8.3. In the city, there are four jewelers to whom the tsar sent 13 bags of gold. The first bag contained one gold ingot, the second - two, the third - three, ..., the

thirteenth - 13 gold ingots. One of the bags was immediately lost somewhere, and
the remaining jewelers distributed them so that each received an equal number of gold ingots and an equal number of bags. In this process, the bag with one ingot went to the first jeweler, with three - to the second, with eleven - to the third. Determine which bags went to the fourth jeweler, if the ingots were not removed from the bags.
Answer: $2,9,10$

Solution: In total, there were $1+2+3+\ldots+13=91$ ingots. When one bag was lost, the remaining number of ingots became a multiple of 4. The number 91 gives a remainder of 3 when divided by 4, so the lost bag must have contained either 3, 7, or 11 ingots. From the condition, we know that the bags with 3 and 11 ingots are still in place, so the bag with 7 ingots was lost, leaving 84 ingots.

Thus, each jeweler received 21 ingots and 3 bags.

The first jeweler needs two bags with a total of 20 ingots (since he already has one bag with 1 ingot). This can be done in one of the following ways: $13+7$ (but 7 ingots were lost), $12+8,11+9$ (but 11 is with another jeweler). Therefore, the first jeweler has bags with 1, 8, and 12 ingots.

The second jeweler is short of 18 ingots to reach 21. This amount can be made with two bags in the following ways: $13+5,12+6$ (but 12 ingots are with the first jeweler), $11+7$ (but 7 ingots were lost), $10+8$ (but 8 ingots are with the first jeweler). Therefore, the second jeweler has bags with 3, 5, and 13 ingots.

The third jeweler is still short of 10 ingots to reach 21. This amount can be made with two bags in the following ways: $9+1$ (but 1 ingot is with the first jeweler), $8+2$ (but 8 ingots are with the first jeweler), $7+3$ (but 7 ingots were lost), $6+4$. Therefore, the third jeweler has bags with 4, 6, and 11 ingots.

Thus, only the bags with 2, 9, 10 ingots remain, which were given to the fourth jeweler.

Criteria: Only the answer - 0 points.

Example where for each jeweler it is specified which bag they received, with no other progress - 1 point. This point does not accumulate with the others.

Proven that the bag with 7 ingots was lost - 1 point.

Proven that a jeweler received specific bags - 2 points for each. These points accumulate with each other and the previous point.