Difference between revisions of "2013 AMC 12A Problems/Problem 17"
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Revision as of 16:00, 3 July 2013
Problem 17
A group of pirates agree to divide a treasure chest of gold coins among themselves as follows. The pirate to take a share takes of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the pirate receive?
Solution
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.
See also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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