The Incredible Puzzle Thread

13468915

Comments

  • edited May 2010
    My method for this was to put the "best" numbers in the middle, then work the rest out from there. I think this is right:

    _64
    2817
    _53

    You got it right, congratulations!
    Avistew, yours is wrong because diagonals also count ;)

    Now, I'm going to get some matches to solve yours, Avistew...
  • edited May 2010
    I already knew A, so just posting B:
    I did a....
    star of david
  • edited May 2010
    Doodinthemood is right :) That's the solution to B.
  • edited May 2010
    I've got one. I don't know how hard or easy it is, since I thought it up myself. I'd probably go with 'easy' though :p
    When is a square not a square?
  • edited May 2010
    When it's in cities, there are stuff called "squares" but they're not always square-shaped?
  • edited May 2010
    A circle.
    0 squared...
  • edited May 2010
    Avistew wrote: »
    When it's in cities, there are stuff called "squares" but they're not always square-shaped?
    Told you it was easy :p
  • edited May 2010
    Remember to keep on trying avistew's part A guys!!!

    Anyway here's quite a nice one:
    I want to be able to show the day of the month at any time using two dice. (showing the first of the month as 01, not just 1.) What numbers do I put on each die?
  • edited May 2010
    Avistew wrote: »
    Yeah, I thought that too, but why specifically from that town? Why is he glad the cousin isn't from there due to that? It doesn't make sense.

    Maybe the town just ahead is where the owner lives, and he'd rather bring more money into the local economy from outsiders?
  • edited May 2010
    Anyway here's quite a nice one:
    I want to be able to show the day of the month at any time using two dice. (showing the first of the month as 01, not just 1.) What numbers do I put on each die?

    Let's see... We need numbers from 0 to 9, but the tens only need to go up to 3. So you put 0-3 on one, but you'll never need 00, so you only need one 0, so the other only needs 1, 2 and 3... Wait, you never need 33 either... However you need 30, so one gets 0, 1, 2 and the other gets 1, 2, 3...

    Then, we need to have 4-9 added. I'd say they can go on either, just put three on each.

    Does that work?
  • edited May 2010
    How would you get 04?
  • edited May 2010
    Can you use the same character for 6 and 9 and just turn the die upside down when needed? That might make things easier?
  • edited May 2010
    awwww, yeah that's the trick that you have to see to solve it :(
  • edited May 2010
    Can you use the same character for 6 and 9 and just turn the die upside down when needed? That might make things easier?

    Oooh, that's the good idea! instead of 9, add 0 to that die too.
    I was thinking in matter of dots, silly me.
  • edited May 2010
    Ok here is a strange puzzle a relative taught me.
    It makes no physical sense but you can still get it I guess.

    You are in a box there is no exits in anyway. All there is is a mirror and a table, neither the mirror nor the table can brake the box and get you out. How do you get out.
  • edited May 2010
    You look in the mirror and remember what you saw. You use that saw to cut the table in two. Two halves make a whole. Shout until your voice is hoarse. Ride that horse out the hole and have a sniff around to acquire the scent. Invest that cent wisely and before long you'll have enough money to write a cheque. Go out with the Czech and make sure she loves you too. Of the two, find out which is father. Ask that father if it's ok to marry the Czech.
    You now are not only free, but are rich and have a foreign wife. Well done.
  • edited May 2010
    You look in the mirror and remember what you saw. You use that saw to cut the table in two. Two halves make a whole. Shout until your voice is hoarse. Ride that horse out the hole and have a sniff around to acquire the scent. Invest that cent wisely and before long you'll have enough money to write a cheque. Go out with the Czech and make sure she loves you too. Of the two, find out which is father. Ask that father if it's ok to marry the Czech.
    You now are not only free, but are rich and have a foreign wife. Well done.

    Dang it:eek:. in reality anything horse onward you didnt need but nice.
  • edited May 2010
    Gman5852 wrote: »
    You are in a box there is no exits in anyway. All there is is a mirror and a table, neither the mirror nor the table can brake the box and get you out. How do you get out.

    I say "the way you got in".

    Also, a hint for my match one: Tabletop RPG gamers have an edge over other people for figuring it out.
  • edited May 2010
    How about this? Hope the ASCII-art gets the point across.
    ._____.
    |\   /|
    | \ / |
    |  x  |
    | / \ |
    |/   \|
    -------
    
  • edited May 2010
    You mean a square and two diagonals? If that's the case, explain to me how you figure that they're equilateral :p
  • edited May 2010
    Doh. Er, nevermind.:o
  • edited May 2010
    Solution to A:
    |><|
    |><|

    Two intertwined equilateral triangles, where the "top" of one triangle touches the "bottom" of the other, and vice versa.

