A king has 100 identical servants, each with a different rank between 1 and 100. At the end of each day, each servant comes into the king's quarters, one-by-one, in a random order, and announces his rank to let the king know that he is done working for the day. For example, servant 14 comes in and says "Servant 14, reporting in."
One day, the king's aide comes in and tells the king that one of the servants is missing, though he isn't sure which one.
Before the other servants begin reporting in for the night, the king asks for a piece of paper to write on to help him figure out which servant is missing. Unfortunately, all that's available is a very small piece that can only hold one number at a time. The king is free to erase what he writes and write something new as many times as he likes, but he can only have one number written down at a time.
The king's memory is bad and he won't be able to remember all the exact numbers as the servants report in, so he must use the paper to help him.
How can he use the paper such that once the final servant has reported in, he'll know exactly which servant is missing?
When the first servant comes in, the king should write down his number. For each other servant that reports in, the king should add that servant's number to the current number written on the paper, and then write this new number on the paper.
Once the final servant has reported in, the number on the paper should equal
(1 + 2 + 3 + ... + 99 + 100) - MissingServantsNumber
Since (1 + 2 + 3 + ... + 99 + 100) = 5050, we can rephrase this to say that the number on the paper should equal
5050 - MissingServantsNumber
So to figure out the missing servant's number, the king simply needs to subtract the number written on his paper from 5050:
MissingServantsNumber = 5050 - NumberWrittenOnThePaper
A woman who lived in Germany during World War II wanted to cross the German/Swiss border in order to escape Nazi pursuers. The bridge which she is to cross is a half mile across, over a large canyon. Every three minutes a guard comes out of his bunker and checks if anyone is on the bridge. If a person is caught trying to escape German side to the Swiss side they are shot. If caught crossing the other direction without papers they are sent back. She knows that it takes at least five minutes to cross the bridge, in which time the guard will see her crossing and shoot her. How does she get across?
She waits until the guard goes inside his hunt, and begins to walk across the bridge. She gets a little more than half way, turns around, and begins to walk toward the german side once more. The guard comes out, sees that she has no papers, and sends her "back" to the swiss side.
Two sisters we are, one is dark and one is fair,
In twin towers dwelling we're quite the pair,
One from land and one from sea,
Tell us truly, who are we?
There are ten people in a house. Everybody wants to make a hand shake with only people shorter than themselves. Assume everybody is different in height. How many hand shakes are made?
0, because a taller person wants to shake hands with a shorter person. But the shorter person doesn't want to shake hands with him.
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)
I drift forever with the current down these long canals they've made.
Tame, yet wild, I run elusive, multitasking to your aid.
Before I came, the world was darker. Colder, sometimes, rougher, true.
But though I might make living easy, I'm good at killing people too.
You have two jugs, one that holds exactly 3 gallons, and one that holds exactly 5 gallons. Using just these two jugs and a fire hose, how can you measure out exactly 4 gallons of water?
Fill the 5-gallon jug to the top, and then pour it into the 3-gallon jug until the 3-gallon jug is full. You now have 2 gallons remaining in the 5-gallon jug. Pour out the 3-gallon jug, and then pour the 2 gallons from the 5-gallon jug into the 3-gallon jug. Finally, fill the 5-gallon jug to the top and pour it into the 3-gallon jug until it's full. Since there was only space left for 1 more gallon in the 3-gallon jug, you now have exactly 4 gallons in the 5-gallon jug.
In the land of Brainopia, there are three races of people: Mikkos, who tell the truth all the time, Kikkos, who always tell lies, and Zikkos, who tell alternate false and true statements, in which the order is not known (i.e. true, false, true or false, true, false). When interviewing three Brainopians, a foreigner received the following statements:
Person 1:
I am a Mikko.
Person 2:
I am a Kikko.
Person 3:
a. They are both lying.
b. I am a Zikko.
Can you help the very confused foreigner determine who is who, assuming each person represents a different race?
Person 1 is a Miko.
Person 2 is a Ziko.
Person 3 is a Kikko.
A horse travels a certain distance each day. Strangely enough, two of its legs travel 30 miles each day and the other two legs travel nearly 31 miles. It would seem that two of the horse's legs must be one mile ahead of the other two legs, but of course this can't be true. Since the horse is normal, how is this situation possible?
The horse operates the mill and travels in a circular clockwise direction. The two outside legs will travel a greater distance than the inside ones.
A man comes to a small hotel where he wishes to stay for 7 nights. He reaches into his pockets and realizes that he has no money, and the only item he has to offer is a gold chain, which consists of 7 rings connected in a row (not in a loop).
The hotel proprietor tells the man that it will cost 1 ring per night, which will add up to all 7 rings for the 7 nights.
"Ok," the man says. "I'll give you all 7 rings right now to pre-pay for my stay."
"No," the proprietor says. "I don't like to be in other people's debt, so I cannot accept all the rings up front."
"Alright," the man responds. "I'll wait until after the seventh night, and then give you all of the rings."
"No," the proprietor says again. "I don't like to ever be owed anything. You'll need to make sure you've paid me the exact correct amount after each night."
The man thinks for a minute, and then says "I'll just cut each of my rings off of the chain, and then give you one each night."
"I do not want cut rings," the proprietor says. "However, I'm willing to let you cut one of the rings if you must."
The man thinks for a few minutes and then figures out a way to abide by the proprietor's rules and stay the 7 nights in the hotel. What is his plan?
The man cuts the ring that is third away from the end of the chain. This leaves him with 3 smaller chains of length 1, 2, and 4. Then, he gives rings to the proprietor as follows:
After night 1, give the proprietor the single ring
After night 2, take the single ring back and give the proprietor the 2-ring chain
After night 3, give the proprietor the single ring, totalling 3 rings with the proprietor
After night 4, take back the single ring and the 2-ring chain, and give the proprietor the 4-ring chain
After night 5, give the proprietor the single ring, totalling 5 rings with the proprietor
After night 6, take back the single ring and give the proprietor the 2-ring chain, totalling 6 rings with the proprietor
After night 7, give the proprietor the single ring, totalling 7 rings with the proprietor
