Given that a, b in {0, 1, 2, ldots, 9} with a | vedclass

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(C) Given that $\left(a+\frac{b}{10}\right)^x = \left(\frac{a}{10}+\frac{b}{100}\right)^y = 1000$. Note that $\frac{a}{10} + \frac{b}{100} = \frac{1}{

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$\frac{1}{3}$

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(C) Given that $\left(a+\frac{b}{10}\right)^x = \left(\frac{a}{10}+\frac{b}{100}\right)^y = 1000$. Note that $\frac{a}{10} + \frac{b}{100} = \frac{1}{10} \left(a + \frac{b}{10}\right)$. Let $k = a + \frac{b}{10}$. Then the given equations are $k^x = 1000$ and $\left(\frac{k}{10}\right)^y = 1000$. From $k^x = 1000$, we have $k = 1000^{1/x} = 10^{3/x}$. From $\left(\frac{k}{10}\right)^y = 1000$, we have $\frac{k}{10} = 1000^{1/y} = 10^{3/y}$. Thus, $k = 10 \cdot 10^{3/y} = 10^{1 + 3/y}$. Equating the two expressions for $k$: $10^{3/x} = 10^{1 + 3/y}$. Comparing the exponents: $\frac{3}{x} = 1 + \frac{3}{y}$. Rearranging the terms: $\frac{3}{x} - \frac{3}{y} = 1$. Dividing by $3$: $\frac{1}{x} - \frac{1}{y} = \frac{1}{3}$.

Given that $a, b \in \{0, 1, 2, \ldots, 9\}$ with $a+b \neq 0$ and that $\left(a+\frac{b}{10}\right)^x = \left(\frac{a}{10}+\frac{b}{100}\right)^y = 1000$. Then, $\frac{1}{x} - \frac{1}{y}$ is equal to

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