HMMT November

Since $v_{p}(\operatorname{gcd}(a, b))=\min \left(v_{p}(a), v_{p}(b)\right)$ and $v_{p}(\operatorname{lcm}(a, b))=\max \left(v_{p}(a), v_{p}(b)\right)$, we may show the following: Claim. For any prime $p$ and non-negative integer $k$, the number of numbers $n$ on the board such that $v_{p}(n)=k$ doesn't change throughout this process. Let the 15 final numbers on the board be $a_{1} \leq a_{2} \leq a_{3} \cdots \leq a_{15}$. Note that $a_{i} \mid a_{j}$ for all $i<j$. For each prime $p$, let $X_{p, i}=v_{p}\left(a_{i}\right)$. Note that by the lemma, we have $$\begin{aligned} \left(X_{2,1}, X_{2,2}, \ldots, X_{2,15}\right) & =(0,0,0,0,0,0,0,0,1,1,1,1,2,2,3) \\ \left(X_{3,1}, X_{3,2}, \ldots, X_{3,15}\right) & =(0,0,0,0,0,0,0,0,0,0,1,1,1,1,2) \\ \left(X_{5,1}, X_{5,2}, \ldots, X_{5,15}\right) & =(0,0,0,0,0,0,0,0,0,0,0,0,1,1,1) \\ \left(X_{7,1}, X_{7,2}, \ldots, X_{7,15}\right) & =(0,0,0,0,0,0,0,0,0,0,0,0,0,1,1) \\ \left(X_{11,1}, X_{11,2}, \ldots, X_{11,15}\right) & =(0,0,0,0,0,0,0,0,0,0,0,0,0,0,1) \\ \left(X_{13,1}, X_{13,2}, \ldots, X_{13,15}\right) & =(0,0,0,0,0,0,0,0,0,0,0,0,0,0,1) \end{aligned}$$ Thus, since $a_{i}=\prod_{p} p^{X_{p, i}}$ for each $i$, so we get the 15 final numbers on the board are $$1,1,1,1,1,1,1,1,2,2,6,6,60,420, \text { and } 360360$$ Adding these up gives 360854 .