2020 MEXT Japanese Government Scholarship Undergraduate Students Natural Sciences Qualifying Examination Mathematics (B): Problem 1(1)

This problem appears at the Qualifying Examinations for Applicants for Japanese Government (MEXT) Scholarships 2020. There are two mathematics exams: one for biology-related natural sciences (Mathematics A), and another for physics- and engineering-related natural sciences (Mathematics B). This problem is from the 2020 Mathematics (B) questionnaire. The official answer key is here.

Problem 1(1)

The largest one among natural numbers that are less than $$ \log_2{3}\cdot\log_3{4}\cdot\log_4{5}\cdots\cdots\log_{2019}{2020} $$ is $ \fbox { A } $.

Solution

The key idea is the following relationship that holds for positive numbers $p, q,$ and $b$. $$ \log_p{q} = \frac{\log_b{q}}{\log_b{p}}, $$ where $p\neq1$ and $b\neq1$.

For convenience, let $$ L \equiv \log_2{3}\cdot\log_3{4}\cdot\log_4{5}\cdots\cdots\log_{2019}{2020}. $$ Selecting $b=e$, we obtain $$\begin{align} L &=\frac{\cancel{\log{3}}}{\log{2}}\cdot \frac{\cancel{\log{4}}}{\cancel{\log{3}}}\cdot \frac{\cancel{\log{5}}}{\cancel{\log{4}}} \cdots\cdots \frac{\log{2020}}{\cancel{\log{2019}}}\\ &=\frac{\log{2020}}{\log{2}}\\ &=\log_2{2020}, \end{align}$$ where $\log$ is the natural logarithm (base $e$). This is equivalent to $$ 2^L = 2020. $$

Noticing that $$\begin{align} 2^{10} &= 1024 \\ 2^{11} &=2048, \end{align}$$ we obtain $$\begin{align} 1024 \leq 2020 \leq 2048 &\Leftrightarrow 2^{10} \leq 2^L \leq 2^{11}\\ &\Leftrightarrow \log_2{2^{10}} \leq \log_2{2^L} \leq \log_2{2^{11}}\\ &\Leftrightarrow 10 \leq L \leq 11. \end{align}$$ because the exponential and logarithmic functions are both monotonic and increasing. Thus, the largest natural number less than $L$ is $10$. ■