Riddle #679

logicmathprobability

An infinite basket

You have a basket of infinite size (meaning it can hold an infinite number of objects). You also have an infinite number of balls, each with a different number on it, starting at 1 and going up (1, 2, 3, etc...). A genie suddenly appears and proposes a game that will take exactly one minute. The game is as follows: The genie will start timing 1 minute on his stopwatch. Where there is 1/2 a minute remaining in the game, he'll put balls 1, 2, and 3 into the basket. At the exact same moment, you will grab a ball out of the basket (which could be one of the balls he just put in, or any ball that is already in the basket) and throw it away. Then when 3/4 of the minute has passed, he'll put in balls 4, 5, and 6, and again, you'll take a ball out and throw it away. Similarly, at 7/8 of a minute, he'll put in balls 7, 8, and 9, and you'll take out and throw away one ball. Similarly, at 15/16 of a minute, he'll put in balls 10, 11, and 12, and you'll take out and throw away one ball. And so on....After the minute is up, the genie will have put in an infinite number of balls, and you'll have thrown away an infinite number of balls. Assume that you pull out a ball at the exact same time the genie puts in 3 balls, and that the amount of time this takes is infinitesimally small. You are allowed to choose each ball that you pull out as the game progresses (for example, you could choose to always pull out the ball that is divisible by 3, which would be 3, then 6, then 9, and so on...). You play the game, and after the minute is up, you note that there are an infinite number of balls in the basket. The next day you tell your friend about the game you played with the genie. "That's weird," your friend says. "I played the exact same game with the genie yesterday, except that at the end of my game there were 0 balls left in the basket." How is it possible that you could end up with these two different results?
Your strategy for choosing which ball to throw away could have been one of many. One such strategy that would leave an infinite number of balls in the basket at the end of the game is to always choose the ball that is divisible by 3 (so 3, then 6, then 9, and so on...). Thus, at the end of the game, any ball of the format 3n+1 (i.e. 1, 4, 7, etc...), or of the format 3n+2 (i.e. 2, 5, 8, etc...) would still be in the basket. Since there will be an infinite number of such balls that the genie has put in, there will be an infinite number of balls in the basket. Your friend could have had a number of strategies for leaving 0 balls in the basket. Any strategy that guarantees that every ball n will be removed after an infinite number of removals will result in 0 balls in the basket. One such strategy is to always choose the lowest-numbered ball in the basket. So first 1, then 2, then 3, and so on. This will result in an empty basket at the game's end. To see this, assume that there is some ball in the basket at the end of the game. This ball must have some number n. But we know this ball was thrown out after the n-th round of throwing balls away, so it couldn't be in there. This contradiction shows that there couldn't be any balls left in the basket at the end of the game. An interesting aside is that your friend could have also used the strategy of choosing a ball at random to throw away, and this would have resulted in an empty basket at the end of the game. This is because after an infinite number of balls being thrown away, the probability of any given ball being thrown away reaches 100% when they are chosen at random.
85.10 %
47 votes

