UkraineMO

Algebra Difficulty 4.8 AIME Prove it Ukraine

Compare the following numbers: $A = 11$ , $B = \log_2 3 \cdot \log_3 4 \cdot \log_4 5 \cdots \log_{2015} 2016$ and $C = \log_3 2 \cdot \log_4 3 \cdot \log_5 4 \cdots \log_{2016} 2015$ .

Solution

Obviously, for any integer $n > 0$ $\log_{n+1} n < 1$ . Hence, $C < 1$ . Also we have that $B \cdot C = 1$ , so $B > 1$ , moreover
$B = \frac{\lg 3}{\lg 2} \cdot \frac{\lg 4}{\lg 3} \cdot \frac{\lg 5}{\lg 4} \cdots \frac{\lg 2016}{\lg 2015} = \frac{\lg 2016}{\lg 2} < 11 = A \Leftrightarrow \lg 2016 < 11 \lg 2 = \lg 2^{11} \Leftrightarrow 2016 < 2^{11} = 2048.$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.