Riddles
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 This topic has 59 replies, 22 voices, and was last updated 8 years, 3 months ago by ☢️ Rand0m3x 🎲.

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August 9, 2012 10:54 pm at 10:54 pm #604497ursketchingMember
I’m in the mood to hear some riddles. I’ll start:
the poor have it, the rich need it, and if you eat it you’ll die.
i’ll give you time to think before i post my answer.
August 9, 2012 10:58 pm at 10:58 pm #1050074CuriosityParticipantLawnmowers?
August 9, 2012 11:05 pm at 11:05 pm #1050075ursketchingMemberexplanation????
August 9, 2012 11:06 pm at 11:06 pm #1050076TheMusicManParticipantI’ve heard that one before.
I’ll just say that no matter how hard you try, nothing meets the criteria.
Have you ever considered the biological meaning of the term “polygon”?
August 9, 2012 11:07 pm at 11:07 pm #1050077DoodleMan™MemberPretty good riddle. I googled it but I’m not giving it away.
August 9, 2012 11:13 pm at 11:13 pm #1050078squeakParticipantNichevo
August 9, 2012 11:27 pm at 11:27 pm #1050079☕️coffee addictParticipantnothing
August 9, 2012 11:27 pm at 11:27 pm #1050080CuriosityParticipantThe people that mow lawns here are poor immigrants and they all have lawnmowers. The rich people need these folks to bring their lawnmowers and mow their lawns, and… Well… Have you ever tried eating a lawnmower?
August 9, 2012 11:38 pm at 11:38 pm #1050081oomisParticipantNada, zip, zilch, efes,zero – an oldie, but goodie.
August 9, 2012 11:51 pm at 11:51 pm #1050082ursketchingMembercoffee addictgood job
curiositynice try but better luck next time
anyone else
August 10, 2012 3:05 am at 3:05 am #1050083557ParticipantA rodef and a nirdaf (each viewed as a single point) are in a closed circle. They both move with the same maximum speed. Can the rodef catch the nirdaf?
If so, how does the rodef catch the nirdaf?
If not, how does the nirdaf avoid capture by the rodef?
August 10, 2012 10:16 am at 10:16 am #1050084CuriosityParticipant557 – guns
August 10, 2012 4:13 pm at 4:13 pm #1050085557ParticipantCuriosity: the rodef and nirdaf are each a single point in a closed circle. A single point can’t carry a gun (since a single point has no dimension, and hence no mass). Alternatively, shooting the nirdaf would violate the given assumption that both move with the same maximum speed.
Here’s a hint for how to visualize the problem:
Let’s assume for a moment that we have a closed square with vertices at (0,0),(0,1), (1,0) and (1,1). This is a simple unit square. Now, assume the rodef begins at one corner of the square, say (1,1), and the nirdaf begins at the opposite corner (0,0). So the nirdaf is sqrt(2) units away from the rodef.
Now, as a conjecture, it would be a reasonable assumption that the rodef and nirdaf would have opposite strategies; namely, the rodef wants to minimize distance between the rodef and nirdaf while the nirdaf wants to maximize distance between the rodef and nirdaf.
In this example, the nirdaf actually has no incentive to move whatsoever. This is because (0,0) is as far as the nirdaf can get from (1,1) where the rodef is. So the nirdaf is maximizing distance from the rodef by remaining at (0,0). The rodef’s optimal strategy from (1,1) is to move in a straight line towards (0,0) because the shortest distance between two points is a straight line (extra credit to whoever proves that the shortest distance between two points is a straight line).
Since the nirdaf will never have an incentive to move given the rodef’s strategy (moving away from (0,0) at any time will shorten the distance with the rodef), and the rodef continually gets closer, then it is clear to see that the rodef will eventually “corner the nirdaf” and catch the nirdaf at (0,0).
So in this contrived example, the rodef catches the nirdaf regardless of what the nirdaf tries to do. The riddle, though, is what happens in a circle?
