HMMT November, 2022

Algebra Difficulty 5.0 AIME, harder Prove it United States

Problem:

Real numbers $x$ and $y$ satisfy the following equations:
$\begin{aligned} x & = \log_{10}\left(10^{y-1}+1\right)-1 \\ y & = \log_{10}\left(10^{x}+1\right)-1 \end{aligned}$
Compute $10^{x-y}$ .

Solution

Solution:

Taking 10 to the power of both sides in each equation, these equations become:
$\begin{aligned} & 10^{x} = \left(10^{y-1} + 1\right) \cdot 10^{-1} \\ & 10^{y} = \left(10^{x} + 1\right) \cdot 10^{-1} \end{aligned}$
Let $a = 10^{x}$ and $b = 10^{y}$ . Our equations become:
$\begin{aligned} 10a & = b/10 + 1 \\ 10b & = a + 1 \end{aligned}$
and we are asked to compute $a / b$ . Subtracting the equations gives
$10a - 10b = b/10 - a \Longrightarrow 11a = 101b/10$
giving an answer of $\frac{101}{110}$ .

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