2017 AIME II Problems/Problem 8
Revision as of 11:36, 23 March 2017 by The turtle (talk | contribs) (Created page with "<math>\textbf{Problem 8}</math> Find the number of positive integers <math>n</math> less than <math>2017</math> such that <cmath>1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4...")
Find the number of positive integers less than such that is an integer.