Another hint:
The optimal strategy for the rodef to minimize distance to the nirdaf is to move along the straight line connecting the rodef’s location on the circle with the nirdaf’s location on the circle.
August 10, 2012 4:26 pm at 4:26 pm #1050086Rough RiderParticipantwhat is a mohel’s favorite drink??
August 11, 2012 6:13 pm at 6:13 pm #1050087CuriosityParticipant557 – I was assuming you were going with the classical interpretation of rodef and nirdaf. How can one 2 dimensional point by a rodef? Furthermore, a nirdaf is allowed to kill a rodef. Assuming they were both people, moving at sprinting pace, a gun would have worked for either of them.
If they’re in a “closed” two dimensional circle (I guess this can be imagined like a circular pair of train tracks), and they both move at light speed, then they can’t catch eachother unless one of them breaks the circle…. Or if the rodef blocks the circle with some object and then corners the nirdaf on it.. your rules aren’t very clear.
August 12, 2012 8:29 am at 8:29 am #1050088Shopping613 🌠ParticipantWasn’t this a riddle thread…not a math thread?
August 12, 2012 1:03 pm at 1:03 pm #1050089557ParticipantCuriosity: Sorry for not being so clear. Perhaps this will clarify…
Consider a closed unit circle centered at the origin (0,0) in the Cartesian coordinate system in the Euclidean plane (i.e. your standard unit circle). Now, consider two different points in the closed unit circle (these points can be inside the circle or on the boundary of the circle, but since the circle is closed, they can never be outside the circle). Let’s call these points A and B, respectively.
Now, assume A and B move freely in the closed circle with equal maximum speed. A tries to “catch” B by getting to the same location as B (i.e. A catches B if both A and B have the same exact coordinates at any given time). B tries to “evade” A.
Assuming A and B play optimal strategies (i.e. A and B make the best possible moves at all times in accordance with their goals), does A “catch” B? If so, how can A be guaranteed to “catch” B? If not, how can B be guaranteed to “evade” A?
To make it more interesting for nonmathoriented people, I had called A the rodef and B the nirdaf since in a literal sense this is the case (i.e. A is the chaser and B is the chased). However, you correctly point out that from a halachic standpoint, the din of rodef and nirdaf is more complex. I apologize for the confusion.
Aside (if it’s confusing, just ignore): This is an example of a pursuitevasion game. There are several interesting properties of these games in general, but there is something particularly interesting about this game.
One of the basic properties of these types of games is to note the path that the chaser takes to catch the chased. This path is called a pursuit curve, which is aptly named because the path of pursuit need not be straight. For example, if a mother is chasing a little child running straight in some direction, chances are that the mother is faster, and she can get onto the child’s path of travel and then follow the child’s path until she catches him. However, if she runs at the same speed as the child, then she cannot just follow the child’s path because she will never catch up. Instead, the mother runs to where the child is heading and intercepts him at some point. If the child changes direction during the chase in response to the mother’s chase, then the mother’s path of pursuit will be curved. Hence the term pursuit curve.
August 12, 2012 10:07 pm at 10:07 pm #1050090Shopping613 🌠ParticipantOk you have a chicken,fox, and chicken feed on one side of the river you want to get them all to the other side eagerly but you can only take one item with you each time. If you leave the chicken with the chicken feed it will eat it, if you leave the fox with the chicken, you will come back to find a fox twice as big and no chicken….how do you do it?
Next, its 1 am and you go to your bathroom to get something, you find a cocariach as big as your hand, what do you do?
August 12, 2012 10:32 pm at 10:32 pm #1050091DoodleMan™MemberOk you have a chicken,fox, and chicken feed on one side of the river you want to get them all to the other side eagerly but you can only take one item with you each time. If you leave the chicken with the chicken feed it will eat it, if you leave the fox with the chicken, you will come back to find a fox twice as big and no chicken….how do you do it?
I think you messed it up a bit. Don’t all three things have to eat one of the others?
