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

Revision as of 20:26, 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)

2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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

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 ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
+==See Also==
+{{AMC12 box|year=2023|ab=B|num-b=6|num-a=8}}
+{{MAA Notice}}

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

Revision as of 20:26, 15 November 2023

Solution

See Also