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.

logicYour friend pulls out a perfectly circular table and a sack of quarters, and proposes a game.
"We'll take turns putting a quarter on the table," he says. "Each quarter must lay flat on the table, and cannot sit on top of any other quarters. The last person to successfully put a quarter on the table wins."
He gives you the choice to go first or second. What should you do, and what should your strategy be to win?

You should go first, and put a quarter at the exact center of the table.
Then, each time your opponent places a quarter down, you should place your next quarter in the symmetric position on the opposite side of the table.
This will ensure that you always have a place to set down our quarter, and eventually your oppponent will run out of space.

logicYou have twelve balls, identical in every way except that one of them weighs slightly less or more than the balls.
You have a balance scale, and are allowed to do 3 weighings to determine which ball has the different weight, and whether the ball weighs more or less than the other balls.
What process would you use to weigh the balls in order to figure out which ball weighs a different amount, and whether it weighs more or less than the other balls?

Take eight balls, and put four on one side of the scale, and four on the other.
If the scale is balanced, that means the odd ball out is in the other 4 balls.
Let's call these 4 balls O1, O2, O3, and O4.
Take O1, O2, and O3 and put them on one side of the scale, and take 3 balls from the 8 "normal" balls that you originally weighed, and put them on the other side of the scale.
If the O1, O2, and O3 balls are heavier, that means the odd ball out is among these, and is heavier. Weigh O1 and O2 against each other. If one of them is heavier than the other, this is the odd ball out, and it is heavier. Otherwise, O3 is the odd ball out, and it is heavier.
If the O1, O2, and O3 balls are lighter, that means the odd ball out is among these, and is lighter. Weigh O1 and O2 against each other. If one of them is lighter than the other, this is the odd ball out, and it is lighter. Otherwise, O3 is the odd ball out, and it is lighter.
If these two sets of 3 balls weigh the same amount, then O4 is the odd ball out. Weight it against one of the "normal" balls from the first weighing. If O4 is heavier, then it is heavier, if it's lighter, then it's lighter.
If the scale isn't balanced, then the odd ball out is among these 8 balls.
Let's call the four balls on the side of the scale that was heavier H1, H2, H3, and H4 ("H" for "maybe heavier").
Let's call the four balls on the side of the scale that was lighter L1, L2, L3, and L4 ("L" for "maybe lighter").
Let's also call each ball from the 4 in the original weighing that we know aren't the odd balls out "Normal" balls.
So now weigh [H1, H2, L1] against [H3, L2, Normal].
-If the [H1, H2, L1] side is heavier (and thus the [H3, L2, Normal] side is lighter), then this means that either H1 or H2 is the odd ball out and is heavier, or L2 is the odd ball out and is lighter.
-So measure [H1, L2] against 2 of the "Normal" balls.
-If [H1, L2] are heavier, then H1 is the odd ball out, and is heavier.
-If [H1, L2] are lighter, then L2 is the odd ball out, and is lighter.
-If the scale is balanced, then H2 is the odd ball out, and is heavier.
-If the [H1, H2, L1] side is lighter (and thus the [H3, L2, Normal] side is heavier), then this means that either L1 is the odd ball out, and is lighter, or H3 is the odd ball out, and is heavier.
-So measure L1 and H3 against two "normal" balls.
-If the [L1, H3] side is lighter, then L1 is the odd ball out, and is lighter.
-Otherwise, if the [L1, H3] side is heavier, then H3 is the odd ball out, and is heavier.
If the [H1, H2, L1] side and the [H3, L2, Normal] side weigh the same, then we know that either H4 is the odd ball out, and is heavier, or one of L3 or L4 is the odd ball out, and is lighter.
So weight [H4, L3] against two of the "Normal" balls.
If the [H4, L3] side is heavier, then H4 is the odd ball out, and is heavier.
If the [H4, L3] side is lighter, then L3 is the odd ball out, and is lighter.
If the [H4, L3] side weighs the same as the [Normal, Normal] side, then L4 is the odd ball out, and is lighter.

