Difference between revisions of "2013 AMC 12A Problems/Problem 17"
Epicwisdom (talk | contribs) m (Created page with "The first pirate takes <math>\frac{1}{12}</math> of the <math>x</math> coins, leaving <math>\frac{11}{12} x</math>. The second pirate takes <math>\frac{2}{12}</math> of the rema...") |
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Note that | Note that | ||
| − | <math>12^{12} = (2^2 * 3)^{12} = 2^{24} | + | <math>12^{12} = (2^2 * 3)^{12} = 2^{24} * 3^{12}</math> |
<math>11! = 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2</math> | <math>11! = 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2</math> | ||
Revision as of 16:04, 10 February 2013
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.
Note that
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.
