be twelve on her next birthday. How is this possible?
Today is her 11th Birthday.animalmath
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?
What number should appear next in this sequence?
1 5 12 34 92 252 ?
688. Add the two previous numbers and multiply by 2.cleanlogicmath
A 400 yard long train, travelling at 30 mph, enters a 4.5 mile long tunnel.
How long will elapse between the moment the front of the train enters the tunnel and the moment the end of the train clears the tunnel?
9 minutes and 27.2727 seconds.cleanlogicmathshort
How many sides does a circle have?
Two. The inside and the outside.interviewlogicmath
The Miller next took the company aside and showed them nine sacks of flour that were standing as depicted in the sketch.
"Now, hearken, all and some," said he, "while that I do set ye the riddle of the nine sacks of flour.
And mark ye, my lords and masters, that there be single sacks on the outside, pairs next unto them, and three together in the middle thereof.
By Saint Benedict, it doth so happen that if we do but multiply the pair, 28, by the single one, 7, the answer is 196, which is of a truth the number shown by the sacks in the middle.
Yet it be not true that the other pair, 34, when so multiplied by its neighbour, 5, will also make 196.
Wherefore I do beg you, gentle sirs, so to place anew the nine sacks with as little trouble as possible that each pair when thus multiplied by its single neighbour shall make the number in the middle."
As the Miller has stipulated in effect that as few bags as possible shall be moved, there is only one answer to this puzzle, which everybody should be able to solve.
The way to arrange the sacks of flour is as follows: 2, 78, 156, 39, 4. Here each pair when multiplied by its single neighbour makes the number in the middle, and only five of the sacks need be moved.
There are just three other ways in which they might have been arranged (4, 39, 156, 78, 2; or 3, 58, 174, 29, 6; or 6, 29, 174, 58, 3), but they all require the moving of seven sacks.