Find the fake diamond in 3

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I apologize if this has been posted before:

This isn't exactly a riddle, but more of a brain teaser. I do not know the answer:

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You have 8 diamonds, 1 of which is actually a fake. You also have a scale with which you can weigh any combination of the diamonds. Note that the scale is NOT a hanging scale--there is just one platter to place objects on. You do not know if the fake diamond is heavier or lighter than the normal diamonds.

Figure out which diamond is actually the fake. You are allowed to perform only 3 weighings of any combination of the diamonds.

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My best answer so far gives me the fake in 4 weighings. I've gotten that several ways, but the way to do it in 3 is still escaping me. Good luck!

Edit: I see someone posted a slightly different version of this question at the same time as me... what are the odds? Hopefully mine's harder Razz

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I'm confused about how the scale works. Typically, it is a balance scale with two pans. That seems not to be the case here. Please clarify.

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Assuming a balance scale:

You can actually solve this in three weighings with up to nine diamonds.

Weigh AAA against BBB with CCC on the side.
If they balance, the fake is in CCC. Weigh CCC agains AAA to see if the fake is heavy or light. Weigh C1 against C2 with C3 on the side. If they balance, C3 is heavy or light depending on the outcome of the previous weighing. Otherwise, C1 or C2 is heavy or light based on the previous and current weighings.

If AAA is heavier (same reasoning applies if lighter) than BBB, weigh AAA against CCC to see if AAA is heavy or BBB is light (assume that AAA is heavy - same reasoning applies if AAA is light). Weigh A1 against A2 with A3 on the side. If they balance, A3 is heavy (based on our assumption). Otherwise, A1 or A2 is heavy.
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If you know if the fake is heavier or lighter, you can determine the fake in ceiling(log3(N)) weighings (ceiling is the smallest integer greater than or equal to its argument, N is the number of objects).

If you don't know if the fake is heavier or lighter, add one weighing.

Objects (N) - Weighings (log3(N))
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1-3 .......... 1
4-9 .......... 2
10-27 ...... 3
etc.

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My understanding of the question is that it is not a hanging scale, so that you cannot balance one "side" against the other. There is only a single place to put the diamonds, and the scale will show a number (whatever it may be) that is the weight of the diamonds.

Let me know if more clarification is needed. Smile

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Do you know what a real diamond weighs?

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Code:

If you do the following three weighings:



1: ABC E

2: AB D F

3: A CD  G



then you'll get the following results (R=weight of real diamond, F=weight of fake diamond):



Fake   Weighing 1   Weighing 2   Weighing 3

A      3R+F         3R+F         3R+F

B      3R+F         3R+F         4R

C      3R+F         4R           3R+F

D      4R           3R+F         3R+F

E      3R+F         4R           4R

F      4R           3R+F         4R

G      4R           4R           3R+F

H      4R           4R           4R



Unfortunately, you can't tell the difference between A being fake and H being fake unless you know the weight of a real diamond.

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Interesting. The problem as I saw it did not specify the weight of a real diamond, so perhaps it was incomplete--or meant to just frustrate the reader. I'll think about it some more.

Thank you!

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I guess this problem is an odd even out kind. seven of whic should be equal in weight but how could you determine the fake one in just 3 combinations, and how does the scale really work?

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Find the fake diamond in 3

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