August 13, 2012 3:32 am at 3:32 am #1050092CuriosityParticipantShopping613 I loled at your second riddle…
August 13, 2012 3:35 am at 3:35 am #1050093CuriosityParticipantTake the chicken,
Go back empty handed,
Take the fox,
On the way back to the seeds return the chicken with you,
take the seeds over to the fox, leaving the chicken at the starting side,
Go back empty handed, and bring the chicken.
August 13, 2012 3:46 am at 3:46 am #1050094CuriosityParticipant557 now I understand your riddle, but assuming they can each anticipate the other’s movements, I don’t have a clue what the answer is, unless the rodef can manipulate the circle.
As for a riddle, an oldy but a goody:
You are in a closed room that has two doors and two other people. One door leads to death, and one leads to life. One person speaks only the truth and one person speaks only lies. You do not know which person is which or which door leads to what. You are allowed to ask one question to either person of your choosing to help you exit through the correct door. What question can you ask to guarantee your survival?
If you’ve already heard this riddle or read the answer elsewhere, please don’t ruin it for others.
August 13, 2012 8:45 am at 8:45 am #1050095Shopping613 🌠ParticipantKnow that one!…curiosity that it correct! And the second riddle…well I screamed silently and decided whatever I needed in there was not important now…..
7 letters are in my name
If you subtract one 12 remains
What am I?
You are trapped in a room with only a mirror and table. There are no doors and windows, how do you escape?
August 13, 2012 2:11 pm at 2:11 pm #1050096DoodleMan™Member7 letters are in my name
If you subtract one 12 remains
What am I?
Shouldn’t it be 8 letters?
August 13, 2012 2:34 pm at 2:34 pm #1050097I can only tryMember557–
Unless Y can move away from X in a straight line, X can either cut the corner or turn inside the radius of the path Y is taking, thereby closing the gap and eventually catching Y.
Even if X and Y are both on the outer edge of the circle, X can still turn inside a smaller radius.
If there was an obstacle (such as an inner circle) within the circle or square, then Y could dodge X forever, since they would both circle that obstacle at the same speed.
Rough Rider–
Slice.
Shopping613–
It depends if the roach is the thing you went to get.
7 letters are in my name
If you subtract one 12 remains
What am I?
Twelves (or drytzen).
August 13, 2012 7:11 pm at 7:11 pm #1050098Shopping613 🌠ParticipantSorry it was six and the answer was dozens
I can only try: very good! Even though it wasn’t the riddle I meant it to be…..and very funny!
Sorry I don’t know how to do italics or bold or the other thingy you can do
But if you use your imagination the words belows actually are in italics!
You are trapped in a room with only a mirror and a table. There are no doors and windows. How do you escape?
Still no one?
Hint : its a play on words
August 14, 2012 9:29 pm at 9:29 pm #1050099Shopping613 🌠ParticipantYou look in the mirror, you see what you saw. You take the saw and cut the table in half, two halves make a hole and you climb out the hole! Very simple!
August 14, 2012 11:41 pm at 11:41 pm #1050100KozovMemberOr, you saw the table, into two halves, and climb out the whole. Forget the mirror.
Or, climb out of the whole of the table/room.
August 15, 2012 3:44 pm at 3:44 pm #1050101557ParticipantICOT: This riddle became popular in the early 1930s, and if it were before 1952, you would not only have the correct answer, but you also noted within your answer the correct strategy X would use to catch Y.
Namely, X could be guaranteed to catch Y by staying on the same radius as Y. In other words, X moves at top speed in such a way that X always lies on the radius vector from the center to Y. If we assume without loss of generality that X stays on the boundary of the circle, then it is easy to check that X does now catch Y in
finite time. Indeed, in the time that it takes Y to run a quartercircle at full speed, X (say starting at the center) performs a semicircle of half the radius of the disc, thus catching Y.
However, in 1952, it was proven that Y need not stay on the boundary and in fact can survive forever in a closed circle. The clever proof is as follows:
First, I establish a winning strategy for Y, and then I will establish that it is feasible, i.e. Y does not leave the closed circle.
