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

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<cmath>\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}</cmath> <cmath>\implies \frac{m^2 - 1}{mn} = \frac{1}{2013}</cmath>
 
<cmath>\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}</cmath> <cmath>\implies \frac{m^2 - 1}{mn} = \frac{1}{2013}</cmath>
  
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A larger value of <math>m</math> will also result in a larger value of <math>n</math> since the numerator increases faster than the denominator, 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>:
+
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>   
 
<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>:
 
Now, we try <math>m=3</math>:

Revision as of 08:52, 2 September 2017

Problem 8

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

We know that the domain of $\arcsin$ 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 $\boxed{371}$.

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

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