Marinada '21

[lang=hr] Neka je $N=(2\cdot 3 \cdot 5\cdot 7\cdot 11)^{100}$. Neka su $a_1, a_2, \ldots, a_{2021}$ prirodni brojevi. Za $j \in \{1, 2, \ldots, 2021\}$, neka je $b_j$ najveći zajednički djelitelj brojeva $a_1, a_2, \ldots, a_{j-1}, a_{j+1}, \ldots, a_{2021}$. Pretpostavimo da je $$b_1+b_2+\ldots+b_{2021}=N.$$ Neka je $X$ najmanja vrijednost koju može poprimiti izraz $a_1+a_2+\ldots+a_{2021}$. Odredi $\frac{X}{N}$. [/lang] [lang=en] Let $N=(2\cdot 3 \cdot 5\cdot 7\cdot 11)^{100}$. Let $a_1, a_2, \ldots, a_{2021}$ be positive integers. For $j \in \{1, 2, \ldots, 2021\}$, let $b_j$ be the greatest common divisor of $a_1, a_2, \ldots, a_{j-1}, a_{j+1}, \ldots, a_{2021}$. Suppose that $$b_1+b_2+\ldots+b_{2021}=N.$$ Let $X$ be the least possible value that $a_1+a_2+\ldots+a_{2021}$ can obtain. Find $\frac{X}{N}$. [/lang]