Posers and Puzzles

Posers and Puzzles

  1. Joined
    08 Dec '06
    Moves
    24226
    09 Apr '19 02:211 edit
    My friend sent me this:
    "There are 42 coins on a table. Of those, 10 are heads-up and the other 32 are tails-up.

    How can you separate the coins into two groups so that each contains the same number of coins that are heads-up?

    Note that you cannot see nor feel the coins, though you can manipulate them. (you are blindfolded and wearing gloves)."

    I solved it and proved that my solution is correct. I'm interested in seeing if other solutions exist.
  2. Standard memberforkedknight
    Defend the Universe
    127.0.0.1
    Joined
    18 Dec '03
    Moves
    16687
    09 Apr '19 02:424 edits
    @damionhonegan said
    My friend sent me this:
    "There are 42 coins on a table. Of those, 10 are heads-up and the other 32 are tails-up.

    How can you separate the coins into two groups so that each contains the same number of coins that are heads-up?

    Note that you cannot see nor feel the coins, though you can manipulate them. (you are blindfolded and wearing gloves)."

    I solved it and proved that my solution is correct. I'm interested in seeing if other solutions exist.
    Reveal Hidden Content
    Take 10 coins from the pile of 42 and move them into a separate pile. Flip all 10 of those over.


    Reveal Hidden Content
    If you happen to pick all 10 of the heads-up coins, both piles have 0. If you pick n of the head-up coins, both piles will have 10-n
  3. Joined
    08 Dec '06
    Moves
    24226
    09 Apr '19 03:471 edit
    @forkedknight
    This is my solution. Are there any other?
  4. Joined
    26 Apr '03
    Moves
    26771
    14 Apr '19 09:22
    @damionhonegan said
    This is my solution. Are there any other?
    Put 32 coins into a new pile and flip the ten coins left in the other pile.
  5. SubscriberAThousandYoung
    Gonzalo de Córdoba
    tinyurl.com/y3aahuno
    Joined
    23 Aug '04
    Moves
    24791
    15 Apr '19 20:22
    Put all the coins on the edge.
  6. Joined
    26 Apr '03
    Moves
    26771
    30 Apr '19 20:03

    Removed by poster

  7. Joined
    26 Apr '03
    Moves
    26771
    03 May '19 07:101 edit
    More seriously
    If there are H heads
    Two piles of coins
    And all coins in one pile, size P, with h heads in it, are flipped
    (You have to flip all coins in a pile. If you don’t, an unflipped coin could be heads or tails, which changes the count in that pile independently of the other pile.)

    Before flip there were h and H-h heads in each pile
    After the flip there are P-h and H-h heads in each pile

    For the piles to have equal numbers of heads
    P-h = H-h
    P = H

    So the size of the pile flipped must be equal to the total number of heads
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