Difference between revisions of "2013 AMC 10A Problems/Problem 21"
Countingkg (talk | contribs) (Created page with "==Problem== A group of <math>12</math> pirates agree to divide a treasure chest of gold coins among themselves as follows. The <math>k^{\text{th}}</math> pirate to take a share...") |
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==Solution== | ==Solution== | ||
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| + | ==See Also== | ||
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| + | {{AMC10 box|year=2013|ab=A|num-b=20|num-a=22}} | ||
| + | {{AMC12 box|year=2013|ab=A|num-b=16|num-a=18}} | ||
Revision as of 22:13, 7 February 2013
Problem
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
See Also
| 2013 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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 10 Problems and Solutions | ||
| 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 | |
