The function $f(n)=\left\lfloor\dfrac{n}{1!}\right\rfloor+\dots+\left\lfloor\dfrac{n}{10!}\right\rfloor$ is non-decreasing and f(1000) < 2012 < f(2000). Since $2000-1000<2^{10}$, you can find n for which f(n) = 2012 in 10 iterations using binary search, or bisection method.