Question
For what value of * the statement $$\left( {\frac{*}{{15}}} \right)$$ $$\left( {\frac{*}{{135}}} \right)$$ = 1 is true ?
Answer: Option D
$$\eqalign{
& {\text{Method 1:}} \cr
& {\text{Let the missing number be }}x \cr
& {\text{Then, }} \cr
& \Leftrightarrow {x^2} = 15 \times 135 \cr
& \Leftrightarrow x = \sqrt {15 \times \left(15 \times 9 \right)} \cr
& \Leftrightarrow x = \sqrt {{{15}^2} \times {3^2}} \cr
& \Leftrightarrow x = 15 \times 3 \cr
& \Leftrightarrow x = 45 \cr
& \cr
& {\text{Method 2:}} \cr
& {\text{Let the missing number be }}x \cr
& {\text{Then, }} \cr
& \Leftrightarrow {x^2} = 15 \times 135 \cr
& \Leftrightarrow {x^2} = 2025 \cr
& \Leftrightarrow x = \sqrt {{2025} } \cr
& \Leftrightarrow x = 45 \cr} $$
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$$\eqalign{
& {\text{Method 1:}} \cr
& {\text{Let the missing number be }}x \cr
& {\text{Then, }} \cr
& \Leftrightarrow {x^2} = 15 \times 135 \cr
& \Leftrightarrow x = \sqrt {15 \times \left(15 \times 9 \right)} \cr
& \Leftrightarrow x = \sqrt {{{15}^2} \times {3^2}} \cr
& \Leftrightarrow x = 15 \times 3 \cr
& \Leftrightarrow x = 45 \cr
& \cr
& {\text{Method 2:}} \cr
& {\text{Let the missing number be }}x \cr
& {\text{Then, }} \cr
& \Leftrightarrow {x^2} = 15 \times 135 \cr
& \Leftrightarrow {x^2} = 2025 \cr
& \Leftrightarrow x = \sqrt {{2025} } \cr
& \Leftrightarrow x = 45 \cr} $$
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