9.1. A father and son are carrying identical cans of preserves. The mass of each can is expressed as an integer number of grams, no less than 300 but no more than 400. The father is carrying 6 kg 500 g, and the son is carrying 2 kg 600 g. How many cans does the father have and how many does the son have?
Answer: The father has 20 cans, and the son has 8 cans.

Solution. The mass $m$ of each can is a common divisor of the numbers 6500 and 2600. Since the GCD $(6500 ; 2600)=2^{2} \cdot 5^{2} \cdot 13=1300$, any common divisor of the numbers 6500 and 2600 is a divisor of 1300. Let's start listing them in pairs, checking the condition $300 \leq m \leq$ 400: 1 and 1300 (both do not fit), 2 and 650 (both do not fit), 4 and 325 (325 fits), 5 and 260 (both do not fit). In the remaining pairs, both divisors are less than 260, so they do not satisfy the condition $300 \leq m \leq 400$. Thus, the father carries $6500: 325=20$ cans, and the son carries $2600: 325=8$ cans.

## Evaluation Criteria.

«+» A complete and well-reasoned solution is provided

«士» A generally correct reasoning and the correct answer are provided, but there are gaps in the justification of its uniqueness (for example, the selection of divisors of the number 1300 is not shown) «士» The mass of the can is correctly and reasonably found, but the question of the problem is not answered «Ғ» Only the correct answer is provided «-» The problem is not solved or is solved incorrectly