Question about York New State Living Environment: Relationships and Biodiversity Lab
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Posted on Aug 17, 2010
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Every associate leaves 4/5(n-1) molecules of a pile of n molecule. This result in an awful formula for the complete process (because every time one molecule must be taken away to make the pile divisible by 5):
4/5(4/5(4/5(4/5(4/5(4/5(p-1)-1)-1)-1)-1)-1), where p is the number of molecule in the original pile, must be a whole number.
The trick is to make the number of molecule in the pile divisible by 5, by adding 4 molecules. This is possible because you can take away those 4 molecules again after taking away one fifth part of the pile: normally, 4/5(n-1) molecules are left of a pile of n molecules; now 4/5(n+4) =4/5(n-1) +4 molecules are left of a pile of n+4 molecules. And because of this, the number of molecule in the pile stays divisible by 5 during the whole process. So we are now looking for a p for which the following holds:
4/5×4/5×4/5×4/5×4/5×4/5× (p+4) = (46/56) × (p+4), where p is the number of molecule in the original pile, must be a whole number.
Therefore, the smallest (p+4) for which the above holds, is 56.
So there were p=5 power 6 - 4 = 15621 molecule in the original pile.
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