Difference between revisions of "2013 AIME I Problems/Problem 8"

Latest revision as of 19:07, 24 January 2021

Problem

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 $m>1$ . Find the remainder when the smallest possible sum $m+n$ is divided by 1000.

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 $\boxed{371}$ .

@@ Line 1: / Line 1: @@
-== Problem 8 ==
+== Problem ==
-The domain of the function f(x) = arcsin(log<math>_{m}</math>(''nx'')) is a closed interval of length <math>\frac{1}{2013}</math> , where ''m'' and ''n'' are positive integers and ''m'' > 1. Find the remainder when the smallest possible sum ''m'' + ''n'' is divided by 1000.
+The domain of the function <math>f(x) = \arcsin(\log_{m}(nx))</math> is a closed interval of length <math>\frac{1}{2013}</math> , where <math>m</math> and <math>n</math> are positive integers and <math>m>1</math>. Find the remainder when the smallest possible sum <math>m+n</math> is divided by 1000.
+== Solution 1==
-== Solution ==
+We know that the domain of <math>\text{arcsin}</math> is <math>[-1, 1]</math>, so <math>-1 \le \log_m nx \le 1</math>. Now we can apply the definition of logarithms:
-The domain of the arcsin function is [-1, 1], so -1 <math>\le</math> log<math>_{m}</math>(''nx'') <math>\le</math> 1.
+<cmath>m^{-1} = \frac1m \le nx \le m</cmath> <cmath>\implies \frac{1}{mn} \le x \le \frac{m}{n}</cmath>
+Since the domain of <math>f(x)</math> has length <math>\frac{1}{2013}</math>, we have that
+<cmath>\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}</cmath> <cmath>\implies \frac{m^2 - 1}{mn} = \frac{1}{2013}</cmath>
-<math>\frac{1}{m} \le nx \le m</math>
+A larger value of <math>m</math> will also result in a larger value of <math>n</math> since <math> \frac{m^2 - 1}{mn} \approx  \frac{m^2}{mn}=\frac{m}{n}</math> meaning <math>m</math> and <math>n</math> increase about linearly for large <math>m</math> and <math>n</math>. So we want to find the smallest value of <math>m</math> that also results in an integer value of <math>n</math>. The problem states that <math>m > 1</math>. Thus, first we try <math>m = 2</math>:
+<cmath>\frac{3}{2n} = \frac{1}{2013} \implies 2n = 3 \cdot 2013 \implies n \notin \mathbb{Z} </cmath>
+Now, we try <math>m=3</math>:
+<cmath>\frac{8}{3n} = \frac{1}{2013} \implies 3n = 8 \cdot 2013 \implies n = 8 \cdot 671 = 5368</cmath>
+Since <math>m=3</math> is the smallest value of <math>m</math> that results in an integral <math>n</math> value, we have minimized <math>m+n</math>, which is <math>5368 + 3 = 5371 \equiv \boxed{371} \pmod{1000}</math>.
-<math>\frac{1}{mn} \le x \le \frac{m}{n}</math>
+==Solution 2==
+We start with the same method as above. The domain of the arcsin function is <math>[-1, 1]</math>, so <math>-1 \le \log_{m}(nx) \le 1</math>.
-<math>\frac{1}{mn} - \frac{m}{n} = \frac{1}{2013}</math>
+<cmath>\frac{1}{m} \le nx \le m</cmath> <cmath>\frac{1}{mn} \le x \le \frac{m}{n}</cmath> <cmath>\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}</cmath> <cmath>n = 2013m - \frac{2013}{m}</cmath>
+For <math>n</math> to be an integer, <math>m</math> must divide <math>2013</math>, and <math>m > 1</math>. To minimize <math>n</math>, <math>m</math> should be as small as possible because increasing <math>m</math> will decrease <math>\frac{2013}{m}</math>, the amount you are subtracting, and increase <math>2013m</math>, the amount you are adding; this also leads to a small <math>n</math> which clearly minimizes <math>m+n</math>.
+We let <math>m</math> equal <math>3</math>, the smallest factor of <math>2013</math> that isn't <math>1</math>. Then we have <math>n = 2013*3 - \frac{2013}{3} = 6039 - 671 = 5368</math>
+<math>m + n = 5371</math>, so the answer is <math>\boxed{371}</math>.
+==Solution 3 (Operation Quadratics)==
+Note that we need <math>-1\le f(x)\le 1</math>, and this eventually gets to <math>\frac{m^2-1}{mn}=\frac{1}{2013}</math>. From there, break out the quadratic formula and note that <cmath>m= \frac{n+\sqrt{n^2+4026^2}}{2013\times 2}.</cmath> Then we realize that the square root, call it <math>a</math>, must be an integer. Then <math>(a-n)(a+n)=4026^2.</math>
+Observe carefully that <math>4026^2 = 2\times 2\times 3\times 3\times 11\times 11\times 61\times 61</math>! It is not difficult to see that to minimize the sum, we want to minimize <math>n</math> as much as possible. Seeing that <math>2a</math> is even, we note that a <math>2</math> belongs in each factor. Now, since we want to minimize <math>a</math> to minimize <math>n</math>, we want to distribute the factors so that their ratio is as small as possible (sum is thus minimum). The smallest allocation of <math>2, 61, 61</math> and <math>2, 11, 3, 3, 11</math> fails; the next best is <math>2, 61, 11, 3, 3</math> and <math>2, 61, 11</math>, in which <math>a=6710</math> and <math>n=5368</math>. That is our best solution, upon which we see that <math>m=3</math>, thus <math>\boxed{371}</math>.
+== See also ==
+{{AIME box|year=2013|n=I|num-b=7|num-a=9}}
+{{MAA Notice}}

2013 AIME I (Problems • Answer Key • Resources)
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