i think this is the answer but i cant quite prove it

My solution: We will show Nora has a winning strategy; that is, she can find a suitable positive integer $n$, so that for any $p_1<p_2<....<p_{2015}$, the numbers $n+p_1, n+p_1p_2, ... n +p_1p_2...p_{2015}$ are each relatively prime to $m=p_1p_2...p_{2015}$. It suffices to determine $n$ mod each $p_i, 1 \le i \le 2015$, and then use CRT. Now modulo each $p_i$, we need $n, n+p_1, n+p_1p_2, ... n+p_1p_2...p_{i-1} \neq 0$ mod $p_i$. Let $X$ be the set of all residues relatively prime to $p_i$ (taken mod $p_i$). Essentially, we need $X \cap \{X-p_1\} \cap \{ X-p_1p_2\} \cap ... \cap \{X-p_1p_2...p_{i-1}\} \neq \emptyset$, where $\{X-a\}$ means subtracting $a$ from each element of $X$ and then take mod $p$. Note that $|\{X-a\}| = p_i-1$ for each $a$. It follows that $|S \cap \{X- a\}| \ge |S|-1$ for each $a,S$ as we can only "lose" one residue each time. Then $|X \cap \{X-p_1\} \cap \{X-p_2\} \cap ... \cap \{X-p_1p_2...p_{i-1} \} | \ge |X| - (i-1) = p_i-i$. Showing this is $\ge 1$ is equivalent to showing $p_i \ge i+1$ which is trivial. Thus we are done. One question- is the Hong Kong TST2 2016 only four questions, or have problems 5/6 not been released yet? If this is the final one out of four problems it seems quite easy...