The Riddle Thread….

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  • #1068720

    Rachel, can you flip the dice over, ie. can 02 and 20 share a side?

    #1068721
    Rachel613
    Member

    Yeah sure you just need to get every combo till 31

    #1068722

    But I assume you cannot flip 13 to become 31, because then it looks backwards. (2 kind of looks the same upside-down)

    #1068723
    Rachel613
    Member

    No i didn’t mean you should flip them over literally, i meant they can switch places. you can move the dice and put any one first or second.

    ICOT – thanx for the welcome! I love this riddle. someone told it to me during a long bus ride in camp and it kept me busy for about an hour….till i chapped the main point 🙂 at least i was distracted!

    #1068724

    Then I give up! Mercy on me?

    #1068725
    Rachel613
    Member

    Tell me what you got so far and where you’re stuck…try to put numbers on and you’ll get to one vital glitch. what have you got for me?

    #1068726
    Rachel613
    Member

    Wow no one could figure this one out since a week ago?!!! It ain’t that hard!

    #1068727
    aussieboy
    Participant

    rachel613: I just saw it. Give me a bit of time to figure it out 🙂

    #1068728
    aussieboy
    Participant

    The problem isnt figuring out which numbers to use. Its figuring out which numbers go where.

    #1068729
    aussieboy
    Participant

    1st die: 1, 2, 3, 4, 5, 9

    2nd die: 6, 11, 13, 19, 25, 29

    I feel like I did to much work though and that there is some kind of shortcut or easier answer.

    #1068730
    aussieboy
    Participant

    Actually you can technically just do this

    1st die: 1, 2, 3, 4, 5, 6

    2nd die: 7, 14, 21, 28, x, x (x = any number [because it doesnt matter])

    I figured this out after like ten minutes but for some reason it didnt make sense to me until I thought about it for another 10 minutes.

    #1068731
    mepal
    Member

    Hey! Look who I see on the board! Nice seeing ya, aussie! Stick around!

    #1068732
    aussieboy
    Participant

    thanks mepal. Its doubtfulk but maybe 🙂

    #1068733
    mepal
    Member

    At least we get a maybe!

    #1068734

    aussieboy-

    The format the puzzle actually wants is as follows:

    ??? ???
    ?0? ?1? <== first of month
    ??? ???
    ??? ???
    ?1? ?1? <== eleventh of month
    ??? ???
    ??? ???
    ?3? ?1? <== thirty-first of month
    ??? ???

    #1068735
    aussieboy
    Participant

    1st die: 1, 2, 3, 4, 5, 6

    2nd die: 0, 1, 2, 7, 8, 9

    This is correct as long as 1, 2, 3… is good and it does not need to be 01, 02, 03…

    Otherwise I dont know how there can be a possible answer.

    #1068736

    aussieboy-

    You’re 99% of the way there.

    You still need something else for 07, 08 and 09 to be possible.

    #1068737
    aussieboy
    Participant

    I give up.

    #1068738

    aussieboy-

    A hint is in the above sentence.

    #1068739
    aussieboy
    Participant

    1st die: 0, 1, 3, 4, 5, 6

    2nd die: 0, 1, 2, 7, 8, 9

    Because 5 and 2 look the same when flipped?

    #1068740

    aussieboy-

    Oooooh, you’re so close.

    It isn’t the 2 that gets flipped…

    #1068741
    aussieboy
    Participant

    1st die: 0, 1, 2, 3, 4, 5,

    2nd die: 0, 1, 2, 7, 8, 9

    ICOT: At this point i didnt really solve the riddle lol

    #1068742
    Jax
    Member

    aussie: long time no see buddy! ok, i won’t ruin your concentration in this thread!

    #1068743
    mepal
    Member

    Yeah jax, its a good thing this threads around. Otherwise….

    #1068744

    aussieboy-

    #1068745
    aussieboy
    Participant

    Hey Jax, and mepal

    ICOT: Its a deal 😉

    #1068746
    #1068747

    Change is good: I

    You will play the following game with one other player and a roll of fifty pennies.

    Each player must take one to six pennies from the roll on alternating turns.

    The player who takes the fiftieth penny loses.

