We do not know where this puzzle originated from. If you have any information, please let us know via email.
I caused my mother’s death and didn’t get convicted.
I married 100 women and never got divorced.
I got born before my father, but I am considered perfectly normal?
Who am I?
A priest, whose mother dies from labor, who marries 100 women to 100 men, and whose father attends his birth.
You are living in a 100-floor apartment block. You know that there is one floor in the block, such that if you drop a light bulb from there or anywhere higher, it will crash upon hitting the ground. If you drop a light bulb from any floor underneath it however, the light bulb will remain intact. If you have two light bulbs at your disposal, how many drop attempts do you need such that you can surely find which the floor in question is?
The answer is 14 drops. You can do this by throwing the first bulb from floors 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100 (notice that the difference decreases always by 1) until it crashes and then start throwing the second bulb from the floors in between. For example, if the first bulb crashes at floor 69, you start throwing the second bulb from floors 61, 62, 63, etc. This way the total number of throws would be always at most 14.
Proving that 14 is optimal is done using the same logic. In order to use at most 13 throws, the first throw should be made from floor 13 or lower. The second throw should be made from floor 13+12 or lower, the third throw should be made from floor 13+12+11 or lower, etc. Continuing with the same argument, we conclude that the 13th drop should be made from floor 13+12+…+2+1=91 or lower. However, if the first light bulb does not crash after the last throw, you will not be able to find out which number among 92-100 is X.
There is a room with a chessboard inside. On each of its 64 squares, there is placed a coin, either heads up or heads down. You enter the room and a person inside points towards one special square on the chessboard and gives you the chance to flip one of the coins (whichever you choose). Then you leave the room, your friend enters and has to guess which was the special square on the chessboard. If you two could devise a plan before entering the room, how would you make sure your friend always guesses correctly which is the special square?
First, you must enumerate the coins with numbers from 1 to 64, locate the mystery coin, and calculate the binary representation of its number, padded with zeros on the left to 6 digits length. For example, if the mystery coin is the 5th one on the 4th row, its number would be 29 and will have a binary representation 011101. Then, consider the following sets of coins:
A1 = {row 1, row 2, row 3, row 4}
A2 = {row 1, row 2, row 5, row 6}
A3 = {row 1, row 3, row 5, row 7}
A4 = {column 1, column 2, column 3, column 4}
A5 = {column 1, column 2, column 5, column 6}
A6 = {column 1, column 3, column 5, column 7}
Now, the strategy is to flip the coin which makes the parity of heads in set Ai odd if and only if the i-th digit in the binary representation of the mystery coin is 1. It is easy to check that this is always a possible thing to do.
In a long list of names, one of the names appears more than half of the time. You will be read the names one at a time, without knowing how many they are, and without being able to write them down. If you have a very weak memory, how can you figure out which is the majority name?
Remember the first name and then keep track of whether it has been repeated more than half of the time. To do that, simply add 1 if you hear the name or subtract one when you hear another name. If the list finishes and your counter is positive, then the first name is the majority. If your counter drops to 0, simply restart the procedure with the next name you hear.
This algorithm, invented by R. Boyer and J. Moore, works, because if the counter ends up at 0, then each of the names up to that moment has been read at most half of the time. Therefore, the majority name appears more than half of the time in the remainder of the list.
I have a 9 letter word, 123456789.
If I lose it, I die.
If I have 234, I can 1234.
If I have 56, I am very sick.
235 is the same as 789.
What is the word?
The answer is HEARTBEAT.
The hands of my alarm clock are indistinguishable. How many times throughout the day their positioning is such that one cannot figure out which is the hour hand, and which is the minute hand?
Remark: AM-PM is not important.
Imagine that you have a third hand which moves 12 times as fast as the minute hand. Then, at any time, if the hour hand moves to the location of the minute hand, the minute hand will move to the location of the imaginary hand. Therefore, our task is to find the number of times during the day when the hour hand and the imaginary hand are on top of each other, and the minute hand is not.
Since the imaginary hand moves 144 times faster than the hour hand, the two hands are on top of each other exactly 143 times between 12AM and 12PM. Out of these 143 times, 11 times all three arrows are on top of each other. Therefore, we have 2 × (143 – 11) = 264 times when we cannot figure out the exact time during the entire 24-hour cycle.
