Petra had written down the natural numbers from 1 to 9. She added two of these numbers, erased them, and wrote down the resulting sum instead of the erased addends. She then had eight numbers, which she managed to divide into two groups with the same product.

Determine the largest possible value of this product.

(E. Novotná)

Hint. It pays to focus on prime numbers.
For comparing the products of the resulting groups of numbers, prime factors will be useful. To have the products of the numbers in both groups be the same, the total number of each individual prime factor must be even. If the number of some prime factors is odd, then dividing them into groups with the same product is not possible.

The prime numbers among the given numbers are only $2,3,5,7$, and 5 and 7 occur exactly once. Given the occurrence of 5 and 7, it suffices to consider three cases that could lead to a solution, which we will analyze step by step:

a) Petra added something to 5 to get a multiple of 7. She could have done this in two ways:

- She added $5+2$ and got the octet of numbers $1,3,4,6,7,7,8,9$. This octet can be divided into groups with the same product:

$$
1 \cdot 7 \cdot 8 \cdot 9=3 \cdot 4 \cdot 6 \cdot 7=2^{3} \cdot 3^{2} \cdot 7=504 \text {. }
$$

- She added $5+9$ and got the octet of numbers $1,2,3,4,6,7,8,14$. This octet can be divided into groups with the same product:

$$
1 \cdot 4 \cdot 6 \cdot 14=2 \cdot 3 \cdot 7 \cdot 8=2^{4} \cdot 3 \cdot 7=336
$$

b) Petra added something to 7 to get a multiple of 5. She could have done this in two ways:

- She added $7+3$ and got the octet of numbers $1,2,4,5,6,8,9,10$. This octet cannot be divided into groups with the same product because 3 appears in the prime factorizations of these numbers a total of three times.
- She added $7+8$ and got the octet of numbers $1,2,3,4,5,6,9,15$. This octet cannot be divided into groups with the same product because 3 appears in the prime factorizations of these numbers a total of five times.

c) Petra added $5+7$ and got the octet of numbers $1,2,3,4,6,8,9,12$. This octet cannot be divided into groups with the same product because 3 appears in the prime factorizations of these numbers a total of five times (or 2 appears nine times).

Petra could have obtained either the product 504 or 336. The largest possible product is therefore 504.

Notes. The division into groups with the same product in case a) is not the only possible one (for example, 1 can be anywhere).

The number of all pairs that can be formed from the given nine numbers is 36. With the initial observation, we have significantly reduced the number of cases to consider.

Part of the problem is also to describe the process leading to the correct answer. Found products without a proper explanation that no other possibilities exist cannot be rated at the highest level.