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King Hiero has 11 metal ingots that are indistinguishable in appearance; the king knows that their weights (in some order) are $1, 2, \ldots, 11$ kg. He also has a bag that will tear if more than 11 kg is placed in it.

Archimedes has learned the weights of all the ingots and wants to prove to Hiero that the first ingot weighs

1 kg. In one step, he can load several ingots into the bag and demonstrate to Hiero that the bag does not tear (the bag cannot be torn!). What is the minimum number of bag loadings Archimedes can use to achieve

the required result?
Let Archimedes first put ingots weighing 1, 2, 3, and 5 kg into the bag, and then put ingots weighing 1, 4, and 6 kg. In both cases, the bag does not tear.

We will prove that this could only happen if the 1 kg ingot was used twice. Indeed, if Archimedes used ingots weighing \( w_{1}, \ldots, w_{6} \) kg instead of ingots weighing 1, ..., 6 kg in these two cases, then these weights would satisfy the system of inequalities

\[ w_{1} + w_{2} + w_{3} + w_{5} \leq 11, \quad w_{1} + w_{4} + w_{6} \leq 11. \]

Adding these inequalities, we get

\[ w_{1} + (w_{1} + w_{2} + \ldots + w_{6}) \leq 22. \]

The sum in parentheses is the sum of six different natural numbers, which is at least \( 1 + 2 + \ldots + 6 = 21 \). Therefore, it follows that

\[ w_{1} \leq 22 - 21 = 1. \]

Thus, \( w_{1} = 1 \), meaning the 1 kg ingot is uniquely determined.

It remains to show that one loading is insufficient. If Archimedes loads only one ingot, the bag will not tear in any case, so no ingot can be identified. Suppose Archimedes loads more than one ingot, and the bag does not tear. If the 1 kg ingot is not loaded into the bag, then replacing it with any ingot in the bag will not change the result; in this case, Hiero will not be able to determine whether this ingot is in the bag. If the required ingot is in the bag, then Hiero will not be able to determine which of the (at least two) loaded ingots is the required one.

## Answer

In two loadings.