You are visiting NYC when a man approaches you.
"Not counting bald people, I bet a hundred bucks that there are two people living in New York City with the same number of hairs on their heads," he tells you.
"I'll take that bet!" you say. You talk to the man for a minute, after which you realize you have lost the bet.
What did the man say to prove his case?
This is a classic example of the pigeonhole principle. The argument goes as follows: assume that every non-bald person in New York City has a different number of hairs on their head. Since there are about 9 million people living in NYC, let's say 8 million of them aren't bald.
So 8 million people need to have different numbers of hairs on their head. But on average, people only have about 100,000 hairs. So even if there was someone with 1 hair, someone with 2 hairs, someone with 3 hairs, and so on, all the way up to someone with 100,000 hairs, there are still 7,900,000 other people who all need different numbers of hairs on their heads, and furthermore, who all need MORE than 100,000 hairs on their head.
You can see that additionally, at least one person would need to have at least 8,000,000 hairs on their head, because there's no way to have 8,000,000 people all have different numbers of hairs between 1 and 7,999,999. But someone having 8,000,000 is an essential impossibility (as is even having 1,000,000 hairs), So there's no way this situation could be the case, where everyone has a different number of hairs. Which means that at least two people have the same number of hairs.
Six jugs are in a row. The first three are filled with coke, and the last three are empty. By moving only one glass, can you arrange them so that the full and the empty glasses alternate?
Move and then pour all coke from second glass to fifth glass.
I sometimes come in a can but I'm not food.
I sometimes come in a bottle but I'm not a beverage.
I come in different colors but I'm not a rainbow.
I'm sometimes used with canvas but I'm not a tent.
I'm used with a brush but I'm not toothpaste.
What Am I?
Four people come to an old bridge in the middle of the night. The bridge is rickety and can only support 2 people at a time. The people have one flashlight, which needs to be held by any group crossing the bridge because of how dark it is.
Each person can cross the bridge at a different rate: one person takes 1 minute, one person takes 2 minutes, one takes 5 minutes, and the one person takes 10 minutes. If two people are crossing the bridge together, it will take both of them the time that it takes the slower person to cross.
Unfortunately, there are only 17 minutes worth of batteries left in the flashlight. How can the four travellers cross the bridge before time runs out?
The two keys here are:
You want the two slowest people to cross together to consolidate their slow crossing times.
You want to make sure the faster people are set up in order to bring the flashlight back quickly after the slow people cross.
So the order is:
1-minute and 2-minute cross (2 minute elapsed)
1-minute comes back (3 minutes elapsed)
5-minute and 10-minute cross (13 minutes elapsed)
2-minute comes back (15 minutes elapsed)
1-minute and 2-minute cross (17 minutes elapsed)