There are 5 houses and each of them has a different color. Their respective owners have different heritages, drink different types of beverages, smoke different brands of cigarettes, and look after different types of pets. It is known that:
The question is, who owns the pet fish?
The German owns the pet fish.
Since the Norwegian lives in the leftmost house (9) and the house next to him is blue (14), the second house must be blue. Since the green house is on the left of the white house (4), the person living in the center house drinks milk (8), and the green house’s owner drinks coffee (5), the fourth house must be green and the fifth one must be white. Since the Brit lives in the red house (1) and the Norwegian lives in the leftmost house (9), the leftmost house must be yellow and the center house must be red. Therefore, the colors of the houses are: YELLOW, BLUE, RED, GREEN, WHITE.
Since the Norwegian from the yellow house smokes Dunhill (7), the man from the blue house must keep a horse (11). The person smoking Blends cannot be in the red house, because this would imply that the person in the green house keeps cats and the Swede keeps dogs in the white house (2, 10). However, in this case the Dane must be drinking tea in the blue house (3) and the person smoking Blends does not have a neighbor drinking water (5), which is a contradiction (15). Also, the person smoking Blends cannot be in the green house, because this would imply that the person in the white house drinks water (15), the Dane lives in the blue house (3), and the German and the Swede live in the last two houses. Since the German smokes Prince (13) and the Swede keeps dogs (2), there is nobody who could smoke Bluemaster and drink beer (12). The person smoking Blends cannot be in the white house either, because this would imply that the person in the green house drinks water (15), when in fact he drinks coffee (5).
Therefore, the person smoking Blends must be in the blue house, and then the German and the Swede must live in the last two houses (2, 13). Since the person who smokes Bluemasters drinks beer (12), this must be the Swede with his dogs in the white house (2). The only option for the person who smokes Pall Mall and raising birds (6) is the red house. Then the Norwegian must keep cats (10) and the German is left with the pet fish in the green house.
Alex and Bob are playing a game. They are taking turns drawing arrows over the segments of an infinite grid. Alex wins if he manages to create a closed loop, Bob wins if Alex does not win within the first 1000 moves. Who has a winning strategy if:
a) Alex starts first (easy)
b) Bob starts first (hard)
Remark: The loop can include arrows drawn both by Alex and Bob.
In both cases, Bob wins. An easy strategy for part a) is the following:
Every time Alex draws an arrow, Bob draws an arrow in such a way that the two arrows form an L-shaped piece and either point towards or away from each other. Since every closed loop must contain a bottom left corner, Alex cannot win.
For part b), Bob should use a modification of his strategy in part a). First, he draws a horizontal arrow. Then, he splits the remaining edges into pairs, as shown in the image below. If Alex draws one arrow on the grid, then Bob draws its paired arrow, such that the two arrows point either towards or away from each other. The only places where a loop can have a bottom left corner are where Bob drew the first arrow or the grid points directly above it. However, if a loop has a bottom left corner there, then it is easy to see that it must have at least one more bottom left corner elsewhere, which is impossible.
The first four moves played by White are 1. f3, 2. Kf2, 3. Kg3, 4. Kh4. If White gets mated in the fourth move, what could be the moves played by Black?
The game proceeds as follows:
1. f3 e5 (or e6)
2. Kf2 Qf6
3. Kg3 Qxf3+
4. Kh4 Be7#
Bob and Jane are taking turns, placing knights and coins respectively on a chessboard. If Bob is allowed to place a knight only on an empty square which is not attacked by another knight, how many pieces at most can he place before running out of moves? Assume that Jane starts second and plays optimally, trying to prevent Bob from placing knights on the board.
Bob can place at most 16 knights. One way to do this is to keep placing knights only on the 32 white squares. In order to see that Jane can prevent Bob from placing more than 16 knights, split the board in four 4×4 grids. Then, group the squares in each grid in pairs, as shown on the image below. If Bob places a knight on any square, then Jane will place a coin on its paired square. This way Bob can place at most one knight on each of the four red squares, one knight on each of the four green squares, one knight on each of the four brown squares, and one knight on each of the four blue squares. Therefore, he can not place more than 64/4 = 16 knights on the board.
Please confirm you want to block this member.
You will no longer be able to:
Please note: This action will also remove this member from your connections and send a report to the site admin. Please allow a few minutes for this process to complete.
