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Soultra
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Posted 1 Year, 4 Months ago #1
Post good puzzles with strong logic for compilation for the group.

A sample to work on:

A heap of ten coins has one defective coin, which has a different weight (could either be light or heavy). In three weighings with a simple pan balance, how would you separate the defective coin from the lot.
JohnBStone
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Posted 1 Year, 4 Months ago #2
What I would do is, I would add two coins, and then look up the solution in the rec.puzzles FAQ where this question is identified as one that gets asked here about once a fortnight.
dagny
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Posted 1 Year, 4 Months ago #3
[SPOILER]

Like Joe Slater suggested: look in the FAQ for the 12 coins solution. If you don't have 2 extra coins, then after the first weighing, label all the coins that certainly are not special: if the first weighing balances, pick 2 coins from the 8 that you just tested. Otherwise, if the weighing did not balance, pick 2 from the remaining 4. In each case, when a conflict arises because you need to balance a double-numbered coin on both sides, or on the same side, just pick another double-labelled coin.

Quite easy

Dirk Vdm
querty
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Posted 1 Year, 4 Months ago #4
Writing main idea down,

1. Divide three groups - a:3 b:3 c:4 , (a,b,c : index)

2. Compare a and b. If equal, c contains a different coin.
imported_baz
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Posted 1 Year, 4 Months ago #5
Not strong enough. You can find the odd coin in as many as 12 coins in 3 weighings.

It is very ironic that the example contradicts the request!
NGR
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Posted 1 Year, 4 Months ago #6
I wonder if this misconception can ever be stamped out. Even the FAQ says you can distinguish among 13 coins, and in fact you can distinguish among 14 (FOURTEEN) coins if you have a 15th known good coin to work with.

Note that we don't need to determine if the defective coin is heavy or light.
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