until you find the answer we will leave this as a possible solution
Teaser #6
- Thread starter ban-all-sheds
- Start date
until you find the answer we will leave this as a possible solution
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JB - surely it is blindingly obvious that a valid solution has to be one that always works, not just 1 time in 6...
You might regard yours as a possible solution, I don't think that many other people will....
You might regard yours as a possible solution, I don't think that many other people will....
i done the test and got it in 2, whats your problem. whats your solution without looking for the answer, Like most teasers, the answer is probably out there on the internet, so if you can't be bothered to work it out, please leave it for others.
Id rather you spoke for yourself rather than " I don't think that many other people will...." I hope you dont speak on my behalf when im not around
Id rather you spoke for yourself rather than " I don't think that many other people will...." I hope you dont speak on my behalf when im not around Sponsored Links
I might do, if only I could learn to speak nonsense...jbonding said:I hope you dont speak on my behalf when im not around
ban-all-sheds said:I might do, if only I could learn to speak nonsense...jbonding said:I hope you dont speak on my behalf when im not around
Read a few of my other posts and im sure you will pick it up quicker
Take 4 groups of 3. ( Called A B C D)
1. Measure A and B.......i....If A = B then Coin is in C or D
ii If A differs from B then coin is in A or B
Now you have 2 choices depending on the outcome above
2i Measure C against A or B......i.... If they equal then the coin is in D
ii If not then the coin is in C as we know that A and B does not hold the fake coin. Also on the scales if C is the highest of the weights then we know its a lighter coin. If its the lowest of the weights we know its heavier.
Third and final weigh.
3i The coin is in C or D so whichever group its in take 2 coins out and measure them. If they equal then the fake coin is the remaining one.
If they differ then look back at whether the coin is lighter or heavier and from the position of the scales you can see which one is the fake one.
Hope that makes sense......
1. Measure A and B.......i....If A = B then Coin is in C or D
ii If A differs from B then coin is in A or B
Now you have 2 choices depending on the outcome above
2i Measure C against A or B......i.... If they equal then the coin is in D
ii If not then the coin is in C as we know that A and B does not hold the fake coin. Also on the scales if C is the highest of the weights then we know its a lighter coin. If its the lowest of the weights we know its heavier.
Third and final weigh.
3i The coin is in C or D so whichever group its in take 2 coins out and measure them. If they equal then the fake coin is the remaining one.
If they differ then look back at whether the coin is lighter or heavier and from the position of the scales you can see which one is the fake one.
Hope that makes sense......
jackpot said:
I'd thought that, but there's a hole, I'm afraid. At you first stage, there is a 1 in 4 chance the fake is going to be in pile D - If that happens then you are going to go into the 3rd weighin without knowing whether the fake is heavier or lighter, and that makes your plan for the 3rd weighing fall down.
Sorry.....
johnny_t said:
Not sure i know what you mean. If at first stage you know the coin is in either A and B or C and D. You would then go on to the second stage. I used an example with the coin being in C and D. If it was in A and B then you would reverse the measure i.e intead of measuring C against A or B you would measure A Aigainst C or D. And only after this 2nd stage will you find out whether coin is heavier or lighter.
I think......
jackpot said:
If the fake is in Pile D, then you only identify that fact by ascertaining that A, B and C all weight the same. You don't actually weigh pile D against anything, so you don't know if it weighs more or less than the others.
It is that knowledge of whether the fake is heavier or lighter that you get from steps 1/2 that allow you to carry out step 3.
(Apologies if I've misread your post - It is possible that I am still too busy brooding on what I did wrong to be able to see the wood for the trees.)
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A 1 2 3
B 4 5 6
C 7 8 9
D 10 11 12
Weigh-1
A v B ..... A equals B (A = B)
Weigh-2
A v C .....A = C therefore group D is odd and heavy xor lightweight
Weigh-3
10 v 11 ..... 10 = 11 therefore 12 is odd BUT heavy or lightweight??
Been there ...
B 4 5 6
C 7 8 9
D 10 11 12
Weigh-1
A v B ..... A equals B (A = B)
Weigh-2
A v C .....A = C therefore group D is odd and heavy xor lightweight
Weigh-3
10 v 11 ..... 10 = 11 therefore 12 is odd BUT heavy or lightweight??
Been there ...
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