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Thread: Daily puzzles/problems.

  1. #1396
    Pièce de Résistance Scheherazade's Avatar
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    15 trees?
    ~
    "It is not that I am mad; it is only that my head is different from yours.”
    ~


  2. #1397
    Registered User billl's Avatar
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    Quote Originally Posted by Scheherazade View Post
    15 trees?
    That was my guess, as well. However, the father had planted even fewer than 15 trees...

  3. #1398
    www.markbastable.co.uk
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    Is this going to involve a Star of David?


    Edit: Yeah, it is. I just drew it. 10.


    Edit again: imagine a tree at each point and each intersection.




    Oops. I miscounted. 11
    Last edited by MarkBastable; 11-14-2011 at 07:20 PM.

  4. #1399
    Something's Gone hoope's Avatar
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    4 trees maybe !
    "He is asleep. Though his mettle was sorely tried,
    He lived, and when he lost his angel, died.
    It happened calmly, on its own,
    The way the night comes when day is done."



  5. #1400
    Registered User billl's Avatar
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    Quote Originally Posted by MarkBastable View Post
    Is this going to involve a Star of David?


    Edit: Yeah, it is. I just drew it. 10.


    Edit again: imagine a tree at each point and each intersection.




    Oops. I miscounted. 11
    Quote Originally Posted by hoope View Post
    4 trees maybe !
    Well, the correct answer is somewhere between Mark and hoope's answers.

  6. #1401
    Pièce de Résistance Scheherazade's Avatar
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    Mark, that one has six rows?

    10 with a five-sided star?
    ~
    "It is not that I am mad; it is only that my head is different from yours.”
    ~


  7. #1402
    Registered User billl's Avatar
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    Quote Originally Posted by Scheherazade View Post
    Mark, that one has six rows?

    10 with a five-sided star?
    That's right.

  8. #1403
    www.markbastable.co.uk
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    Yeah, you're right.

    I envisaged a Star of David (first message).

    Drew a five-pointed star and counted the nodes (second message)

    Then Googled 'Star of David' and linked it (third message).


    My excuse is that I was distracted by Steve Jobs buggering up my iPhone with an update from beyond the grave.

    God, I hate Apple. If my company didn't insist on me using an iPhone, I'd toss this thing out of the window,
    Last edited by MarkBastable; 11-14-2011 at 07:59 PM.

  9. #1404
    Pièce de Résistance Scheherazade's Avatar
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    Have to admit, I was considering a triangle (after my initial pentagol failed) until Mark's post.

    A quick one:

    You are on your way to the market with some gold coins in your pocket; however, you need to cross nine bridges before you reach your destination and there is a troll sitting under each bridge... They greedily ask for half of each passer's coin possession but, being somewhat good-hearted trolls still, they return one coin back.

    How many coins do you need to have at the beginning of your journey so that you are left with at least two coins when you arrive at the market?

    PS: No word games. Straight forward maths.
    ~
    "It is not that I am mad; it is only that my head is different from yours.”
    ~


  10. #1405
    www.markbastable.co.uk
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    While my phone is in triage, I'll think about this in print.

    So, in order to have two after the ninth bridge, you have to have three after the eighth bridge, because the troll will take....hang on...how's he going to take half of three?

    So you have to have four. He takes two, gives one back, so you've got three after the ninth bridge, which fulfils the 'at least two' requirement.

    To have four after the eighth bridge, you have to have six after the seventh bridge, because the troll will take three and give you one back.

    To have six after the seventh bridge, you have to have ten after the sixth bridge, because the troll will take five and give you one back.

    So - 18 after the fifth.

    Okay, now i've either got to actually work it out, or - which would be much more fun - come up with a formula that works for any number of bridges.

    Essentially, the number of coins you need for any bridge is 2 to the power of the number of the bridge (taking the last as bridge 0, and the penultimate bridge as bridge 1), plus two more coins.

    I think.

    That would mean it's (two to the power of eight) plus two.

    So I think it's 258. It would be cowardly of me now to actually work out whether I'm right by going through the bridges one by one.
    Last edited by MarkBastable; 11-14-2011 at 08:46 PM.

  11. #1406
    Something's Gone hoope's Avatar
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    Quote Originally Posted by billl View Post
    That's right.
    Well I thought the trick is how u read it ...
    But, Bill when u said in the hint " five rows and four trees each " that confused ! Coz there is no EACH mentioned in the puzzle words when u read it !

    Good one ... And good job Mark & Scher
    "He is asleep. Though his mettle was sorely tried,
    He lived, and when he lost his angel, died.
    It happened calmly, on its own,
    The way the night comes when day is done."



  12. #1407
    Registered User billl's Avatar
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    Quote Originally Posted by hoope View Post
    Well I thought the trick is how u read it ...
    But, Bill when u said in the hint " five rows and four trees each " that confused ! Coz there is no EACH mentioned in the puzzle words when u read it !

    Good one ... And good job Mark & Scher
    It's true there's lots of puzzles like this that have a trick reading--it might've been better if I had put "each" in both places (I almost left it out entirely, glad I thought to put it in...!)

  13. #1408
    Registered User prendrelemick's Avatar
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    Quote Originally Posted by MarkBastable View Post
    While my phone is in triage, I'll think about this in print.

    So, in order to have two after the ninth bridge, you have to have three after the eighth bridge, because the troll will take....hang on...how's he going to take half of three?

    So you have to have four. He takes two, gives one back, so you've got three after the ninth bridge, which fulfils the 'at least two' requirement.

    To have four after the eighth bridge, you have to have six after the seventh bridge, because the troll will take three and give you one back.

    To have six after the seventh bridge, you have to have ten after the sixth bridge, because the troll will take five and give you one back.

    So - 18 after the fifth.

    Okay, now i've either got to actually work it out, or - which would be much more fun - come up with a formula that works for any number of bridges.

    Essentially, the number of coins you need for any bridge is 2 to the power of the number of the bridge (taking the last as bridge 0, and the penultimate bridge as bridge 1), plus two more coins.

    I think.

    That would mean it's (two to the power of eight) plus two.

    So I think it's 258. It would be cowardly of me now to actually work out whether I'm right by going through the bridges one by one.
    Perhaps you have to have 2 coins at the last bridge. He takes one and gives you one back. (A bit like VAT. ) In fact if that's right, you only need two coins when you set out, as each troll will take one and give you one back. Remember Trolls are fiscally inept.
    Last edited by prendrelemick; 11-15-2011 at 02:50 AM.
    ay up

  14. #1409
    Registered User billl's Avatar
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    HA!

    Just like old times.

  15. #1410
    www.markbastable.co.uk
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    Quote Originally Posted by prendrelemick View Post
    Perhaps you have to have 2 coins at the last bridge. He takes one and gives you one back. (A bit like VAT. ) In fact if that's right, you only need two coins when you set out, as each troll will take one and give you one back. Remember Trolls are fiscally inept.

    Oh, *^$£*.

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