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There's an episode of Columbo with a similar problem, expressed as: there are a number of bags. Each bag contains many gold coins. The gold coins weigh 1 ounce
by geenew 4y ago
There's an episode of Columbo with a similar problem, expressed as: there are a number of bags. Each bag contains many gold coins. The gold coins weigh 1 ounce each. However, one bag contains fake coins, which appear the same as the real coins, but have a slightly different weight - 1.1 ounces per coin. You have a scale, which you can only use once, and you must find the bag with the fake coins.
The solution is: Take one coin from the first bag, two coins from the second bag, three coins from the third bag, etc. Weigh all of the selected coins at the same time to get a total weight. Subtract from the total weight the triangular number of the number of bags[1], in ounces. The remainder, divided by 0.1, will be the number of the bag with the fake coins.
[1] eg, for three bags, 1 + 2 + 3 = 6
- recursive 4y agoI'm amazed that there was an episode of a popular TV drama aired that had such a detailed puzzle in it.
- hgsgm 4y agoColumbo episodes were unusually long. Each was the length of a short movie: ~1 to ~1.5hrs. One of the Die Hard movies had the "extract 1 gallon from a 3 gallon and a 5 gallon jug" puzzle in it.
- terminal_d 4y agoIsn't that overcomplicating the solution? The only requirement is to have a unique number of coins from the fake coin pouch. Since we don't know which one that is, every pouch gets a different number of coins, and not necessarily a sequence starting from 1. Then, you weigh them all assuming they're all real, but wait, they're not! So you take the excess weight, divide it by the excess weight per fake coin, and the bag from which you took out that many coins is the bag with the fake coins.