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Fun fun riddles/problems

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Comments

  • JonSnowIsAliveJonSnowIsAlive Member Posts: 196
    Found the one using {1, 7, 7, 7, 7}:


    It's (7 + 1/7) * (7 + 7) = 100! Yay me!

  • JonSnowIsAliveJonSnowIsAlive Member Posts: 196
    Fifth riddle:

    There's a wolf, a chicken and some corn:

    image

    They all belong to a farmer who would very much like to transport them across a river using a boat. The problem is the boat is so small he can only bring one "passenger" at a time (either wolf, chicken of corn). There is no limit to the number of crossings.
    1. If the wolf stays with the chicken, its gonna eat it;
    2. If the chicken stays with the corn, its gonna eat it.
    Is there a way for the farmer to make this work?

  • Montresor_SPMontresor_SP Member Posts: 2,139
    Yes. :)

    - Sail the chicken to the other side.
    - Sail back alone.
    - Sail the corn to the other side.
    - Sail back with the chicken.
    - Sail the wolf to the other side.
    - Sail back alone.
    - Sail the chicken to the other side. Done.

    JonSnowIsAlive
  • JonSnowIsAliveJonSnowIsAlive Member Posts: 196
    Riddle #6

    You are facing a closed door standing next to 3 light switches:

    image

    Behind that door stands a lamp. Exactly one of the 3 light switches controls that lamp. How can you identify that switch without opening the door (and looking at the lamp...) more than once?

  • zurathanzurathan Member Posts: 53
    wow nice, @JonSnowIsAlive has correctly solved my second puzzle,
    and here is yours answer
    ok there is 3 swiches and i want to name them "1", "2" and "3" then you need to open switch 1 w8 some time then close it and open 2 and go check lamp, if its open your switch is 2 and if its closed you just check heat of it, if its hot then the switch is 1 but if its cold then swich might be 3 or something broken whatever

    here is a smilar riddle just for you JonSnowIsAlive
    there is pipe just next to the swichs room, the pipe starts on one wall and ends another wall, no other room has pipes in them, and you can hear water discharge in it but you dont know which way.
    is it possible to find out which way this water go without breaking it???

  • zurathanzurathan Member Posts: 53
    edited August 2013

    Fifth riddle:

    There's a wolf, a chicken and some corn:

    image

    They all belong to a farmer who would very much like to transport them across a river using a boat. The problem is the boat is so small he can only bring one "passenger" at a time (either wolf, chicken of corn). There is no limit to the number of crossings.

    1. If the wolf stays with the chicken, its gonna eat it;
    2. If the chicken stays with the corn, its gonna eat it.
    Is there a way for the farmer to make this work?
    dude this is famous and well known puzzle but i recently found a book that mentions the origins of this puzzle, aperently its a 1240 years old british puzzle basicly.
    image
    try to solve the original one. i gurantee you hate mankind immedietly. i disgusted myself for being human alone. why not elf or dwarf? why did i pick human?

  • FinneousPJFinneousPJ Member Posts: 6,456
    That original one is quite controversial I see :D

    Fredjo
  • JonSnowIsAliveJonSnowIsAlive Member Posts: 196
    @zurathan I have an answer which I would never ever use in real life... You could heat up the left end of the pipe. If the right end of the pipe become warmer, it means the water travels left-to-right, else its right-to-left.

    Will try to find a less what-the-hell solution!

  • zurathanzurathan Member Posts: 53
    actually this is the answer. if you are an electrical engineer or simply a gnome, you could make a device to measure magnetic field to determine the discharge. i heard this way you could figure out the amount of water discharge too, but simple torch is enough if not you could try your body heat too if you are too desperate to learn it. nevermind thx

  • FinneousPJFinneousPJ Member Posts: 6,456
    This is actually used to measure blood flow. You don't measure magnetic field though. More info from my course book Medical Instrumentation - Application and Design, John G. Webster, 2010

    image

    If you're interested.

    JonSnowIsAlive
  • JonSnowIsAliveJonSnowIsAlive Member Posts: 196
    @zurathan Cool solution. Requires some tools, but definitely less damaging and dangerous that mine!

    @FinneousPJ Nice article! In what field are studying exactly? Saw the keyword "medical", then was pleasantly surprised to see an integral...? Math is so cool.

  • FinneousPJFinneousPJ Member Posts: 6,456
    Biomedical engineering, very cool.

  • Morte50Morte50 Member Posts: 161

    @Morte50

    You answer is not only right, it is very insightful. Damn this thread is cool! Cheers!

    Also, to be completely honest, I don't like the normal distribution that much, its tail is way too thin. Try the Gumbel distribution for a change!

