## Dog chasing red articles of clothing

What kind of dog chases red articles of clothing?

A Bulldog.

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What kind of dog chases red articles of clothing?

A Bulldog.

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Jay escaped from jail and headed to the country. While walking along a rural road, he saw a police car speeding towards him. Jay ran toward it for a short time and then fled into the woods. Why did he run toward the car?

Jay was just starting to cross a bridge when he saw a police car. He ran toward the car to get off the bridge before running into the woods.

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I had no father, only a grandfather. I shall bear no sons, only daughters. What am I?

A bee.

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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.

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One morning an airline president is leaving on a business trip and finds he left some paperwork at his office. He runs into his office to get it and the night watchman stops him and says, "Sir, don't get on the plane. I had a dream last night that the plane would crash and everyone would die!" The man takes his word and cancels his trip. Sure enough, the plane crashes and everyone dies. The next morning the man gives the watchman a $1,000 reward for saving his life and then fires him. Why did he fire the watchman that saved his life?

He was fired from sleeping on his job.

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What is a frog’s favorite year?

Leap Year.

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A guard is stationed at the entrance to a bridge. He is tasked to shoot anyone who tries to cross to the other side of the bridge, and to turn away anyone who comes in from the opposite side of the bridge. You are on his side of the bridge and want to escape to the other side.
Because the bridge is old and rickety, anyone who tries to cross it does so at a constant speed, and it always takes exactly 10 minutes to cross.
The guard comes out of his post every 6 minutes and looks down the bridge for any people trying to leave, and at all other times he sits in his post and snoozes. You know you can sneak past him when he's sleeping, but the problem is that you won't be able to make it all the way to the other side of the bridge before he sees you (since he comes out every 6 minutes, but it takes 10 minutes to cross).
One day a brilliant idea comes to you, and soon you've successfully crossed to the other side of the bridge without being shot. How did you do it?

Right after the guard goes back to his post after checking the bridge, you sneak by and make your way down the bridge. After a little bit less than 6 minutes, you turn around and start walking back toward the guard. He will come out and see you, and assume that you are a visitor coming from the other side of the bridge, since you're only about 4 minutes from the end of the other side of the bridge. He will go back into his post since he doesn't plan to turn you away until you reach him, and then you turn back around and make your way the rest of the way to the other side of the bridge.

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Bill and Stacie are delighted when their new baby, Patrick, is born on February 29th, 2008. They think it's good luck to for him to be born on the special day of the leap year. But then they start thinking about when to celebrate his next birthday. After some thought, they decide that they want to celebrate Patrick's next birthday (when he turns 1) exactly 365 days after he was born, just like most people do.
What will be the date of this birthday?

The date of the birthday will be February 28th, 2009.
At first it might seem like his birthday should be March 1st, 2009, since February 29th is the day after February 28th in the leap year, while March 1st is the day after February 28th in non-leap years. But this is the wrong way to think about it.
The right way to think about it is that 365 days after the day before March 1st is always February 28th, regardless of whether it's a leap year or not. So Patrick's birthday will be February 28th.

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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.

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Pronounced as one letter,
And written with three,
Two letters there are,
And two only in me.
I’m double, I’m single,
I’m black blue and gray,
I’m read from both ends,
And the same either way.

Eye.

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