How do I solve without using calculator How many times [math]f(n)[/math] is an integer for [math] 2001\leq n\leq 2099[/math] where [math]f(n)=\frac{n^6}{6!}+\frac{n^5}{5!}+\frac{n^4}{4!}+\frac{n^3}{3!}+\frac{n^2}{2!}+n+1[/math] ?
valid question