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Auto: EEvoookioov M.A.

The tsar has 7 bags with gold coins, each containing 100 coins. The tsar remembers that in one bag all the coins weigh 7 g, in the second 8 g, in the third 9 g, in the fourth 10 g, in the fifth 11 g, in the sixth 12 g, and in the seventh 13 g, but he does not remember which is which.

The tsar informed the court sage and pointed to one of the bags. The sage can take any number of coins from this and other bags, but they all look the same. However, the sage has large two-pan scales without weights (they will accurately show if the weights on the pans are equal, and if not, which pan is heavier). Can the sage determine which coins are in the indicated bag with no more than two weighings?
Note that if we take one coin from each bag, their total weight will be $7+8+\ldots+13=70$ grams. Let's call such a set of coins a set.

Suppose the coins in the bag indicated by the king weigh $x$ grams each. If we take 70 such coins, their weight will be $70 x$ - the same as $x$ sets. That is, if the sage finds out how many sets are needed to balance these

70 coins, he will answer the king's question.

Let the first weighing compare the weight of these 70 coins and 10 sets. If the scales are in balance, then $x=10$ and the problem is solved; if the coins outweigh, then $x>10$; if the sets outweigh, then $x<10$. If $x>10$, then in one weighing he can easily find out whether the coins weigh 11, 12, or 13 grams each. Indeed, now he can compare the weight of 70 coins and 12 sets. If the coins outweigh, then $x=13$; if the scales are in balance, then $x=12$; if the sets outweigh, then $x=11$.

Similarly, the sage can proceed in the case of $x<10$: the coins then may weigh 7, 8, or 9 grams each, so it remains to compare the weight of 70 coins and 8 sets.

Note that in each weighing, the sage uses no more than $70+12$ coins from the bag indicated by the king, and no more than 12 coins from each of the other bags. So there will be enough coins in the bags.

## Answer

Yes, he can.