Create a number using only the digits 4,4,3,3,2,2,1 and 1.
So I can only be eight digits.
You have to make sure the ones are separated by one digit, the twos are separated by two digits the threes are separated with three digits and the fours are separated by four digits.
You can easily "tile" an 8x8 chessboard with 32 2x1 tiles, meaning that you can place these 32 tiles on the board and cover every square.
But if you take away two opposite corners from the chessboard, it becomes impossible to tile this new 62-square board.
Can you explain why tiling this board isn't possible?
Color in the chessboard, alternating with red and blue tiles. Then color all of your tiles half red and half blue. Whenever you place a tile down, you can always make it so that the red part of the tile is on a red square and the blue part of the tile is on the blue square.
Since you'll need to place 31 tiles on the board (to cover the 62 squares), you would have to be able to cover 31 red squares and 31 blue squares. But when you took away the two corners, you can see that you are taking away two red spaces, leaving 30 red squares and 32 blue squares. There is no way to cover 30 red squares and 32 blue squares with the 31 tiles, since these tiles can only cover 31 red squares and 31 blue squares, and thus, tiling this board is not possible.
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.
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.
Frank and some of the boys were exchanging old war stories. James offered one about how his grandfather (Captain Smith) led a battalion against a German division during World War I. Through brilliant maneuvers he defeated them and captured valuable territory. Within a few months after the battle he was presented with a sword bearing the inscription: "To Captain Smith for Bravery, Daring and Leadership, World War One, from the Men of Battalion 8." Frank looked at James and said, "You really don't expect anyone to believe that yarn, do you?"
What is wrong with the story?
It wasn't called World War One until much later. It was called the Great War at first, because they did not know during that war and immediately afterward that there would be a second World War (WW II).
A boy goes and buys a fishing pole that is 6' 3" long. As he goes to get on the bus, the driver stops him. The driver tells him that he can't take anything longer than 6' onto the bus. The boy goes back into town, purchases one more thing, and the driver allows the boy on the bus. What did the boy buy, and what did he do with it?
The boy bought 6' long box. He put the fishing pole in diagonally and the entire package was only 6'!
One day, Emperor Akbar posed a question to Birbal. He asked him what Birbal would choose if he offered either justice or a gold coin.
"The gold coin," said Birbal without hesitation.
On hearing this, Akbar was taken aback. "You would prefer a gold coin to justice?" he asked, not believing his own ears.
"Yes," said Birbal.
The other courtiers were amazed by Birbal's display of idiocy. They were full of glee that Birbal had finally managed himself to do what these courtiers had not been able to do for a long time - discredit Birbal in the emperor's eyes!
"I would have been disappointed if this was the choice made even by my lowliest of servants," continued the emperor. "But coming from you it's not only disappointing, but shocking and sad. I did not know you were so debased!"
How did Birbal justify his answer to the enraged and hurt Emperor?
"One asks for what one does not have, Your Majesty." said Birbal, smiling gently and in quiet tones.
"Under Your Majesty´s rule, justice is available to everybody. But I am a spendthrift and always short of money and therefore I said I would choose the gold coin."
The answer immensely pleased the emperor and respect for Birbal was once again restored in the emperor's eyes.
You're walking down a path and come to two doors. One of the doors leads to a life of prosperity and happiness, and the other door leads to a life of misery and sorrow. You don't know which door is which.
In front of the door is ONE man. You know that this man either always lies, or always tells the truth, but you don't know which. The man knows which door is which.
You are allowed to ask the man ONE yes-or-no question to figure out which door to go through. To make things more difficult, the man is very self-centered, so you are only allowed to ask him a question about what he thinks or knows; your question cannot involve what any other person or object (real or hypothetical) might say.
What question should you ask to ensure you go through the good door?
You should ask: "If I asked you if the good door is on the left, would you say yes?"
Notice that this is subtly different than asking "Is the good door on the left?", in that you are asking him IF he would say yes to that question, not what his answer to the question would be. Thus you are asking a question about a question, and if it ends up being the liar you are talking to, this will cause him to lie about a lie and thus tell the truth. The four possible cases are:
The man is a truth-teller and the good door is on the left. He will say "yes".
The man is a truth-teller and the good door is on the right. He will say "no".
The man is a liar and the good door is on the left. He will say "yes" because if you asked him "Is the good door on the left?", he would lie and say "no", and so when you ask him if he would say "yes", he will lie and say "yes".
The man is a liar and the good door is on the right. Similar to the previous example, he'll say "no".
So regardless of whether the man is a truth-teller or a liar, this question will get a "yes" if the door on the left is the good door, and a "no" if it's not.
You 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.
"Welcome back to the show. Before the break, Mr Ixolite here made it to our grand finale! How do you feel Mr.Ix?"
"Nervous."
"Okay, now to win the star prize of one million pounds all you have to do is answer the following question in 90 seconds."
"Okay, I'm ready."
"Right. In 90 seconds name 100 words that do NOT contain the letter 'A'. Start the clock!"
Can you help?
One, two, three, four, five...one hundred! I just counted from 1 to 100 in ninety seconds (it is possible).
