Difficult riddles

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.
94.24 %
44 votes

logic

Four big houses

There are 4 big houses in my home town. They are made from these materials: red marbles, green marbles, white marbles and blue marbles. Mrs Jennifer's house is somewhere to the left of the green marbles one and the third one along is white marbles. Mrs Sharon owns a red marbles house and Mr Cruz does not live at either end, but lives somewhere to the right of the blue marbles house. Mr Danny lives in the fourth house, while the first house is not made from red marbles. Who lives where, and what is their house made from ?
From, left to right: #1 Mrs Jennifer - blue marbles #2 Mrs Sharon - red marbles #3 Mr Cruz - white marbles #4 Mr Danny - green marbles If we separate and label the clues, and label the houses #1, #2, #3, #4 from left to right we can see that: a. Mrs Jennifer's house is somewhere to the left of the green marbles one. b. The third one along is white marbles. c. Mrs Sharon owns a red marbles house d. Mr Cruz does not live at either end. e. Mr Cruz lives somewhere to the right of the blue marbles house. f. Mr Danny lives in the fourth house g. The first house is not made from red marbles. By (g) #1 isn't made from red marbles, and by (b) nor is #3. By (f) Mr Danny lives in #4 therefore by (c) #2 must be red marbles, and Mrs Sharon lives there. Therefore by (d) Mr Cruz must live in #3, which, by (b) is the white marbles house. By (a) #4 must be green marbles (otherwise Mrs Jennifer couldn't be to its left) and by (f) Mr Danny lives there. Which leaves Mrs Jennifer, living in #1, the blue marbles house.
94.24 %
44 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.
93.70 %
40 votes

logic

Zen Master

A new student met the Zen Master after traveling hundreds of miles by yak cart. He was understandably pleased with himself for being selected to learn at the great master's feet . The first time they formally met, the Zen Master asked, "May I ask you a simple question?" "It would be an honor!" replied the student. "Which is greater, that which has no beginning or that which has no end?" queried the Zen Master. "Come back when you have the answer and can explain why." After the student made many frustrated trips back with answers which the master quickly cast off with a disapproving negative nod, the Zen Master finally said, "Perhaps I should ask you another question?" "Oh, please do!" pleaded the exasperated student. The Zen Master then asked, "Since you do not know that, answer this much simpler riddle. When can a pebble hold back the sea?" Again the student was rebuffed time and again. Several more questions followed with the same result. Each time, the student could not find the correct answer. Finally, completely exasperated, the student began to weep, "Master, I am a complete idiot. I can not solve even the simplest riddle from you!" Suddenly, the student stopped, sat down, and said, "I am ready for my second lesson." What was the Zen Master's first lesson?
The student's first lesson was that in order to learn from the Zen Master, the student should be asking the questions and not the Zen Master.
93.70 %
40 votes

logic

Fancy restaurant

A fancy restaurant in New York was offering a promotional deal. A married couple could eat at the restaurant for half-price on their anniversary. To prevent scams, the couple would need proof of their wedding date. One Thursday evening, a couple claimed it was their anniversary, but didn't bring any proof. The restaurant manager was called to speak with the couple. When the manager asked to hear about the wedding day, the wife replied with the following: "Oh, it was a wonderful Sunday afternoon, birds were chirping, and flowers were in full bloom." After nearly 10 minutes of ranting, she comes to tell him that today was their 28th wedding anniversary. "How lovely", the manager said, "However, you do not qualify for the discount. Today is not your anniversary, you are a liar". How did the manager know that it wasn't their anniversary?
The calendar repeats itself every 28 years. So, if they were married on a Sunday 28 years ago, the day they were at the restaurant would also have to be a Sunday. Since it was a Thursday, the manager knew they were lying, and abruptly kicked them out of his restaurant.
93.70 %
40 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
93.39 %
38 votes

MORE RIDDLES >
<<<234
Page 2 of 4.