5. Given the natural numbers $1,2,3 \ldots, 10,11,12$. Divide them into two groups such that the quotient of the product of all numbers in the first group divided by the product of all numbers in the second group is an integer, and takes the smallest possible value. What is this quotient?
Solution. Let's factorize the given numbers into prime factors and find their product.

$$
1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 10 \cdot 11 \cdot 12=2^{10} \cdot 3^{5} \cdot 5^{2} \cdot 7 \cdot 11
$$

The factors 7 and 11 do not have pairs. One of the factors 3 does not have a pair. Therefore, to make the quotient an integer, these prime factors without pairs must be placed in the numerator. The quotient cannot be less than what these three irreducible factors provide. The remaining threes, fives, and twos should be paired in the numerator and denominator so that the pairs cancel out. One way to do this is written below

$$
\frac{7 \cdot 11 \cdot 1 \cdot 2 \cdot 3 \cdot 8 \cdot 9 \cdot 10}{4 \cdot 5 \cdot 6 \cdot 12}=231
$$

Answer. 231.