    Jon

    Hmm, but the normal distribution is so enormously convenient to work with. With the more esoteric distributions, things tend to go sideways in a hurry. Especially when you're working with Big Data. Can't say I've ever had a need to use the Gumbel anyway, though I did play around the generalized Pareto a while back. Ahh, the fun one can have with probability distributions...

  • JonSnowIsAliveJonSnowIsAlive Member Posts: 196
    @Morte50 I guess you may be talking about closed-form expression. Generelly speaking, that is not a requierment for me. Also, there are well documented phenomenons that have nothing to do with the normal distribution.

    I for one frequently model claims frequency in a car insurance context. The common basis of such models is the Poisson distribution which for very small values of its parameter cannot be correctly approximated by the normal distribution.

  • Morte50Morte50 Member Posts: 161
    Closed forms usually aren't strictly a requirement I agree. But it does help if at least the conditionals are well-behaved ('cause when in doubt, run a Gibbs sampler!), hence my preference for the normal. But then again I'm slightly biased by my own typical applications, which is typically large-scale data with a horrible signal-to-noise ratio that invariably needs permutation/bootstrapping at some point. So it helps if the other parts of the model are cheap to compute. Hence, even my binary outcome variables tend to get the normal treatment which, though violating at least three assumptions of the linear regression model still seems to work splendidly :-). In other contexts I would probably be more conscientious about the accuracy of my distributional assumptions, but here that'd just get in the way too much.

  • ElectricMonkElectricMonk Member Posts: 599
    edited August 2013
    zurathan said:

    Fifth riddle:

    There's a wolf, a chicken and some corn:

    image

    They all belong to a farmer who would very much like to transport them across a river using a boat. The problem is the boat is so small he can only bring one "passenger" at a time (either wolf, chicken of corn). There is no limit to the number of crossings.

    1. If the wolf stays with the chicken, its gonna eat it;
    2. If the chicken stays with the corn, its gonna eat it.
    Is there a way for the farmer to make this work?
    dude this is famous and well known puzzle but i recently found a book that mentions the origins of this puzzle, aperently its a 1240 years old british puzzle basicly.
    image
    try to solve the original one. i gurantee you hate mankind immedietly. i disgusted myself for being human alone. why not elf or dwarf? why did i pick human?
    1st crossing: sister 1 and sister 2 cross from side A to side B of the river; sister 2 stays on side B.
    2nd crossing: sister 1 returns to side A.
    3rd crossing: sister 1 and brother 1 cross from side A to side B; sister 1 and brother 1 stays on side B.
    4th crossing: sister 2 returns to side A.
    5th crossing: sister 2 and brother 2 cross from side A to side B; brother 2 stays on side B.
    6th crossing: sister 2 returns to side A.
    7th crossing: sister 2 and sister 3 cross from side A to side B; sister 2 stays on side B.
    8th crossing: sister 3 returns to side A.
    9th crossing: sister 3 and brother 3 cross from side A to side B and stay on side B.

    alternatively:
    7th crossing: sister 3 and brother 3 cross from side A to side B; brother 3 stays on side B.
    8th crossing: sister 3 returns to side A.
    9th crossing: sister 2 and sister 3 cross from side A to side B and stay on side B.


    ...but, the picture says that 11 crossings are required at a minimum, so I assume that I've misunderstood one of the rules of the problem.

  • Morte50Morte50 Member Posts: 161
    edited August 2013
    @jaysl659
    I think the problem is that none of the men can be in the presence of any of the women unless her brother is present as well. After your third crossing, S1, S2 and B1 are all on the same side, however briefly. So naturally, B1 will immediately rape S2 before she has a chance to get back to the boat.

    A solution avoiding this, I believe, goes thusly:

    Crossing
    (Side 1 || Side 2)

    1: S1,S2
    (B1,B2,B3,S3 || S1,S2)

    2: S2
    (B1,B2,B3,S2,S3 || S1)

    3: S2,S3
    (B1,B2,B3 || S1,S2,S3)

    4: S3
    (B1,B2,B3,S3 || S1,S2)

    5: B1,B2
    (B3,S3 || B1,B2,S1,S2)

    6: B2,S2
    (B2,S2,B3,S3 || B1,S1)

    7: B2,B3
    (S2,S3 || B1,B2,B3,S1)

    8: S1
    (S1,S2,S3 || B1,B2,B3)

    9: S1,S2
    (S3 || B1,B2,B3,S1,S2)

    10: S2
    (S2,S3 || B1,B2,B3,S1)

    11: S2,S3
    ( || B1,B2,B3,S1,S2,S3)

    ElectricMonk
  • ElectricMonkElectricMonk Member Posts: 599
    Morte50 said:

    @jaysl659
    I think the problem is that none of the men can be in the presence of any of the women unless her brother is present as well. After your third crossing, S1, S2 and B1 are all on the same side, however briefly. So naturally, B1 will immediately rape S2 before she has a chance to get back to the boat.