    On B:
    Wouldn't a Star of David make six equilateral triangles like you originally posted?
  • edited May 2010
    Bamse wrote: »
    Solution to A:
    |><|
    |><|

    Two intertwined equilateral triangles, where the "top" of one triangle touches the "bottom" of the other, and vice versa.

    On B:
    Wouldn't a Star of David make six equilateral triangles like you originally posted?
    No, six small ones and two big ones, so eight altogether. I'm not sure if your solution is indeed equilateral triangles but I'm guessing if they are there are six of them. Either way, that's not the solution I was looking for, but that's definitely an interesting suggestion.
  • edited May 2010
    Avistew wrote: »
    No, six small ones and two big ones, so eight altogether. I'm not sure if your solution is indeed equilateral triangles but I'm guessing if they are there are six of them. Either way, that's not the solution I was looking for, but that's definitely an interesting suggestion.
    Ah, I didn't even think of the bigger ones. Yeah, they are equilateral, but like you said, there are six of them when you count the two big ones.
  • edited May 2010
    Okay, everyone, bamse didn't find the answer to A, but the answer to C (6 equilateral triangles with 6 matches), which deserves kudos too.
  • edited May 2010
    Create the rhombus
    A___B
    / /
    /___/
    D C

    where ADC = ABC = 60 degrees. Join corners A and C. Join lines AB and CD with a match parallel to AD and BC. Call the point where the last two matches meet X, the point where the last match meets AB Y and the point where the last match meets CD Z.
    We have equilateral triangles ACD, ACB, AXY and CXZ.
  • edited May 2010
    That looks super complicated and I have no clue if they're equilateral or not, but that's not the answer I'm looking for.

    Just like the answer to B, the answer to A can be given with words, without any drawing or anything involved. If that helps. I'll even accept both the tabletop RPG term and the geometry one.
  • edited May 2010
    Here's an attempt at some more descriptive ASCII art of my solution:
    ...____
    ../\/.../
    ././\../
    /_/_\/

    It consists of two large equilateral triangles, where one is upside down, and they share one match. The sixth match can be anywhere, touching the top and bottom line, but must be parallel to the left and right side.
  • edited May 2010
    Ah, I see. Looks good to me. Congrats then! Wasn't the solution I was looking for (mine has 4 equilateral triangles that are the same size, and doesn't require any calculating) but if it works you get points. Want to try and figure out the one I had in mind or should I give the solution?

    Here is a hint:
    Every side of every triangle is one match. That's why you don't need to calculate anything, since all matches are the same size.
  • edited May 2010
    This is what bamse has done:
    backv.png
    I followed his instructions XD It does work. Just not especially nice, and the triangles are different sizes, and two matches cross.
  • edited May 2010
    The answer to A (4 equilateral triangles using 6 matches) is
    a tetrahedron (triangular pyramid with 4 equal sides). The trick is thinking beyond the 2D
  • edited May 2010
    Well, bamse gets credit for figuring out a new solution.
  • edited May 2010
    I've got one:
    How do you get seven as the result of dividing twelve by two?
  • edited May 2010
    Oh, I know! But I figured it out so fast, I think I knew it already, so I won't say.
  • edited June 2010
    I've got one:
    How do you get seven as the result of dividing twelve by two?

    When you suck at math and don't have a calculator?
  • edited June 2010
    Just keep dividing by 2 until you reach 0.75?
  • edited June 2010
    cheat?
  • edited June 2010
    You only divide once, and it's an exact (and -somehow- correct) division
  • edited June 2010
    Well, since I don't want to hold the whole thread up, the answers I would have accepted are
    d4
    ,
    tetrahedron
    and, while inaccurate,
    pyramid
    (because I would have assumed you just didn't know the right word).
  • edited June 2010
    Avistew wrote: »
    Well, since I don't want to hold the whole thread up, the answers I would have accepted are
    d4
    ,
    tetrahedron
    and, while inaccurate,
    pyramid
    (because I would have assumed you just didn't know the right word).
    I would've said that if Javi hadn't said it first in post 232.
Sign in to comment in this discussion.