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Posers and Puzzles

Joined 08 Dec '06 Moves 24006 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.

127.0.0.1

Joined 18 Dec '03 Moves 16177 @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 ContentTake 10 coins from the pile of 42 and move them into a separate pile. Flip all 10 of those over.

Reveal Hidden ContentIf 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

Joined 08 Dec '06 Moves 24006 @forkedknight This is my solution. Are there any other?

Joined 26 Apr '03 Moves 26735 @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.

tinyurl.com/y3czz7bk

Joined 23 Aug '04 Moves 24791 Put all the coins on the edge.

Joined 26 Apr '03 Moves 26735

Joined 26 Apr '03 Moves 26735 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