2013 AMC 12A Problems/Problem 17
The first pirate takes of the coins, leaving .
The second pirate takes of the remaining coins, leaving .
Continuing this pattern, the eleventh pirate must take of the remaining coins after the first ten pirates have taken their share, which leaves . The twelfth pirate takes all of this.
All the 2s and 3s cancel out of , leaving
in the numerator.
We know there were just enough coins to cancel out the denominator in the fraction. So, at minimum, is the denominator, leaving coins for the twelfth pirate.