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Difference between revisions of "2023 AMC 12B Problems/Problem 7"

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

Revision as of 17:27, 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)

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