    A solution avoiding this, I believe, goes thusly:


    Crossing
    (Side 1 || Side 2)

    1: S1,S2
    (B1,B2,B3,S3 || S1,S2)

    2: S2
    (B1,B2,B3,S2,S3 || S1)

    3: S2,S3
    (B1,B2,B3 || S1,S2,S3)

    4: S3
    (B1,B2,B3,S3 || S1,S2)

    5: B1,B2
    (B3,S3 || B1,B2,S1,S2)

    6: B2,S2
    (B2,S2,B3,S3 || B1,S1)

    7: B2,B3
    (S2,S3 || B1,B2,B3,S1)

    8: S1
    (S1,S2,S3 || B1,B2,B3)

    9: S1,S2
    (S3 || B1,B2,B3,S1,S2)

    10: S2
    (S2,S3 || B1,B2,B3,S1)

    11: S2,S3
    ( || B1,B2,B3,S1,S2,S3)
    Ah, yes, I was working under the assumption that a woman could not be left with a man (w/out her brother present) under the three following conditions: waiting on side 1, waiting on side 2, or traveling in the boat.

    I clearly underestimated the depravity of these men.

    Montresor_SP
  • zurathanzurathan Member Posts: 53
    edited August 2013
    i m glad that my riddles and puzzles have been solved one by one. but i realized that i didnt answer them compactly or at all.

    the pipe with water flow. how do we findout which way it flows?
    we heat midle of the pipe, according to thermodynamics the pipe and water distribute the heat differently, so the flowing side will be alitle bit hotter. oh yea i forgot to mention that we can use some tools for it like a lighter or torch. and since we can do it you may solve this with ultra high teck equipments or alien technology too, that solution will be equally correct!

    find 100 with 1,7,7,7,7 and elementary school math. which solved correctly by @JonSnowIsAlive
    (7+7)x(7+7^-1)=100 is also same

    find 100 with 7,7,7,7 and any sicentific calculator function. which hasn't been solved yet so i decide to add some hints here.
    you can use square root with invisible 2 in it but if you need to use another 7 on that weird sigh inorder to achieve X^(-1/7) function, and any logarithmic function if it has invisible 10 or number e in it, factorial !,power as ^, trigonometrical functions. and i m hoping that you will figure out this hint is designed to misnavigate you completely! ;)
    all the hint you will get is this don't open the rest. there is the answer.
    seriously? dont spoil this man this puzle has so much quality
    you are still not ready for this mind blowing solution
    last warning man! sorry.
    ok you have to use the invisible 0 zeroes. 100= 7x7/.7/.7 is the answer. so any calculator will understand that you mean 0.7 with that. have fun

    and the brothers and their sisters puzzle. which solved correcty by @Morte50.
    this puzle has more than one vay to solv it but morte's solution so perfect that i wont try to improve it. i m glad that the decency of those fictional sisters is in safe hands, nevertheless they are still screwed for having such brothers.

    thanks for allowing me to contribute in such a quality forum. my (stolen from unknown places) puzzles will continue.

  • TeflonTeflon Member, Translator (NDA) Posts: 517
  • Montresor_SPMontresor_SP Member Posts: 2,139
    Here's an oldie:

    You have nine identical gold coins. Except of them is counterfeit and weighs a bit less than the others. You also have an old-fashioned balance scale (see below). You can only use the scale two times.

    Using the scale, how do you identify which coin is counterfeit?

    image

  • Awong124Awong124 Member Posts: 2,643

    Here's an oldie:

    You have nine identical gold coins. Except of them is counterfeit and weighs a bit less than the others. You also have an old-fashioned balance scale (see below). You can only use the scale two times.

    Using the scale, how do you identify which coin is counterfeit?

    image

    - Split the 9 coins into 3 groups of 3.
    - Weigh any two of them against each other. If one group is lighter, then the counterfeit coin is in that group. If they are balanced, then the counterfeit coin is in the unweighed group.
    - Take any two of the coins in the unweighed group and weigh them against each other. If one of them is lighter, then that is the counterfeit coin. If they are balanced, then the unweighed coin is the counterfeit.

    Montresor_SP
  • zurathanzurathan Member Posts: 53
    i love those balance scale puzzles, and knew a modified version of it. and glad that i know its origin too this time. in my country we have a famous puzzle dude which presently turned to polithics, named emrehan halıcı. he published some puzzle books and this puzzle is at most 20 years old, don't know for sure.

    you have 12 identical golden coins (or something really valuable) and one of them is fake.
    fake one is differ with weight only, but you dont know lighter or heavier.
    you have only 3 use with a scale. (i really can't make up a proper story for this rediculous limitation)(as a bonus answer you might try fixing the question, as filling with a background, i don't know maybe scale will be broken if you try using it 4th time)

    how could you possibly pinpoint the fake one?
    as answer you need the whole possibilities diagram. i mean you must be able to figure out the fake one %100 not %99.
    i will try my best for explaning the solution with my rubbish english. just give me a couple of days.
    have fun and be warned this puzle is really hard and long!

