Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
 
(13 intermediate revisions by 10 users not shown)
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 <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.
+
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 <math>[-1, 1]</math>, so <math>-1 \le log_{m}(nx) \le 1</math>.
+
<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{m}{n} - \frac{1}{mn} = \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>
  
−
<math>n = 2013m - \frac{2013}{m}</math>
+
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>.
  
−
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>m</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>
  
−
We let <math>m</math> equal 3, the smallest factor of <math>2013</math> that isn't <math>1</math>.. <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>.
  
−
<math>m + n = 5371</math>, so the answer is <math>\boxed{371}</math>.
+
== See also ==
 +
{{AIME box|year=2013|n=I|num-b=7|num-a=9}}
 +
{{MAA Notice}}

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 $\boxed{371}$.

See also

2013 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_8&oldid=143132"