When I was younger I used to spend a pretty strange amount of time reading through brain teaser books. As such, I occasionally hear a brain teaser and know the answer without being able to remember how in the world you get there. That’s what happened with today’s teaser…. I heard it on the Skeptics Guide to the Universe podcast this morning, answered immediately, then spent the rest of the train ride working out why I was right. I got there somewhere near Hyde Park.

Lets see how you do: A jeweler has 9 pearls, all identical shape and feel. He knows one weighs slightly more than the other 8, but all he has to measure with is a balance scale (one with two arms that compares weights to each other). What is the minimum number of times he needs to use the scale in order to figure out for certain which is the heavy pearl?

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2 weighings. Divide the 9 pearls into 3 groups of 3; weigh 2 groups to find the heavy one (if they're equal, it's the group you didn't weigh). Repeat with 3 groups of 1 pearl.

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It should be 3.

Let's see if I guessed right…the jeweler splits the 9 pearls into two sets of 4, with one left over.

1. He puts the two sets of 4 on the opposite sides of the balance scale.

1.a. If the two sets are the same weight, then the heavier pearl is the one left out. He can stop.

1.b. If the two sets are not the same weight, then the heavier pearl is in the heavier set. Clear the scale; keep the heavier set of pearls.

2. Put 2 of the pearls from the last set on one side of the scale, and 2 on the other. The heavier pearl will make one pair of pearls heavier. Clear the scale; keep the heavier set of pearls.

3. Put one pearl from the last set on each side of the scale. The heavier pearl will be obvious.

So the jeweler either gets the answer after the first attempt, or he has to use 3 attempts.

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I remember this one from long ago as well. Probably learned it in one of my comp-sci classes.

Jeweler divides the pearls up into three groups of three A, B, C

1) He weighs group A vs Group B

a) if they're the same, choose group C, and go to step 2.

b) if they're different, take the group that's heavier, and go to step 2.

2) He takes any two of the three in this final group, and weighs them against each other.

a) If one is heavier, that's the answer

b) if they're both the same, the remaining, unweighed one is heavier.

2 weighings.

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I guessed 3, and tried to justify it.

Didn't realize that 2 would work, also.

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I recall a variation in which it is known that one is Different, but unknown whether heavier or lighter, and the answer was three weighings. Or something. I'll see if that's available on the intertubes somewhere.

3 v 3, if same then 1 v 1 from unweighed group, then sub in the last unweighed 1…

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Nice job! You got caught in the spam filter for a bit…not sure why.

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Yeah, I knew the answer was two but I went down the same road you did at first. It was only my distant memory of the answer that kept me going.

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That's always the tough part about having it in the hazy part of my memory. There's always a chance there's some nuance I'm forgetting that makes the answer easier/harder/different.

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Good job! You get co-credit with Eric. His got caught in the spam filter for a while, so in a way you're both first.

Everyone's a winner and all must have prizes!

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