1. We will divide the integers from 1 to 9 into three groups of three and then multiply the numbers within each group.

a) Determine these three products, given that two of them are equal and smaller than the third product.

b) Assume that one of the three products, denoted as $S$, is smaller than the other two products (which may be the same). Find the largest possible value of $S$.
1. First, we express the product of all nine numbers using its prime factorization:

$$
1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9=2^{7} \cdot 3^{4} \cdot 5 \cdot 7
$$

a) Let's denote two of the considered (different) products as $S$ and $Q$, with $S<Q$. From the equality

$$
S \cdot S \cdot Q=2^{7} \cdot 3^{4} \cdot 5 \cdot 7
$$

we can see that the prime numbers 5 and 7 must be present in the product $Q$, so $Q=5 \cdot 7 \cdot x = 35x$, where $x$ is one of the remaining numbers $1, 2, 3, 4, 6, 8,$ and $9$. Furthermore, we see that in the factorization of the relevant $x$, the prime number 2 must have an odd exponent and the prime number 3 must have an even exponent - this is satisfied only by the numbers 2 and 8. For $x=2$, however, we get $Q=35 \cdot 2=70 < S=2^{3} \cdot 3^{2}=72$, which contradicts the assumption $S<Q$. Therefore, it must be $x=8$, for which we get $Q=35 \cdot 8=280$ and $S^{2}=2^{4} \cdot 3^{4}$, or $S=2^{2} \cdot 3^{2}=36$. The triplet of products is thus $(36, 36, 280)$.

It remains to show that the obtained triplet actually corresponds to a partition of the given nine numbers into triplets:

$$
S=1 \cdot 4 \cdot 9=36, \quad S=2 \cdot 3 \cdot 6=36, \quad Q=5 \cdot 7 \cdot 8=280
$$

b) Let's denote the considered products as $S, Q,$ and $R$, where $S<Q$ and $S<R$ (it is not excluded that $Q=R$). In the solution to part a), we found that the equality

$$
S \cdot Q \cdot R=70 \cdot 72 \cdot 72
$$

holds. If we show that there exists a partition of the numbers such that $S=70$ and $R=Q=72$, then $S=70$ will be the largest value sought, because if for some partition $S \geq 71$, it must be $R \geq 72$ and $Q \geq 72$, and thus $S \cdot Q \cdot R \geq 71 \cdot 72 \cdot 72$, which clearly contradicts the previous equality. Finding the required partition is easy:

$$
S=2 \cdot 5 \cdot 7=70, \quad Q=1 \cdot 8 \cdot 9=72, \quad R=3 \cdot 4 \cdot 6=72
$$

For a systematic and complete solution to part a), award 3 points, of which only 1 point for a random, unexplained finding of the result. For a systematic and complete solution to part b), award 3 points, of which only 1 point for a random, unexplained finding of the result.