12 coins by engiin!!

0Maths? Probability?:P0-

0 01font00there are 12 coins. = 12 coins02font02br
00one of them is false; so 1-12 = 11. 02br
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0011 coins of same weight02br
001 is not 02br
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00how to find the false coin by three weighs on a simple scale? (i don't understand)0-

0 Maybe he wants to know the probability of taking that false coin in three attempts:) 0-

0Yes; you have 12 coins, one of which has a different weight from the other 11. You also have a balance. How do you detect the odd coin out, using the balance only 3 times?02br
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00MrP0-

0 But we still need something else! Do you return the coins you take out to the "box" or not?:) 0-

0 wouldn't the odd coin look different? it would be thinner or fatter than the rest if it's lighter/heavier.0-

0I think to do it in three, you need to know if it's heavier or lighter, not just "different."02br
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00Let's assume lighter.02br
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00Take any four and put it them on one side of the balance, and any other four and but them on the other.02br
00 - If it balances, the odd coin is one of the four not yet weighed.02br
00 - If it do

0You don't need to know if it's lighter or heavier, to do it in three; but it is a horribly complicated procedure, with several "if this, do X, but if that, do Y" branchlets.02br
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00(In other words, it isn't one of those "Why didn't I think of that?" solutions.)02br
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00MrP0-

0 Hello all:)!02br
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00I don't think we should weight them all together. Wouldn't it be easier to take three coins and see if there is any possibility for the "different" coin to be one of those? If it is different, it should weight more or less than the others, this way would be much easier..050010id38

0 (But what if the "false" coin isn't among those three?) 0-