You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
You need to conduct 7 races.
First, separate the horses into 5 groups of 5 horses each, and race the horses in each of these groups. Let's call these groups A, B, C, D and E, and within each group let's label them in the order they finished. So for example, in group A, A1 finished 1st, A2 finished 2nd, A3 finished 3rd, and so on.
We can rule out the bottom two finishers in each race (A4 and A5, B4 and B5, C4 and C5, D4 and D5, and E4 and E5), since we know of at least 3 horses that are faster than them (specifically, the horses that beat them in their respective races).
This table shows our remaining horses:
A1 B1 C1 D1 E1
A2 B2 C2 D2 E2
A3 B3 C3 D3 E3
For our 6th race, let's race the top finishers in each group: A1, B1, C1, D1 and E1. Let's assume that the order of finishers is: A1, B1, C1, D1, E1 (so A1 finished first, E1 finished last).
We now know that horse D1 cannot be in the top 3, because it is slower than C1, B1 and A1 (it lost to them in the 6th race). Thus, D2 and D3 can also not be in the to 3 (since they are slower than D1).
Similarly, E1, E2 and E3 cannot be in the top 3 because they are all slower than D1 (which we already know isn't in the top 3).
Let's look at our updated table, having removed these horses that can't be in the top 3:
A1 B1 C1
A2 B2 C2
A3 B3 C3
We can actually rule out a few more horses. C2 and C3 cannot be in the top 3 because they are both slower than C1 (and thus are also slower than B1 and A1). And B3 also can't be in the top 3 because it is slower than B2 and B1 (and thus is also slower than A1). So let's further update our table:
A1 B1 C1
A2 B2
A3
We actually already know that A1 is our fastest horse (since it directly or indirectly beat all the remaining horses). So now we just need to find the other two fastest horses out of A2, A3, B1, B2 and C1. So for our 7th race, we simply race these 5 horses, and the top two finishers, plus A1, are our 3 fastest horses.
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.
