Here's another interesting coin puzzle. If you already know the answer to
this, please don't give it away, since this is a fun problem for people to
figure out on their own.
You are kidnapped, blindfolded and placed in front of a pile of coins of
unknown quantity. You are told that 2003 of these coins are heads-up, and
that you will be released unharmed if you can sort these coins into two
piles such that each pile has exactly the same number of coins with heads
showing. How can you do this?
+ You cannot distinguish heads from tails, be it by sight, touch, etc.
+ You may manipulate the coins in any manner you wish; however, you may not
remove any coins, or balance the coins on their sides. ;-)
At first when I read this problem, I wondered how it could even be
possible! But it turns out the solution is quite simple (unlike your
previous problem :-) ).
The solution: Just flip any 2003 coins! Make these 2003 coins one pile,
and put the rest of the coins in the other pile.
Here's why it works.
Let N = total number of coins, N > 2003. (The N=2003 case is trivial).
Split the coins into 2 piles, pile 1 with 2003 coins, pile 2 with the rest.
Let x = number of heads-up coins in pile 1 (before flipping). So there are
2003-x heads-up coins in pile 2. Now flip every coin in pile 1, so pile 1
now has x tails-up coins, which means 2003-x coins are heads-up. That's
now the same as pile 2.