Find the smallest positive integer $j \geq 1$ such that for every polynomial $p(x)$ with integer coefficients and every integer $a$, the integer $p^{(j)}(a)$ is divisible by $2016$.
(Here $p^{(j)}(a)$ denotes the $j$-th derivative of $p$ evaluated at $a$.)
We have $j = 8$. We have that $2016 = 2048 − 32 = 32(64 − 1) = 2^5 3^2 5$.
Given a polynomial $p(x)$, if the monomial $a_kx^k$ appears in $p$ and $k \geq 8$, then after taking $8$ derivatives it will become the monomial $a_k \times k \times (k-1) \times \dots \times (k-7)x^{k-8}$. Considering $k, k-1, \dots, k-7$ modulo $3$, $7$, and $8$, we conclude that $2^5 3^2 7$ divides $a_k \cdot k \cdot (k-1) \cdot \dots \cdot (k-7) \cdot n^{k-8}$. Any term of $p$ of degree less than $8$ will disappear after taking $8$ derivatives. Thus $2016$ divides $p^{(8)}(n)$ for all $n$, as desired.
For $k \leq 7$, we have that $(x^k)^{(k)} = k!$ and the number of times $k!$ is divisible by $2$ is at most that of $7!$. But $7!$ is divisible by $2$ only $4$ times, so $2016$ does not divide $7!$. So by choosing $p(x) = x^k$, we may show that no smaller integer works. Thus $j=8$.
Special thanks to Olivier Massicot, Brian Zhang, Ritoprovo Roy, Charles D.G. Burns, Tanay Saha, Chandan Relekar, Andrew Glatten, and Roei Levy Israel for also submitting solutions for this challenge problem.