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#1 Re: Help Me ! » Twelve Coins Puzzle » 2012-05-26 08:24:04
Thanks for the help guys. Hey, so do think that a formula can be extrapolated from this method?
For three coins, you have three to the third, minus one (the FFF) distinct scale possibilities. So you have 26, but then that needs to be divided by two, because each coin has two scale possibilities, one if its heavy, one if its light. So you then have 13, and this needs to be made 12 because, using this method, you need a number of coins which is divisible by three. If we had four weighs, though, could we find a fake out of 39 coins?
3 to the 4th is how many distinct scale possibilities there are. Minus one (FFFF, which doesnt tell us if the coin is heavy or light), gives us 80. Divide that by two because of light/heavy and you have 40. And the next lowest number divisible by three is 39. I'm not sure if that formula is right, or if there even is a formula. I may be missing something here. If that formula is true, though, it means that if you had 10 weighs, you could find a fake coin in 29,523 coins!
#2 Re: Help Me ! » Twelve Coins Puzzle » 2012-05-26 06:02:00
Thanks Bob! 
#3 Re: Help Me ! » Twelve Coins Puzzle » 2012-05-26 04:33:15
Okay, so I've changed a few of the series around. They are now
1. 123
2. 113
3. 122
4. 111
5. 212
6. 231
7. 232
8. 233
9. 331
10. 313
11. 321
12. 322
Now the position of the scale is observed so that if the first plate is down, it is marked D, if the first plate is up, it is marked U, and if its flat its marked F. I now have the list so that each series gives two possibilities. The first is if that coin were the fake coin AND it was heavy. The second is that if the coin were fake coin AND light. That gives
1. DUF, UDF
2. DDF, UUF
3. DUU, UDD
4. DDD, UUU
5. UDU, DUD
6. UFD, DFU
7. UFU, DFD
8. UFF, DFF
9. FFD, FFU
10. FDF, FUF
11. FUD, FDU
12. FUU, FDD
I think that tells you which coin is fake and whether its heavy or light. What do you think?
#4 Re: Help Me ! » Twelve Coins Puzzle » 2012-05-26 04:03:54
Thanks for the help Bob! I'll try to tweak it so it finds heavier or lighter as well.
#5 Help Me ! » Twelve Coins Puzzle » 2012-05-26 03:25:09
- BCtrain
- Replies: 12
The other day I was looking for some math puzzles for a school project and came across the "Twelve Coins" problem. The problem is: You have twelves coins, and one of them is fake. It weighs more or less than the other coins. Whether its heavier or lighter than the other coins is not known. With a simple scale, you have three weighs to figure out which coin is the fake.
Okay, so, I came up with a method which (I hope) works. After I developed my method, I looked up the "classic" method and found it was simpler than mine, but there are multiple ways to solve it and I would like to know if mine works as well.
My method: You number each coin 1-12, and give each one a distinct series of three numbers (1-3). These are the places where the coin will be on each weigh. 1 the first pan, 2 is the other pan, 3 is off the scale. So, a coin with a 112, would be on one pan, then the same pan, then the other pan. The coins are each numbered so that during each weigh, there are four coins in one pan, four coins in the other pan, and four coins off the scale. Here are the numbers.
1. 123
2. 112
3. 122
4. 111
5. 212
6. 231
7. 232
8. 223
9. 331
10. 313
11. 321
12. 333
So, you weight the coins like this, and take note of the scale on each weigh. If the scale is uneven, the fake coin must be in pan one or pan two.
If the scale is uneven AND has switched which side is up (its gone from one uneven position to other) the fake coin must have switched pans (from one to two or from two to one). If the scale is even, the coin must be off the scale (in the three position). Okay, so, the three positions of the pans then must (I hope) tell you which coin is the fake, because each 1,2,3 position series gives you a distinct series of positions for the scale. Remember that, if the coin is fake, the scale should be uneven when the coin is in the one or two position, should switch from previously being uneven whenever the coin moves from one to two or two to one, and should be even whenever the coin is in the three position. The coin, then, can be found by matching the the viewed positions of the scale with the predicted positions for each given scale.
Here, U = uneven scale, S = a scale that's uneven, but in a different uneven position than the one it started in in the first round, F= a flat scale
1. USF
2. UUS
3. USS
4. UUU
5. USU
6. UFS
7. UFU
8. UUF
9. FFU
10.FUF
11.FUS
12.FFF
Judged on the three positions of the scale, you can tell which coin is the fake. Just match the positions to the coin's predicted positions. This method does not tell you if the coin is heavy or light (the problem doesn't say you have to), but I think it does tell if the coin is fake.
Are there any problems with it? Or does this convoluted method actually work?
Thanks for the help!
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