You have been given the task of transporting 3,000 apples 1,000 miles from Appleland to Bananaville. Your truck can carry 1,000 apples at a time. Every time you travel a mile towards Bananaville you must pay a tax of 1 apple but you pay nothing when going in the other direction (towards Appleland). What is highest number of apples you can get to Bananaville?
833 apples.
Step one: First you want to make 3 trips of 1,000 apples 333 miles. You will be left with 2,001 apples and 667 miles to go.
Step two: Next you want to take 2 trips of 1,000 apples 500 miles. You will be left with 1,000 apples and 167 miles to go (you have to leave an apple behind).
Step three: Finally, you travel the last 167 miles with one load of 1,000 apples and are left with 833 apples in Bananaville.
How to measure exactly 4 gallon of water from 3 gallon and 5 gallon jars, given, you have unlimited water supply from a running tap.
Step 1. Fill 3 gallon jar with water. ( 5p – 0, 3p – 3)
Step 2. Pour all its water into 5 gallon jar. (5p – 3, 3p – 0)
Step 3. Fill 3 gallon jar again. ( 5p – 3, 3p – 3)
Step 4. Pour its water into 5 gallon jar untill it is full. Now you will have exactly 1 gallon water remaining in 3 gallon jar. (5p – 5, 3p – 1)
Step 5. Empty 5 gallon jar, pour 1 gallon water from 3 gallon jar into it. Now 5 gallon jar has exactly 1 gallon of water. (5p – 1, 3p – 0)
Step 6. Fill 3 gallon jar again and pour all its water into 5 gallon jar, thus 5 gallon jar will have exactly 4 gallon of water. (5p – 4, 3p – 0)
There are 1 million closed school lockers in a row, labeled 1 through 1,000,000.
You first go through and flip every locker open.
Then you go through and flip every other locker (locker 2, 4, 6, etc...). When you're done, all the even-numbered lockers are closed.
You then go through and flip every third locker (3, 6, 9, etc...). "Flipping" mean you open it if it's closed, and close it if it's open. For example, as you go through this time, you close locker 3 (because it was still open after the previous run through), but you open locker 6, since you had closed it in the previous run through.
Then you go through and flip every fourth locker (4, 8, 12, etc...), then every fifth locker (5, 10, 15, etc...), then every sixth locker (6, 12, 18, etc...) and so on. At the end, you're going through and flipping every 999,998th locker (which is just locker 999,998), then every 999,999th locker (which is just locker 999,999), and finally, every 1,000,000th locker (which is just locker 1,000,000).
At the end of this, is locker 1,000,000 open or closed?
Locker 1,000,000 will be open.
If you think about it, the number of times that each locker is flipped is equal to the number of factors it has. For example, locker 12 has factors 1, 2, 3, 4, 6, and 12, and will thus be flipped 6 times (it will end be flipped when you flip every one, every 2nd, every 3rd, every 4th, every 6th, and every 12th locker). It will end up closed, since flipping an even number of times will return it to its starting position. You can see that if a locker number has an even number of factors, it will end up closed. If it has an odd number of factors, it will end up open.
As it turns out, the only types of numbers that have an odd number of factors are squares. This is because factors come in pairs, and for squares, one of those pairs is the square root, which is duplicated and thus doesn't count twice as a factor. For example, 12's factors are 1 x 12, 2 x 6, and 3 x 4 (6 total factors). On the other hand, 16's factors are 1 x 16, 2 x 8, and 4 x 4 (5 total factors).
So lockers 1, 4, 9, 16, 25, etc... will all be open. Since 1,000,000 is a square number (1000 x 1000), it will be open as well.
There are 100 ants on a board that is 1 meter long, each facing either left or right and walking at a pace of 1 meter per minute.
The board is so narrow that the ants cannot pass each other; when two ants walk into each other, they each instantly turn around and continue walking in the opposite direction. When an ant reaches the end of the board, it falls off the edge.
From the moment the ants start walking, what is the longest amount of time that could pass before all the ants have fallen off the plank? You can assume that each ant has infinitely small length.
The longest amount of time that could pass would be 1 minute.
If you were looking at the board from the side and could only see the silhouettes of the board and the ants, then when two ants walked into each other and turned around, it would look to you as if the ants had walked right by each other.
In fact, the effect of two ants walking into each other and then turning around is essentially the same as two ants walking past one another: we just have two ants at that point walking in opposite directions.
So we can treat the board as if the ants are walking past each other. In this case, the longest any ant can be on the board is 1 minute (since the board is 1 meter long and the ants walk at 1 meter per minute). Thus, after 1 minute, all the ants will be off the board.
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.
