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# Treebeard's Stumper AnswerSeptember 19, 1997

We're starting the year in science studying Oology, so here's a classic stumper slightly modified to involve eggs. Suppose you have a dozen eggs that look and feel identical. All are equal in weight except one. You also have an old fashioned double arm balance. You can put any number of eggs on each pan of the balance to see which side tips or if they are the same. Can you find which egg is different in only 3 attempts if you know that the odd egg is heavier than the others? What if you only know that the odd egg is different (heavier or lighter), but you don't know which?

Can you find the heavy egg using a balance just 3 times? Katherine Koch (Jordan's mom) explains: "Divide the eggs in half, 6 on each of the scale's pans. One side will weigh more. Then divide the heavier in half again, now you have 3 eggs on each pan. One side will weigh more, so take any 2 of those 3 eggs. If they balance, the one left over is the odd one, otherwise the scale will reveal it." It's MUCH harder if you only know that the odd egg is different. John Norris almost got it.

Ode to an Egg by John Norris

We've been told to ponder an egg,
So I sit, crossing over a leg.
I look and I look,
I could write a whole book!
Well... I just hope the darn thing don't breg!

Thanks John, that last line took real courage! It's much harder if you only know that the odd egg is different, but not if it's heavier or lighter. The scale may tip, but is the one side heavier or the other lighter? It's interesting how such a small difference in wording makes a huge difference in difficulty!

If you divide the eggs in half on the balance each time, then you can't get enough information to solve the general case. One trick is to divide the eggs into three groups at each step, but only weigh two of them, as in the last step of Katherine's solution. Now in 3 weighings we can distinguish 3^3=27 cases, and not just 2^3=8 cases. There are actually 24 cases to distinguish, 12 eggs, each heavier and lighter, so it's important to get maximum information from each weighing. The other trick is to use eggs known to be normal weight in future weighings. I think this works:

1. Number the eggs 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

2. Put 1, 2, 3, 4 on the left side of the scale and 5, 6, 7, 8 on the right side. If it remains balanced go to [3], otherwise go to [4].

3. The different egg is one of 9, 10, 11, 12. Put 9 and 10 on the left side, and 11 with any good egg, say 1, on the right side. If it is balanced, then the different egg is 12, and you can find out if it is heavier or lighter by comparing it with any other egg. If the scale is unbalanced, say to the left (the other case is analogous), then the different egg is one of 9 or 10 and heavier, or 11 and lighter. Next, compare 9 with 10. In case of balance, the different egg is 11 (and lighter). Otherwise the different egg is the heaviest one.

4. Assume the scale was unbalanced to the left (the other case is analogous). Then the egg is one of 1, 2, 3, 4 and heavier, or one of 5, 6, 7, 8 and lighter. Next put 1, 2, 5 on the left and 3, 4, 6 on the right. Now there are three possibilities:

• If the scale remains unbalanced to the left, the odd egg is 1 or 2 and heavier, or 6 and lighter. Then compare 1 with 2.

• If the scale gets unbalanced to the right, then the odd egg is 5 and lighter, or 3 or 4 and heavier. Then compare 3 with 4.

• If the scale gets balanced, then the odd egg is 7 or 8 and lighter. Compare them with each other.
Check out http://einstein.et.tudelft.nl/~arlet/puzzles/logic.html for a more general solution.

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