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

A bee.

logicA man was to be sentenced, and the judge told him, "You may make a statement. If it is true, I'll sentence you to four years in prison. If it is false, I'll sentence you to six years in prison." After the man made his statement, the judge decided to let him go free.What did the man say?

He said, "You'll sentence me to six years in prison." If it was true, then the judge would have to make it false by sentencing him to four years. If it was false, then he would have to give him six years, which would make it true. Rather than contradict his own word, the judge set the man free.

crazyfunnyWhy shouldn’t you tell an Easter egg a joke?

It might crack up.

logicA deliveryman comes to a house to drop off a package. He asks the woman who lives there how many children she has.
"Three," she says. "And I bet you can't guess their ages."
"Ok, give me a hint," the deliveryman says.
"Well, if you multiply their ages together, you get 36," she says. "And if you add their ages together, the sum is equal to our house number."
The deliveryman looks at the house number nailed to the front of her house. "I need another hint," he says.
The woman thinks for a moment. "My youngest son will have a lot to learn from his older brothers," she says.
The deliveryman's eyes light up and he tells her the ages of her three children. What are their ages?

Their ages are 1, 6, and 6. We can figure this out as follows:
Given that their ages multiply out to 36, the possible ages for the children are:
1, 1, 36 (sum = 38)
1, 2, 18 (sum = 21)
1, 3, 12 (sum = 16)
1, 4, 9 (sum = 14)
1, 6, 6 (sum = 13)
2, 2, 9 (sum = 13)
2, 3, 6 (sum = 11)
3, 3, 4 (sum = 10)
When the woman tells the deliveryman that the children's ages add up to her street number, he still doesn't know their ages. The only way this could happen is that there is more than one possible way for the children's ages to add up to the number on the house (or else he would have known their ages when he looked at the house number). Looking back at the possible values for the children's ages, you can see that there is only one situation in which there are multiple possible values for the children's ages that add up to the same sum, and that is if their ages are either 1, 6, and 6 (sums up to 13), or 2, 2, and 9 (also sums up to 13). So these are now the only possible values for their ages.
When the woman then tells him that her youngest son has two older brothers (who we can tell are clearly a number of years older), the only possible situation is that their ages are 1, 6, and 6.

logic You've been placed on a course of expensive medication in which you are to take one tablet of Sildenafil and one tablet of Citrate daily. You must be careful that you take just one of each because taking more of either can have serious side effects. Taking Sildenafil without taking Citrate, or vice versa, can also be very serious, because they must be taken together in order to be effective. In summary, you must take exactly one of the Sildenafil pills and one of the Citrate pills at one time. Therefore, you open up the Sildenafil bottle, and you tap one Sildenafil pill into your hand. You put that bottle aside and you open the Citrate bottle. You do the same, but by mistake, two Citrates fall into your hand with the Sildenafil pill. Now, here's the problem. You weren't watching your hand as the pills fell into it, so you can't tell the Sildenafil pill apart from the two Citrate pills. The pills look identical. They are both the same size, same weight (10 micrograms), same color (Blue), same shape (perfect square), same everything, and they are not marked differently in any way. What are you going to do? You cannot tell which pill is which, and they cost $300 a piece, so you cannot afford to throw them away and start over again. How do you get your daily dose of exactly one Sildenafil and exactly one Citrate without wasting any of the pills?

Carefully cut each of the three pills in half, and carefully separate them into two piles, with half of each pill in each pile. You do not know which pill is which, but you are 100% sure that each of the two piles now contains two halves of Cirate and half of Sildenafil. Now go back into the Sildenafil bottle, take out a pill, cut it in half, and add one half to each stack. Now you have two stacks, each one containing two halves of Sildenafil and two halves of Citrate. Take one stack of pills today, and save the second stack for tomorrow.

logicOne 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.

logicWhen Manish was three years old he carved a nail into his favorite tree to mark his height. Six years later at age nine, Manish returned to see how much higher the nail was. If the tree grew by five centimeters each year, how much higher would the nail be.

The nail would be at the same height since trees grow at their tops.