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 "2023 AMC 12B Problems/Problem 7"

(Created blank page)
 
Line 1: Line 1:
 +
==Solution==
  
 +
We have
 +
<cmath>
 +
\begin{align*}
 +
\sqrt{\frac{\log \left( n^2 \right) - \left( \log n \right)^2}{\log n - 3}}
 +
& = \sqrt{\frac{2 \log n - \left( \log n \right)^2}{\log n - 3}} \\
 +
& = \sqrt{\frac{\left( \log n \right) \left( 2 - \log n\right)}{\log n - 3}} .
 +
\end{align*}
 +
</cmath>
 +
 +
Because <math>n</math> is an integer and <math>\log n</math> is well defined, <math>n</math> must be a positive integer.
 +
 +
Case 1: <math>n = 1</math> or <math>10^2</math>.
 +
 +
The above expression is 0. So these are valid solutions.
 +
 +
Case 2: <math>n \neq 1, 10^2</math>.
 +
 +
Thus, <math>\log n > 0</math> and <math>2 - \log n \neq 0</math>.
 +
To make the above expression real, we must have <math>2 < \log n < 3</math>.
 +
Thus, <math>100 < n < 1000</math>.
 +
Thus, <math>101 \leq n \leq 999</math>.
 +
Hence, the number of solutions in this case is 899.
 +
 +
Putting all cases together, the total number of solutions is
 +
\boxed{\textbf{(E) 901}}.
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Revision as of 18:27, 15 November 2023

Solution

We have \begin{align*} \sqrt{\frac{\log \left( n^2 \right) - \left( \log n \right)^2}{\log n - 3}} & = \sqrt{\frac{2 \log n - \left( \log n \right)^2}{\log n - 3}} \\ & = \sqrt{\frac{\left( \log n \right) \left( 2 - \log n\right)}{\log n - 3}} . \end{align*}

Because $n$ is an integer and $\log n$ is well defined, $n$ must be a positive integer.

Case 1: $n = 1$ or $10^2$.

The above expression is 0. So these are valid solutions.

Case 2: $n \neq 1, 10^2$.

Thus, $\log n > 0$ and $2 - \log n \neq 0$. To make the above expression real, we must have $2 < \log n < 3$. Thus, $100 < n < 1000$. Thus, $101 \leq n \leq 999$. Hence, the number of solutions in this case is 899.

Putting all cases together, the total number of solutions is \boxed{\textbf{(E) 901}}.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12B_Problems/Problem_7&oldid=203359"