    You go first.

    How many pennies should you remove on your first turn? Why?

    (from a game magazine)

    Change is good: II

    The director of the U.S. mint challenges you to the following game:

    1) No quarter may touch any other quarter.

    Should you go first or second?

    What should your strategy be?

    (from an online puzzle site)

    #1068748

    Change is good: I

    Correction: The player who takes the fiftieth penny wins.

    #1068749
    Dr. Pepper
    Participant

    I can only try-

    OK I gave it an hour and no one answered.

    Change is good: I

    I would remove 1 on my first turn.

    Each subsequent turn I would pick the amount that when added to the amount he took would equal 7. (If he picked 2 I would pick 5…)

    Therefore 1 => 8 => 15 => 22 => 29 => 36 => 43.

    There are now seven pennies left.

    Any number he chooses to remove now will leave as little as one and as much as six left. I choose the rest and win!

    #1068750

    Dr. Pepper-

    Bingo!

    I solved it by realizing that you needed to put down the 43rd penny, which meant you needed to put down the 36th penny, etc. This method works, but is cumbersome.

    #1068751
    Dr. Pepper
    Participant

    I can only try-

    Change is good: II

    I’ve been racking my brain all day on this one but can’t come up with the correct answer. I think this is probably the one you are looking for (although it is wrong).

    I would go first.

    To decide where to put the first quarter I would draw two straight, non-parallel lines on the table (with my trusty pencil of course) where each end of the line touches the circumference of the table. I’d then rip out my protractor that I always carry in my pocket protector and use it to find the midpoint of each line. Next I would draw a perpendicular bisector for each line towards the other end of the table. The point where the two lines meet is the exact center of the table.

    The center of the first quarter is placed on the center of the table.

    After Mr. Director places a quarter I would place a quarter on the exact opposite side of the table. Therefore if he has an empty spot to place a quarter there should be an empty spot on the exact opposite side. Hence I will be the last on to place a quarter on the table.

    This will not work once the diameter of the table reaches a certain length which I can definitely not calculate while in the state of mind I am currently in. (It has to do with my favorite constant.)

    #1068752

    Dr. Pepper-

    Fantastic!

    You got the solution, and even described how to locate the exact center of the table, rather than just saying “put the first quarter in the middle”.

    I didn’t get the answer to the puzzle, but the “chap” was so clever that I stored it in the vault.

    As far as I can tell, this will work for any diameter table, except one so small that a second quarter can’t fit. If this is wrong, I’m interested in how/why.

    (As an aside, I think one of New York’s professional sports teams is regularly enjoying celebrations which involve the use of your favorite constant – kain yirbu.)

    #1068753
    Dr. Pepper
    Participant

    I can only try-

    Take seven Nickels (or any coin for that matter but preferably without ridges), put one in the center and the other six around it. Look closely and you will see a space between two of the coins. That space is twice the decimal of PI (multiplied by a constant).

    Let’s say Mr. Director starts a pattern that is a family of concentric circles centered at the center of the table. In each subsequent circle the little space will get bigger and bigger until there is room for another quarter. Once there is an odd amount of quarters in a circle there will be a space for Mr. Director to place a quarter without a corresponding spot on the “other” side of the table.

    QED

    #1068754

    Dr. Pepper-

    If I understand you correctly:

    1) Put a nickel on a flat table.

    3) If the six outer nickels are all touching the inner one, there is no way that each outer nickel can touch both adjacent ones, although they will come very close.

    4) The total gap between the surrounding nickels is a function of pi. (That space is twice the decimal of PI [multiplied by a constant]).

    Is the above correct?

    Eventually there will be space(s) large enough for a nickel within the circle, and if the circle is completed by an odd number of coins the director will win.

    You can ensure that there is never a circle with an odd number of coins or an unbalanced gap.

    By putting your coin exactly opposite his, the gap in the circle must remain divided evenly between the two sides.

    #1068755

    Attention-Getting Quiz

    2) The only difference in the green, yellow, red and purple variety of these vegetables is their stage of ripeness.

    3) Capsaicin is an irritant in this device, used by law-enforcement agencies.

