Problem:
The domain of the function f(x)=arcsin(logm(nx)) is a closed interval of length 20131, where m and n are positive integers and m>1. Find the remainder when the smallest possible sum m+n is divided by 1000.
Solution:
The domain of f(x) is the solution of the inequality −1≤logm(nx)≤1, or equivalently, mn1≤x≤nm. Thus the domain is a closed interval of length 20131=nm−mn1=mnm2−1. Hence n=m2013(m2−1). Because m and m2−1 are relatively prime, m must be a factor of 2013=3⋅11⋅61. The smallest possible value of n is 32013(32−1)=5368, and the smallest possible sum m+n is 5368+3=5371. The requested remainder is 371.
The problems on this page are the property of the MAA's American Mathematics Competitions