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Strange Miles

You are somewhere on Earth. You walk due south 1 mile, then due east 1 mile, then due north 1 mile. When you finish this 3-mile walk, you are back exactly where you started. It turns out there are an infinite number of different points on earth where you might be. Can you describe them all? It's important to note that this set of points should contain both an infinite number of different latitudes, and an infinite number of different longitudes (though the same latitudes and longitudes can be repeated multiple times); if it doesn't, you haven't thought of all the points.
One of the points is the North Pole. If you go south one mile, and then east one mile, you're still exactly one mile south of the North Pole, so you'll be back where you started when you go north one mile. To think of the next set of points, imagine the latitude slighty north of the South Pole, where the length of the longitudinal line around the Earth is exactly one mile (put another way, imagine the latitude slightly north of the South Pole where if you were to walk due east one mile, you would end up exactly where you started). Any point exactly one mile north of this latitude is another one of the points you could be at, because you would walk south one mile, then walk east a mile around and end up where you started the eastward walk, and then walk back north one mile to your starting point. So this adds an infinite number of other points we could be at. However, we have not yet met the requirement that our set of points has an infinite number of different latitudes. To meet this requirement and see the rest of the points you might be at, we just generalize the previous set of points. Imagine the latitude slightly north of the South Pole that is 1/2 mile in distance. Also imagine the latitudes in this area that are 1/3 miles in distance, 1/4 miles in distance, 1/5 miles, 1/6 miles, and so on. If you are at any of these latitudes and you walk exactly one mile east, you will end up exactly where you started. Thus, any point that is one mile north of ANY of these latitudes is another one of the points you might have started at, since you'll walk one mile south, then one mile east and end up where you started your eastward walk, and finally, one mile north back to where you started.
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There are 100 ants on a board that is 1 meter long, each facing either left or right and walking at a pace of 1 meter per minute. The board is so narrow that the ants cannot pass each other; when two ants walk into each other, they each instantly turn around and continue walking in the opposite direction. When an ant reaches the end of the board, it falls off the edge. From the moment the ants start walking, what is the longest amount of time that could pass before all the ants have fallen off the plank? You can assume that each ant has infinitely small length.
The longest amount of time that could pass would be 1 minute. If you were looking at the board from the side and could only see the silhouettes of the board and the ants, then when two ants walked into each other and turned around, it would look to you as if the ants had walked right by each other. In fact, the effect of two ants walking into each other and then turning around is essentially the same as two ants walking past one another: we just have two ants at that point walking in opposite directions. So we can treat the board as if the ants are walking past each other. In this case, the longest any ant can be on the board is 1 minute (since the board is 1 meter long and the ants walk at 1 meter per minute). Thus, after 1 minute, all the ants will be off the board.
87.96 %
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87.88 %
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4 gallon of water

How to measure exactly 4 gallon of water from 3 gallon and 5 gallon jars, given, you have unlimited water supply from a running tap.
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87.77 %
58 votes

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Equation riddle

If, Fernando + Alonso + McLaren = 6 Fernando x Alonso = 2 Alonso x McLaren = 6 Then, McLaren x Fernando =?
3 Explanation: Fernando + Alonso + McLaren = 6 Fernando x Alonso = 2 Alonso x McLaren = 6 Given. Fernando x Alonso = 2 Alonso x McLaren = 6 Rewriting Both equations in terms of Alonso, Fernando = 2/Alonso McLaren = 6/Alonso Replacing above values in equation "Fernando + Alonso + McLaren = 6" 2/Alonso + Alonso + 6/Alonso =6 (2 + Alonso^2 + 6)/Alonso = 6 8 + Alonso^2 = 6Alonso Alonso^2 - 6Alonso + 8 = 0 (Alonso - 4) (Alonso - 2) = 0 Therefore; Alonso = 4 or 2 Let's take value of Alonso as 2 Fernando = 2/2 = 1 McLaren = 6/2 = 3 Therefore; McLaren x Fernando = 3 x 1 = 3
87.71 %
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Tiling Without Corners

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.
87.71 %
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Half of two plus two

Is half of two plus two equal to two or three?
Three. It seems that it could almost be either, but if you follow the mathematical orders of operation, division is performed before addition. So... half of two is one. Then add two, and the answer is three.
87.71 %
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2 Player and N Coin – Strategy Puzzle

There are n coins in a line. (Assume n is even). Two players take turns to take a coin from one of the ends of the line until there are no more coins left. The player with the larger amount of money wins. Would you rather go first or second? Does it matter? Assume that you go first, describe an algorithm to compute the maximum amount of money you can win. Note that the strategy to pick maximum of two corners may not work. In the following example, first player looses the game when he/she uses strategy to pick maximum of two corners. Example 18 20 15 30 10 14 First Player picks 18, now row of coins is 20 15 30 10 14 Second player picks 20, now row of coins is 15 30 10 14 First Player picks 15, now row of coins is 30 10 14 Second player picks 30, now row of coins is 10 14 First Player picks 14, now row of coins is 10 Second player picks 10, game over. The total value collected by second player is more (20 + 30 + 10) compared to first player (18 + 15 + 14). So the second player wins.
Going first will guarantee that you will not lose. By following the strategy below, you will always win the game (or get a possible tie). (1) Count the sum of all coins that are odd-numbered. (Call this X) (2) Count the sum of all coins that are even-numbered. (Call this Y) (3) If X > Y, take the left-most coin first. Choose all odd-numbered coins in subsequent moves. (4) If X < Y, take the right-most coin first. Choose all even-numbered coins in subsequent moves. (5) If X == Y, you will guarantee to get a tie if you stick with taking only even-numbered/odd-numbered coins. You might be wondering how you can always choose odd-numbered/even-numbered coins. Let me illustrate this using an example where you have 6 coins: Example 18 20 15 30 10 14 Sum of odd coins = 18 + 15 + 10 = 43 Sum of even coins = 20 + 30 + 14 = 64. Since the sum of even coins is more, the first player decides to collect all even coins. He first picks 14, now the other player can only pick a coin (10 or 18). Whichever is picked the other player, the first player again gets an opportunity to pick an even coin and block all even coins.
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Flipping Lockers

