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Well, the last puzzle brought up quite a storm. Congratulations to FoieGras for his solution!
Today, let's take a shot at the common coin weighing puzzle. Of course, just like yesterday's common true/false puzzle, there is going to be a difficult twist to it.
Here we go!
You have an arbitrarily accurate scale at your disposal, and 52 coins. These coins are divided into 13 piles of 4. All coins are identical and each weigh X grams, except for one pile, where the four coins are fake and each weigh X+d grams. More precisely, X is a positive integer and d is a non-zero (real) number strictly between -5 and +5.
Given 2 weighings on your scale, find which pile has the fake coins. Also find the values of X and d.
GL!
 
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I love your puzzles JeeJee keep it up, will work on this one when i can. Gosh I wish I could keep up withe everyone :-)
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+ Show Spoiler +Make a pile with coins chosen from piles 1-13 like so:
4 4 4 4 3 3 3 3 2 2 1 1 0
You will get 34x + (up to 4) d
If your answer is within 14 of a multiple of 34, then you know what X is (closest multiple).
Then pile them on like this:
0 1 2 3 1 2 3 4 3 4 3 4 0
Since you know X, you can figure out (up to 4) d. Then depending on if the ratio between these two numbers (up to 4) d is 0, 1/4, 1/2, 3/4, 1/3, 2/3, 1, 4/3, 3/2, 2, 3, 4, or infinity, you can figure out which pile it is in and also d.
If your answer is not within 14 of a multiple of 34, you know that the bad pile is in piles one through eight. Weigh this pile:
0 1 2 3 1 2 3 4 4 4 4 4 4
You now have 36x + (up to 4) d. If your answer is within 15 of a multiple of 36, then you know X, and hence (up to 4) d, and determine the bad pile and d depending if the ratio is 0, 1/4, 1/2, or 3/4, 1/3, 2/3, or 1. This will be the case if the bad pile is in piles one through seven. If your answer is not, then you know it was the eighth pile and now you know both 34X + 4d and 36X + 4d so you can find both X and d this way.
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On February 14 2010 05:41 EmeraldSparks wrote:+ Show Spoiler +Make a pile with coins chosen from piles 1-13 like so:
4 4 4 4 3 3 3 3 2 2 1 1 0
You will get 34x + (up to 4) d
If your answer is within 14 of a multiple of 34, then you know what X is (closest multiple).
Then pile them on like this:
0 1 2 3 1 2 3 4 3 4 3 4 0
Since you know X, you can figure out (up to 4) d. Then depending on if the ratio between these two numbers (up to 4) d is 0, 1/4, 1/2, 3/4, 1/3, 2/3, 1, 4/3, 3/2, 2, 3, 4, or infinity, you can figure out which pile it is in and also d.
If your answer is not within 14 of a multiple of 34, you know that the bad pile is in piles one through eight. Weigh this pile:
0 1 2 3 1 2 3 4 4 4 4 4 4
You now have 36x + (up to 4) d. If your answer is within 15 of a multiple of 36, then you know X, and hence (up to 4) d, and determine the bad pile and d depending if the ratio is 0, 1/4, 1/2, or 3/4, 1/3, 2/3, or 1. This will be the case if the bad pile is in piles one through seven. If your answer is not, then you know it was the eighth pile and now you know both 34X + 4d and 36X + 4d so you can find both X and d this way.
Intense solution! I would like to add that I believe 0 1 2 3 1 2 3 4 3 4 3 4 4 makes more sense and would be correct rather than 0 1 2 3 1 2 3 4 3 4 3 4 0, when doing the second pile.
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For the coins in the fake pile, are all the fake coins identical?
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On February 14 2010 18:13 Cambium wrote:
For the coins in the fake pile, are all the fake coins identical?
they're all identically fake, yes (i.e. d is constant)
I should mention I haven't solved this puzzle so if you're looking for confirmation from me, look elsewhere  I am looking over EmeraldSparks's solution now screw that TSL is on.. after that although I can tell you that InFdude's is not correct unless I am mistaken -- how do you determine which pile is the fake one?
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No I don't have the answer, and no I didn't make this stuff up =)
Third possibility: it's a puzzle for which I don't know the solution.
Or, to be fair, didn't know. I just now had a chance to look at EmeraldSparks's solution and it seems to be correct, although I haven't checked it extremely rigorously. I am impressed though, that's one hell of a solution, GJ!
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Ok, we could be a little more polite..
here's a proof of concept solve:
ok
let's say X is 3
and d is 2.5
and pile 11 is fake (aka 5.5 each)
weighing one: 4 4 4 4 3 3 3 3 2 2 1* 1 0
48+36+12+5.5+3=104.5
within 14 of a multiple of 34 (which is 102, 2.5 diff), we know X is 3
weighing two: 0 1 2 3 1 2 3 4 3 4 3* 4 0 = 30+xd
0+3+6+9+3+6+9+12+9+12+5.5*3+12+0=97.5
3*30=90. 97.5-90=7.5
7.5/2.5 = 3, .: pile 11 is the fake pile, .: d=2.5
Which part are you uncertain about?
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South Africa4316 Posts
Once again, excuse the lack of mathematical know how, I'll have to explain things in words once or twice. Also, I'm not sure what an arbitrarily accurate scale means, but I assume it means that they give a weight, but that it's not necessarily in grams. Maybe I missunderstood that, that could explain why.
+ Show Spoiler +The first step is to find the weights of the items. Since the weight of X is always an integer, it means the final weight will always be 52X + 4d. However, since the limits of d are -5 and 5, it means you can always find out what both X and d are, since the total will only be reachable in one way. For example, if the final weight is 124gs, then X must be 2, and d must be 5. If X was 3, (154), then the final weight couldn't get to 124 even if d was -5. So the first step is to have group 1 be split 2-2, and the rest evenly between 4-0 and 0-4. If the scale is even, then obviously group 1 is the fake group. However, if it is not then you add the two weights to find the total weight of the groups, allowing you to calculate the weight of X and d.
Once you have that, it's a simple process of splitting the groups up. The groups can be divided as follows:
1-0
2-0
3-0
4-0
2-1
3-1
0-1
0-2
0-3
0-4
1-2
1-3
2-2
The total weight will now be Full Total Weight - 14X - nd. Since you know what X and d is, you can calculate n, narrowing down the results. if n = 3, then the answer is either 1-0 or 0-1, and you can see which one by looking at which side is the heaviest. If n = 0, it can be either 4-0, 3-1, 1-3, or 0-4. You can now calculate what the difference is between the two sides. The difference will either be 2d or 4d. If it is 2d, then its either 3-1 or 1-3, if it is 4d it is 4-0 or 0-4, depending on which side weighs the most at the end.
I'm fairly sure this solution works if I understand the idea of an arbitrarily accurate scale correctly. If not, then I will have to try again!
EDIT: for some reason I thought the scale had two sides, and gave a weight. I'll come redo it once I get back from watching Alice 
EDIT2: EmeraldSparks's solution is damn elegant. Very nice!
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glad you're enjoying the puzzles <3
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South Africa4316 Posts
In the humanities, theres not much that forces you to think, so these puzzles make a nice change
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EPT![[image loading]](http://imgs.xkcd.com/comics/labyrinth_puzzle.png)
