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wiki:donuts:fall2014:answers:pirates
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wiki:donuts:fall2014:answers:pirates [2014/11/25 13:44] jmb438 created |
wiki:donuts:fall2014:answers:pirates [2014/12/01 11:09] (current) jmb438 |
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| + | Solution: (adapted from [[http://www.puzzle.dse.nl/teasers/index_us.html#coconut_chaos|THE ULTIMATE PUZZLE SITE]]) | ||
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| + | Each pirate leaves //4/5(n-1)// coins of a pile of //n// coins. This results in an awful formula for the complete process (because every time one coin must be taken away to make the pile divisible by 5): | ||
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| + | //4/5(4/5(4/5(4/5(4/5(p-1)-1)-1)-1)-1)//, where //p// is the integer number of coins in the original pile. | ||
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| + | The trick is to make the number of coins in the pile divisible by 5, by adding 4 coins. This is possible because you can take away those 4 coins again after taking away one fifth part of the pile: normally, //4/5(n-1)// coins are left of a pile of //n// coins; now //4/5(n+4)=4/5(n-1)+4// coins are left of a pile of //n+4// coins. Because of this, the number of coins in the pile stays divisible by 5 during the whole process. So we are now looking for a //p// for which the following holds: | ||
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| + | //4/5×4/5×4/5×4/5×4/5×(p+4)=(4<sup>5</sup>/5<sup>5</sup>)×(p+4)//, where p is the number of coins in the original pile. | ||
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| + | The smallest //(p+4)// for which the above holds, is //5<sup>5</sup>//. So there were //p=56-4=3121// coins in the original pile. | ||
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wiki/donuts/fall2014/answers/pirates.1416948299.txt.gz · Last modified: 2014/11/25 13:44 by jmb438