100 men are in a room, each wearing either a white or black hat. Nobody knows the color of his own hat, although everyone can see everyone else's hat. The men are not allowed to communicate with each other at all (and thus nobody will ever be able to figure out the color of his own hat).
The men need to line up against the wall such that all the men with black hats are next to each other, and all the men with white hats are next to each other. How can they do this without communicating? You can assume they came up with a shared strategy before coming into the room.
The men go to stand agains the wall one at a time. If a man goes to stand against the wall and all of the men already against the wall have the same color hat, then he just goes and stands at either end of the line. However, if a man goes to stand against the wall and there are men with both black and white hats already against the wall, he goes and stands between the two men with different colored hats. This will maintain the state that the line contains men with one colored hats on one side, and men with the other colored hats on the other side, and when the last man goes and stands against the wall, we'll still have the desired outcome.
A women walks into a bank to cash out her check.
By mistake the bank teller gives her dollar amount in change, and her cent amount in dollars.
On the way home she spends 5 cents, and then suddenly she notices that she has twice the amount of her check.
How much was her check amount?
The check was for dollars 31.63.
The bank teller gave her dollars 63.31
She spent .05, and then she had dollars 63.26, which is twice the check.
Let x be the dolars of the check, and y be the cent.
The check was for 100x + y cent
He was given 100y + x cent
Also
100y + x - 5 = 2(100x + y)
Expanding this out and rearranging, we find:
98y = 199x + 5
Which doesn't look like enough information to solve the problem except that x and y must be whole numbers, so:
199x ≡ -5 (mod 98)
98*2*x + 3x ≡ -5 (mod 98)
3x ≡ -5 ≡ 93 (mod 98)
This quickly leads to x = 31 and then y = 63
Alternative solution by substitution:
98y = 199x + 5
y = (199x + 5)/98 = 2x + (3x + 5)/98
Since x and y are whole numbers, so must be (3x + 5)/98.
Call it z = (3x+5)/98
so 98z = 3x + 5, or 3x = 98z - 5 or x = (98z - 5)/3
or x = 32z-1 + (2z-2)/3.
Since everything is a whole number, so must be (2z-2)/3.
Call it w = (2z-2)/3, so 3w = 2z-2 so z = (3w+2)/2 or
z = w + 1 + w/2. So w/2 must be whole, or w must be even.
So try w = 2. Then z = 4. Then x = 129. Then y = 262.
if you decrease y by 199 and x by 98, the answer is the same:
y = 63 and x = 31.
You've been placed on a course of expensive medication in which you are to take one tablet of Plusin and one tablet of Minusin daily. You must be careful that you take just one of each because taking more of either can have serious side effects. Taking Plusin without taking Minusin, 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 Plusin pills and one of the Minusin pills at one time.
Therefore, you open up the Plusin bottle, and you tap one Plusin pill into your hand. You put that bottle aside and you open the Minusin bottle. You do the same, but by mistake, two Minusins fall into your hand with the Plusin pill.
Now, here's the problem. You weren't watching your hand as the pills fell into it, so you can't tell the Plusin pill apart from the two Minusin 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 $500 a piece, so you cannot afford to throw them away and start over again. How do you get your daily dose of exactly one Plusin and exactly one Minusin 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 Minusin and half of Plusin. Now go back into the Plusin 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 Plusin and two halves of Minusin. Take one stack of pills today, and save the second stack for tomorrow.
A monk leaves at sunrise and walks on a path from the front door of his monastery to the top of a nearby mountain. He arrives at the mountain summit exactly at sundown. The next day, he rises again at sunrise and descends down to his monastery, following the same path that he took up the mountain.
Assuming sunrise and sunset occured at the same time on each of the two days, prove that the monk must have been at some spot on the path at the same exact time on both days.
Imagine that instead of the same monk walking down the mountain on the second day, that it was actually a different monk. Let's call the monk who walked up the mountain monk A, and the monk who walked down the mountain monk B. Now pretend that instead of walking down the mountain on the second day, monk B actually walked down the mountain on the first day (the same day monk A walks up the mountain).
Monk A and monk B will walk past each other at some point on their walks. This moment when they cross paths is the time of day at which the actual monk was at the same point on both days. Because in the new scenario monk A and monk B MUST cross paths, this moment must exist.
A bad king has a cellar of 1000 bottles of delightful and very expensive wine. A neighboring queen plots to kill the bad king and sends a servant to poison the wine.
Fortunately (or say unfortunately) the bad king's guards catch the servant after he has only poisoned one bottle.
Alas, the guards don't know which bottle but know that the poison is so strong that even if diluted 100,000 times it would still kill the king. Furthermore, it takes one month to have an effect.
The bad king decides he will get some of the prisoners in his vast dungeons to drink the wine. Being a clever bad king he knows he needs to murder no more than 10 prisoners – believing he can fob off such a low death rate – and will still be able to drink the rest of the wine (999 bottles) at his anniversary party in 5 weeks time.
Explain what is in mind of the king, how will he be able to do so?
Think in terms of binary numbers. (now don’t read the solution, give a try).
Number the bottles 1 to 1000 and write the number in binary format.
bottle 1 = 0000000001 (10 digit binary)
bottle 2 = 0000000010
bottle 500 = 0111110100
bottle 1000 = 1111101000
Now take 10 prisoners and number them 1 to 10, now let prisoner 1 take a sip from every bottle that has a 1 in its least significant bit. Let prisoner 10 take a sip from every bottle with a 1 in its most significant bit. etc.
prisoner = 10 9 8 7 6 5 4 3 2 1
bottle 924 = 1 1 1 0 0 1 1 1 0 0
For instance, bottle no. 924 would be sipped by 10,9,8,5,4 and 3. That way if bottle no. 924 was the poisoned one, only those prisoners would die.
After four weeks, line the prisoners up in their bit order and read each living prisoner as a 0 bit and each dead prisoner as a 1 bit. The number that you get is the bottle of wine that was poisoned.
1000 is less than 1024 (2^10). If there were 1024 or more bottles of wine it would take more than 10 prisoners.
A man walks into a bar and asks the bartender for a glass of water. The bartender reaches under the bar and brings out a gun and aims it at the man. The man says thank you and leaves. What happened?
The man had the hiccups and the water helped him stop it, and the gun scared him which also help stop his hiccups as well.
You 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.
