If you know xmodc1,xmodc2,...,xmodcn then you can find out xmodlcm(c1,c2,...,cn) if the given values are not self-contradictory(eg. xmod2=0 and xmod4=1). Conversely, if you know xmodlcm(c1,c2,...,cn) you can find each of the given n values. So first find xmodlcm(c1,c2,...,cn) and this would retain all the information. Now, if k divides lcm(x1,x2,...,xn) you can easily find xmodk otherwise you can't.

The editorial proves that all you have to do is check whether $$$k \mid \text{lcm}(c_1, \dots, c_n)$$$. This is equivalent to saying $$$\gcd(k, \text{lcm}(c_1, \dots, c_n)) = k$$$. But we can calculate the left side very easily since $$$\gcd(k, \text{lcm}(c_1, \dots, c_n)) = \text{lcm}(\gcd(k, c_1), \dots, \gcd(k, c_n))$$$. The time complexity will be $$$O(n(\log(C) + \log(K)))$$$. Also there will not be overflow because the intermediate value will always be $$$\leq k$$$.

These identities hold because if you think of an integer's prime factorization as a multiset, $$$\mid$$$ corresponds to $$$\subseteq$$$, $$$\gcd$$$ corresponds to $$$\cap$$$, and $$$\text{lcm}$$$ corresponds to $$$\cup$$$