Difficult riddles

logicmath

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.
Step 1. Fill 3 gallon jar with water. ( 5p – 0, 3p – 3) Step 2. Pour all its water into 5 gallon jar. (5p – 3, 3p – 0) Step 3. Fill 3 gallon jar again. ( 5p – 3, 3p – 3) Step 4. Pour its water into 5 gallon jar untill it is full. Now you will have exactly 1 gallon water remaining in 3 gallon jar. (5p – 5, 3p – 1) Step 5. Empty 5 gallon jar, pour 1 gallon water from 3 gallon jar into it. Now 5 gallon jar has exactly 1 gallon of water. (5p – 1, 3p – 0) Step 6. Fill 3 gallon jar again and pour all its water into 5 gallon jar, thus 5 gallon jar will have exactly 4 gallon of water. (5p – 4, 3p – 0) We are done !
87.36 %
56 votes

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.
87.19 %
44 votes

logic

Teenage boy

In a small town in the United States, a teenage boy asked his parents if he could go to a friend's party. His parents agreed, provided that he was back before sunrise. He left the house that evening clean-shaven and when he returned just before the following sunrise his parents were amazed to see that he had a fully grown beard. What happened?
The small town was Barrow in Alaska, the northernmost town in the United States. When the sun sets there in the middle of November, it does not rise again for 65 days. That allowed plenty of time for the boy to grow a beard before the next sunrise.
87.14 %
55 votes

logic

Unusual paragraph

This is an unusual paragraph. I’m curious as to just how quickly you can find out what is so unusual about it. It looks so ordinary and plain that you would think nothing was wrong with it. In fact, nothing is wrong with it! It is highly unusual though. Study it and think about it, but you still may not find anything odd. But if you work at it a bit, you might find out. Try to do so without any coaching.
The letter "e", which is the most common letter in the English language, does not appear once in the long paragraph.
86.69 %
53 votes

logicshort

Sequence of numbers

Finish the sequence: 7 8 5 5 3 4 4 ?
6 - the number of letters in the month august; (January has 7 letters, February has 8 etc.)
86.62 %
42 votes

logicmath

Camel and Banana

The owner of a banana plantation has a camel. He wants to transport his 3000 bananas to the market, which is located after the desert. The distance between his banana plantation and the market is about 1000 kilometer. So he decided to take his camel to carry the bananas. The camel can carry at the maximum of 1000 bananas at a time, and it eats one banana for every kilometer it travels. What is the most bananas you can bring over to your destination?
First of all, the brute-force approach does not work. If the Camel starts by picking up the 1000 bananas and try to reach point B, then he will eat up all the 1000 bananas on the way and there will be no bananas left for him to return to point A. So we have to take an approach that the Camel drops the bananas in between and then returns to point A to pick up bananas again. Since there are 3000 bananas and the Camel can only carry 1000 bananas, he will have to make 3 trips to carry them all to any point in between. When bananas are reduced to 2000 then the Camel can shift them to another point in 2 trips and when the number of bananas left are <= 1000, then he should not return and only move forward. In the first part, P1, to shift the bananas by 1Km, the Camel will have to Move forward with 1000 bananas – Will eat up 1 banana in the way forward Leave 998 banana after 1 km and return with 1 banana – will eat up 1 banana in the way back Pick up the next 1000 bananas and move forward – Will eat up 1 banana in the way forward Leave 998 banana after 1 km and return with 1 banana – will eat up 1 banana in the way back Will carry the last 1000 bananas from point a and move forward – will eat up 1 banana Note: After point 5 the Camel does not need to return to point A again. So to shift 3000 bananas by 1km, the Camel will eat up 5 bananas. After moving to 200 km the Camel would have eaten up 1000 bananas and is now left with 2000 bananas. Now in the Part P2, the Camel needs to do the following to shift the Bananas by 1km. Move forward with 1000 bananas – Will eat up 1 banana in the way forward Leave 998 banana after 1 km and return with 1 banana – will eat up this 1 banana in the way back Pick up the next 1000 bananas and move forward – Will eat up 1 banana in the way forward Note: After point 3 the Camel does not need to return to the starting point of P2. So to shift 2000 bananas by 1km, the Camel will eat up 3 bananas. After moving to 333 km the camel would have eaten up 1000 bananas and is now left with the last 1000 bananas. The Camel will actually be able to cover 333.33 km, I have ignored the decimal part because it will not make a difference in this example. Hence the length of part P2 is 333 Km. Now, for the last part, P3, the Camel only has to move forward. He has already covered 533 (200+333) out of 1000 km in Parts P1 & P2. Now he has to cover only 467 km and he has 1000 bananas. He will eat up 467 bananas on the way forward, and at point B the Camel will be left with only 533 Bananas.
85.53 %
58 votes

logic

Three Childrens' Ages

A deliveryman comes to a house to drop off a package. He asks the woman who lives there how many children she has. "Three," she says. "And I bet you can't guess their ages." "Ok, give me a hint," the deliveryman says. "Well, if you multiply their ages together, you get 36," she says. "And if you add their ages together, the sum is equal to our house number." The deliveryman looks at the house number nailed to the front of her house. "I need another hint," he says. The woman thinks for a moment. "My youngest son will have a lot to learn from his older brothers," she says. The deliveryman's eyes light up and he tells her the ages of her three children. What are their ages?
Their ages are 1, 6, and 6. We can figure this out as follows: Given that their ages multiply out to 36, the possible ages for the children are: 1, 1, 36 (sum = 38) 1, 2, 18 (sum = 21) 1, 3, 12 (sum = 16) 1, 4, 9 (sum = 14) 1, 6, 6 (sum = 13) 2, 2, 9 (sum = 13) 2, 3, 6 (sum = 11) 3, 3, 4 (sum = 10) When the woman tells the deliveryman that the children's ages add up to her street number, he still doesn't know their ages. The only way this could happen is that there is more than one possible way for the children's ages to add up to the number on the house (or else he would have known their ages when he looked at the house number). Looking back at the possible values for the children's ages, you can see that there is only one situation in which there are multiple possible values for the children's ages that add up to the same sum, and that is if their ages are either 1, 6, and 6 (sums up to 13), or 2, 2, and 9 (also sums up to 13). So these are now the only possible values for their ages. When the woman then tells him that her youngest son has two older brothers (who we can tell are clearly a number of years older), the only possible situation is that their ages are 1, 6, and 6.
83.96 %
52 votes

logicmathshort

Really hard algebra puzzle

2+3=8, 3+7=27, 4+5=32, 5+8=60, 6+7=72, 7+8=? Solve it?
98 2+3=2*[3+(2-1)]=8 3+7=3*[7+(3-1)]=27 4+5=4*[5+(4-1)]=32 5+8=5*[8+(5-1)]=60 6+7=6*[7+(6-1)]=72 therefore 7+8=7*[8+(7-1)]=98 x+y=x[y+(x-1)]=x^2+xy-x
80.86 %
43 votes

logicshort

The smallest number

Which is the smallest number that you can write using all the vowels exactly once?
fIvE thOUsAnd (5000).
80.62 %
56 votes