There are 1 million closed school lockers in a row, labeled 1 through 1,000,000. You first go through and flip every locker open. Then you go through and flip every other locker (locker 2, 4, 6, etc...). When you're done, all the even-numbered lockers are closed. You then go through and flip every third locker (3, 6, 9, etc...). "Flipping" mean you open it if it's closed, and close it if it's open. For example, as you go through this time, you close locker 3 (because it was still open after the previous run through), but you open locker 6, since you had closed it in the previous run through. Then you go through and flip every fourth locker (4, 8, 12, etc...), then every fifth locker (5, 10, 15, etc...), then every sixth locker (6, 12, 18, etc...) and so on. At the end, you're going through and flipping every 999,998th locker (which is just locker 999,998), then every 999,999th locker (which is just locker 999,999), and finally, every 1,000,000th locker (which is just locker 1,000,000). At the end of this, is locker 1,000,000 open or closed?
Locker 1,000,000 will be open. If you think about it, the number of times that each locker is flipped is equal to the number of factors it has. For example, locker 12 has factors 1, 2, 3, 4, 6, and 12, and will thus be flipped 6 times (it will end be flipped when you flip every one, every 2nd, every 3rd, every 4th, every 6th, and every 12th locker). It will end up closed, since flipping an even number of times will return it to its starting position. You can see that if a locker number has an even number of factors, it will end up closed. If it has an odd number of factors, it will end up open. As it turns out, the only types of numbers that have an odd number of factors are squares. This is because factors come in pairs, and for squares, one of those pairs is the square root, which is duplicated and thus doesn't count twice as a factor. For example, 12's factors are 1 x 12, 2 x 6, and 3 x 4 (6 total factors). On the other hand, 16's factors are 1 x 16, 2 x 8, and 4 x 4 (5 total factors). So lockers 1, 4, 9, 16, 25, etc... will all be open. Since 1,000,000 is a square number (1000 x 1000), it will be open as well.
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Two words are anagrams if and only if they contain the exact same letters with the exact same frequency (for example, "name" and "mean" are anagrams, but "red" and "deer" are not). Given two strings S1 and S2, which each only contain the lowercase letters a through z, write a program to determine if S1 and S2 are anagrams. The program must have a running time of O(n + m), where n and m are the lengths of S1 and S2, respectively, and it must have O(1) (constant) space usage.
First create an array A of length 26, representing the counts of each letter of the alphabet, with each value initialized to 0. Iterate through each character in S1 and add 1 to the corresponding entry in A. Once this iteration is complete, A will contain the counts for the letters in S1. Then, iterate through each character in S2, and subtract 1 from each corresponding entry in A. Now, if the each entry in A is 0, then S1 and S2 are anagrams; otherwise, S1 and S2 aren't anagrams. Here is pseudocode for the procedure that was described: def areAnagrams(S1, S2) A = new Array(26) A.initializeValues(0) for each character in S1 arrayIndex = mapCharacterToNumber(character) //maps "a" to 0, "b" to 1, "c" to 2, etc... A[arrayIndex] += 1 end for each character in S2 arrayIndex = mapCharacterToNumber(character) A[arrayIndex] -= 1 end for (i = 0; i < 26; i++) if A[i] != 0 return false end end return true end
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