Problem 5. Consider four consecutive numbers $n, n+1, n+2, n+3$. For which $n$ is the LCM of the first three numbers greater than the LCM of the last three?
Answer. Any odd number not less than 5.

Solution. Consider the triplet of numbers $n, n+1, n+2$. Any two adjacent numbers are coprime. If the numbers $n$ and $n+2$ have a common divisor, then it is a divisor of their difference, 2. Thus, if the number $n$ is odd, then all three numbers $n, n+1, n+2$ are pairwise coprime, so $\operatorname{HOK}(n, n+1, n+2) = n(n+1)(n+2)$. If $n$ is even, then $\operatorname{HOK}(n, n+1, n+2) = n(n+1)(n+2) / 2$. Clearly, $\operatorname{HOK}(n+1, n+2, n+3)$ can be less than $\operatorname{HOK}(n, n+1, n+2)$ only if $n$ is odd. The inequality takes the form

$$
\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)>\frac{(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)}{2}
$$

or after simplification $2 n > n + 3$. Finally, $n > 3$, hence due to the oddness of $n \geq 5$.

Criteria. 2 points for calculating the LCM for the case of odd $n$, 3 points for the case of even $n$. 7 points for a complete solution. 0 points for individual examples.