There are nine silver coins marked 1, 2, 3, 4, 5, 6, 7, 8, and 9.
They are supposed to weigh the number of grams that is written on them.
One of the coins is fake and is lighter than it should be.
Find the fake coin using the balance scale twice.
As in the classic problem of detecting the one light coin out of 9 in only 2 weighings, the first weighing must be 3 coins vs 3 coins.
http://perplexus.info/show.php?pid=76
So we need two subsets of 3 coins that should be equal:
One possibility is {1,4,8} vs {2,5,6}
if one pan is lighter it contains the fake, if they are the same the fake is among {3,7,9}
Once you have it narrowed down to one of three you can use a known fair coin to help you weight two of them. For example if {1,4,8} is known to contain a light coin then weigh {1,3} vs. {4} since we know the 3 is fair if the left side is light the 1 is the fake.
The pair of triplets I chose for the first weighing is not the only possibility. I may try to enumerate the others.
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Posted by Jer
on 2016-01-18 12:17:21 |
A solution