Let T = {t_1, t_2, t_3, …} be a sequence of time periods of equal length. At the ith time period, Y runs in a straight line that is perpendicular to Y’s radius vector at the start of the period. Y runs into the halfplane that does not contain X (if X is on the radius then Y may choose either halfplane). Certainly, X does not catch Y in this time period.
Let Y repeat this procedure for all time periods. As long as T is an infinite sequence, Y is never caught.
Now, I must show that this strategy is feasible. Note that if R_i is the distance of Y from the center at the start of the ith time period, then (R_i+1)^2 = (R_i)^2 + (t_i)^2. (This is just a simple application of the famous Pythagorean Theorem.) If the sum of (t_i)^2 is finite, then the R_i’s are bounded so that (multiplying by a constant if necessary) Y does not leave the closed circle. Let t_i = 1/i.
Then, Y is never caught and never leaves the closed circle.
August 15, 2012 5:54 pm at 5:54 pm #1050102frummy in the tummyParticipantI’ve always loved this one:
3 good friends (they know each other very well) of absolutely equal intelligence, Huey, Dewey, and Louie are captured by Bin Laden and given an interesting opportunity for one of them to save his own life (ol’ Laden is in a good mood today).
Bin Laden tells them: There are a total of 3 red hats and 2 black hats. I will blindfold all of you and place one hat on each of your heads. I will then remove the other 2 hats from the room and then remove the blindfolds. The first one who correctly states the color of his own hat will be spared; the other two will be summarily executed. If you answer incorrectly, are caught looking in someone else’s eye’s reflection, or any other form of cheating, you will be killed immediately. Good Luck! MUAHAHAHAHA!!!!
Bin Laden goes through the process and places red hats on all 3 of their heads. After about 2 minutes, Louie confidently yells out “It’s red!” and enjoys his freedom as he watches Huey and Dewey being burned alive in vats of excrement (coincidentally, Bin Laden was found the next week and guess what he’s burning in now for eternity?).
How did Louie know? (It could have been Huey or Dewey who figured it out as well, but Louie happened to be faster today.)
August 15, 2012 7:06 pm at 7:06 pm #1050103I can only tryMember557
Please let me know if the following are correct:
2) Do you mean half the diameter of the disc?
Please provide the following:
2) The values of t1, t2.
3) What direction (if any) is Y moving in during the nonnth time periods?
I realize the values are variable, but applying specific numbers may help me understand.
Thank you.
August 16, 2012 12:28 am at 12:28 am #1050104I can only tryMember557
A Google search for your puzzle shows that the question was first posed by Richard Rado, and the solution you described was given by Abram Besicovitch.
Google Books allowed me to see a sketch of the solution in action, so you can disregard my request for clarification.
I did notice that the solution has X getting closer to Y as well as Y getting closer to the circle’s edge with every move, so in the “real world”, even if X and Y were microscopic, X’s edge would eventually touch Y’s – it’s just that their exact centers wouldn’t ever be in exact alignment.
I’d still appreciate an answer to my “semicircle” and “diameter” questions.
Also, since maintaining distance isn’t what’s important, but rather just preventing X and Y from occupying the same spot, couldn’t Y use a similar or identical plan to evade X inside a square?
Just for your amusement, here are terms I don’t remember hearing:
radius vector
{t_1, t_2, t_3, …}
ith
halfplane
Furthermore, your application of the Pythagorean theorem and R_i’s being bounded was way over my head – oh, well.
August 16, 2012 12:45 am at 12:45 am #1050105I can only tryMemberfrummy in the tummy –
Let’s name the three people X, Y and Z.
August 16, 2012 2:27 pm at 2:27 pm #1050106frummy in the tummyParticipantICOT –
Great Job! Heard it before or you figured it out?
