and one of them is fake. The fake could be either heavier or lighter than the others, you don't know. Could you find out which one is fake just with 3 tries on the balance scale?
Got this brain-teaser from Advait, who heard it from Brandon. Baby solved it in half an hour!!! My kids'd better be smart.
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label coins a-l
measure 4 (a-d) coins against another 4 (e-h)
if the same then the spare 4 (i-l) contains the odd one out, measure 1(i) v 1(j) of these
if i and j are the same take one of the spares (k) and measure against one of those you know to be true (i or j), if the same, the unmeasured one (l) is the fake, if different k must be fake
if the i and j are different you know any of the remaining coins are real so measure i against any of the others to know if it is fake, if it isn't j is.
Note: this take 3 steps.
if a-d are different to e-h (remember which is heavier)
take two from each set a,b,e,f and measure the against i-l if they are different you can work out whether the fake is heavier or lighter; which means you know whether a or b are fake or e or f are fake, you can then eliminate which one in the same method as i or j above
Note: this take 3 steps
if they are the same take c and g and measure against i and j
if they are different, you know whether the fake is heavier or lighter so can work out whether it is c or g that is the fake
if they are the same mesure one of them against a real coin and find which one is fake in the same way as i and j above
Note: this take 4 steps
Please let me know if I'm correct, took me 1 hour.
Hi Felix!
You're almost there, but it's possible to find out in just 3 steps... :) jia you~~
Haiya, I've think a lot but still can't find a solution. Guess my IQ is not catching up with my age. Please tell me the answer. Thank you.
Label coins as A-L.
Step 1: compare 4 coins (ABCD) with another 4 (EFGH)
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*If they are equal, then the fake is one of IJKL. ABCDEFGH are not fake.
Step 2: compare IJ with either 2 of ABCDEFGH.
**If they are equal, then the fake is one of KL. IJ are not fake.
Step 3: compare K with either one of ABCDEFGHIJ.
If they are equal, then L is fake.
Else, then K is fake.
**If IJ and 2 of ABCDEFGH are not equal, then the fake is one of IJ. KL are not fake.
Step 3: compare I with either one of ABCDEFGHKL.
If they are equal, J is fake.
Else, I is fake.
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*If ABCD and EFGH are not equal, then the fake is one of ABCD or EFGH. At this point, we know whether ABCD is heavier or lighter than EFGH.
->Suppose ABCD is heavier than EFGH.
Step 2: Compare {AB(possibly heavy fake)+E(possibly light fake)+I(definitely not fake)} and {CD(possibly heavy fake)+F(possibly light fake)+J(definitely not fake)}.
-->If AB+E+I is lighter than CD+F+J, then E or C or D is fake.
Step 3: compare C with D.
If they are equal, E is fake.
If C is heavier than D, C is fake.
Else, D is fake.
-->If AB+E+I is heavier than CD+F+J, then A or B or F is fake.
Step 3: compare A with B.
If they are equal, F is fake.
If A is heavier than B, A is fake.
Else, B is fake.
-->If AB+E+I is equal to CD+F+J, then G or H is fake.
Step 3: compare G with I.
If they are equal, H is fake.
Else, G is fake.
->Suppose ABCD is lighter than EFGH. The reasoning is similar to the case when ABCD is heavier than EFGH (replace A with E, B with F, C with G, D with H and proceed accordingly).
I think we've covered all possible cases. Did I miss anything? :)
Thanks very much. Your baby is smart enough to think this out in 30 minutes time. So don't be greedy, hehe. Happy Sunday.
I have better solution for the first case, where ABCDEFGH are equal. (Step 1)
We know that IJKL has 1 fake coin, not knowing it's heavier or lighter.
Step 2:
So, take I aside; and compare JKL with either 3 coins from ABCDEFGH.
If JKL is heavier, we know that one of them is fake and the fake is heavier.
Step 3:
Compare K & L, if they are equal; J must be fake.
If K is heavier, then K is fake.
The same principal apply if the fake is lighter.
If JKL is equal with either 3 of ABCDEFGH, then the fake is I.
But we still have 1 more chance to find out the fake is heavier or lighter, so just compare I with any of those coins and you know the result.
Using this method, combine with your method, we managed to find out the fake coins in 3 weighing steps, and know if the fake is heavier or lighter.
You're right! Thanks for improving the solution. :)
Any other puzzles? hehee~~
Wow, good puzzle. But with the second case, where ABCD and EFGH are not equal, in step 2 you said:
Compare {AB(possibly heavy fake)+E(possibly light fake)+I(definitely not fake)} and {CD(possibly heavy fake)+F(possibly light fake)+J(definitely not fake)}.
What's the point of having I(definitely not fake) and J(definitely not fake) in there? They will only cancel each other out, so why not just compare ABE with CDF?
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