animallogicmathThere 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.

## Similar riddles

See also best riddles or new riddles.

cleanlogicmathshortHow do you make the number one disappear by adding to it?

Add the letter 'G' and it becomes Gone.

animalfunnylogicshortIf 20 blackbirds are on a fence and you shoot one, how many remain?

None, they would all fly away from the sound of the shot.

logicmathYou 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.

logicmathThe owner of a banana plantation has a camel. He wants to transport his 3000 bananas to the market, which is located after the desert. The distance between his banana plantation and the market is about 1000 kilometer. So he decided to take his camel to carry the bananas. The camel can carry at the maximum of 1000 bananas at a time, and it eats one banana for every kilometer it travels.
What is the most bananas you can bring over to your destination?

First of all, the brute-force approach does not work. If the Camel starts by picking up the 1000 bananas and try to reach point B, then he will eat up all the 1000 bananas on the way and there will be no bananas left for him to return to point A.
So we have to take an approach that the Camel drops the bananas in between and then returns to point A to pick up bananas again.
Since there are 3000 bananas and the Camel can only carry 1000 bananas, he will have to make 3 trips to carry them all to any point in between.
When bananas are reduced to 2000 then the Camel can shift them to another point in 2 trips and when the number of bananas left are <= 1000, then he should not return and only move forward.
In the first part, P1, to shift the bananas by 1Km, the Camel will have to
Move forward with 1000 bananas – Will eat up 1 banana in the way forward
Leave 998 banana after 1 km and return with 1 banana – will eat up 1 banana in the way back
Pick up the next 1000 bananas and move forward – Will eat up 1 banana in the way forward
Leave 998 banana after 1 km and return with 1 banana – will eat up 1 banana in the way back
Will carry the last 1000 bananas from point a and move forward – will eat up 1 banana
Note: After point 5 the Camel does not need to return to point A again.
So to shift 3000 bananas by 1km, the Camel will eat up 5 bananas.
After moving to 200 km the Camel would have eaten up 1000 bananas and is now left with 2000 bananas.
Now in the Part P2, the Camel needs to do the following to shift the Bananas by 1km.
Move forward with 1000 bananas – Will eat up 1 banana in the way forward
Leave 998 banana after 1 km and return with 1 banana – will eat up this 1 banana in the way back
Pick up the next 1000 bananas and move forward – Will eat up 1 banana in the way forward
Note: After point 3 the Camel does not need to return to the starting point of P2.
So to shift 2000 bananas by 1km, the Camel will eat up 3 bananas.
After moving to 333 km the camel would have eaten up 1000 bananas and is now left with the last 1000 bananas.
The Camel will actually be able to cover 333.33 km, I have ignored the decimal part because it will not make a difference in this example.
Hence the length of part P2 is 333 Km.
Now, for the last part, P3, the Camel only has to move forward. He has already covered 533 (200+333) out of 1000 km in Parts P1 & P2. Now he has to cover only 467 km and he has 1000 bananas.
He will eat up 467 bananas on the way forward, and at point B the Camel will be left with only 533 Bananas.

logicmathHow 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)
We are done !

animalcleanlogicA horse is tied to a fifteen-foot rope and there is a bale of hay 25 feet away from him. The horse however is still able to eat from the hay. How is this possible?

The rope wasn't tied to anything.

logicmathshortIf you're 8 feet away from a door and with each move you advance half the distance to the door. How many moves will it take to reach the door?

You will never reach the door! If you only move half the distance, then you will always have half the distance remaining no matter, how small is the number.

logicmathprobabilityWhat is the least number of people that need to be in a room such that there is greater than a 50% chance that at least two of the people have the same birthday?

Only 23 people need to be in the room.
Our first observation in solving this problem is the following:
(the probability that at least 2 people have the same birthday + the probability that nobody has the same birthday) = 1.0
What this means is that there is a 100% chance that EITHER everybody in the room has a different birthday, OR at least two people in the room have the same birthday (and these probabilities don't add up to more than 1.0 because they cover mutually exclusive situations).
With some simple re-arranging of the formula, we get:
the probability that at least 2 people have the same birthday = (1.0 - the probability that nobody has the same birthday)
So now if we can find the probability that nobody in the room has the same birthday, we just subtract this value from 1.0 and we'll have our answer.
The probability that nobody in the room has the same birthday is fairly straightforward to calculate. We can think of this as a "selection without replacement" problem, where each person "selects" a birthday at random, and we then have to figure out the probability that no two people select the same birthday. The first selection has a 365/365 chance of being different than the other birthdays (since none have been selected yet). The next selection has a 364/365 chance of being different than the 1 birthday that has been selected so far. The next selection has a 363/365 chance of being different than the 2 birthdays that have been selected so far.
These probabilities are multiplied together since each is conditional on the previous. So for example, the probability that nobody in a room of 3 people have the same birthday is (365/365 * 364/365 * 363/365) =~ 0.9918
More generally, if there are n people in a room, then the probability that nobody has the same birthday is (365/365 * 364/365 * ... * (365-n+2)/365 * (365-n+1)/365)
We can plug in values for n. For n=22, we get that the probability that nobody has the same birthday is 0.524, and thus the probabilty that at least two people have the same birthday is (1.0 - 0.524) = 0.476 = 47.6%.
Then for n=23, we get that the probability that nobody has the same birthday is 0.493, and thus the probabilty that at least two people have the same birthday is 1.0 - 0.493) = 0.507 = 50.7%. Thus, once we get to 23 people we have reached the 50% threshold.

logicmathYou are standing in a pitch-dark room. A friend walks up and hands you a normal deck of 52 cards. He tells you that 13 of the 52 cards are face-up, the rest are face-down. These face-up cards are distributed randomly throughout the deck.
Your task is to split up the deck into two piles, using all the cards, such that each pile has the same number of face-up cards. The room is pitch-dark, so you can't see the deck as you do this.
How can you accomplish this seemingly impossible task?

Take the first 13 cards off the top of the deck and flip them over. This is the first pile. The second pile is just the remaining 39 cards as they started.
This works because if there are N face-up cards in within the first 13 cards, then there will be (13 - N) face up cards in the remaining 39 cards. When you flip those first 13 cards, N of which are face-up, there will now be N cards face-down, and therefore (13 - N) cards face-up, which, as stated, is the same number of face-up cards in the second pile.

funnylogicmathAn infinite number of mathematicians are standing behind a bar. The first asks the barman for half a pint of beer, the second for a quarter pint, the third an eighth, and so on. How many pints of beer will the barman need to fulfill all mathematicians' wishes?

Just one.