August 16, 2012 4:00 pm at 4:00 pm #1050107frummy in the tummyParticipant557/ICOT –
Although it’s certainly an interesting problem as stated, the one problem with describing it as pursuer (X) and pursued (Y) is that it does not factor in reaction time. In order for Y to travel in the perpendicular direction of travel of X for any infinitesimal movement (during your time t_i), Y must know X’s direction of travel before X actually performs said movement. Assuming that this is impossible, X can randomly preempt the direction of travel of Y and cut him off, and after enough guesses should eventually catch him. I’m with ICOT in not understanding your entire logic (I’m an engineer, not a mathematician :)), so please correct me if I’m wrong.
August 16, 2012 8:48 pm at 8:48 pm #1050108ursketchingMemberthese riddles take too much thinking. i have a good one:
you peel the outside, cook the inside, eat the outside, throwout the inside its a food.good luck
August 17, 2012 2:06 am at 2:06 am #1050109DoodleMan™Memberursketching: I think corn.
August 17, 2012 2:52 pm at 2:52 pm #1050110I can only tryMemberCuriosity–
You are in a closed room that has two doors and two other people. One door leads to death, and one leads to life. One person speaks only the truth and one person speaks only lies. You do not know which person is which or which door leads to what. You are allowed to ask one question to either person of your choosing to help you exit through the correct door. What question can you ask to guarantee your survival?
Select either of the other two people at random and ask him “Which door leads to safety?”
Wait until that person falls asleep (this may take a while), then quickly drag him to the door that he pointed to, open the door, and throw him out. If he lives, he told the truth, and you can follow him. If he dies, serves him right for lying to you, and you can quickly slam the door and exit via the other door.
frummy in the tummy–
August 17, 2012 4:26 pm at 4:26 pm #1050111557ParticipantICOT:
1) Yes.
2) Yes. Also, I meant to say that Y stays on the boundary of the circle.
Yes also to your square question – in fact, this strategy works in any compact metric space. (My square visualization hint at the beginning was correct given the assumption on X and Y’s optimal strategies. It turns out, though, that Y has an optimal strategy that does not necessarily maximize the distance from X.)
Radius vector – this is just the line drawn from the center of the circle to Y.
{t_1, t_2, t_3, …} – I don’t know if it’s possible to do math type here, but the “_number” is supposed to mean subscript. The number just identifies the time period, so t_1 is the first period of movement and so on.
ith – “i” just means that it refers to any number. It could refer to t_1 or t_2 or t_557 or anything you want. So using “i” instead of a number is a generalization. So if the ith time period happens to be t_557, then the (i+1)th time period is t_558. But “i” can be any number.
halfplane – in our example of a closed circle, we were using R^2 space, i.e. your standard 2dimensional metric space. R^2 is the metric space that children do math in for all of grade school; when you draw a graph with an xaxis and yaxis, this is R^2 space. The whole R^2 space makes up a plane. If I only consider everything above the xaxis, then I am considering a “halfplane.”
So in the context of the problem, we could consider Y’s radius vector as dividing the “world” in half. By imagining that the vector actually is a line rather than a line segment, everything to one side of the line would constitute a halfplane whereas everything to the other side of the line would constitute the other halfplane.
The Pythagorean Theorem is actually quite easy; draw a line from the center of the circle to where Y is located. This is Y’s radius vector. Now Y is moving perpendicular to his radius vector, so draw a small line segment perpendicular to the radius vector that begins at Y’s current location. This line segment represents Y’s movement. Now connect the other end of the segment to the center of the circle. This will be Y’s new radius vector after moving. Note that since the original radius vector is perpendicular to the line segment of movement, we now have a right triangle, where the new radius vector after moving is the hypotenuse of the triangle.
The boundedness of the radii is a bit more difficult, but basically the idea is that in order for the strategy to work, Y cannot make “too big” of moves or else the strategy will require him to leave the circle, which he is not allowed to do.
frummy in the tummy: If I understand your kasha (which I might not), you are concerned that Y does not know where X is heading necessarily, so he cannot know which halfplane to choose.
