Question

The domain of the function $f(x) = \arcsin(\log_m(nx))$ is a closed interval of length $\frac{1}{2013},$ where $m$ and $n$ are positive integers and $m \gt 1.$ Find the remainder when the smallest possible sum $m + n$ is divided by $1000.$

Answer

Solution:

The function is defined when $-1 \le \log_m(nx) \le 1,$ that is $\frac{1}{m} \le nx \le m,$ so the domain is $\left[\frac{1}{mn}, \frac{m}{n}\right],$ with length $\frac{m}{n} - \frac{1}{mn} = \frac{m^2 - 1}{mn} = \frac{1}{2013}.$ Hence $n = \frac{2013(m^2 - 1)}{m}.$ Since $m$ is relatively prime to $m^2 - 1,$ $m$ must divide $2013 = 3 \cdot 11 \cdot 61.$

Because $n \approx 2013m,$ the sum $m + n$ grows with $m,$ so take the smallest factor $m = 3:$ then $n = \frac{2013 \cdot 8}{3} = 5368$ and $m + n = 5371.$

The remainder upon division by $1000$ is $371.$

2013 AIME I Problem 8

Solution:

Problem 8 in Other Years