Marinada '24

[lang=hr] Suma $$\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+\frac{5}{3!+4!+5!}+...+\frac{n}{(n-2)!+(n-1)!+(n)!}$$ može se zapisati kao $1/2-1/f(n)$, gdje je $f(n)$ funkcija u $n$. Pokazuje se da je $f(n)$ cjelobrojan, a mi vas pitamo da napišete najmanji $n$ takav da je $f(n)$ djeljiv s $2024$. Ako takav ne postoji, vratite $-1$. [/lang] [lang=en] The sum $$\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+\frac{5}{3!+4!+5!}+...+\frac{n}{(n-2)!+(n-1)!+(n)!}$$ can be written as $1/2-1/f(n)$, where $f(n)$ is a function of $n$. It has been shown that $f(n)$ is an integer, and we ask you to find the smallest $n$ such that $f(n)$ is divisible by $2024$. If no such $n$ exists, return $-1$. [/lang]