  • TJ_HookerTJ_Hooker Member Posts: 2,438
    edited August 2013
    Alright, I think I got it. That being said, I could easily have messed up somewhere without realizing it. Hopefully I've worded it well enough for people to understand.
    Split the coins into 3 groups of 4

    1) Weigh 2 groups. If scales tip, go to 2), if not go to 3)

    2) Remove 2 coins from one side of scale and 1 coin from the other. Swap 1 coin from each side of the scale to the other. Add an unweighed coin to the side that you removed 2 coins from. If the scales tip the same way as in 1), go to 4). If the scale tips the opposite way, go to 5). If the scales do not tip, go to 6)

    3) Weigh 2 of the unweighed coins. If the scales tip, go to 7), if not go to 8)

    4) Remove all the coins that were added or swapped last step. You will now have 1 coin on one side and 2 coins on the other. Remove the single coin and move one of the move one of the remaining coins to the opposite side (so there's 1 coin on each side). If the side that had 2 coins was high, then whichever side is now high is the fake coin. If the side that had 2 coins was low, whichever side is now low is the fake coin. If the scales are now balanced the 3rd coin is fake.

    5) We now know that the fake is one of the 2 coins that were swapped (let's call them A and B). Weigh A against any other coin (besides B). If the scales tip then A is fake, if not B is fake.

    6) We now know that the fake coin is 1 of the 3 removed last step. Take the 2 coins that were removed from the same side (let's call them A and B), and weigh them. If A and B were on the low side after 1), then whichever coin is now low is fake. If A and B were on the high side after 1), then whichever coin is now high is fake. If the scales are now balanced, the 3rd coin is fake.

    7) We now know 1 of the 2 coins just weighed (let's call them A and B) is fake. Weigh A against any other coin (besides B). If the scales tip, A is fake. If not, B is fake.

    8) We now know 1 of the 2 remaining un-weighed coins (let's call them A and B) is fake. Weigh A against any other coin (besides B). If the scales tip, A is fake. If not, B is fake.

  • zurathanzurathan Member Posts: 53
    edited August 2013
    @TJ_Hooker impressive speed!
    i think you got it. but i think i couldn't recognize the answer if i didn't know the answer for sure, i must say.
    it may because of my bad english or too complicated demostration.

    first weight 4to4 and leave the 4
    there is two possibilities, sides stay same or different.
    if they same this means we know the 8 of them genuine and you have 2 wheighting left to analyse rest4,
    you need to leave 1 posible fake here, and weight 3 posible fake with 3 genuine, this time we have 3 possibilities. either one of 3 possible fake is light, heavy or genuine, last weighting should clear all possibilities with prowing the fake one is light or heavy.
    now if in first weighting sides may different and a scale could be heavier then the rest. this means 4 of them possibly heavy and 4 of them possibly light and 4 of them genuine.
    this time we need to label the coins as what are they, may be we have a pen or something. and mark on them as H,L,G. then you have to put one side H,H,L and other one same H,H,L and leaving 2 possible light and genuine ones.
    again 2 possibilities either sides same or different.
    same sides means all 6 of them genuine and you use 3rd weighting with two L to check which one is lighter.
    and if the sides different this means either H's are heavy or the L is light, again the rest is genuine.
    3rd scale must be weighting remaining possible heavy ones to check which one is heavier.
    as a bonus answer for backround story for this puzzle is: you are near graduation of jewelcrafting school. and this puzzle is one of the tests. and someone allready solved this with 4 weighting. you are desperatly trying to impress your teachers. ok lame background i could only come up whit this lame story in this time... i realy wanna hear your imagination about this.

    i hope its understandable enough for first time encounterers. i know its a wall of writing but the solution is rewarding imho. have fun

    TJ_Hooker
  • OlvynChuruOlvynChuru Member Posts: 2,415
    After over half a decade, here's a new riddle!

    A bard equips a long sword that should deal 1d8 damage; the bard has proficiency in all weapons. The bard whacks a goblin with the long sword; the goblin takes 472 slashing damage and is obliterated. The bard hits several more goblins with the long sword; each one takes exactly 472 damage.

    The bard then equips a bastard sword that should deal 2d4 damage. The bard whacks a goblin with this sword and the goblin takes 724 damage; its guts fly all the way to the moon. Several other goblins are slashed by this bastard sword, and each of them takes exactly 724 damage.

    The bard then equips a two-handed sword that should deal 1d10 damage. The bard whacks the last goblin in the encounter. How much damage does this goblin take? (Note: It is a specific amount of damage, not a range)

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