A horse travels a certain distance each day. Strangely enough, two of its legs travel 30 miles each day and the other two legs travel nearly 31 miles. It would seem that two of the horse's legs must be one mile ahead of the other two legs, but of course this can't be true. Since the horse is normal, how is this situation possible?
The horse operates the mill and travels in a circular clockwise direction. The two outside legs will travel a greater distance than the inside ones.
A boy goes and buys a fishing pole that is 6' 3" long. As he goes to get on the bus, the driver stops him. The driver tells him that he can't take anything longer than 6' onto the bus. The boy goes back into town, purchases one more thing, and the driver allows the boy on the bus. What did the boy buy, and what did he do with it?
The boy bought 6' long box. He put the fishing pole in diagonally and the entire package was only 6'!
Two workmen were repairing a roof. Suddenly, they both fell down the chimney and found themselves in the large fireplace.
One man`s face was smeared with soot and one wasn`t. The one with the clean face washed his and the dirty man did not and went back to work. Why?
When the two men looked at each other, the clean man thought his face was dirty as well. The dirty man, looking at the first's place, thought his was clean.
A man and woman run through a field holding hands. They bound toward the sunset, happy as can be. Suddenly, the man moves off of his straight-line course and starts veering to his left. At the same time, the woman begins running off to her right.
They continue this for a full minute, but never let go of each others' hands. How is this possible?
The man was facing forward, but the woman was running backwards. The man's right hand was holding the woman's right hand. They both veered in the same geographic direction, but it was the man's left and the woman's right because the woman was running backwards.
A new student met the Zen Master after traveling hundreds of miles by yak cart. He was understandably pleased with himself for being selected to learn at the great master's feet .
The first time they formally met, the Zen Master asked, "May I ask you a simple question?" "It would be an honor!" replied the student.
"Which is greater, that which has no beginning or that which has no end?" queried the Zen Master. "Come back when you have the answer and can explain why."
After the student made many frustrated trips back with answers which the master quickly cast off with a disapproving negative nod, the Zen Master finally said, "Perhaps I should ask you another question?"
"Oh, please do!" pleaded the exasperated student.
The Zen Master then asked, "Since you do not know that, answer this much simpler riddle. When can a pebble hold back the sea?" Again the student was rebuffed time and again. Several more questions followed with the same result. Each time, the student could not find the correct answer. Finally, completely exasperated, the student began to weep, "Master, I am a complete idiot. I can not solve even the simplest riddle from you!"
Suddenly, the student stopped, sat down, and said, "I am ready for my second lesson."
What was the Zen Master's first lesson?
The student's first lesson was that in order to learn from the Zen Master, the student should be asking the questions and not the Zen Master.
You have twelve balls, identical in every way except that one of them weighs slightly less or more than the balls.
You have a balance scale, and are allowed to do 3 weighings to determine which ball has the different weight, and whether the ball weighs more or less than the other balls.
What process would you use to weigh the balls in order to figure out which ball weighs a different amount, and whether it weighs more or less than the other balls?
Take eight balls, and put four on one side of the scale, and four on the other.
If the scale is balanced, that means the odd ball out is in the other 4 balls.
Let's call these 4 balls O1, O2, O3, and O4.
Take O1, O2, and O3 and put them on one side of the scale, and take 3 balls from the 8 "normal" balls that you originally weighed, and put them on the other side of the scale.
If the O1, O2, and O3 balls are heavier, that means the odd ball out is among these, and is heavier. Weigh O1 and O2 against each other. If one of them is heavier than the other, this is the odd ball out, and it is heavier. Otherwise, O3 is the odd ball out, and it is heavier.
If the O1, O2, and O3 balls are lighter, that means the odd ball out is among these, and is lighter. Weigh O1 and O2 against each other. If one of them is lighter than the other, this is the odd ball out, and it is lighter. Otherwise, O3 is the odd ball out, and it is lighter.
If these two sets of 3 balls weigh the same amount, then O4 is the odd ball out. Weight it against one of the "normal" balls from the first weighing. If O4 is heavier, then it is heavier, if it's lighter, then it's lighter.
If the scale isn't balanced, then the odd ball out is among these 8 balls.
Let's call the four balls on the side of the scale that was heavier H1, H2, H3, and H4 ("H" for "maybe heavier").
Let's call the four balls on the side of the scale that was lighter L1, L2, L3, and L4 ("L" for "maybe lighter").
Let's also call each ball from the 4 in the original weighing that we know aren't the odd balls out "Normal" balls.
So now weigh [H1, H2, L1] against [H3, L2, Normal].
-If the [H1, H2, L1] side is heavier (and thus the [H3, L2, Normal] side is lighter), then this means that either H1 or H2 is the odd ball out and is heavier, or L2 is the odd ball out and is lighter.
-So measure [H1, L2] against 2 of the "Normal" balls.
-If [H1, L2] are heavier, then H1 is the odd ball out, and is heavier.
-If [H1, L2] are lighter, then L2 is the odd ball out, and is lighter.
-If the scale is balanced, then H2 is the odd ball out, and is heavier.
-If the [H1, H2, L1] side is lighter (and thus the [H3, L2, Normal] side is heavier), then this means that either L1 is the odd ball out, and is lighter, or H3 is the odd ball out, and is heavier.
-So measure L1 and H3 against two "normal" balls.
-If the [L1, H3] side is lighter, then L1 is the odd ball out, and is lighter.
-Otherwise, if the [L1, H3] side is heavier, then H3 is the odd ball out, and is heavier.
If the [H1, H2, L1] side and the [H3, L2, Normal] side weigh the same, then we know that either H4 is the odd ball out, and is heavier, or one of L3 or L4 is the odd ball out, and is lighter.
So weight [H4, L3] against two of the "Normal" balls.
If the [H4, L3] side is heavier, then H4 is the odd ball out, and is heavier.
If the [H4, L3] side is lighter, then L3 is the odd ball out, and is lighter.
If the [H4, L3] side weighs the same as the [Normal, Normal] side, then L4 is the odd ball out, and is lighter.
Marty and Jill want to copy three 60 minute tapes. They have two tape recorders that will dub the tapes for them, so they can do two at a time. It takes 30 minutes for each side to complete; therefore in one hour two tapes will be done, and in another hour the third will be done. Jill says all three tapes can be made in 90 minutes. How?
Jill will rotate the three tapes. Let's call them tapes 1,2, and 3 with sides A and B. In the first 30 minutes they will tape 1A and 2A, in the second 3 minutes they will tape 1B and 3A (Tape 1 is now done). Finally, in the last 30 minutes, they will tape 2B and 3B.
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.