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Viewing 15 posts - 1 through 15 (of 35 total)
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  • #69255
    Harry
    Keymaster

    Do you have a great puzzle to challenge our community? Why not publish it here to give them something a little different to try between challenges.

    #84945
    GreenReble
    Participant

    Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are da and ja, in some order. You do not know which word means which.

    I tried but i don’t think i am even close to the level to do this.If you find the answer please explain to me :3

    #84989
    Harry
    Keymaster

    @GreemReble: we don’t usually post puzzles unless the poster provides a solution for the Elves to check, but we will make an exception for this one. I haven’t had time to think about whether it is even possible to answer. Can you say where you got it from?

    Harry

    #85707
    GreenReble
    Participant

    Sorry for the late reply, i got the puzzle from Wikipedia but the first time i heard about the puzzle was from YouTube.

    #85757
    BreakTheCipher
    Participant

    It is quite an odd puzzle – particularly with the uncertainty of ‘da’ and ‘ja’, and the god of random. I think you would have to ask self-referring questions like ‘Is the answer to are you a god of truth ‘da’?’ but I haven’t gotten anything that works yet.

    #85763
    Axel5
    Participant

    I think this bears some resemblance to the three gods riddle from Ted-ed. The video (including solution) is available on YouTube. It’s totally solvable if it is a permutation of this riddle.

    #86128
    PuttPutt86
    Participant

    I have got as far as working out that you need to formulate the questions in such a way that it doesn’t matter which of “da” and “ja” means “yes” (and which means “no) – you just don’t have enough questions.

    So I considered the following question: “Are you FALSE?”

    – TRUE will reply “no”
    – FALSE will reply “no”
    – RANDOM could reply either “yes” or “no”

    Now let’s consider the question “If I asked you “Are you FALSE?” would you respond “da”?”

    If “da” means “yes”:
    – TRUE will reply “no” (or “ja”)
    – FALSE will reply “yes” (or “da”)
    – RANDOM could reply either “yes” or “no” (“da” or “ja”)

    If “da” means “no”:
    – TRUE will reply “yes” (or “ja”)
    – FALSE will reply “no” (or “da”)
    – RANDOM could reply either “yes” or “no” (“da” or “ja”)

    So you can see that we now have a question to which TRUE will ALWAYS reply “ja” and FALSE will ALWAYS reply “da” (irrespective of which of “ja” and “da” means “yes” and which means “no”).

    I feel like we just need to get another question out of the way first, one that helps us identify at least one god that is not RANDOM, so we can ask the above question to it, to work out if it’s TRUE or FALSE…

    Hope this helps!

    #85978
    GreenReble
    Participant

    Yes that’s where i originally got the puzzle, i love that channel.

    #86762
    cribbage
    Participant

    In The Times quick crossword this morning;

    Reluctance to proceed: code left to be cracked. (4,4)

    I hope nobody has these with regards to the codes so far on the Cipher Challenge!

    #86775
    Harry
    Keymaster

    Wouldn’t be too surprised if some are feeling an extreme lack of energy at this stage.

    #86777
    The-Letter-Wriggler
    Participant

    Brrr icy – put some socks on them.
    It could leave you clef toed !

    #87324
    Rhydwen
    Participant

    The following question is taken from ‘Amusements in Mathematics’ by Henry Ernest Dudeney. This book was first published in 1917. Can you solve it without the use of a computer?

    127. SIMPLE DIVISION

    Sometimes a very simple question in elementary arithmetic will cause a good deal of perplexity. For example, I want to divide the four numbers, 701, 1,059, 1,417, and 2,312, by the largest number possible that will leave the same remainder in every case. How am I to set to work? Of course, by laborious system of trial one can in time discover the answer, but there is quite a simple method of doing it if you can only find it.

    #87336
    The-Letter-Wriggler
    Participant

    =======================================================
    Some jokes I came across . . .

    The square-root said to minus one “why can’t we be together?”
    Minus-one sighed and said “it’s complex.”

    Well i overheard and came to their rescue.
    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    Ha ha, is that a real or an imaginary joke?

    =======================================================
    Why didn’t the Romans find algebra very challenging?

    Because X was ALWAYS ten!!!
    =======================================================

    Chuckle over here are some fun math to work on.

    WORK THOSE NUMBERS
    ===================
    ============================================================ #1
    IT ALL ADDS UP
    ===============
    Using addition only, arrange the numerals 1 through 9 so that they will equal 100.

    Concatenation allowed. (hint you can use the equals symbol more than once)

    ============================================================ #2
    A SYMBOL IN PLACE
    ==================
    Without changing the order of these digits, place the fewest
    possible mathematical symbols in order to make the equation true:

    1 2 3 4 5 6 7 8 9 = 100

    It can of course, be done quite easy by putting a line through the equals
    making it a ‘not equal to’ sign, but that is the easy way out!

    ============================================================ #3
    DIVIDING THE DIGITS
    ====================
    Rearrange the digits and place a dividing line so as to make it equal 1/11.

    1 1 2 3 3 4 4 5 7

    ============================================================ #4
    A SURPRISE IN COUNTING
    =======================
    Amusing as a general question.
    Although ‘a’ is the third most common letter in English after e and t you would be surprised how far you have to count before it is used!

    Using the correct way of verbal counting in English were the ‘and’ is left out,
    one hundred one, one hundred two, etc.
    (actually if you say un like one hundred’un one, one hundred’un two that is ok here)

    What is the first number that ‘a’ appears in?
    What is the first number that ‘b’ appears in?
    What is the first number that ‘c’ appears in?

    #87217
    The-Letter-Wriggler
    Participant

    ◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►
    CAN YOU CRACK OPEN THE SAFE?
    =============================

    HERE IS THE BUTTON-PAD OF a ‘DIAMOND’ MANUFACTURED SAFE.

    ░░░░░░░░░░░
    ░░DIAMOND░░
    ░░░░░░░░░░░
    ░░░░░3░░░░░
    ░░░4░░░5░░░
    ░6░░░░░░░4░
    ░░2░░░░░6░░
    ░░░0░░░3░░░
    ░░░░░5░░░░░
    ░░░░░░░░░░░
    ░MODEL░No.░
    ░░░4CAC0░░░
    ░░░░░░░░░░░

    WHAT IS THE CODE USED TO OPEN THE SAFE?

    WILL YOU BE THE FIRST TO POST THE CORRECT CODE?

    WHAT WILL YOU FIND INSIDE?

    ◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►◄●►

    #87338
    The-Letter-Wriggler
    Participant

    TLW THREE CARDS
    ===============
    I have taken 3 cards from a standard deck and placed them face down.
    Maybe you can tell me what the three cards are if I tell you the following,

    To the left of the Queen, are one or two Jacks.
    To the right of the Jack, are one or two Jacks.
    To the right of the club, are one or two diamonds.
    To the left of the diamond, are one or two diamonds.

    Now, name the three cards from left to right.

    By the way, there are three solutions that satisfies the four statements
    so I will accept any of the three different but correct solutions.
    (Maybe you can give all three solutions?)

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