AIME 2013 I · 第 8 题

Solution 1

We know that the domain of $\text{arcsin}$ is $[-1, 1]$, so $-1 \le \log_m nx \le 1$. Now we can apply the definition of logarithms:

$$m^{-1} = \frac1m \le nx \le m$$ $$\implies \frac{1}{mn} \le x \le \frac{m}{n}$$ Since the domain of $f(x)$ has length $\frac{1}{2013}$, we have that

$$\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}$$ $$\implies \frac{m^2 - 1}{mn} = \frac{1}{2013}$$ A larger value of $m$ will also result in a larger value of $n$ since $\frac{m^2 - 1}{mn} \approx \frac{m^2}{mn}=\frac{m}{n}$ meaning $m$ and $n$ increase about linearly for large $m$ and $n$. So we want to find the smallest value of $m$ that also results in an integer value of $n$. The problem states that $m > 1$. Thus, first we try $m = 2$:

$$\frac{3}{2n} = \frac{1}{2013} \implies 2n = 3 \cdot 2013 \implies n \notin \mathbb{Z}$$ Now, we try $m=3$:

$$\frac{8}{3n} = \frac{1}{2013} \implies 3n = 8 \cdot 2013 \implies n = 8 \cdot 671 = 5368$$ Since $m=3$ is the smallest value of $m$ that results in an integral $n$ value, we have minimized $m+n$, which is $5368 + 3 = 5371 \equiv \boxed{371} \pmod{1000}$.

Solution 2

We start with the same method as above. The domain of the arcsin function is $[-1, 1]$, so $-1 \le \log_{m}(nx) \le 1$.

$$\frac{1}{m} \le nx \le m$$ $$\frac{1}{mn} \le x \le \frac{m}{n}$$ $$\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}$$ $$n = 2013m - \frac{2013}{m}$$ For $n$ to be an integer, $m$ must divide $2013$, and $m > 1$. To minimize $n$, $m$ should be as small as possible because increasing $m$ will decrease $\frac{2013}{m}$, the amount you are subtracting, and increase $2013m$, the amount you are adding; this also leads to a small $n$ which clearly minimizes $m+n$.

We let $m$ equal $3$, the smallest factor of $2013$ that isn't $1$. Then we have $n = 2013*3 - \frac{2013}{3} = 6039 - 671 = 5368$

$m + n = 5371$, so the answer is $\boxed{371}$.

Solution 3 (Operation Quadratics)

Note that we need $-1\le f(x)\le 1$, and this eventually gets to $\frac{m^2-1}{mn}=\frac{1}{2013}$. From there, break out the quadratic formula and note that

$$m= \frac{n+\sqrt{n^2+4026^2}}{2013\times 2}.$$ Then we realize that the square root, call it $a$, must be an integer. Then $(a-n)(a+n)=4026^2.$

Observe carefully that $4026^2 = 2\times 2\times 3\times 3\times 11\times 11\times 61\times 61$! It is not difficult to see that to minimize the sum, we want to minimize $n$ as much as possible. Seeing that $2a$ is even, we note that a $2$ belongs in each factor. Now, since we want to minimize $a$ to minimize $n$, we want to distribute the factors so that their ratio is as small as possible (sum is thus minimum). The smallest allocation of $2, 61, 61$ and $2, 11, 3, 3, 11$ fails; the next best is $2, 61, 11, 3, 3$ and $2, 61, 11$, in which $a=6710$ and $n=5368$. That is our best solution, upon which we see that $m=3$, thus $\boxed{371}$.

Solution 4 (Factor! and slightly cheese)

By considering the range of $\arcsin$, we obtain $-1 \leq\log_{m} (nx) \leq 1 \Rightarrow \frac{1}{mn} \leq x \leq \frac{m}{n}$. Hence we consider the difference: $\frac{m}{n} - \frac{1}{mn} = \frac{m^2 - 1}{mn} = \frac{1}{2013}$. Now just cross multiply to obtain $2013(m-1)(m+1) = mn$, and that $2013 = 3 \cdot 11 \cdot 61$. We could try a few cases: $m = 3$, for instance, yields $n = 671 \cdot 8 = 5368$. Adding $3$ gives $5371$. Now, if we consider greater $m$, we see that the term $(m-1)(m+1)$ grows very fast: this is also due to the fact that $61,11 \gg 3$. Hence, we may just conclude that the answer is $5371 \Rightarrow \boxed{371}$.