To simplify the expression \((7^{-3})^7 \times \frac{7^{20}}{6^{-2}}\) while applying the properties of integer exponents, let's break it down step by step.
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Apply the Power of a Power Property: \[ (7^{-3})^7 = 7^{-3 \cdot 7} = 7^{-21} \]
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Combine the Exponents with \(7^{20}\): \[ 7^{-21} \times 7^{20} = 7^{-21 + 20} = 7^{-1} \]
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Simplify the Division by a Negative Exponent: Dividing by \(6^{-2}\) is equivalent to multiplying by \(6^2\): \[ \frac{7^{-1}}{6^{-2}} = 7^{-1} \times 6^2 = \frac{6^2}{7^1} \]
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Calculate \(6^2\): \[ 6^2 = 36 \]
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Combine the Expressions: So, we have: \[ \frac{6^2}{7^1} = \frac{36}{7} \]
The final expression, with only positive exponents, is \(\frac{36}{7}\).
Therefore, the answer is: \[ \frac{36}{7} \] (as an improper fraction).