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

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(C) Given that $\left(a+\frac{b}{10}\right)^x = \left(\frac{a}{10}+\frac{b}{100}\right)^y = 1000$. We can rewrite the second term as $\left(\frac{1}{1

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$\frac{1}{3}$

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(C) Given that $\left(a+\frac{b}{10}ight)^x = \left(\frac{a}{10}+\frac{b}{100}ight)^y = 1000$. We can rewrite the second term as $\left(\frac{1}{10}(a+\frac{b}{10})ight)^y = 1000$. Let $K = a+\frac{b}{10}$. Then $K^x = 1000$ and $(\frac{K}{10})^y = 1000$. From $K^x = 1000$,we have $K = 1000^{1/x} = 10^{3/x}$. From $(\frac{K}{10})^y = 1000$,we have $\frac{K}{10} = 1000^{1/y} = 10^{3/y}$. Thus,$K = 10 	imes 10^{3/y} = 10^{1 + 3/y}$. Equating the two expressions for $K$: $10^{3/x} = 10^{1 + 3/y}$. Therefore,$\frac{3}{x} = 1 + \frac{3}{y}$. Rearranging the terms,we get $\frac{3}{x} - \frac{3}{y} = 1$. Dividing by $3$,we obtain $\frac{1}{x} - \frac{1}{y} = \frac{1}{3}$.

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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