    4) 1885 is the year a pharmacist invented this. After 1950 it was still produced, although it no longer had a point.

    5) A Scoville scale will tell you this (not your weight).

    6) A variety of spiced meat uses these for flavoring (not the seemingly eponymous food).

    #1068756
    squeak
    Participant

    Dr. Pepper

    Member

    a space between two of the coins. That space is twice the decimal of PI (multiplied by a constant).

    Wouldn’t it also be 100 times the decimal of PI, multiplied by a different constant?

    #1068758
    hiijacker
    Participant

    Ok, do not attempt this one unless you have a few days time to kill. This riddle took me over 3 days to solve. But you can solve it in pieces.

    There are two unknown whole numbers, x and y, both greater than 1, and less than 100. One mathematician, Mr. Product is given the product of these two numbers, while another mathematician, Mr. Sum is given the sum of these two numbers.

    The following conversation takes place:

    Mr. Product: I do not know the numbers.

    Mr. Sum: I knew you didn’t knew the numbers.

    Mr. Product: Now I know the numbers

    Mr. Sum: Now I know the numbers, too.

    What are the numbers?

    #1068760
    Dr. Pepper
    Participant

    I haven’t solved it just yet but I think I’m getting there.

    Does anyone mind if I give hint?

    #1068761

    Dr. Pepper-

    Does anyone mind if I give hint?

    No hints or answers for another hour, please.

    #1068762
    Dr. Pepper
    Participant

    I don’t have the answer yet, I only did as much as I could think of mentally while on the train this morning. The hint I wanted to give is just to notify others that a not-so-famous conjecture may be needed.

    #1068763

    Still not there.

    One more hour, please.

    #1068764
    Dr. Pepper
    Participant

    OK but it’s taking lots of self control.

    (I have the whole answer now.)

    #1068765

    It’ll take me too long.

    Go ahead with the answer, and I won’t look.

    #1068766
    hiijacker
    Participant

    This riddle was posted in Scientific American in the 80’s or 90’s. You can solve the first 2/3 of it in your head, but the last third it helps to have a pen and paper.

    I solved this by brute force, without any formulas.

    #1068767
    Dr. Pepper
    Participant

    Here it goes.

    Mr. Product: I do not know the numbers.

    => Either none of the numbers are prime or one of the numbers is prime but not both. (If both numbers are prime then Mr. Product would just factor the product and have the number.)

    Mr. Sum: I knew you didn’t knew [sic] the numbers.

    => The sum can’t be even. (Goldbach’s conjecture – Every even integer greater than 2 can be written as the sum of two primes. Therefore the sum can’t be even or it can be written as the sum of two primes which we know the numbers are not two primes. Sorry everyone but the proof is beyond the scope of this thread.)

    => One number must be odd and one number must be even.

    => The product must be even.

    Mr. Product: Now I know the numbers

    => There is one unique way to attain that product using an odd number and an even number. (There is at least one other way, possibly more, to attain that product using two even numbers since one number must be composite.)

    Mr. Sum: Now I know the numbers, too.

    => All of us in the CR also do!

    4 & 13

    There aren’t as many possibilities as you may think once you start throwing out possible pairs based on criteria. The upper bound is “red herring” as it will be the same answer regardless on whether it is 15 or 1,000,000. I tried an upper bound of 10 and found no solution. When I raised it to 15 I found this solution which I believe is the only solution regardless of the upper bound.

    #1068768
    hiijacker
    Participant

    While you are working on that, here is one of my favorite riddles that can be solved in your head.

    You have two one-hour fuses: lighting one end of a fuse will cause it to burn down to the other end in exactly one hour’s time. You know nothing else about the fuses; in particular you don’t know how long any segment of a fuse will burn, only that an entire fuse takes one hour. How can you tell when exactly 45 minutes have passed?

    #1068769
    Dr. Pepper
    Participant

    Light one in the middle (or light both ends), it will take one half hour to burn.

    When that is finished burning light each end and the middle of the next one and it will take 15 minutes to burn.

    That should take 45 minutes.

    #1068770
    hiijacker
    Participant

    Good job, here is the full explanation, with each exchange, and how to derive the possibilities:

    P says: I do not know the numbers.

