# Riddle #846

## 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.
72.84 %

## Similar riddles

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## King and Wind Bottles

A bad king has a cellar of 1000 bottles of delightful and very expensive wine. A neighboring queen plots to kill the bad king and sends a servant to poison the wine. Fortunately (or say unfortunately) the bad king's guards catch the servant after he has only poisoned one bottle. Alas, the guards don't know which bottle but know that the poison is so strong that even if diluted 100,000 times it would still kill the king. Furthermore, it takes one month to have an effect. The bad king decides he will get some of the prisoners in his vast dungeons to drink the wine. Being a clever bad king he knows he needs to murder no more than 10 prisoners – believing he can fob off such a low death rate – and will still be able to drink the rest of the wine (999 bottles) at his anniversary party in 5 weeks time. Explain what is in mind of the king, how will he be able to do so?
Think in terms of binary numbers. (now don’t read the solution, give a try). Number the bottles 1 to 1000 and write the number in binary format. bottle 1 = 0000000001 (10 digit binary) bottle 2 = 0000000010 bottle 500 = 0111110100 bottle 1000 = 1111101000 Now take 10 prisoners and number them 1 to 10, now let prisoner 1 take a sip from every bottle that has a 1 in its least significant bit. Let prisoner 10 take a sip from every bottle with a 1 in its most significant bit. etc. prisoner = 10 9 8 7 6 5 4 3 2 1 bottle 924 = 1 1 1 0 0 1 1 1 0 0 For instance, bottle no. 924 would be sipped by 10,9,8,5,4 and 3. That way if bottle no. 924 was the poisoned one, only those prisoners would die. After four weeks, line the prisoners up in their bit order and read each living prisoner as a 0 bit and each dead prisoner as a 1 bit. The number that you get is the bottle of wine that was poisoned. 1000 is less than 1024 (2^10). If there were 1024 or more bottles of wine it would take more than 10 prisoners.
80.87 %

## 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.
80.74 %

## 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?
80.15 %

## How long was he walking

Every day, Jack arrives at the train station from work at 5 pm. His wife leaves home in her car to meet him there at exactly 5 pm, and drives him home. One day, Jack gets to the station an hour early, and starts walking home, until his wife meets him on the road. They get home 30 minutes earlier than usual. How long was he walking? Distances are unspecified. Speeds are unspecified, but constant. Give a number which represents the answer in minutes.
The best way to think about this problem is to consider it from the perspective of the wife. Her round trip was decreased by 30 minutes, which means each leg of her trip was decreased by 15 minutes. Jack must have been walking for 45 minutes.
80.15 %

## Cards in the dark

You are standing in a pitch-dark room. A friend walks up and hands you a normal deck of 52 cards. He tells you that 13 of the 52 cards are face-up, the rest are face-down. These face-up cards are distributed randomly throughout the deck. Your task is to split up the deck into two piles, using all the cards, such that each pile has the same number of face-up cards. The room is pitch-dark, so you can't see the deck as you do this. How can you accomplish this seemingly impossible task?
Take the first 13 cards off the top of the deck and flip them over. This is the first pile. The second pile is just the remaining 39 cards as they started. This works because if there are N face-up cards in within the first 13 cards, then there will be (13 - N) face up cards in the remaining 39 cards. When you flip those first 13 cards, N of which are face-up, there will now be N cards face-down, and therefore (13 - N) cards face-up, which, as stated, is the same number of face-up cards in the second pile.
79.99 %

79.77 %

## Ramanujan Magic number

What is the smallest number, that can be expressed as the sum of the cubes of two different sets of numbers?
Hardy-Ramanujan discovered 1729 as a magic number. Why 1729 is a magic number? 10^3 + 9^3 = 1729 and 12^3 + 1^3 = 1729 Taxicab number Ta(2)
79.77 %

## Whoever...

Whoever makes it, tells it not. Whoever takes it, knows it not. Whoever knows it, wants it not.
Counterfeit money.
79.46 %

## Course of expensive medication

You've been placed on a course of expensive medication in which you are to take one tablet of Plusin and one tablet of Minusin daily. You must be careful that you take just one of each because taking more of either can have serious side effects. Taking Plusin without taking Minusin, or vice versa, can also be very serious, because they must be taken together in order to be effective. In summary, you must take exactly one of the Plusin pills and one of the Minusin pills at one time. Therefore, you open up the Plusin bottle, and you tap one Plusin pill into your hand. You put that bottle aside and you open the Minusin bottle. You do the same, but by mistake, two Minusins fall into your hand with the Plusin pill. Now, here's the problem. You weren't watching your hand as the pills fell into it, so you can't tell the Plusin pill apart from the two Minusin pills. The pills look identical. They are both the same size, same weight (10 micrograms), same color (Blue), same shape (perfect square), same everything, and they are not marked differently in any way. What are you going to do? You cannot tell which pill is which, and they cost \$500 a piece, so you cannot afford to throw them away and start over again. How do you get your daily dose of exactly one Plusin and exactly one Minusin without wasting any of the pills?
Carefully cut each of the three pills in half, and carefully separate them into two piles, with half of each pill in each pile. You do not know which pill is which, but you are 100% sure that each of the two piles now contains two halves of Minusin and half of Plusin. Now go back into the Plusin bottle, take out a pill, cut it in half, and add one half to each stack. Now you have two stacks, each one containing two halves of Plusin and two halves of Minusin. Take one stack of pills today, and save the second stack for tomorrow.
79.46 %