Difference between revisions of "2013 AMC 12A Problems/Problem 17"
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| − | We know there were just enough coins to cancel out the denominator in the fraction. So, at minimum, <math>x</math> is the | + | We know there were just enough coins to cancel out the denominator in the fraction. So, at minimum, <math>x</math> is the denominator, leaving <math>1925</math> coins for the twelfth pirate. |
== See also == | == See also == | ||
{{AMC12 box|year=2013|ab=A|num-b=16|num-a=18}} | {{AMC12 box|year=2013|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 22:59, 13 October 2017
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 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
