Archimedes knows this, Hiero doesn’t. At first Archimedes puts A, B, C, D in the bag then he exchanges D and E

Since the bags helds both times and (1l, 2l, 3l, 4l) and (1l, 2l, 3l, 5l) are the only sets of four which held, Hiero

knows that A, B, C weigh (1l, 2l, 3l) i(n) s(ome) o(rder), D and E weigh (4l, 5l) i.s.o. and F, G, H, I, J, K weigh

(6l, 7l, 8l, 9l, 10l, 11l) i.s.o.

The third try is A, D and F and the bag holds. Since D weighs at least 4g and F at least 6g Hiero knows that these

must be their exact weights and he now knows the weight of A, D, E and F

Bonus question:

There is no need for another weighing because Hiero already knows the weight of 4 ingots

At first I thought wording wasn’t precise, but it is: to identify four, it’s one more weighing *in addition* to what was done to identify one ingot. Not a separate set of completely new N+1 weighings.

But yeah, cute puzzle, I can ID ingot 1 in two weighings, and then 4,5,6 in one more weighing. ]]>

First try: as before, we put four ingots in the bag. As the bag survives, we have either (1,2,3,4) or (1,2,3,5).

Second try :Archimedes removes 3 ingots from the bag, which he knows to be the three heaviest, and he replaces them by just one ingot: the 10 libra ingot. Thus the bag contains 2 ingots (x,10).

Since the bag survives, x=1: Archimedes draws the bag x=1 from the bag and presents it successfully to the King as the 1 libra ingot.

Only 2 tries have been required to identify the 1 libra ingot. ]]>

1st weighing:1,2,3,4. The bag is safe.

Archimedes removes one of the previous four ingots, and replaces it by 5: that is the second weighing.

As the bag is still alive, this proves that the ingot removed has a weight of 4 libras.

Indeed, if 3 were removed the second weighing would sum up to 1+2+4+5=12, and the bag would break up. Same conclusion if ingot 2 libras or 1 libra were removed after 1st weighing.

So the 4 libra ingot has been identified.

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