Y does not need to know X’s direction; Y only needs to know X’s location.
August 17, 2012 8:27 pm at 8:27 pm #1050112I can only tryMember557
Thank you for the answers and for taking the time to write detailed explanations of all the terms I wasn’t familiar with.
I’m still not 100% percent clear on the “bounded” calculation, but I’ll b”n take another shot at “getting it” after Shabbos.
August 18, 2012 7:42 pm at 7:42 pm #1050113CuriosityParticipantICOT that’s a creative solution! The original riddle is with a king who is watching over this scenario. He executes you if you walk through the wrong door, and it’s not some sort of mechanical or natural death that is physically connected by the door. Thus, your answer wouldn’t really work because the man you threw through the door wouldn’t be executed. Also, the question you are asking him doesn’t really guarantee anything, and he just kind of serves as a guinnea pig of sorts. Good effort, but try again!
August 19, 2012 1:02 am at 1:02 am #1050114ursketchingMembermoskidoodleyou are right.
August 20, 2012 6:26 pm at 6:26 pm #1050115I can only tryMemberCuriosity
Thank you.
I already knew the “real” answer, but didn’t want to steal it from someone else who could solve it.
There are two possibilities. Let’s name the two people “X” and “Y” (such imaginative names, I know):
1) Ask X “which door would Y tell me is safe?”, then use the other door. If X is the honest one, he will point to the door that Y (the liar) would have indicated, while if X is the liar, he would dishonestly point to the door Y would not indicate.
2) Ask X “Hypothetically, if I asked you which door was safe, which door would you point to?”, then use the indicated door. If X was the truthteller, he’d honestly point to the safe door that he’d point to if asked. If X is the liar, he’d dishonestly point to the safe door that he wouldn’t have pointed to if he’d actually (not hypothetically) been asked.
June 4, 2013 12:21 am at 12:21 am #1050116Shticky GuyParticipantRead each sentence in sequence. Please don’t skip any steps or the mathematics will be thrown off.*****************************
Take the following test mentally. Don’t write down answers and don’t shout them out. *****************************
1. Pick a number from 2 to 9. It can be 2 or 9, or any number inbetween. *****************************
2. Take your number and multiply it by 9. *****************************
3. That should give you a 2 digit number. Take these 2 digits and add them together. *****************************
4. Take the resulting number and subtract 5 from it. *****************************
5. Take that number and correspond it to the alphabet, numbering the letters. A=1, B=2, C=3, D=4 and so on. *****************************
6. Take your letter, and think of a country that begins with that letter. *****************************
7. Take the last letter in the name of that country, and think of an animal that starts with this letter. *****************************
8. Now, take the last letter in the name of that animal, and think of a color that starts with this letter. *****************************
That’s it. You have finished the test. Now before you reveal your final answer, . . . . . * * * * * * * * * * * * * *
December 22, 2013 4:43 pm at 4:43 pm #1050117☕ DaasYochid ☕ParticipantWhat do yeshivish’e cars and analogies have in common?
For the answer, see here:
http://www.theyeshivaworld.com/coffeeroom/topic/classicemunahquestion#post503915
December 23, 2013 9:01 pm at 9:01 pm #1050118ronrsrMemberWhat can you sit on, sleep on, and brush your teeth with?
December 23, 2013 9:14 pm at 9:14 pm #1050119🐵 ⌨ GamanitParticipantit
December 23, 2013 10:10 pm at 10:10 pm #1050120LevAryehMemberThis is a pointless thread, because when you know the answers you aren’t supposed to post them, and when you don’t know the answers you have nothing useful to say.
December 25, 2013 12:29 am at 12:29 am #1050121ronrsrMemberanswer: A chair, a bed, and a toothbrush.
December 23, 2014 1:45 pm at 1:45 pm #1050122☕️coffee addictParticipanteveryone knows you take out three sifrei torah when rosh chodesh teves falls out on shabbos, when do we take out 5 sifrei torah (in a shul that only has 5) there are two answers

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