    So, we can be sure that the product that P knows, can be analysed in more than one way. We name such a product a “fuzzy product”.

    Here is an example ( P = 50 = 5 * 10 = 2 * 25). P can’t really know whether 5, 10 or 2, 25 are the numbers X, Y. But if P was 55, then the numbers X, Y would definitely be 5, 11.

    S says: I knew that you didn’t know them. I do not know them either.

    From the first part of S’s answer, we understand that S has such a sum that every possible X, Y pair corresponding to this sum gives a fuzzy product.

    So the Possible Sums are : 11, 17, 23, 27, 29, 35, 37, 41, 47, 53

    So, let’s name those sums that have only fuzzy product solutions, good sums.

    One example to make that more clear: If you take any of the above sums, and check all possible combinations which yield that sum, you will find out that all of them have a fuzzy product.

    11 = 2+9 = 3+8 = 4+7 = 5+6

    * 2*9 = 18 = 3*6 (so, 18 is fuzzy)

    * 3*8 = 24 = 4*6 (so, 24 is fuzzy)

    * 4*7 = 28 = 2*14 (so, 28 is fuzzy), and…

    * 5*6 = 30 = 2*15 (so, 30 is fuzzy), thus, all combinations for 11 give fuzzy products.

    So, it is now obvious that S has one of the above 10 sums, or else he wouldn’t be sure that P has a fuzzy number, and thus wouldn’t have said that “I knew that you did not know them”.

    P says: But now I know them!

    FGS! (For God’s Sake) How did P found those numbers?!

    Well, when P heard S saying that he (S) knew that he (P) didn’t know the numbers in the first place, he made the calculations above and found out that S had one of those 10 sums(which produce only fuzzy products).

    But thou shall not forget that P, being P, knows P 🙂 (the product)

    So, P knows P and those 10 possible sums. Hence, the only thing he can do now is to solve those 10 equation systems with 2 unknowns and 2 equations and hope that only one of those equation systems has a valid solution.

    { X + Y = S

    { X * Y = P

    with S = {11, 17, 23, 27, 29, 35, 37, 41, 47, 53}

    and P known (to our friend P)

    But we know from P’s answer that he found the numbers, and that means one thing: only one of those equation systems has a valid solution.

    S says: Now, I know them too!

    Of course, S does not have the privilege to know the product P. The only thing that he knows (except from S 🙂 are the ten equation systems from which P found the numbers.

    Now that the two math geniuses S, P learned the numbers, it is right time that we learn them too.

    From the fact that his friend P could find the numbers, he draws the conclusion that only one of those equation systems has a valid solution. So, S is dying to find the product P, which, by the way, let’s name “rare product” implying that it has the following property: only one of all pairs of numbers which have this product, has a good sum. If this sounds complicated, read until clear.

    But S says that he found out the numbers. That can only mean that he has a sum that of all possible pairs which yield that sum, only one has a rare product.

    What is therefore left for us is to check all of the 10 possible sums to find the one that has a unique pair of numbers which give a rare product.

    We are then presented with:

    SUM = 17

    PRODUCT = 52

    which reassures us that 17 is the one and only of all the 10 possible sums that of all pairs of numbers which sum to it (e.g. 5, 12 or 8, 9) only one has a rare product, i.e. a product which of all pairs of numbers which have that product only one has a good sum, i.e. a sum which all pairs of numbers which sum to it have fuzzy products.

    So which numbers have a product of 52 and a sum of 17?

    They are 4 and 13…

    That’s all folks, I hope you had a nice time!

    #1068771
    hiijacker
    Participant

    So it seems I need to stump the Pepper:

    Here goes.

    You are blindfolded and enter into a room. The room has a table with 100 quarters scattered. Your are told that 20 of these quarters are tails and 80 are heads. You are also told that if you can split the coins into 2 piles where the number of tails is the same in both piles, then you get to keep all of the quarters. You are allowed to move the coins, but you can never tell what state a quarter is currently in (the blindfold prevents you from seeing, and you cannot tell by feeling it).

    How do you go about splitting the quarters so that you can win?

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