2013 AIME I Problem 8

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

The domain requires

-1 \le \log_m(nx) \le 1,

so

x

runs over

\left[\frac{1}{mn}, \frac{m}{n}\right],

which has length

\frac{m^2 - 1}{mn}

Then

mn = 2013(m^2 - 1).

Since

\gcd(m, m^2 - 1) = 1,

m

must divide

2013,

and the smallest

m \gt 1

minimizes the sum.

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.

Problem 8 in other years